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Definition of "momentum operator" in quantum theory

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Stephen Parrott

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Mar 10, 2010, 3:13:50 AM3/10/10
to
Recently there were some posts discussing the definition of "momentum
operator" in quantum mechanics. One poster asked about it, and another
replied that the answer could be found in any textbook.


In fact, this definition is not obvious and not treated in a
sufficiently fundamental manner by most textbooks.


Most books define the momentum operator as the infinitesimal generator
of spatial translations. For definiteness and simplicity, consider the
case of a single non-relativistic particle in one dimension. All
treatments known to me take its Hilbert space to be the space of all
square-integrable functions on the real line, and the position operator
X to be "multiplication by the coordinate x": (Xf)(x) := xf(x) , where
f is a function.


In this framework, the unitary operator U of spatial translation by a
number b is supposed to satisfy


(1) UXU* = X-b ,


where I use the mathematicians' notation * for adjoint instead of the
physicists' "dagger" because it is available in ASCII.


If we define the "usual" momentum operator


P := -i d/dx, i.e., (Pf)(x) := -i df/dx ,


then U := exp(-ibP) satisfies (1). However, P is not unique; if we
defined


(Qf(x)) := -i df/dx + A(x)f(x), where A is any given function,


then U := exp(-ibQ) would also satisfy (1), so Q would have just as
much claim to being the "official" momentum operator as P.


In fact, for a particle moving in an electromagnetic field, something
like Q is generally taken to be the "correct" momentum operator,
contrary to the claims of one poster, who ridiculed
another poster for asking if the momentum operator might depend on the
potential. In this situation, it does, at least in "standard" treatments.


This new momentum operator arises via a Lagrangian treatment. The
classical Lorentz force law (the "high school" force law, not including
radiation) can be "derived" from a Lagrangian L of the general form


(2) L(x,v) = mechanical Lagrangian + interaction Lagrangian


Here x and v are vectors in four-dimensional Minkowski space, with x
referring to a charged particle's position and v its four-velocity.
(Sometimes an "electromagnetic Lagrangian" is added to the right side,
but it will not contribute in the following discussion so I omit it.)


The "mechanical Lagrangian" is the Lagrangian for a free particle in the
absence of electromagnetic fields. The "interaction Lagrangian" will be
the part responsible for the Lorentz force law F = E + vxB, where F
denotes force, E electric field, and B magnetic field. The interaction
Lagrangian is of the form v.A = v.A(x), where A is the four-potential,
and "." denotes the Minkowski inner product.


In Lagrangian theory, the "momentum" is taken to be the partial
derivative of L with respect to v, which decomposes into a sum of two
terms corresponding to the two terms on the right of (2): a "mechanical
momentum" which turns out to be the usual mv (m:= mass) plus an
"interaction momentum" which turns out to be A:


(3) p := partial L with respect to v = mv + A


These are four-dimensional vectors. Since v has four components,
"partial L with respect to v" denotes a four-component object.
Eliminating the "time" component gives a three-dimensional expression
which looks the same in ASCII.


The three-dimensional A is the "magnetic potential" which gives the
magnetic field B via the usual relation


B = del x A .


Since A is a function of the coordinate x = (x1, x2, x3), texts
usually write (3) as


(4) p = mv + A(x) .


If we identify the function A = A(x) with a multiplication operator
(actually, it's three multiplication operators since A is a 3-vector
A(x) = (A1(x), A2(x), A3(x)) ), then (4) suggests that the
quantum-mechanical momentum might be represented by the corresponding
operator, where the mechanical term mv is the "usual" momentum operator
which for a particle in one dimension would be -id/dx. (In three
dimensions, which we have to use here to talk about magnetic potentials,
it will have three components, each of the form -i d/dx^j.)


In typical textbooks, this might be called a "derivation" of (4) as a
momentum operator for charged particles, but all we have done is to play
with some symbols. I can see no reason to imagine, a priori, that there
might be some physical content to (4).


What is amazing is that if we *do* use (4) as a momentum operator and
write down a corresponding Schroedinger equation, it has a profound
physical consequence which can be experimentally verified! You can find
the details in any text which discusses the "Aharonov-Bohm effect". This
effect states that under certain conditions, electrons can be affected
by an electromagnetic field even though THEY NEVER ENTER THE FIELD. And
you can see this in the laboratory!


I sympathize with one of the posters who is struggling to obtain a
fundamental understanding of quantum mechanics. If he ever does
succeed, I hope he will share the secret with the rest of us who are
still mystified after years of study!


Juan R.

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Mar 10, 2010, 7:44:22 AM3/10/10
to
Stephen Parrott wrote on Wed, 10 Mar 2010 08:13:50 +0000:

> Recently there were some posts discussing the definition of "momentum
> operator" in quantum mechanics. One poster asked about it, and
> another replied that the answer could be found in any textbook.

And a third poster is patiently giving to him sections of textbooks
with the relevant stuff and he is also patiently correcting his
mistakes, even those recent about dimensional analysis.

In this new thread you add more mistakes which are corrected below.

> In fact, this definition is not obvious and not treated in a
> sufficiently fundamental manner by most textbooks.

Nope. Let me show how the responses are found in textbooks :-D

> Most books define the momentum operator as the infinitesimal
> generator of spatial translations. For definiteness and simplicity,
> consider the case of a single non-relativistic particle in one
> dimension. All treatments known to me take its Hilbert space to be
> the space of all square-integrable functions on the real line, and
> the position operator X to be "multiplication by the coordinate x":
> (Xf)(x) := xf(x) , where f is a function.
>
>
> In this framework, the unitary operator U of spatial translation by
> a number b is supposed to satisfy
>
>
> (1) UXU* = X-b ,
>
>
> where I use the mathematicians' notation * for adjoint instead of the
> physicists' "dagger" because it is available in ASCII.
>
>
> If we define the "usual" momentum operator
>
>
> P := -i d/dx, i.e., (Pf)(x) := -i df/dx ,
>
>
> then U := exp(-ibP) satisfies (1). However, P is not unique; if we
> defined
>
>
> (Qf(x)) := -i df/dx + A(x)f(x), where A is any given
> function,
>
>
> then U := exp(-ibQ) would also satisfy (1), so Q would have just as
> much claim to being the "official" momentum operator as P.

The expression for P is unique. For any f(x) the next identity is
easily checked

f(x+a) = exp(a d/dx) f(x) [1]

if a is infinitesimal

exp(a d/dx) = 1 + a d/dx

and from here one obtains the value for P. See

http://en.wikipedia.org/wiki/Momentum_operator#Derivation

or, better, the page 190 in Volume I of the well-known textbook Quantum
Mechanics by Cohen-Tannoudji, Diu, and Lalöe

P = -ihbar d/dx

Here in tereafter I will refer to this book as [QM].

Your Q gives an extra exp(A(x)), that violates the identity [1] This
is why textbooks give P.

> In fact, for a particle moving in an electromagnetic field, something
> like Q is generally taken to be the "correct" momentum operator,
> contrary to the claims of one poster, who ridiculed another poster
> for asking if the momentum operator might depend on the potential.
> In this situation, it does, at least in "standard" treatments.

Your claims are plain wrong and violate any textbook in the matter.
As said above any textbook in quantum mechanics gives the expression
for the momentum as P = -ihbar d/dx, not as your Q. Details below.

> This new momentum operator arises via a Lagrangian treatment. The
> classical Lorentz force law (the "high school" force law, not
> including radiation) can be "derived" from a Lagrangian L of the
> general form
>
>
> (2) L(x,v) = mechanical Lagrangian + interaction Lagrangian
>
>
> Here x and v are vectors in four-dimensional Minkowski space, with x
> referring to a charged particle's position and v its four-velocity.

In quantum mechanical [x,p], x is not a four-position and the v
associated to p is not a four-velocity but ignore this detail now.

> (Sometimes an "electromagnetic Lagrangian" is added to the right side,
> but it will not contribute in the following discussion so I omit it.)
>
>
> The "mechanical Lagrangian" is the Lagrangian for a free particle in
> the absence of electromagnetic fields. The "interaction Lagrangian"
> will be the part responsible for the Lorentz force law F = E + vxB,
> where F denotes force, E electric field, and B magnetic field.

Here E and B are three-vectors, but above you defined v as a
four-velocity. It may be noted that in the Lorentz force law of above
v is a three vector

http://en.wikipedia.org/wiki/Lorentz_force

If you want to give a covariant expression of the four-force using a
four- velocity u then you may use the tensor F_ab, not three-vectors E
and B.

http://en.wikipedia.org/wiki/Lorentz_force#Covariant_form_of_the_Lorentz_force

> The interaction
> Lagrangian is of the form v.A = v.A(x), where A is the
> four-potential, and "." denotes the Minkowski inner product.

However, here v is the four-velocity, which is neither the
three-vector v nor the four-vector u used in the spatial and covariant
versions of the Lorentz force given above.

> In Lagrangian theory, the "momentum" is taken to be the partial
> derivative of L with respect to v, which decomposes into a sum of two
> terms corresponding to the two terms on the right of (2): a
> "mechanical momentum" which turns out to be the usual mv (m:= mass)
> plus an "interaction momentum" which turns out to be A:
>
>
> (3) p := partial L with respect to v = mv + A

That is not the definition. This is only valid for particles moving to
low speeds. The definition contains an extra gamma factor

p = m gamma v + eA

For low speeds v<<c follows that gamma -> 1 and you recover your
approx. expression of above.

> These are four-dimensional vectors. Since v has four components,
> "partial L with respect to v" denotes a four-component object.
> Eliminating the "time" component gives a three-dimensional expression
> which looks the same in ASCII.
>
>
> The three-dimensional A is the "magnetic potential" which gives the
> magnetic field B via the usual relation
>
>
> B = del x A .
>
>
> Since A is a function of the coordinate x = (x1, x2, x3), texts
> usually write (3) as
>
>
> (4) p = mv + A(x) .
>
>
> If we identify the function A = A(x) with a multiplication operator
> (actually, it's three multiplication operators since A is a 3-vector
> A(x) = (A1(x), A2(x), A3(x)) ), then (4) suggests that the
> quantum-mechanical momentum might be represented by the corresponding
> operator, where the mechanical term mv is the "usual" momentum
> operator which for a particle in one dimension would be -id/dx.
> (In three dimensions, which we have to use here to talk about
> magnetic potentials, it will have three components, each of the form
> -i d/dx^j.)

The definition of momentum p= -ihbar d/dx is the definition of
momentum of the particle, it is then

p = m gamma v + eA = -ihbar d/dx

wich contradicts your early claim about

"In fact, for a particle moving in an electromagnetic field, something
like Q is generally taken to be the "correct" momentum operator"

with your (wrong) definition Q := -i d/dx + A(x)

In presence of fields the momentum is

p = m gamma v + eA = -ihbar d/dx

for a free particle it is

p = m gamma v = -ihbar d/dx

for a free low velocity particle

p = m v = -ihbar d/dx

Etc.

In all the cases the commutation relationship [x,p] is exactly
satisfied both in presence or absence of A.

*This is all said in textbooks*. For example, in the Chapter III "The
postulates of quantum mechanics" of the book [QM], the authors
explicitely say:

"However, it is the conjugate momentum p and not the mechanical
momentum mv which becomes in quantum mechanics the operator P wich
satisfies the canonical commutation relations (B-33)"

Note that in that section they only consider a Hamiltonian (B-46)
valid for low speed particles, thus their expression for the momentum
(B-45) does not contain the extra gamma factor reported above.

(...)



> I sympathize with one of the posters who is struggling to obtain a
> fundamental understanding of quantum mechanics. If he ever does
> succeed, I hope he will share the secret with the rest of us who are
> still mystified after years of study!

As showed the above questions are found in textbooks and the mistakes
reported here and in the other thread would be avoided if the textbooks
were opened. Moreover, I share here the well-known Gell-Mann view
about the existence of a tendency towards mistifying quantum mechanics.

As he likes to say, quantum mechanics is quantum mechanics just that.

It may be remarked that the quantum mechanical [x,p] = ihbar is the
quantum version of the well-known classical result {x,p} = 0, where
{,} are the Poisson brackets.

Claiming that the quantum mechanical relation is only valid for free
particles and that we may modify it in presence of interactions would
be such a nonsensical statement as claiming that {x,p} would be also
modified, because in presence of interaction the p is not that of a
free particle. Of course {x,p} still holds in presence of magnetic
fields.


--
http://www.canonicalscience.org/

BLOG:
http://www.canonicalscience.org/publications/canonicalsciencetoday/canonicalsciencetoday.html

Igor Khavkine

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Mar 10, 2010, 12:17:38 PM3/10/10
to
=========== Moderator's note ==================================

It's definitely the canonical momentum which has to be identified
with translations in position space. All this follows from a careful
analysis of the representation theory of the Galileo group in the non-
relativistic case. In the relativistic case it's nearly the same except
for massless particles with spin >=1 where there does not exist a position
operator.

===============================================================

On Mar 10, 9:13�am, Stephen Parrott <postn...@email.toast.net> wrote:

> Most books define the momentum operator as the infinitesimal generator
> of spatial translations. �For definiteness and simplicity, consider the
> case of a single non-relativistic particle in one dimension. �All
> treatments known to me take its Hilbert space to be the space of all
> square-integrable functions on the real line, and the position operator
> X to be "multiplication by the coordinate x": �(Xf)(x) := xf(x) , where
> f is a function.
>
> In this framework, the �unitary operator U of spatial translation by a
> number b is supposed to satisfy
>
> (1) � � � � � �UXU* = X-b ,
>
> where I use the mathematicians' notation * for adjoint instead of the
> physicists' "dagger" because it is available in ASCII.
>
> If we define the "usual" momentum operator
>
> � � � � �P := -i d/dx, � i.e., �(Pf)(x) := -i df/dx ,
>
> then �U := exp(-ibP) satisfies (1). �However, P is not unique; �if we
> defined
>
> � � � � �(Qf(x)) := -i df/dx + A(x)f(x), where A is any given function,
>
> then �U := exp(-ibQ) would also satisfy (1), so Q would have just as
> much claim to being the "official" momentum operator as P.

Of course, you are right that operators P and Q are equivalent if (1)
is taken as the defining property of the momentum operator. However,
to the best of my knowledge, the momentum operator is defined as the
generator of infinitesimal translations on wave functions:

(Uf)(x) = f(x-b)

Then the only possibility is U=exp(ibP), which is not equal to
V=exp(ibQ). BTW, I'm probably mixing up signs somewhere. The operators
exp(ibQ) correspond to a translation followed by multiplication by a
certain position dependent phase. The operators U and V are physically
distinguishable because we can compare their matrix elements <f|U|g>
and <f|V|g>. These matrix elements should be experimentally accessible
just as the matrix elements <f|exp(-itH)|g> of the unitary time
evolution operator are.

The freedom to chose different pairs of observables X and P or X and Q
satisfying the canonical commutation relation is there, but it is the
same freedom as choosing different coordinates on the classical phase
space, which are related to each other by a canonical transformation
(i.e., that preserves the canonical Poisson brackets).

> [...] �The


> classical Lorentz force law (the "high school" force law, not including
> radiation) can be "derived" from a Lagrangian L of the general form
>
> (2) � L(x,v) = mechanical Lagrangian + interaction Lagrangian

[...]


> In Lagrangian theory, the "momentum" is taken to be the partial
> derivative of L with respect to v, which decomposes into a sum of two
> terms corresponding to the two terms on the right of (2): a "mechanical
> momentum" which turns out to be the usual mv (m:= mass) plus an
> "interaction momentum" which turns out to be A:
>
> (3) � �p := partial L with respect to v = mv + A

[...]


> Since �A �is a function of the coordinate x = (x1, x2, x3), texts
> usually write (3) as
>
> (4) � �p = mv + A(x) .
>
> If we identify the function �A = A(x) with a multiplication operator
> (actually, it's three multiplication operators since A is a 3-vector
> A(x) = (A1(x), A2(x), A3(x)) ), then (4) suggests that the
> quantum-mechanical momentum might be represented by the corresponding
> operator, where the mechanical term mv is the "usual" momentum operator
> which for a particle in one dimension would be -id/dx. �(In three
> dimensions, which we have to use here to talk about magnetic potentials,
> it will have three components, each of the form -i d/dx^j.)

If I understand what you wrote correctly, you are suggesting that mv
is -id/dx and that equation (4) defines the canonical quantum
mechanical momentum operator as -id/dx+A(x). Unfortunately, I believe
that's incorrect. The canonical momentum p must be identified with -id/
dx. Then, if you are interested in the operator mv, it will be given
by mv = p-A(x) = -id/dx-A(x). I hope the discussion below will clarify
why.

> In typical textbooks, this might be called a "derivation" of (4) as a
> momentum operator for charged particles, but all we have done is to play
> with some symbols. �I can see no reason to imagine, a priori, that there
> might be some physical content to (4).

Perhaps at this point I can make some remarks that will clarify the
situation and give some "a priori" reasons.

First, some matters of terminology. You've made reference to the
"momentum" (which I'll call the "canonical momentum"), the "mechanical
momentum", and the "interaction momentum". The relation between them
is

canonical momentum = mechanical momentum + interaction momentum (*)

Now, in accord with my previous discussion with Jay, which you've
alluded to, I maintain that the definition of the canonical momentum
is that it generates infinitesimal translations. From my previous
comments, it should be clear that this fixes the quantum mechanical
canonical momentum operator uniquely. It immediately follows that the
commutation relation

[position,canonical momentum] = i

holds independently of the presence or absence of any interactions,
since no mention has been made of them so far. Classically, the
infinitesimal translations should be generated via Poisson brackets.
In that case, as you've pointed out, for any canonical pair P and X of
coordinates on the classical phase space, one can apply a symplectic
transformation and obtain another canonical pair Q and X, with X the
same, Q different from P, yet still {X,P}={X,Q}=1. This is perhaps not
too surprising, since the quantum mechanical operators P and Q were
different from each other by a multiplicative phase, and phase
information gets lost in the classical limit. So, classically, the
canonical momentum is not uniquely fixed, but can be associated with a
well defined equivalence class.

The question of explaining (*) now can be sharpened to this: what is
the relation between the canonical momentum and the velocity? From
your discussion of electromagnetic forces, the answer to that question
is obviously interaction dependent. But what is the precise
connection? The traditional answer, as you've already pointed out, is
via the Legendre transform, which transforms the Lagrangian picture
into the canonical one.

However, lets take a slightly deeper look here. Through centuries of
observation, we've concluded that any physical system can be modeled
extremely well by specifying its (a) states, (b) observables and (c)
dynamical evolution law. These components of a model are not
arbitrary, but exhibit specific structure (depending on the level of
approximation). On the fundamental quantum level, we have (a) vectors
in a Hilbert space (or density matrices), (b) linear, self-adjoint
operators on the same Hilbert space, (c) unitary evolution. In the
classical limit, the remnants of that structure are (a) points in a
phase space manifold (or probability distributions on it), (b)
functions on the phase space, (c) a symplectic form preserved by the
equations of motion (a vector field on phase space). Note that this
description is not in full generality, as for instance absence of time-
translation invariance complicates the structures involved.

In the quantum case, continuity and unitarity of the dynamical
evolution law imply that observables satisfy the Heisenberg equations
of motion dO/dt = -i[O,H], where H is the self-adjoint generator of
infinitesimal time translations, usually called the Hamiltonian. If X
is the position operator and P is the canonical momentum operator
(uniquely defined), the commutation relation [X,P]=i implies that H
will be some expression in X and P (irreducibility, Schur's lemma).
Combining these pieces, we find that

dX/dt = -i[X,H] = some expression in X and P.

In other words, knowledge of the dynamical evolution law (basically
knowledge of H) uniquely fixes the relationship between the canonical
momentum P and the velocity dX/dt.

In the classical case, instead of the quantum commutator, we have the
Poisson bracket or the symplectic structure. The symplectic structure
is there whether we ask for it or not. It so happens that classical
equations of motions always preserve a symplectic 2-form on the
system's phase space (for appropriately idealized, isolated systems).
>>From this it follows that the phase space vector field associated to
the equations of motion is Hamiltonian, that is observables evolve
according to the canonical equations dO/dt = {O,H}, where H is some
observable, again called the Hamiltonian. The Lagrangian variational
principle is just a convenient way to package the equations of motion
together with the symplectic structure. Once again, knowledge of the
Hamiltonian fixes the precise relationship between dX/dt and P. A
canonical transformation may pick out another representative from the
same equvalence class of observables as P, say Q. Then dX/dt will have
a different expression in terms of Q, but so will H.

The lesson I like to take from these observations is the following.
The time evolution of physical systems exhibits special structure.
This structure can be split into two parts: a kinematical part
(Hilbert space or symplectic manifold) and a dynamical part
(Hamiltonian observable). The canonical momentum can be defined
whenever the kinematical structure is present, independent of the
presence or absence of interactions (dynamics). However, it has no
fixed relationship with the velocity (time derivative of the
position). Once the dynamics (including interactions, if any) is
specified, the relationship between the canonical momentum and
velocity observables is fixed.

Finally, about the the nomenclature "mechanical momentum" or
"interaction momentum". They never really entered the preceding
discussion. They happen to be specific expressions in terms of the
velocity, like mv, and have no a priori relation to the canonical
momentum. They are obviously related to the canonical momentum for
specific choices of the dynamics. However, the fact that all three
terms have the word "momentum" in them, I believe, is no more than a
historical artifact.

> I sympathize with one of the posters who is struggling to obtain a
> fundamental understanding of quantum mechanics. �If he ever does
> succeed, I hope he will share the secret with the rest of us who are
> still mystified after years of study!

Hope this helps. :-)

Igor

Stephen Parrott

unread,
Mar 10, 2010, 2:45:17 PM3/10/10
to
The original post implicitly assumed units in which the charged particle
had charge 1, so that the Lorentz equation was F = E + v x B and
the interaction Lagrangian was v.A. (I do that without thinking in
private calculations.) In arbitrary units, for charge q, those should
be F = q(E + vxB) and qv.A for the interaction Lagrangian.

Igor Khavkine

unread,
Mar 10, 2010, 3:52:59 PM3/10/10
to
On Mar 10, 6:17�pm, Igor Khavkine <igor...@gmail.com> wrote:
> =========== Moderator's note ==================================
>
> It's definitely the canonical momentum which has to be identified
> with translations in position space. All this follows from a careful
> analysis of the representation theory of the Galileo group in the non-
> relativistic case. In the relativistic case it's nearly the same except
> for massless particles with spin >=1 where there does not exist a position
> operator.
>
> ===============================================================

In the case of relativistic field theory, though a particle
interpretation of states and observables is sometimes available, it by
no means always is (as strong field effects like Klein's paradox,
Hawking and Unruh's effects evince). On the other hand, all states,
observables and equations of motion can be unambiguously formulated in
terms of field degrees of freedom. Then, the notion of a canonical
momentum density is still available defined by the property of
generating local translations of the field amplitude (phi -> phi+delta-
phi). That is the field analog of the canonical momentum that I
discussed in my previous post.

Of course, if one is interested in a Poincare invariant theory and the
conserved quantities associated to space-time translations via
Noether's theorem, these quantities are commonly called momenta as
well, even in relativistic field theory. However, I believe, the fact
that the latter kind of momentum and canonical momentum share the word
"momentum" in their names is also a historical artefact.

Igor

Stephen Parrott

unread,
Mar 11, 2010, 12:43:15 PM3/11/10
to
[ Mod. note: I would encourage everyone to keep the focus of the
conversation on the relevant physics, rather than, sometimes
unfortunate, personal remarks. -ik ]

Your post is so intemperate that I'm not motivated to reply in detail to
every point you raise. I'll just say initially that I disagree with
almost all of them. I will give brief replies to a few of them.


>
> f(x+a) = exp(a d/dx) f(x) [1]
>
> if a is infinitesimal
>
> exp(a d/dx) = 1 + a d/dx
>
> and from here one obtains the value for P.

Yes, if you start from [1], but translation by *a* need not take
f(x) into f(x+a), but could take f(x) into exp(ig(x,a))f(x+a), where g
is a real function which has to satisfy a so-called "cocycle equation"
to guarantee that translating by *a*, and then by *b*, is the same as
translating by a+b.

This is because both f(x+a) and exp(ig(x,a))f(x+a) define the same
probability distribution |f(x+a)|^2 on position space. The infinitesimal
generator for the action just described is of the form I gave:

(Ph)(x) = -i dh/dx + A(x)h(x) ,

where A is an arbitrary real function.


>
>> In fact, for a particle moving in an electromagnetic field, something
>> like Q is generally taken to be the "correct" momentum operator,
>> contrary to the claims of one poster, who ridiculed another poster
>> for asking if the momentum operator might depend on the potential.
>> In this situation, it does, at least in "standard" treatments.
>
> Your claims are plain wrong and violate any textbook in the matter.
> As said above any textbook in quantum mechanics gives the expression
> for the momentum as P = -ihbar d/dx, not as your Q. Details below.

This is an example of what I meant by your post being intemperate. My
"claims" are not claims, but correct conclusions from my assumptions,
which happen to be different from your assumptions. For example, you
assume that the operator of translation by a must take f(x) into f(x+a),
but I used a more general assumption.

A more extensive explanation is given in my reply to Igor Khavkine.


>>
>> The "mechanical Lagrangian" is the Lagrangian for a free particle in
>> the absence of electromagnetic fields. The "interaction Lagrangian"
>> will be the part responsible for the Lorentz force law F = E + vxB,
>> where F denotes force, E electric field, and B magnetic field.
>
> Here E and B are three-vectors, but above you defined v as a
> four-velocity.

You didn't read carefully what I wrote. *v* could be a three-vector or
a four-vector, depending on context. Obvious, in the context
F = E + v x B, v must represent a three-vector.

It may be noted that in the Lorentz force law of above
> v is a three vector
>
> http://en.wikipedia.org/wiki/Lorentz_force
>
> If you want to give a covariant expression of the four-force using a
> four- velocity u then you may use the tensor F_ab, not three-vectors E
> and B.
>
> http://en.wikipedia.org/wiki/Lorentz_force#Covariant_form_of_the_Lorentz_force

You are obviously assuming and implying that I am making very stupid
mistakes. Had you inquired into my background (by Google, for example),
you would have learned I am the author of the book , "Relativistic
Electrodynamics and Differential Geometry", Springer, 1987 It is
probably out of print now, but when published it was very well-reviewed,
and sold unusually well for a book that technical.

Of course, I can make mistakes like everybody else, but the chance that
an author of a book on relativistic electrodynamics published by a good
publisher would make such elementary mistakes would seem quite small.


>
>> The interaction
>> Lagrangian is of the form v.A = v.A(x), where A is the
>> four-potential, and "." denotes the Minkowski inner product.
>
> However, here v is the four-velocity, which is neither the
> three-vector v nor the four-vector u used in the spatial and covariant
> versions of the Lorentz force given above.
>
>> In Lagrangian theory, the "momentum" is taken to be the partial
>> derivative of L with respect to v, which decomposes into a sum of two
>> terms corresponding to the two terms on the right of (2): a
>> "mechanical momentum" which turns out to be the usual mv (m:= mass)
>> plus an "interaction momentum" which turns out to be A:
>>
>>
>> (3) p := partial L with respect to v = mv + A
>
> That is not the definition. This is only valid for particles moving to
> low speeds. The definition contains an extra gamma factor
>
> p = m gamma v + eA
>
> For low speeds v<<c follows that gamma -> 1 and you recover your
> approx. expression of above.

You didn't carefully read what I wrote.The second paragraph of the quote
above states that v is a four-velocity. The "gamma" factor is contained
in the four-velocity v, not in the rest mass m.

Again, you are very quick to assume that you are dealing with some kind
of incompetent. Your previous post was also of that nature, but
referring to a different person. I should imagine that he would be
offended and/or hurt.

> It may be remarked that the quantum mechanical [x,p] = ihbar is the
> quantum version of the well-known classical result {x,p} = 0, where
> {,} are the Poisson brackets.
>
> Claiming that the quantum mechanical relation is only valid for free
> particles and that we may modify it in presence of interactions would
> be such a nonsensical statement as claiming that {x,p} would be also
> modified, because in presence of interaction the p is not that of a
> free particle. Of course {x,p} still holds in presence of magnetic
> fields.
>
>

???

Your posts make quite a few negative judgments about the competence of
others. May I ask what are your qualifications to make such judgments?


Do you have a Ph.D., and if so, from what university? Do you have a
position at any university? Do you have any publications? I could find
nothing in the arXiv.


Stephen Parrott

Stephen Parrott

unread,
Mar 11, 2010, 12:43:15 PM3/11/10
to
Oh, my!


Had I realized that my post would have produced such impassioned
responses, I would never have posted it. Not because I think that there
is anything wrong with it, but because I would prefer not to feel
obligated to spend time on replies. But I guess I should [sigh].


My notation and terminology seems to have caused some confusion. I was
trying to write something which would resonate even in ASCII, so that
readers could get a general sense of what I was trying to say without
getting bogged down in details.


For example, I used *v* to represent a velocity, which was to be
interpreted as either a normal 3-dimensional velocity, or a relativistic
four-velocity, according to the context in which only one or the other
would make sense. This resulted in being taken to task (in a "gotcha!"
manner), for referring to the "high school [Lorentz] force law" F = q(E
+ v x B) without stating that v is a 3-velocity, rather than a
4-velocity. Since the context was high school (or maybe freshman)
physics in which all the symbols are standard (E := electric field, a
3-vector, etc.), I thought that readers could not possibly interpret v
as a four-velocity in this context. But, sadly, I was wrong.


Similar, I thought that the terms "mechanical" and "interaction"
momentum would communicate the idea that the mechanical momentum was
what we normally think of as momentum---for example for an electron the
mechanical momentum would be the momentum it would have if it were an
uncharged particle of the same mass, and the "interaction momentum"
would be something additional arising from the electron's charge and the
surrounding electromagnetic potentials. I hoped that readers who might
not be familiar with Lagrangian formulations might at least get the
general idea, and if further curious would consult any standard text for
details. I cannot write a text in the ASCII format available here,
without using mathematical symbols.


I wrote the post partly to express wonder that playing with symbols in
such a formal and almost meaningless way could lead to the Aharonov-Bohm
effect, and to express further wonder at that effect. I realize that
not everyone feels such wonder, but there *are* those who do, and the
post was addressed to them.


And, on the barely conscious level, I was probably distressed that good
questions of one poster would be ridiculed. As one person commented,
that poster does often ask elementary questions whose answers can be
found in textbooks. It seems to me that the appropriate response to
that is simply to ignore them, if one doesn't feel like answering them.
That is what I usually do, this current mess being an unfortunate exception.


Many of his questions seem to me naive, but that is to be expected when
one is learning with very little background a field as complicated and
controversial as physics. (I think he is largely self-taught.) And,
almost by definition, to suggest that a person should not ask naive
questions is tantamount to telling him that he should not ask anything
at all. If a person is asking about things he does not understand, how
can he tell if a question is naive?


I thought that some of his questions about the definition of momentum
were good ones which deserved respect and serious consideration. Of
course, I'm not implying that any particular person is obligated to
answer them. In particular, I feel no such obligation, but sometimes if
I have some free time, I might choose to answer anyway.

Stephen Parrott

Lou Pecora

unread,
Mar 14, 2010, 8:44:16 AM3/14/10
to
In article <pan.2010.03...@canonicalscience.com>,

"Juan R." Gonz�lez-�lvarez <now...@canonicalscience.com> wrote:

> It may be remarked that the quantum mechanical [x,p] = ihbar is the
> quantum version of the well-known classical result {x,p} = 0, where
> {,} are the Poisson brackets.

Minor point: I assume you really meant {x,p} = 1 or {x_i,p_j} = delta_ij

--
-- Lou Pecora

Juan R. Gonzàlez-Àlvarez

unread,
Mar 14, 2010, 8:44:16 AM3/14/10
to
Stephen Parrott wrote on Thu, 11 Mar 2010 17:43:15 +0000:

(Previous names, and dates sniped by Parrot)

> [ Mod. note: I would encourage everyone to keep the focus of the
> conversation on the relevant physics, rather than, sometimes
> unfortunate, personal remarks. -ik ]
>
> Your post is so intemperate that I'm not motivated to reply in detail
> to every point you raise. I'll just say initially that I disagree
> with almost all of them. I will give brief replies to a few of them.
>>
>> f(x+a) = exp(a d/dx) f(x) [1]
>>
>> if a is infinitesimal
>>
>> exp(a d/dx) = 1 + a d/dx
>>
>> and from here one obtains the value for P.
>
> Yes, if you start from [1], but translation by *a* need not take f(x)
> into f(x+a), but could take f(x) into exp(ig(x,a))f(x+a), where g is
> a real function which has to satisfy a so-called "cocycle equation"
> to guarantee that translating by *a*, and then by *b*, is the same
> as translating by a+b.
>
> This is because both f(x+a) and exp(ig(x,a))f(x+a) define the same
> probability distribution |f(x+a)|^2 on position space. The
> infinitesimal generator for the action just described is of the form
> I gave:
>
> (Ph)(x) = -i dh/dx + A(x)h(x) ,
>
> where A is an arbitrary real function.

In my another more detailed reply I did many remarks. Now I will add
another remark.

In your original message you defined A to be the electromagnetic
four-potential. Now, you restrict your definition of 'momentum' to
expressions where A is a *real* function, but the electromagnetic
potential A used in the quantum theory is *not*.

(...)

Stephen Parrott

unread,
Mar 14, 2010, 8:44:16 AM3/14/10
to
Igor Khavkine wrote:
> =========== Moderator's note ==================================
>
> It's definitely the canonical momentum which has to be identified
> with translations in position space. All this follows from a careful
> analysis of the representation theory of the Galileo group in the non-
> relativistic case. In the relativistic case it's nearly the same except
> for massless particles with spin >=1 where there does not exist a position
> operator.
>
> ===============================================================
> ...

> First, some matters of terminology. You've made reference to the
> "momentum" (which I'll call the "canonical momentum"), the "mechanical
> momentum", and the "interaction momentum". The relation between them
> is
>
> canonical momentum = mechanical momentum + interaction momentum (*)
> ...

OK, let's take this as our definition of "canonical momentum". In a
classical context, for a charged particle of mass m and velocity v in a
magnetic potential A = A(x), the "mechanical momentum" will be defined
as mv, and the "interaction momentum" as A (using units in which the
particle has charge 1; otherwise it would be qA).

In usual physics notation, v = v^i = (v^1, v^2, v^3) In the present
three-dimensional context, the magnetic potential A is usually called
the "vector potential", and it satisfies

(1) B = del x A , where B is the magnetic field.

Given B, equation (1) does not uniquely define A. In a simply connected
region, it defines A up to the addition of the gradient of a scalar
function. In a region which is not simply connected, further ambiguity
is possible. (This is at the heart of the Aharonov-Bohm effect.)

Now we have to decide what we should (or must?) take as the
quantum-mechanical analog P of the canonical momentum. This P=(P^1, P^2,
P^3) will be an triple of operators on the Hilbert space of states.
Everyone seems to agree that for a spinless particle of nonzero mass,
this Hilbert space should be taken as the space of square-integrable
functions on R^3 (three-dimensional real Euclidean space).

[Because (1) only makes sense in three dimensions, we have to use
R^3 instead of R^1, but it may be useful to sometimes pretend we are
using R^1 to make certain points in a simple manner.]

Everyone seems to agree that a component P^i of the canonical momentum
operator should be an infinitesimal generator of translations in the
i'th direction, but there seems to be disagreement on whether this means
that it should satisfy (pretending for notational simplicity that R^3 is
R^1 with coordinate denoted "x" and corresponding multiplication
operator (Xf)(x) := xf(x))

(2a) exp(isP) X exp(-isP) = X - sI (s an arbitrary
scalar, I the identity operator), or

(2b) (exp(isP)f)(x) = f(x-s) (as you suggest), or

something else. If we use your suggestion (2b), then, as you say, the
canonical momentum P has to be (Pf)(x) := -i df/dx.

I am accustomed to using (2a). I have only two elementary quantum
mechanics books in my library, George Mackey's "Mathematical Foundations
of Quantum Mechanics" (which is foundational, but not mathematically
elementary) and Asher Peres' "Quantum Theory: Concepts and Methods"
(mathematically elementary but physically fairly sophisticated).

[When I retired from teaching (though not from research) and moved
from Boston to rural Nevada, I gave away almost all my books because
there wasn't room in my small SUV. It costs half a day to visit the
nearest library which might possibly have a serious physics book, and
two days to the nearest good research library. So I can't discuss
other books.]

Mackey's book (from which I learned the subject) does it exactly as I
did, using (2a).

Peres does it in a way that looks slightly different, but amounts to
(2a). He points out that your (2b) should be

(2c) (exp(isP)f)(x) = exp(isg(x))f(x-s)

with g an arbitrary real function. (He writes exp(ig(x)) instead of
exp(isg(x)), which I assume is an error or typo.) That is because (2b)
and (2c) define the same position-space probability function |f(x-s)|^2.
Either (2a) or (2c) gives

(3) (Pf)(x) := -i df/dx + h(x)f(x) ,

for our one-dimensional simplification, where h = g' is an arbitrary
real function. In three dimensions,

(4) (P^i f)(x) = -i df/dx^i + A^i(x)f(x),

where A is an arbitrary vector field. I will write this as

(5) P = -iD + A,

where I am introducing a 3-component differentiation operator
D = (d/dx^1, d/dx^2, d/dx^3). Thus -iD is the "usual" momentum operator
as presented in the more elementary quantum mechanics texts.

For an uncharged spinless, elementary particle of mass m > 0 in
three-dimensions in a potential V, I hope everyone can agree that that
the physically correct Hamiltonian H is

(6) H = -D^2/2m + V = (-iD)^2/2m + V ,

where for simplicity of ASCII notation, V=V(x) will stand both for a
scalar function on R^3 and for the operator of multiplication by that
function. Also, D^2 stands for the Laplacian, which is the "inner
product" of the "vector" D (of operators) with itself. If P = -iD as
you think, then this is the more familiar H = -P^2/2m + V.

The question we need to address is: WHAT IS THE CORRESPONDING
HAMILTONIAN FOR A CHARGED PARTICLE IN A POTENTIAL A? Please try to
answer this before continuing.

I do not have at hand a book which answers this question. I have only
my memory, which tells me that one obtains the Hamiltonian for the
charged particle by making the so-called "minimal coupling" replacement
of -iD in (6)

(7) replace -iD by -iD + A ,

where A is the magnetic potential ("vector potential") and use the
electric potential for V. This results in a new Hamiltonian H' given by

(8) H' = (-iD + A)^2/2m + V .

This is the Hamiltonian used to derive the Aharonov-Bohm effect. If the
canonical momentum P is given by (5) as suggested by Lagrangian theory,
then (8) assumes the familiar form

(9) H' = P^2/2m + V.

We can think of this in two ways: either (a) we have changed the
momentum from your -iD to -iD + A but used the usual expression (9) for
the Hamiltonian or (b) we have retained your expression for the
momentum, but changed the Hamiltonian. Obviously, which is "correct" is
a matter of semantics only, without physical significance.

Note also that changing the Hamiltonian is equivalent to CHANGING THE
SCHROEDINGER EQUATION. My impression is that most people would prefer
to change the expression for momentum (since this is routinely done in
the classical context anyway) than change the Schroedinger equation.

[I can't pass this point without mentioning a remarkable fact which
was one of the main points of my original post, though it isn't
particularly relevant to the present discussion. The new Hamiltonian
H' depends only on the electromagnetic POTENTIALS A and V and not on
the electromagnetic FIELDS B and E as one would expect. Even if the
fields vanish in a region which is not simply connected, the
potentials need not vanish, nor be of the trivial form A = grad phi.
Thus even if the electron stays in such a region and never sees any
fields, its behavior according to (8) can differ from the behavior
of a similar uncharged particle. This is the theoretical basis
for the Aharonov-Bohm effect, and I understand that it has been
amply experimentally verified.]

I hope you can see how an intelligent beginning student might get
confused! Why does one sometimes use a different "momentum" in the
presence of an electromagnetic potential? Or, from the other point of
view, why does one change the Schroedinger equation in the presence of
a magnetic "potential" (which does not affect what freshmen think of as
"potential energy")? These are not stupid questions.

Stephen Parrott

Stephen Parrott

unread,
Mar 14, 2010, 8:44:16 AM3/14/10
to
>> If we define the "usual" momentum operator
>>
>> P := -i d/dx, i.e., (Pf)(x) := -i df/dx ,
>>
>> then U := exp(-ibP) satisfies (1). However, P is not unique; if we
>> defined
>>
>> (Qf(x)) := -i df/dx + A(x)f(x), where A is any given function,
>>
>> then U := exp(-ibQ) would also satisfy (1), so Q would have just as
>> much claim to being the "official" momentum operator as P.
>
> Of course, you are right that operators P and Q are equivalent if (1)
> is taken as the defining property of the momentum operator. However,
> to the best of my knowledge, the momentum operator is defined as the
> generator of infinitesimal translations on wave functions:
>
> (Uf)(x) = f(x-b)
>
> Then the only possibility is U=exp(ibP), which is not equal to
> V=exp(ibQ). BTW, I'm probably mixing up signs somewhere. The operators
> exp(ibQ) correspond to a translation followed by multiplication by a
> certain position dependent phase. The operators U and V are physically
> distinguishable because we can compare their matrix elements <f|U|g>
> and <f|V|g>. These matrix elements should be experimentally accessible
> just as the matrix elements <f|exp(-itH)|g> of the unitary time
> evolution operator are.

Two days ago, I sent an extensive reply to this. Meanwhile, I noticed
an error in the reply. Though it didn't affect the conclusions, it
seemed important enough to correct, so I prepared a correction to append
to it when it appeared.

But it hasn't appeared. Twice before, recently, articles which I sent
*never* appeared, so I assume that there is a good chance that it will
never appear. If it does, I will append the correction. If it doesn't,
I will revise the correction and submit it as a new posting titled
something like "Cocycles in the definition of momentum", because it
contains some new material which I think interesting.

[Moderator's note: I am posting this message in its entirety, just to
make sure nothing is lost. In general, one needs to allow a couple of
days for a post to show up, due to the moderation process.
Occasionally, it can take longer. The moderators are only human. If
your messages have gotten through in the past, then they are probably
getting through now. If a message isn't posted, it is rejected, and a
valid (or easily decipherable) email address should get a rejection
method. If you post to a news server, you can use alt.test to make sure
your posts get that far. Moderated groups send the posts behind the
scenes via email to the moderation address. A news server can be
misconfigured in that respect. You can always send directly to the
moderation address, bypassing a news server completely. Of course, even
this won't work if none of your email gets out, but you can check this
in other ways. For up-to-date information with moderation addresses etc
see http://www.astro.multivax.de:8000/spr/spr.html. -P.H.]

Meanwhile, I'll just add a brief comment. It's not clear to me *how*
one can "compare their matrix elements <f|U|g> and <f|V|g>". These are
complex numbers, whose real and imaginary parts seem to have no physical
significance. Only |<f|U|g>|^2 seems to have experimental significance,
as a transition probability.

If we agree that only the modulus squared of the matrix elements is
experimentally accessible, then shouldn't the defining condition for an
operator T to implement spatial translation by *b* be

(1) |(Tf)(x)|^2 = |f(x-b)|^2

instead of your less general

(2) Tf(x) = f(x-b) ?

I agree that many, perhaps the vast majority, of elementary quantum
mechanics books do use condition (2), but that may be only for
simplicity. (Anyone who has taught will probably attest that hardly any
students will notice the difference!)

If we are brave enough to ignore the "conventional wisdom" of the
elementary texts and use (1) instead of (2), we end up with a momentum
operator P of the form I specified,

(3) Pf(x) = -idf/dx + A(x)f(x),

where A is an arbitrary function.

That is not obvious, but has to be worked out. It is not difficult, but
requires care. Indeed, the book of Asher Peres, "Quantum Theory:
Concept and Methods" attempts to work it out and gets the wrong answer,
in my opinion. The "cocycles" come in when one attempts to work it out.

Readers may enjoy working it out for themselves before I post my
"solution". I'll give a hint to get you started. It is immediate that
an operator satisfying (1) must be of the form

(4) (Tf)(x) = exp(ig(x,b))f(x-b) ,

where g is a real function which is yet to be determined. Since T
depends on b, there really should be a subscript b on T, but I didn't
want to complicate the ASCII in the above introduction. From here on it
may be helpful to introduce it, writing T_b for the "T" above.

If we translate our system by an amount *b*, and then again by *a*, for
consistency we should get the same result as translating it by *a+b*:

(5) T_a T_b = T_{a+b}

That will give you a so-called "cocycle equation" for the unknown
function g. Generally, it's hard to solve cocycle equations, but
solutions for this one can be written down by inspection. (Readers
unfamiliar with cocycles may find that the more challenging part of the
problem; those familiar with cocycles will immediately think of looking
for "coboundaries".)

As I write this, I realize that (5) is actually too special. Since
states (wave functions) are only determined "up to a phase", (5) really
should be

(6) T_{a+b} = c(a,b) T_a T_b,

where for fixed a,b, c(a,b) is a constant of modulus 1. Note that *c*
is *not* a function of the coordinate *x*, i.e., its not (T_{a+b}f)(x) =
c(a,b)(x) (T_a (T_b f))(x). That's because c(a,b)(x) (T_a (T_b f))(x)
is not the same *state* as c(a,b)(x) (T_a (T_b f))(x) when c(a,b)(x) is
a nonconstant function of x, though it does define the same probability
distribution.

Anyway, I used the simpler (5), and ended up with (3), P = -id/dx + A .
I doubt that (6) will lead to anything more general, but it might. It
would be neat if it did! Then we might even have publishable collective
research on s.p.r. (except that the referees would probably say that it
is "trivial" anyway, as it likely will look *in retrospect*).
(Assuming, that is, that *this* post doesn't get stuck in the s.p.r
"black hole".)

'

M

Juan R. Gonzàlez-Àlvarez

unread,
Mar 14, 2010, 8:44:16 AM3/14/10
to
"Juan R." Gonzàlez-Àlvarez wrote on Wed, 10 Mar 2010 13:44:22 +0100:

> Stephen Parrott wrote on Wed, 10 Mar 2010 08:13:50 +0000:

(...)

> It may be remarked that the quantum mechanical [x,p] = ihbar is the
> quantum version of the well-known classical result {x,p} = 0, where
> {,} are the Poisson brackets.

Sorry, my glaring mistake. Of course, the conmmutator [,] for
*classical* x and p is clearly zero, but the Poisson bracket for
classical x and p is

{x_i,p_j} = delta_ij [1]

> Claiming that the quantum mechanical relation is only valid for free
> particles and that we may modify it in presence of interactions
> would be such a nonsensical statement as claiming that {x,p} would
> be also modified, because in presence of interaction the p is not
> that of a free particle. Of course {x,p} still holds in presence of
> magnetic fields.

I.e. for a free non-relativistic classical particle p=mv and [1] holds

For a free non-relativistic classical particle p=mv+eA and [1] holds.

Juan R.

unread,
Mar 14, 2010, 2:47:04 PM3/14/10
to
Stephen Parrott wrote on Sun, 14 Mar 2010 12:44:16 +0000:

(...)

> where A is the magnetic potential ("vector potential") and use the
> electric potential for V. This results in a new Hamiltonian H'
> given by
>
> (8) H' = (-iD + A)^2/2m + V .
>
> This is the Hamiltonian used to derive the Aharonov-Bohm effect. If
> the canonical momentum P is given by (5) as suggested by Lagrangian
> theory, then (8) assumes the familiar form
>
> (9) H' = P^2/2m + V.

This is not the familiar Hamiltonian. The usual (and correct but valid
only for low velocities) Hamiltonian is

H = (P - eA)^2/2m + V

This has been remarked several times here. And also in a last message
by moderator Hendrik van Hees in the other thread "Two questions about
quantum momentum and position".

The Hamiltonian H is easily find in textbooks on quantum mechanics and
field theory. I already cited the relevant pages from the textbook
in Quantum Mechanics by Cohen-Tannoudji, Diu, and Lalöe.

Now I will add that H is (1.59) in Mandl and Shaw textbook
about QFT.

Jay R. Yablon

unread,
Mar 15, 2010, 4:01:01 PM3/15/10
to
"Stephen Parrott" <post...@email.toast.net> wrote in message
news:4B98A9B6...@email.toast.net...
I have been away for a few days, and have perused but not yet had
time to study the many posts in the two threads on this topic. So with
that caveat, and with Dr. Parrott's reference to the EM potential which
is sort of where I started this discussion (with V(q-hat) which is often
conflated with the time component A^0 = V of the vector potential A^u
whose reality independent of electric and magnetic fields is amply
confirmed by Aharonov-Bohm), let me assume the role of "naive student"
for which it seems I may be well-suited ;-) and try to hone in some more
on what bothered me about the momentum operator and got me asking
questions in the first place.

In the other thread, HvH remarked that "a lot of misconceptions are
introduced when one goes over to Dirac's final and most appropriate
representation free formulation in terms of the abstract Hilbert space
and an operator algebra representing the observables." I'd like to
discuss what it means to discuss the momentum operator "free of
representation," but to minimize misunderstanding or diverse
interpretations, do so not with words, but equations.

Let's start by positing a dimensionless, non-Hermitian creation operator
a-hat and annihilation operator a-hat* (* = Hermitian conjugate).

Now, I've seen in many places that the momentum and position operators
p-hat and x-hat are constructed from these according to:

p-hat = A (a-hat - a-hat*) (1)

x-hat = B (a-hat + a-hat*) (2)

All I will say about A and B for now is that A is a coefficient of
proportionality which must have dimensions of momentum
(mass*distance/time) and B is a coefficient of proportionality which
must have dimensions of distance.

It is also true that (1) and (2) commute with one another in all
circumstance according to:

[x-hat, p-hat] = h-bar (3)

All of the above, are what I think of as "representation-free"
relationships, in the same way, for example, that in Dirac's equation (i
gamma^u d_u -m) psi=0, one can put the gamma^u matrices into the Dirac
representation to study non-relativistic situations and the Weyl
representation to study extreme relativistic situations. There are two
different sets of gamma matrices, but they both can be used in the same
equation which does not change based on which set of matrices we use.

Now, let me embellish (1)-(3) a little more. Suppose I next tell
you that by virtue of (1)-(3) we also have:

[a-hat,a-hat*] = 1 (5)

It seems to me that this is free of representation; it is part of what
makes (3) work properly.

Suppose then I tell you that there is a number operator:

N = a-hat* a-hat (6)

Now, this operator is a matrix with 0,1,2,3,4...oo along its diagonal.
Because it is an explicit matrix, I might be inclined to say that (6) is
"representation-dependent," but what then is the representation?
Numbers are numbers. 0,1,2,3,4...oo do not become 3, 7, 6.25, 7001,
etc. in some other representation. So, is (6) in some representation
already, or is it independent of representation? That is not a
rhetorical question. It is just a question. I am trying to find out
when we transition from being representation-free, to having explicit
matrix operators in a specific representation? And, what that really
means? And, what causes that transition to occur?

Suppose I now tell you that (Sakurai (2.3.23)):

<n'|a|n> = sqrt(n) delta_n',n-1 (7)

<n'|a*|n> = sqrt(n+1) delta_n',n-1 (8)

Have I now finally chosen a "number" representation? Did I make this
choice already in (6), or did I only do so in (7) and (8) once I
bra-keted a-hat and a-hat* with <n'| and |n>? Given (6), can (7) and
(8) even be any different than what is shown above? Again, I am not
being rhetorical, I am asking a question.

Now, let's say I tell you for the coefficients in (1) and (2) that:

A = -i sqrt(h-bar m omega / 2) (9)

B = sqrt (h-bar/2m omega) (10)

Here, I don't think I have chosen a representation, but I for sure think
I have chosen a Hamiltonian and an interaction operator V(x-hat) for a
harmonic oscillator, namely that in (2.3.1) of Sakurai, because omega
only comes from that particular choice of V(x-hat). It is a "bicycle
for a fish" to use Igor's colorful characterization.

So, I started out this thread trying to sort out which of the
foregoing relationships are "universal" or "independent of
representation" (and I don't know if these two quoted expressions means
the same thing), which are dependent on a particular choice of
representation, and which are dependent not only on a representation
choice, but on even further choices such as a choice of Hamiltonian
including a choice of V(x-hat). And, I stated out trying in particular
to understand what A and B are in (1) and (2) and whether these can only
be determined once a Hamiltonian and its V(x-hat) has been chosen, or
whether these is an expression for each of A and B that is independent
of representation and independent of Hamiltonian and independent of
anything and everything. And, trying to understand, if the latter, then
what are A and B generally, without the "fish bicycles" in (9) and (10)?
And, in general, what determines A and B, and therefore, the explicit
operator matrix for both the momentum and position? And, how
independent are A and B of certain specific choices that we make of
representation and of Hamiltonian?

Finally, perhaps I am confused, but I take "choice of
representation" in the context of Sakurai section 1 to mean using an
operator on particular bras and kets. But, I also keep also in mind
that eigenvectors psi and eigenvalues lambda are obtained from a given
operator O, and not the other way around, which is to say the
eigenvectors of an operator O are a "blank slate" of sorts which are
filled in, up to an overall coefficient, by finding the lambda and psi
*for which* (O-lambda)psi=0. So I still have a little confusion over
how the selection of bras and kets affects anything of substance, other
than the overall coefficient. And, of course, A and B are nothing more
than overall coefficients.

Jay



Hendrik van Hees

unread,
Mar 17, 2010, 2:46:51 AM3/17/10
to
Jay R. Yablon wrote:

> In the other thread, HvH remarked that "a lot of misconceptions
> are
> introduced when one goes over to Dirac's final and most appropriate
> representation free formulation in terms of the abstract Hilbert space
> and an operator algebra representing the observables." I'd like to
> discuss what it means to discuss the momentum operator "free of
> representation," but to minimize misunderstanding or diverse
> interpretations, do so not with words, but equations.

This is a total misunderstanding of what I tried to say. The opposite is
true! I think the clearest way to formulate quantum theory is in fact
Diracs representation-free formulation.

Confusing is what is written in this thread not what's in Dirac's famous
textbook which is one of the best books on quantum theory ever written!

The only way to clarify questions like the one discussed here, namely
how to obtain the operator algebra for a given system, is to use
symmetry principles. Canonical quantization, e.g., works only in very
few examples like non-relativistic QT for a point particle and some
simple cases of relativistic QFT.

The symmetry analysis in the case of the Galileo Group, which can be
found in the excellent textbook by Ballentine for various cases, clearly
leads to the definition the momentum operator which corresponds to the
canonical momentum in classical mechanics, not mechanical momentum, and
thus for a particle in an external electromagnetic field with vector
potential A and scalar Potential phi reads

H=(p-q A)^2/(2m)+q phi

where A is a three-vector field and phi a scalar field.

The velocity is then defined as the covariant derivative of the position
operator

v=1/i [x,H]=(p-q A)/m

and thus in the correspondence principle, understood in this unique
group-theoretical way, indeed the momentum operator is revealed to
correspond to canonical momentum and not to m v!

How important this careful analysis of the observable algebra is,
reveals, e.g., the case of massless relativistic particles with spin
>=1. What comes out there is that there is no position operator in the
usual sense. A photon by itself is not localizable, and a physically
sensible position measurement can only be defined in the extended sense
of positive operator valued measures (POVMs). See Arnold's nice FAQ
entry on this isue:

http://www.mat.univie.ac.at/~neum/physfaq/topics/S2g

--
Hendrik van Hees
Institut f�r Theoretische Physik
Justus-Liebig-Universit�t Gie�en
http://theorie.physik.uni-giessen.de/~hees/

Stephen Parrott

unread,
Mar 17, 2010, 4:39:10 AM3/17/10
to

----------------Moderator's note---------------------------------------

(8) is not "another convention", but simply wrong, because you defined
the Hamiltonian by (6). Then necessarily in the position representation
(the usual name of it in any useful textbook I know, by the way) it must
read as you say in (7). Also (8) is uniquely determined by your algebra
and thus must read H=(2N+1)/2.

This whole thing is well represented in the majority of textbooks (just
look for the simple harmonic oscillator). There are very many textbooks
on elementary quantum theory and as the 2nd law of thermodynamics tells
us there are many bad ones, but there are also plenty very good texts
and not too much need for another one :-).

HvH.
-----------------------------------------------------------------------


I thought of another way of answering your question (to the extent that
I understand it) which is closer to the way things are typically done in
physics texts. It should be read along with my previous answer.


In algebra, one manipulates symbols according to certain rules, without
thinking about what the symbols might represent. For example, in high
school algebra, one can write x+y=y+x without thinking about whether x
represents the number 3, or the number 5, etc.


The same is true for operator algebra, except that the rules of
manipulation are rules which operators would satisfy. Our toy operator
algebra will be constructed from symbols x, p, and a. In your
presentation, a and a* were considered the primary symbols in terms of
which x and p were constructed, but it is more usual to consider x and p
as primary, so I'll do it that way.


If these symbols did represent operators, we could take adjoints of
those operators, so we also have symbols x*, p*, a*. We will eventually
want x and p to represent Hermitian operators, so two of our algebraic
rules will be x=x* and p=p*. Another algebraic rule will be


(1) [p,x] := px - xp = -i, where "i" is the imaginary unit.


Then we define a by


(2) a := (x + ip)/2 , so that

a* = (x-ip)/2 .


If we wanted to start with "a" as a primary symbol, we could invert this
relation and then define "x" and "p" by


(3) x = a + a*

p = -i(a - a*)

It follows that


(4) [a,a*] := aa*-a*a = 1.


Next define


(5) n := a*a . (This "n" is an abstract symbol, which does not stand
for a number.)


A "representation" of this algebra is a way to assign specific operators
(on some Hilbert space) to these symbols so that the above rules hold,
along with the usual algebraic rules for operators. Since I want to
distinguish between the algebraic symbols a, x, p, etc. and the
corresponding operators in a representation, I will use capital letters
for the operators. For example, I will assign operator X to symbol x,
operator A to symbol a, operator N to symbol n.

My previous answer spoke of a "usual" representation. Actually, there
are two "usual" representations which look quite different, though they
turn out to be essentially the "same":


(1) This representation seems to have no standard name, so I will call
it the "Position Representation". Its Hilbert space is taken to be the
set of all square-integrable complex-valued functions f on the real
line, and X and P are defined by: Xf(x) := xf(x) and Pf(x) := -idf/dx.


(2) The "Number Representation".


We may take for its Hilbert space any separable,
infinite-dimensional Hilbert space. We could use the Hilbert space of
square-integrable functions discussed above, but it's easier to think
more abstractly. Such a Hilbert space can be constructed by taking a
countably infinite set of symbols to represent an orthonormal basis, and
using Hilbert space algebra to represent arbitrary vectors as infinite
linear combinations of them. For the orthonormal basis, I will use the
symbols

|0>, |1>, |2>, ... , |k>, ... ,


indexed by non-negative integers k. To define an operator A, it is
sufficient to define A|k> for all k, because the rules of operator and
Hilbert space algebra then define A on all other vectors. (Technically,
A has to be "bounded" for this to be true, but I'm sliding over
technicalities to rapidly present the big picture.) In the "Number
Representation, one defines A|0> := 0 (note: |0> is not the same as the
"zero vector" 0, one reason why mathematicians dislike this common "ket"
notation), and for k > 0,


A|k> := c_k |k-1> ,


where the complex coefficients C_k (which can be taken to be real, it
turns out) are chosen to make the commutation relation (4), [A,A*] = 1,
hold. It is a routine exercise to find such c_k, but their precise
values are not important to us. It is then routine to show that


N|k> := A*A|k> = k|k> for all non-negative integers k.


That is, with respect to the orthonormal basis {|k>}, N has a diagonal
matrix with diagonal elements the non-negative integers 0, 1, 2, ... .


Because the matrix of N with respect to this basis so simple, this
Number Representation is convenient for problems for which N is a
central figure. But the matrices of the operators X and P with respect
to this basis are complicated, much more complicated than the
definitions of X and P in the "Continuous Representation".


Conversely, if we use the "Position Representation" to obtain simple
forms for X and P, then N will look complicated. It is an enlightening
but nontrivial exercise to find an explicit form for the number basis
|k> as square-integral functions on the line; this exercise is solved
in many beginning quantum texts. It turns out that the so-called
"ground state" |0> is a Gaussian in the "Position Representation".


The algebra element


(6) h := (p^2 + x^2)/2


is considered to be the Hamiltonian for a particular "simple harmonic
oscillator". Usually texts throw in a positive parameter w^2 and some
hbars, writing for example, h := (p^2 + w^2 x^2)/2, but this just
complicates the algebra without adding any essential conceptual element,
so I'll use (6). In the "Position Representation", the Hamiltonian is
represented by the operator


(7) H_{continuous} = (- d^2/dx^2 + X^2)/2 .


In the "Number Representation", h is represented by a much simpler
operator,


(8) H_{number} = 2N + 1


[Note to certain readers: Please don't flame me for writing (8) instead
of the H = N + 1/2 usually seen in textbooks. I probably started with
slightly different conventions.]


Since the "Position Representation" looks so different from the "Number
Representation", one might be tempted to think that they could lead to
different physics. But there is a nontrivial theorem that states that
they are related by a unitary equivalence, so that mathematicians
consider them as essentially the "same". That is, they are the "same"
except for the particular names that we give to the Hilbert space
elements. Unless the names can affect the physics, we will get the same
physics from either, so we may use whichever representation is more
convenient.


More precisely, the theorem states that any two so-called "irreducible"
representations of this particular algebra are related by a unitary
transformation. It is a good exercise to show that both the Position
and Number representations are irreducible.


My previous post pointed out that there are representations that are not
equivalent to these, but the example I gave was rather trivial. It
consisted of simply "doubling" the Number Representation. That is also
called the "spin 1/2" representation, which is not irreducible for the
algebra we are considering, but becomes irreducible if you add some
additional algebra elements representing "spin".


All of these things are treated systematically in mathematics texts, but
usually in a more haphazard way in physics texts. The mathematics texts
take a long time to read and will likely contain more material than a
physicist may need. Unfortunately, it is not easy to learn this stuff
from books alone.


Why doesn't someone write a good book? That is a natural question. I
think it is because writing a good book requires so much dedication and
time, and is unlikely to be sufficiently rewarded.


The best I have seen from the standpoint of fundamental understanding
are the books of Mackey, but they assume a mathematical level higher
than typically possessed by physics graduate students. Also, I think
that Mackey was mainly self-taught in physics, so that there are
probably a lot of gaps in his books from a physics standpoint.


Stephen Parrott

Igor Khavkine

unread,
Mar 17, 2010, 5:45:00 AM3/17/10
to
On Mar 17, 9:39 am, Stephen Parrott <postn...@email.toast.net> wrote:

> (1)     [p,x] := px - xp = -i, where "i" is the imaginary unit.

> (2)     a := (x + ip)/2 , so that
>
>          a* = (x-ip)/2 .

> It follows that


>
> (4)     [a,a*] := aa*-a*a = 1.

> (5)     n := a*a .   (This "n" is an abstract symbol, which does not stand


> for a number.)
>
> A "representation" of this algebra is a way to assign specific operators
> (on some Hilbert space) to these symbols so that the above rules hold,
> along with the usual algebraic rules for operators.  Since I want to
> distinguish between the algebraic symbols a, x, p, etc. and the
> corresponding operators in a representation, I will use capital letters
> for the operators.  For example, I will assign operator X to symbol x,
> operator A to symbol a, operator N to symbol n.

> (6)      h := (p^2 + x^2)/2

> (8)      H_{number} = 2N + 1


>
> [Note to certain readers:  Please don't flame me for writing (8) instead
> of the H = N + 1/2 usually seen in textbooks. I probably started with
> slightly different conventions.]

To temper the flames, if any, I just want to point out where your
conventions differ from the usual ones.

According to (1) and (2), (4) should actually be [a,a*]=1/2, which
means that your versions of (1) and (2) are off by a factor of sqrt(2)
from the usual ones. Restoring this factor in the appropriate places
brings your formulas in agreement with the standard ones. This is an
innocuous mistake and nothing to flame about.

Igor

P.S.: Stephen, apologies for the delay in replying to your recent
longer post. It may still take a few days before I have time to
compose a reply.

Stephen Parrott

unread,
Mar 17, 2010, 6:18:30 PM3/17/10
to
Not all of your questions are clear to me, but one seems clear enough to
answer, and maybe the answer will help with the rest. If it doesn't, no
harm is done.


>
> Let's start by positing a dimensionless, non-Hermitian creation operator
> a-hat and annihilation operator a-hat* (* = Hermitian conjugate).
>
> Now, I've seen in many places that the momentum and position operators
> p-hat and x-hat are constructed from these according to:
>
> p-hat = A (a-hat - a-hat*) (1)
>
> x-hat = B (a-hat + a-hat*) (2)

> All I will say about A and B for now is that A is a coefficient of
> proportionality which must have dimensions of momentum
> (mass*distance/time) and B is a coefficient of proportionality which
> must have dimensions of distance.
>
> It is also true that (1) and (2) commute with one another in all
> circumstance according to:
>
> [x-hat, p-hat] = h-bar (3)

You're getting ahead of yourself.
To justify this, you have to specify the commutation relations [a,a*]=
*something* between a and a*, as you did in (5) below. Also, the right
side of (3) must be anti-Hermitian (as is the commutator of any two
Hermitian operators, so you probably want to insert an "i" there.

> All of the above, are what I think of as "representation-free"

> relationships, ...

Right.

>
> Now, let me embellish (1)-(3) a little more. Suppose I next tell
> you that by virtue of (1)-(3) we also have:
>
> [a-hat,a-hat*] = 1 (5)
>
> It seems to me that this is free of representation; it is part of what
> makes (3) work properly.
>
> Suppose then I tell you that there is a number operator:
>
> N = a-hat* a-hat (6)
>
> Now, this operator is a matrix with 0,1,2,3,4...oo along its diagonal.
> Because it is an explicit matrix, I might be inclined to say that (6) is
> "representation-dependent," but what then is the representation?

That's essentially right, too, except for the statement: "Now, this
operator is a matrix with 0,1,2,3,4...oo along its diagonal." That's an
unjustified statement which could be false. It truth or falsity depends
on how you "represent" a-hat as an operator on that Hilbert space.

To "represent" a-hat means to choose a Hilbert space and an operator A
on that space satisfying [A,A*]=1. There is a "usual" way of doing that
for which "this operator is a matrix with 0,1,2,3,4...oo along its
diagonal". And there are other ways for which that statement is false.

For example, if the "usual" representation uses a Hilbert space H, then
we can make a new representation on the direct sum of H with itself, in
which the "new" A is the direct sum of the "old" A with itself. Then
the "new" N := (New A)*(New A) will have a matrix (with respect to an
appropriate basis) which is diag{0,0,1,1,2,2,3,3, ...}. This "new" N is
not unitarily equivalent to the "old" N. It could conceivably lead to
different physics.

There is a moral here. Your question was mainly about what people mean
when they talk about a "representation" of an operator algebra. If you
realized that that was what you were actually asking, chances are that
you could answer it yourself by looking up the definition in some
mathematics book.

You might not find it in many physics books. Physics books often don't
pay proper attention to definitions, instead merely throwing around
jargon like "representation" in the expectation that readers will pick
up the meaning by hearing it enough in different contexts, the way that
children learn a language. Most people learn this kind of thing in
courses where they can get additional clues by asking questions and
discussions with other students. Learning only from books, you are
working under a handicap.

If there is a decent university near you, you might learn faster by
auditing some courses. You may theoretically have to pay a fee for
that, but most instructors will probably let anyone sit in their classes
for free so long as they don't create additional work. I have done it
many times and even graded exams for informal "auditors". Instructors
will usually go out of their way to help anyone who sincerely wants to
learn.

Stephen Parrott

Stephen Parrott

unread,
Mar 18, 2010, 4:38:58 AM3/18/10
to
Igor Khavkine wrote:

> On Mar 17, 9:39 am, Stephen Parrott <postn...@email.toast.net> wrote:
>
>> (1) [p,x] := px - xp = -i, where "i" is the imaginary unit.
>
>> (2) a := (x + ip)/2 , so that
>>
>> a* = (x-ip)/2 .
>
>> It follows that
>>
>> (4) [a,a*] := aa*-a*a = 1.
>
>> (5) n := a*a . (This "n" is an abstract symbol, which does not stand
>> for a number.)
>>
>> A "representation" of this algebra is a way to assign specific operators
>> (on some Hilbert space) to these symbols so that the above rules hold,
>> along with the usual algebraic rules for operators. Since I want to
>> distinguish between the algebraic symbols a, x, p, etc. and the
>> corresponding operators in a representation, I will use capital letters
>> for the operators. For example, I will assign operator X to symbol x,
>> operator A to symbol a, operator N to symbol n.
>
>> (6) h := (p^2 + x^2)/2
>
>> (8) H_{number} = 2N + 1
>>
>> [Note to certain readers: Please don't flame me for writing (8) instead
>> of the H = N + 1/2 usually seen in textbooks. I probably started with
>> slightly different conventions.]
>
> To temper the flames, if any, I just want to point out where your
> conventions differ from the usual ones.
>
> According to (1) and (2), (4) should actually be [a,a*]=1/2, which
> means that your versions of (1) and (2) are off by a factor of sqrt(2)
> from the usual ones. Restoring this factor in the appropriate places
> brings your formulas in agreement with the standard ones. This is an
> innocuous mistake and nothing to flame about.
>
> Igor
>

If anybody cares, I did, in fact, drop a factor of 1/2 in (4), which
should be [a,a*] = 1/2 as Igor says. I haven't gone through the rest to
see how it should be changed, and I don't intend to. There are more
interesting and important things to think about.

Stephen Parrott

Jay R. Yablon

unread,
Mar 18, 2010, 5:21:11 AM3/18/10
to
"Stephen Parrott" <post...@email.toast.net> wrote in message
news:4BA03155...@email.toast.net...

> Not all of your questions are clear to me, but one seems clear enough
> to
> answer, and maybe the answer will help with the rest. If it doesn't,
> no
> harm is done.
>
Thanks. My first reply was to your second post. This second reply is
to your first post. Such are the vagaries of the SPR moderation, as
Jonathan just pointed out.

>
>
>
>>
>> Let's start by positing a dimensionless, non-Hermitian creation
>> operator
>> a-hat and annihilation operator a-hat* (* = Hermitian conjugate).
>>
>> Now, I've seen in many places that the momentum and position
>> operators
>> p-hat and x-hat are constructed from these according to:
>>
>> p-hat = A (a-hat - a-hat*) (1)
>>
>> x-hat = B (a-hat + a-hat*) (2)
>
>> All I will say about A and B for now is that A is a coefficient of
>> proportionality which must have dimensions of momentum
>> (mass*distance/time) and B is a coefficient of proportionality which
>> must have dimensions of distance.
>>
>> It is also true that (1) and (2) commute with one another in all
>> circumstance according to:
>>
>> [x-hat, p-hat] = h-bar (3)
>
> You're getting ahead of yourself.
> To justify this, you have to specify the commutation relations [a,a*]=
> *something* between a and a*, as you did in (5) below.

That is just a question of ordering of presentation, but I agree that
(5) logically precedes (3).

>Also, the
> right
> side of (3) must be anti-Hermitian (as is the commutator of any two
> Hermitian operators, so you probably want to insert an "i" there.

A typo.

>> All of the above, are what I think of as "representation-free"
>> relationships, ...
>
> Right.

Good.

I guess based on my experience with Lie groups (such as 2x2 = 3+1 for
SU(2), to pick the simplest example), I would have to say that
diag{0,1,2,3...} is the *irreducible* representation of the number
operator?

> For example, if the "usual" representation uses a Hilbert space H,

"usual" = "irreducible"?

> then
> we can make a new representation on the direct sum of H with itself,
> in
> which the "new" A is the direct sum of the "old" A with itself. Then
> the "new" N := (New A)*(New A) will have a matrix (with respect to an
> appropriate basis) which is diag{0,0,1,1,2,2,3,3, ...}. This "new" N
> is
> not unitarily equivalent to the "old" N. It could conceivably lead to
> different physics.

Which is a "composite" representation? Just as 2x2 = 3+1 for SU(2) give
you spin 1 and spin 0 states compose out of spin 1/2 states?

> There is a moral here. Your question was mainly about what people
> mean
> when they talk about a "representation" of an operator algebra. If
> you
> realized that that was what you were actually asking, chances are that
> you could answer it yourself by looking up the definition in some
> mathematics book.

Yes and no. I think my question about representations came up along the
way in my effort to answer a deeper underlying question. As I thought
about my prior study of Lie groups, I started answering my questions
about "representations" for myself.

My real question, which I tried to highlight in my first reply which was
to your second reply, and which I will summarize here in my second reply
which is to your first reply ;-) is essentially this:

If we use the creation and annihilation operators to construct position
and momentum operators:

x = A (a + a*) (1)
p = -i B (a - a*)

A must have length dimension and B must have momentum dimension. My
bottom line question is this: What, in general, what determines A and B?
The representation? The Hamiltonian? Something else? What? My other
reply goes into a bit more detail, but I do not want to be redundant so
I just refer to that by reference, with the above summary. I want to
know as a general rule, what determines the A and B coefficients in the
above.

As a separate, though intertwined question, I would like to understand
how general or how limited are the specific choices:

<n'|a|n> = sqrt(n) delta_n',n-1 (2)


<n'|a*|n> = sqrt(n+1) delta_n',n-1

for the creation and annihilation operators. That is, on what
assumptions do (2) above depend? I presently think of these as the
non-relativistic creation and annihilation operators in an irreducible
number representation. Am I using stating this correctly, or is there a
better way to state / understand this?

> You might not find it in many physics books. Physics books often
> don't
> pay proper attention to definitions, instead merely throwing around
> jargon like "representation" in the expectation that readers will pick
> up the meaning by hearing it enough in different contexts, the way
> that
> children learn a language. Most people learn this kind of thing in
> courses where they can get additional clues by asking questions and
> discussions with other students. Learning only from books, you are
> working under a handicap.

Hendrik did suggest Dirac's book, and since I have not for a moment
regretted taking his advice to obtain Sakurai, that may be next up on my
Amazon purchases.

> If there is a decent university near you, you might learn faster by
> auditing some courses. You may theoretically have to pay a fee for
> that, but most instructors will probably let anyone sit in their
> classes
> for free so long as they don't create additional work. I have done it
> many times and even graded exams for informal "auditors". Instructors
> will usually go out of their way to help anyone who sincerely wants to
> learn.

Oh, Stephen,

There is nothing I can think of, working as a 56 year old patent
attorney, that I would rather do than sit through some good physics and
/ or physical mathematics courses, but the practicalities of my life
unfortunately do not permit that at this time. So, I do the next best
thing in the 2-3 hours per day on average that I can devote to physics:
I read, I calculate out things that seem interesting, and I ask
questions and float trial balloons of understanding (or
misunderstanding -;) on Usenet. And, I audit what both of my kids are
doing in their studies.

I guess I am not quite in the boondocks of Nevada as you are, and back
in the 1980s, I did audit Hans Ohanian's class on General Relativity and
Gravitation at RPI. I decided that GR should be my *first* physics
class (while those who do formal physics study at a university such as
my son who received his physics degree from Cornell in 2007 almost
always do relativity toward the end of their studies), because the
geometric beauty of gravitational theory IMHO is the "gold standard" of
physical theory. It was my belief that knowing GR *before* studying
that which most other people study well before GR would give me an
internal mental "aesthetic" standard for not only studying, *but
critically evaluating*, other theories, such as particle physics, Yang
Mills theory, quantum field theory, and quantum mechanics, all of which
I subsequently studied (more or less in that order) and continue to
study. Thus, I can look at things such as the canonical commutation
relationships [x,p]=i hbar and have the reaction that "that is some
canonical rule that I am told to use because it happens to work but
nobody really understands *why* nature gives us such a rule." Then,
even when I use that rule to "shut up and calculate," I know that nobody
really understands quantum (as Feynman said) in the same way that GR
just makes perfect intuitive sense. The physics which seem most
aesthetically natural, are things such as "the particle moves this way
because that is the natural way to move through the geometry." Or, the
nineteenth century version of this, which is that physical processes
occur in the "laziest" way possible, so as to minimize action. The zen
that underlies this, is to be able to say that nature does what it does
because it has to do so, yet not have that be a tautology. When I study
quantum, I look hard for whatever principles I can find which have the
same feel of naturalness to them. One of the reasons I studied path
integrals before canonical formalism (again reversing the more
conventional order of study), is because path integrals have a natural
(though as I have been told here, not completely rigorous) connection to
the classical action first pointed out by Dirac and then given calculus
limit expression by Feynman, and so simply place any classical theory
which can be represented through a Lagrangian, into a larger context
with a classical limit. Then, the action principle itself, which is a
close cousin of the beauty of geodesic motion through geometry, ends up
being subsumed as well into this larger context. Unfortunately, those
sorts of simple, intuitive principles often get hidden behind
mathematical apparatus, and one has to be very careful understanding and
balancing the symbiotic relationship between physics and mathematics,
and never lose sight of simple (though hopefully not not simplistic)
physical understanding. One cannot do physics without math, but neither
should physics get lost in the math if what gets sacrificed is the
intuitive understanding of nature. In this enterprise, we are, at
bottom, trying to read "the mind of God," which is what makes physics
such a beautiful and passionate pursuit for those of us who have "the
bug," but we use mathematics and experimental apparatus to tell us if we
really are reading that mind correctly, or just making things up in our
own minds, which is always a danger even for the best and brightest of
us.

But I have digressed from replying to your much-appreciated practical
advice. While taking classes is not a practical possibility at the
moment, (and given that my daughter is herself an undergraduate at RPI,
my auditing classes there at this time would be even more mortifying for
her than when I volunteered a few years back to chaperone a high school
dance -- "you mean that old guy sitting in on our class is your dad?"
;-) I am not without resources. Aside from all these Usenet discussions
which are extremely valuable and for which I deeply appreciate the
insights I am able to gain by reading what others here have to say, my
son, who is now pursuing a doctoral degree at NorthWestern in quantum
optics (and involved with his advisor in a project using negative
dispersion optical gyroscopes to detect gravity waves) is an invaluable
resource. He walks me through the standard pedagogy that he is
experiencing as a university grad student, and I clue him in about the
stuff that they don't teach in the early classes but save for the end,
like GR and Yang Mills. And, my daughter often has me review her
homework in physical chemistry, so between my two like-minded kids and
all the discussion here, I have kind of scratched together the most
practical system of "auditing" that I can muster at this time of my
life. Now, when I hopefully retire in a few years as you have already
done . . . nearby physics professors, watch out! ;-)

Thanks,

Jay
>
> Stephen Parrott

Jay R. Yablon

unread,
Mar 22, 2010, 11:23:23 AM3/22/10
to
[Note: This is a second attempt to post this. I first sent this early
Wednesday morning but it has not yet posted. Jay.]

"Stephen Parrott" <post...@email.toast.net> wrote in message

news:4BA0722E...@email.toast.net...

Let me stop here. Because this allows me to ask my underlying question
most directly.

Let us indeed us "a" as the primary symbol underlying our operator
algebra. And let us take a to be physically dimensionless. As a
consequence of the definitions (3), x and p are also dimensionless.

Now let us say we want x to have position dimension and p to have
momentum dimension. So, we need coefficients A and B for each of (3)
such that:

(3a) x = A (a + a*)

p = -i B (a - a*)

A must have length dimension and B must have momentum dimension. One
*example* of A and B comes from the Hamiltonian H-hat=p-hat^2/2m +
V(x-hat) for a harmonic oscillator potential V(x-hat)=m omega^2 x-hat^2
/ 2, and is given by (within Sakurai (2.3.25))

A = sqrt (h-bar/2m omega) (9)

B = sqrt(h-bar m omega / 2) (10)

Here is the underlying question(s) I had when I started the first
thread:

Whether the answer is "the representation" or "the Hamiltonian" or
something else: ***In general, what determines A and B?***

Further, can you (anyone) write some generalized expression for each of
A and B that is *independent* of the choice of V(x-hat)? If you cannot,
then must I conclude that the *physical* momentum and position operators
are dependent on the interaction term V(x-hat)?

And moving beyond my original question, is the distinction that Igor
made elsewhere:

"canonical momentum = mechanical momentum + interaction momentum (*)"

part of this answer, that is, is one of these types of momentum
*dependent* on the interaction term V(x-hat) in the Hamiltonian and
another *not dependent* thereon? If so, what depends upon what?

Jay


[[Mod. note -- 190 unnecessarily-quoted lines snipped here. As another
moderator recently said in another thread in this newsgroup, please help
to keep the newsgroup readable by quoting (only) *judiciously*! -- jt]]

Igor Khavkine

unread,
Mar 22, 2010, 1:23:26 PM3/22/10
to
On Mar 22, 4:23 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
> "Stephen Parrott" <postn...@email.toast.net> wrote in message
> news:4BA0722E...@email.toast.net...

> > (3)     x = a + a*


> >
> > p = -i(a - a*)
>
> Let me stop here.  Because this allows me to ask my underlying question
> most directly.
>
> Let us indeed us "a" as the primary symbol underlying our operator
> algebra.  And let us take a to be physically dimensionless.  As a
> consequence of the definitions (3), x and p are also dimensionless.
>
> Now let us say we want x to have position dimension and p to have
> momentum dimension.  So, we need coefficients A and B for each of (3)
> such that:
>
> (3a)     x = A (a + a*)
>
>  p = -i B (a - a*)
>
> A must have length dimension and B must have momentum dimension.  One
> *example* of A and B comes from the Hamiltonian H-hat=p-hat^2/2m +
> V(x-hat) for a harmonic oscillator potential V(x-hat)=m omega^2 x-hat^2
> / 2, and is given by (within Sakurai (2.3.25))
>
> A = sqrt (h-bar/2m omega)    (9)
>
> B = sqrt(h-bar m omega / 2)   (10)
>
> Here is the underlying question(s) I had when I started the first
> thread:
>
> Whether the answer is "the representation" or "the Hamiltonian" or
> something else:  ***In general, what determines A and B?***

Nothing determines A and B except an arbitrary choice, which can be
made or changed at any time.

Taken abstractly, the operators a and a* do not correspond to any
observables of a physical system. Equally abstractly, there is a 1-
parameter family of real coefficients A and B which combine a and a*
into x and p, as in your (3a). These operators are still mathematical
abstractions not related to any physical system. Only when some choice
of x and p (and hence A and B) is arbitrarily made to correspond to
operationally defined observables in some particular physical system,
do a and a* acquire by extension some connection to physics. All of
this is done without reference to whatever interaction this physical
system exhibits.

> Further, can you (anyone) write some generalized expression for each of
> A and B that is *independent* of the choice of V(x-hat)?

Any A and B will do. None of them depend on V(x-hat).

>  If you cannot,
> then must I conclude that the *physical* momentum and position operators
> are dependent on the interaction term V(x-hat)?

No you must not and cannot. If you wish to make such a statement, you
must prove it (or find a verifiable proof of it somewhere). In
mathematics, absence of proof does not constitute disproof. Logic is a
hard mistress.

Igor

Arnold Neumaier

unread,
Mar 22, 2010, 2:02:18 PM3/22/10
to
Igor Khavkine wrote:
> On Mar 22, 4:23 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>> "Stephen Parrott" <postn...@email.toast.net> wrote in message
>> news:4BA0722E...@email.toast.net...
>
>>> (3) x = a + a*
>>>
>>> p = -i(a - a*)
>> Let me stop here. Because this allows me to ask my underlying question
>> most directly.
>>
>> Let us indeed us "a" as the primary symbol underlying our operator
>> algebra. And let us take a to be physically dimensionless. As a
>> consequence of the definitions (3), x and p are also dimensionless.
>>
>> Now let us say we want x to have position dimension and p to have
>> momentum dimension. So, we need coefficients A and B for each of (3)
>> such that:
>>
>> (3a) x = A (a + a*)
>>
>> p = -i B (a - a*)
>>
>> A must have length dimension and B must have momentum dimension. One
>> *example* of A and B comes from the Hamiltonian H-hat=p-hat^2/2m +
>> V(x-hat) for a harmonic oscillator potential V(x-hat)=m omega^2 x-hat^2
>> / 2, and is given by (within Sakurai (2.3.25))
>>
>> A = sqrt (h-bar/2m omega) (9)
>>
>> B = sqrt(h-bar m omega / 2) (10)
>>
>
> Any A and B will do. None of them depend on V(x-hat).

But the product of A and B must have a fixed value, to get the CCR.

Jay R. Yablon

unread,
Mar 22, 2010, 3:10:19 PM3/22/10
to

"Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
news:4BA7B03...@univie.ac.at...

Yes, exactly. And that product must be h-bar/2.

It seems to me that this is why Igor referred to A and B as a
"1-parameter family of real coefficients." While A and B, a priori, are
two parameters, the constraint AB=h-bar/2 reduces the freedom of choice
for A and B down to but a single parameter. And if I read him
correctly, Igor's use of the word "family" suggests that, e.g., (9) and
(10) are but a single example of a whole family of possible A and B
choices constrained such that AB=h-bar/2.

Also key is "Only when some choice of x and p (and hence A and B) is

arbitrarily made to correspond to operationally defined observables in
some particular physical system, do a and a* acquire by extension some

connection to physics." This tells me that one must *begin* with
observables and an understanding of what it means operationally to
observe them, which observables must then be represented as eigenvalues
of some operator(a) related in a definite specified way to operators x,
p, a and a*, in some chosen representation of these operators.

I am starting to think that part of my own misconception at the outset
of this thread, was in asking what is *the* momentum operator, as if A
and B individually must be some definite parameters in all cases, beyond
the fact that their product must be h-bar/2, and beyond the fact that
the momentum operator must generate translations as in Sakurai 1.6.
Above, (9) and (10) used in (3a) yield "a" momentum operator and "a"
position operator, which obey "*the* CCR." But, there is no such thing,
objectively speaking, as "the" position and momentum operators.

If what I have just stated if accurate, then Igor's last post answered
my original question. Along the way, I have been motivated to go back
through sections 1 and 2 of Sakurai to reach a deeper understanding than
I had the first time I studied them. Maybe not yet perfect, but better
the second time around. ;-)

Jay

Jay R. Yablon

unread,
Mar 24, 2010, 11:22:19 PM3/24/10
to
"Igor Khavkine" <igo...@gmail.com> wrote in message
news:4b10bb3e-34eb-45d1...@j21g2000yqh.googlegroups.com...

I think I finally got it! Let me refer to Sakurai 1.6:

We define the infinitesimal translation operator as:

T(dx')|x'> = |x' + dx'> (1.6.12)

After the expansion (1.6.13) and (1.6.14) we impose four properties
developed in (1.6.15)-(1.6.19): probability conservation which requires
unitarity; linear composition of successive translations; identity under
inversion; reduction to identity as dx'->0. These are all satisfied by:

T(dx') = 1-iKdx' (1.6.20)

with K Hermitian. Once (1.6.20) is shown to have these four sensible
properties, one then shows that in relation to the x operator of x|x'> =
x'|x>, one must have the commutator:

[x,K] = i (1.6.27 for a single dimension)

Then, we just use dimensional analysis to find that p = h-bar K and get
to the canonical commutation relationship

[x,p] = i h-bar (1.6.33)

In sum: If

T(dx') = 1-iKdx = 1 - (i/h-bar) p dx (1)

is the generator of translations

T(dx')|x'> = |x' + dx'> (1.6.12)

as it must be based on the four sensible requirements from
(1.6.15)-(1.6.19), then we must necessarily have:

[x,p] = i h-bar (1.6.33)

This "forward" development I have understood for awhile. What now
occurs to me is that the logic also works in the "reverse" direction
also:

That is, if the CCR (1.6.33) is true, then the p operator in the CCR
will of necessity generate translations of x' eigenvalues related to the
x operator according to x|x'> = x'|x> via (1). That is:

(1) if and only if (1.6.33)

What this means in plain English is that so long as two operators x and
p satisfy (1.6.33), then p will be a generator of translations for the
associated x with which is commutes according to (1.6.33). If a
particular p does NOT commute with a particular x, then that p will NOT
generate proper translations for that x. Thus, a particular p will be
tied to a particular x, and the CCR is the relationship which these two
must have in order for p to generate translations for x, on an "if and
only if" basis.

So, I can choose whatever A and B suit my fancy in

(3a) x = A (a + a*)
p = -i B (a - a*)

and so long as (1.6.33) are satisfied, then p will be the translation
generator for x.

Even more simply: the CCR is a "litmus test" we use to determine if
a given p operator will properly generate translations for a given x
operator. If so, then the p and the x are "good to go" as momentum and
position operators. If not, then back to the drawing board for those
operators.

Did I finally "get it"?

Thanks,

Jay

[[Mod. note -- 31 excessively-quoted lines snipped here. -- jt]]

Arnold Neumaier

unread,
Mar 26, 2010, 7:07:53 PM3/26/10
to
Jay R. Yablon wrote:
> we must necessarily have:
>
> [x,p] = i h-bar (1.6.33)
>
> and so long as (1.6.33) are satisfied, then p will be the translation
> generator for x.

No. This is far from correct. For (1.6.33) does not determine p
uniquely. Tf p satisfies this, and we redefine p to be p'=p+P(x)
with an arbitrary function P(x) then p' also satisfies this.

But only one p satisfying (1.6.33) is the true translation generator.

The 1-parameter family of creation/annihilation operators comes from
the fact that when one starts with the canonical x and the corresponding
translation generator p then a and a^* are determined by these only
after one has chosen a frequency that defines the free Hamiltonian
whose excitations are the modes which a/a^* are annihilating/creating.

This is explained in more detail in Section 14.2 of
Arnold Neumaier and Dennis Westra
Classical and Quantum Mechanics via Lie algebras
http://lanl.arxiv.org/pdf/0810.1019v1

Jay R. Yablon

unread,
Mar 28, 2010, 5:15:04 AM3/28/10
to
"Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
news:4BAB1D72...@univie.ac.at...

My son is home for the weekend and I asked him about all of this
last night. He pointed out that with a quadratic Hamiltonian:

H-hat = p-hat/2m + .5 m omega^2 q-hat^2 (1)

we are simply selecting a and a* (* = conjugate transpose) to take
advantage of the fact that any quadratic in x and y can be written as:

x^2 + y^2 = (x+iy)(x-iy) (2)

and so are defining:

a = sqrt(m omega/2 h-bar) (q-hat + i p-hat/m omega)
a*= sqrt(m omega/2 h-bar) (q-hat - i p-hat/m omega) (3)

so as to make use of the quadratic nature of (1) via (2). And, he said
that in general, for some different V(q-hat) other than the:

V(q-hat) = .5 m omega^2 q-hat^2 (4)

which is implicit in (1), we would want to use a different definition of
a, a* rather than those in (3).

He also explained to me when I told him that it bothered me that
everyone seems to just posit a harmonic oscillator and an omega and move
on (you do that right away in section 14.1: omega "is the frequency of
the harmonic oscillator"), and that it seemed arbitrary and limiting to
me to do so, he said that omega as it relates to the spring constant k
via omega = sqrt(k/m) represents a "stiffness" of the potential against
moving out of equilibrium (which you also say). And, he pointed out
that in the series expansion that for a potential with a minimum at
zero, the zero and first order terms drop out and the coefficient of the
leading term is (1/2!) d^2V/dx^2 |_x=0, which means that the harmonic
oscillator will be the leading order for most or all physical potentials
of interest. (Is this also how "perturbation theory" starts out?) (We
talked a bit about how a Coulomb potential fits into this and I also
thought about the Higgs which has a less-than-zero ground state and
maybe relates in some way to your V_0 = -.5 h-bar omega, but that goes a
bit further afield from the immediate inquiry.)

In light of all that, let me now refer to your (14.6) and the
"simple formula":

H = omega a*a (5)

Is this a general formula that we can use for any Hamiltonian, and not
just the Harmonic Hamiltonian (1)? That is, in the general case, might
we start with:

H-hat = p-hat/2m + V(q-hat) (6, your 14.1)

(with or without the V_0 you use to focus on the difference between
energy levels) and use (2) to write a more general relationship in place
of (3) in which we invert (4) into:

sqrt(2V(q-hat) / m omega^2) = q-hat (7)

and then substitute (7) into (3) to write:

a= sqrt(m omega/2 h-bar) (sqrt(2V(q-hat) / m omega^2) + i p-hat/m omega)
a*=sqrt(m omega/2 h-bar) (sqrt(2V(q-hat) / m omega^2)- i p-hat/m omega)
(8) ?

This gives us a more general set of relations a(p-hat, V(q-hat)),
a*(p-hat, V(q-hat)) which reproduce the general (6, your 14.1), rather
than a(p-hat, q-hat), a(p-hat, q-hat) for the simpler and more specific
case of the harmonic oscillator that reproduces (1, your unnumbered
equation just after (14.1))?

Jay


Arnold Neumaier

unread,
Mar 28, 2010, 7:58:57 AM3/28/10
to
Jay R. Yablon wrote
[in: Second reply to question about momentum operators]:

The point is that this depends on a choice of omega.
For different omega one gets different a and a^*.


> so as to make use of the quadratic nature of (1) via (2). And, he said
> that in general, for some different V(q-hat) other than the:
>
> V(q-hat) = .5 m omega^2 q-hat^2 (4)
>
> which is implicit in (1), we would want to use a different definition of
> a, a* rather than those in (3).
>
> He also explained to me when I told him that it bothered me that
> everyone seems to just posit a harmonic oscillator and an omega and move
> on (you do that right away in section 14.1: omega "is the frequency of
> the harmonic oscillator"), and that it seemed arbitrary and limiting to
> me to do so, he said that omega as it relates to the spring constant k
> via omega = sqrt(k/m) represents a "stiffness" of the potential against
> moving out of equilibrium (which you also say).

The point here is that a harmonic oscillator has a distinguished
frequency, and only with this choice, eveything becomes simple.

On the other hand, an anharmonic oscillator has no simple distinguished
expansion frequency, Indeed, approximations made by truncating the
expansion (and in perturbative field theory this sort of thing _must_
be done) depend on the choice of omega, and the best approximation
is usually not obtained with the omega that comes from the expansion of
the Hamiltonian, but a different one. (Google ''variational perturbation
theory'' for details.)

> And, he pointed out
> that in the series expansion that for a potential with a minimum at
> zero, the zero and first order terms drop out and the coefficient of the
> leading term is (1/2!) d^2V/dx^2 |_x=0, which means that the harmonic
> oscillator will be the leading order for most or all physical potentials
> of interest. (Is this also how "perturbation theory" starts out?)

This is how ''bare'' perturtbation theory starts out. The fact that
one has the freedom to choose omega is the key to understanding therole
of the counterterms in renormalization theory. They are the field
version of adding to omega an extra term.


> In light of all that, let me now refer to your (14.6) and the
> "simple formula":
>
> H = omega a*a (5)
>
> Is this a general formula that we can use for any Hamiltonian, and not
> just the Harmonic Hamiltonian (1)?

This is the general formula for a harmonic oscillator with frequency
omega. Since it has no nonlinearities, it is not suitable for any
anharmonic dynamics. But
H = omega a*a + c(a+a^*)+d
is for any real c,d also a harmonic oscillator, with frequency different
from omega if c is nonzero. To bring it into the canonical form (5),
one needs a Bogoliubov transformation of the form (14.11).
(Exercise: Figure out which one does the job!)


> That is, in the general case, might
> we start with:
>

> H-hat = p-hat^2/2m + V(q-hat) (6, your 14.1)

My (14.1) had an extra square, so I added this.

You can use any of the Hermitian representations of phat,qhat in terms
of a and a^* that reproduce the CCR and insert it into this equation
to get a Hamiltonian in the form H=H(a^*,a). Any of these is an equally
valid form of the anharmonic oscillator, and their spectra compuuted by
perturbation theory are identical. But if you truncate the series then
the spectrum depends on which representation you chose, and choosing the
bare representation that gets the frequewncy from the curvature of V
at the minimum of V is usually not the best choice.

One gets a Hamiltonian that is a linear combination of a^*a, a+a^* and 1
if and only iff V is quadratic, and one gets in this case the canonical
form (5) only if the represenation is thas bare one.

This is the reason why the bare free field representations in QFT
always have the form
H= integral dp omega(p) a^*(p)a(p),
which is the infinite-dof version of the Harmonic oscillator.
And the mass counterterm is just the deviation form the bare
representation in which everything is ill-defined and a better
representation with which renormalization is possible.


Arnold Neumaier

Jay R. Yablon

unread,
Mar 30, 2010, 12:57:00 PM3/30/10
to
"Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
news:4BAF2495...@univie.ac.at...
I agree. But this gets to what has been giving me discomfort about all
of this, some of which I have overcome, and some of which remains.

A couple of weeks ago I spoke with Igor at some length about the
definition of a and a* and by implication p and x operators being
dependent on omega. He pointed out at the time how these operators need
omega like a "fish needs a bicycle." Now you are saying that the choice
of omega is important. I am trying to raise my own comfort that in
choosing an omega, we are not doing something which is arbitrary. Now,
there are places in physics where we do make arbitrary choices: for
example, a choice of coordinates or a choice of gauge. But our theories
are constructed to be invariant under these choices, that is, to give us
the same answers about what we will observe irrespective of these
choices. So, if omega is a "fish bicycle," how do we become comfortable
that in the end, the way we choose omega is not going to change the
physics that comes out in the wash at the end of the day? Or, if
different omegas do result in different physics or serve some other
purpose, I'd like to better understand this, and what goes into the
determination of what "omega" to use in any given context. More to
follow below...

>
>> so as to make use of the quadratic nature of (1) via (2). And, he
>> said
>> that in general, for some different V(q-hat) other than the:
>>
>> V(q-hat) = .5 m omega^2 q-hat^2 (4)
>>
>> which is implicit in (1), we would want to use a different definition
>> of
>> a, a* rather than those in (3).
>>
>> He also explained to me when I told him that it bothered me that
>> everyone seems to just posit a harmonic oscillator and an omega and
>> move
>> on (you do that right away in section 14.1: omega "is the frequency
>> of
>> the harmonic oscillator"), and that it seemed arbitrary and limiting
>> to
>> me to do so, he said that omega as it relates to the spring constant
>> k
>> via omega = sqrt(k/m) represents a "stiffness" of the potential
>> against
>> moving out of equilibrium (which you also say).
>
> The point here is that a harmonic oscillator has a distinguished
> frequency, and only with this choice, eveything becomes simple.

I will not get into the details because it will divert from the main
questions I have, but suffice it to say that after doing some
calculation based on writing, in stripped down form:

V(x) + p^2 - i [p,sqrt(V(x))] = (sqrt(V(x)-ip) (sqrt(V(x)+ip) (1),

I am 100% convinced that we should be developing our operators based on
a harmonic oscillator, so long doing so is physically reasonable. My
discomfort now shifts to omega itself and its meaning. My concern now,
is that I do not want to be like the drunk who looks under the lamppost
for a quarter because "that is where the light is." I know that the
harmonic oscillator is the best "light." I now must convince myself
that what we find underneath that light of the harmonic oscillator, with
a suitably-selected omega, is something of value that can be relied upon
for obtaining physically accurate results, and why. More to follow...

> On the other hand, an anharmonic oscillator has no simple
> distinguished
> expansion frequency, Indeed, approximations made by truncating the
> expansion (and in perturbative field theory this sort of thing _must_
> be done) depend on the choice of omega, and the best approximation
> is usually not obtained with the omega that comes from the expansion
> of
> the Hamiltonian, but a different one. (Google ''variational
> perturbation
> theory'' for details.)

Agreed. And the last thing I want to get involved in if it can be
avoided is truncating an expansion because then we start to forego exact
results.

>> And, he pointed out
>> that in the series expansion that for a potential with a minimum at
>> zero, the zero and first order terms drop out and the coefficient of
>> the
>> leading term is (1/2!) d^2V/dx^2 |_x=0, which means that the harmonic
>> oscillator will be the leading order for most or all physical
>> potentials
>> of interest. (Is this also how "perturbation theory" starts out?)
>
> This is how ''bare'' perturtbation theory starts out. The fact that
> one has the freedom to choose omega is the key to understanding
> therole
> of the counterterms in renormalization theory. They are the field
> version of adding to omega an extra term.

Let's now get down to brass tacks, as they say. We base our a, a*, p, q
operators on a harmonic oscillator with a choice of omega because that
is clearly the simplest way to do things, so long as by choosing
simplicity we are not sacrificing accuracy or doing something arbitrary
because the"harmonic light" happens to arbitrarily be where it is. You
have made a statement in the paragraph above, and added more discussion
in the remaining paragraphs below, that contain information that I am
probably not discerning at this juncture, so permit me please to ask
some more naive questions to see if I can understand all of this.

I realize that omega, or alternatively, the stiffness, is a key
parameter that cannot be avoided. So, here is the most direct question
I can ask about all of this: "What is the *physical* meaning of omega?"
Maybe this is naive, but I would like for you to be able to say "omega
is the frequency of such and such." Particularly since you tell me
there is a freedom in choosing omega, how am I assured that the physics
which is predicted will not change based on one choice of omega over the
other? How do I best choose an omega? And, again, what does omega
represent, *physically*? Or, is omega just part of the "bag of tricks"
that is used in renormalization? Or, does this omega end up being some
sort of variable of integration, like the omega in the Fourier
transform, in which case it does not matter much what omega is chosen to
be because it a dummy variable?

Here, I somewhat go back to what my son told me about the stiffness of
the potential, and in particular, the "curvature" of the potential at
its minimum / equilibrium point. The best I can make of all of this at
the moment, is that we assume, reasonably, that V(q) has a single
minimum, or at least some minimum which "predominates" over all the
other minima, and that omega is chosen to in some way relate to the
"curvature" at that minimum, whereby a sharper curvature corresponds to
a higher stiffness and so a more sharply "bounded" set of states for
particle in that potential.

If this is roughly accurate, then the question becomes one of
information loss, if any. That is, if I have a generalized anharmonic
potential V(x), and I am choosing a particular omega based on the
stiffness / curvature of V(q) at the minimum, how am I assured that I
will not forego the effects of all the other harmonics that are
discarded with my particular choice of omega, and that they are retained
somewhere in my operator algebra, to "magically" reappear and have these
effects accounted for when I extract a (continuous or discrete) spectral
distribution of energies at the end of the day?

Igor says that omega is a fish bicycle. You point out that omega "the
freedom to choose omega is the key to understanding the role of the
counterterms in renormalization theory." Based on what I know about
renormalization, which is one of the "ickiest" things we do in physics
these days, my gut tells me that the questions we are discussing are the
types of questions to which we must gain satisfactory answers if we are
to ever "punch our way out of the paper bag" of renormalization and find
a more pleasing way to do physics.

So, Arnold, please help me (any any other readers who may have the same
concerns) to become comfortable that choosing a particular omega for a
harmonic oscillator, even when I have an anharmonic potential V(x) of
any form whatsoever, is still going to get me the right answers, that
even if I am making an arbitrary choice that that choice will make no
difference in the end (i.e., that the physical results will be invariant
with regard to that choice), and that I am not losing any information,
even at high orders, by making a particular choice of omega or by
"looking under the light" of the harmonic oscillator for my physics
because that is the simplest pace to look. And, finally, please help me
understand why something (omega) that Igor has called a "fish bicycle"
in another context is still so darned important in order to get any
physics done.

Jay.

[Note to moderators: please do not snip the remainder below, because
they are pertinent and referred to in the discussion above.]

Igor Khavkine

unread,
Mar 31, 2010, 2:46:36 PM3/31/10
to
On Mar 30, 6:57�pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:

> A couple of weeks ago I spoke with Igor at some length about the
> definition of a and a* and by implication p and x operators being
> dependent on omega. �He pointed out at the time how these operators need
> omega like a "fish needs a bicycle." �Now you are saying that the choice
> of omega is important.

Arnold and I were talking about different things and our statements
are not in conflict or contradiction. If you find this confusing,
without more information about where your confusion lies, I can only
direct you to re-read the relevant discussion.

> So, Arnold, please help me (any any other readers who may have the same
> concerns) to become comfortable that choosing a particular omega for a
> harmonic oscillator, even when I have an anharmonic potential V(x) of

> any form whatsoever, is still going to get me the right answers [...]

I do not see how your question can have an answer. What do you
consider a "right answer"? You've not stated any particular problem
that you are trying to solve yet. Both Arnold and I have restricted
our comments to specific contexts (though different ones at different
times; canonical commutation relations, infinitesimal symmetry
generators, perturbative field theory, etc.). However, you do not
appear to be sticking to any of these contexts. Without making your
questions mathematically precise or referring to a specific physical
system they cannot be answered any better than the question of the
number of angels that can fit on the head of a pin.

Igor

Arnold Neumaier

unread,
Apr 1, 2010, 12:44:44 PM4/1/10
to
Jay R. Yablon wrote:
> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
> news:4BAF2495...@univie.ac.at...

> So, here is the most direct question


> I can ask about all of this: "What is the *physical* meaning of omega?"
> Maybe this is naive, but I would like for you to be able to say "omega
> is the frequency of such and such." Particularly since you tell me
> there is a freedom in choosing omega, how am I assured that the physics
> which is predicted will not change based on one choice of omega over the
> other? How do I best choose an omega? And, again, what does omega
> represent, *physically*?

Omega has no physical meaning, except if the system under study is
already harmonic. Since an anharmonic oscillator is no harmonic,
it is impossible to assign to it a single frequency whose overtones give
the spectrum (which is at the heart of what quantum mechanics wants to
compute). But the harmonic oscillator is very widely used as an
approximative tool. Now ''the'' harmonic oscillator does not exist,
since there is a harmonic oscillartor for each frequency, and these
differ in their spectrum. Depending on what one wants to calculate,
different harmonic oscillators produce in perturbation theory
approximations of different quality. Thus one tries to choose the
frequency in a way that the errors remain reasonable. In quantum
field theory, one actually needs an infinite frequency, to get
renormalization working. This removes the arbitrariness in this case,
but it reappears in the freedom to choose an energy scale in the
renormalization scheme. Customarily, this is chosen in an ad hoc way,
reflecting the energies relevant for the things to predict.

The best choice is an art, not a science, except in simple cases.
I recommended alreadty that you look up variational perturbation theory,
which is the most important way to try to find an optimal omega
from the problem itself.


> Or, is omega just part of the "bag of tricks"
> that is used in renormalization? Or, does this omega end up being some
> sort of variable of integration, like the omega in the Fourier
> transform, in which case it does not matter much what omega is chosen to
> be because it a dummy variable?

The art of approximation is always (and will always be) based on a
large bag of tricks. A good choice of omega matters very much once
you make later approximations based on this choice. And you have to
make approximations if you want to compute anything for a problem
involving QED, or even for much simpler problems such as an x^4
potential well.


> Here, I somewhat go back to what my son told me about the stiffness of
> the potential, and in particular, the "curvature" of the potential at
> its minimum / equilibrium point. The best I can make of all of this at
> the moment, is that we assume, reasonably, that V(q) has a single
> minimum, or at least some minimum which "predominates" over all the
> other minima, and that omega is chosen to in some way relate to the
> "curvature" at that minimum, whereby a sharper curvature corresponds to
> a higher stiffness and so a more sharply "bounded" set of states for
> particle in that potential.

One can choose omega based on the quadratic Taylor series around
a minimum, and naive perturbation theory does exactly this.
But in QFT this is a very poor choice, and leands to meaningless
results except in lowest order of perturbation theory.


> If this is roughly accurate, then the question becomes one of
> information loss, if any. That is, if I have a generalized anharmonic
> potential V(x), and I am choosing a particular omega based on the
> stiffness / curvature of V(q) at the minimum, how am I assured that I
> will not forego the effects of all the other harmonics that are
> discarded with my particular choice of omega,

No matter how you choose omega, the harmonic approximation distorts the
spectrum severely, unless your problem is almost harmonic.


> and that they are retained
> somewhere in my operator algebra,

The information is not lost - it reappears faithfully in the resulting
interaction potential V = H(a^*,a) - omega a^*a. The problem is that
when omega is poorly chosen then V is by no means a small perturbation
in any sense, but perturbation theory can be successful only if it is.


> Igor says that omega is a fish bicycle.

You misunderstood him. Omega is a cyclop's eye for the harmonic
oscillator, essential for describing harmonic oscillations.

It is a fish bicycle only for creation or annihilation operators,
since these come without any relation to frerquencies. They are simply
characterized by given commutator relations, and depending on the
problem, these can be generated in many ways.

A harmonic oscillator (or a nonharmonic one) is characterized by the
Hamiltonian, not by the operators used to define it. But the simplest
expression for the Hamiltonian, and the one in terms of which one can
find it spectrum most easily, is the standard one.

> You point out that omega "the
> freedom to choose omega is the key to understanding the role of the
> counterterms in renormalization theory." Based on what I know about
> renormalization, which is one of the "ickiest" things we do in physics
> these days, my gut tells me


As long as you do not even understand the anharmonic oscillator,
you are very far from understanding renormalization. So in your place
I wouldn't give more weight to what your guts tell you than to what
a blind man can see.


> So, Arnold, please help me (any any other readers who may have the same
> concerns) to become comfortable that choosing a particular omega for a
> harmonic oscillator, even when I have an anharmonic potential V(x) of
> any form whatsoever, is still going to get me the right answers, that
> even if I am making an arbitrary choice that that choice will make no
> difference in the end (i.e., that the physical results will be invariant
> with regard to that choice), and that I am not losing any information,
> even at high orders,

You'll make very big errors at high order even when your choice of
omega is only slightly off the best choice. Only at the lowest order
is a straightforward choice - the stiffness frequency - a good one.

You'll have to become comfortable with this, since this is the
mathematical nature of all asymptotic expansions! The optimal order
to use is usually quite low.


Arnold Neumaier

Bob_for_short

unread,
Apr 2, 2010, 1:23:55 PM4/2/10
to
On 1 avr, 18:44, Arnold Neumaier <Arnold.Neuma...@univie.ac.at> wrote:

>> Jay R. Yablon wrote: ...


>> Particularly since you tell me
> > there is a freedom in choosing omega, how am I assured that the physics
> > which is predicted will not change based on one choice of omega over the
> > other? �How do I best choose an omega? �And, again, what does omega
> > represent, *physically*?
>
> Omega has no physical meaning, except if the system under study is

> already harmonic. ...
...


> Thus one tries to choose the
> frequency in a way that the errors remain reasonable.

...


> This is how ''bare'' perturtbation theory starts out. The fact that

> one has the freedom to choose omega is the key to understanding the
> role of the counterterms in renormalization theory. They are the field


> version of adding to omega an extra term.

...


> In quantum
> field theory, one actually needs an infinite frequency, to get
> renormalization working. This removes the arbitrariness in this case,
> but it reappears in the freedom to choose an energy scale in the
> renormalization scheme. Customarily, this is chosen in an ad hoc way,
> reflecting the energies relevant for the things to predict.
>

Many years ago I worked with theoretical description of H2 molecule.
It is two Hydrogen atoms bound with a potential of finite depth. Of
course, near the equilibrium (ground) state the potential is nearly
quadratic. The first (vibrational) levels are nearly equidistant but
by the 14-th level they get somewhat closer and anyway their number is
limited - the fifth level corresponds to a dissociated molecule H + H.

The harmonic frequency omega has a clear physical meaning there.
Anyway we can directly solve the problem with numerical methods and
get numbers comparable with the experimental data.

An oscillator solution like A*sin(omega*t) may be somewhat imprecise
in some non-linear problems and sometimes one may search a solution if
the form: A(t)*sin[omega(t)*t], i.e., with a time-dependent omega and
amplitude. Often omega(t) is expanded in a series itself so it
contains the �bare� frequency omega(0) and perturbative corrections to
it. In QFT some perturbative corrections to the solution can also be
represented as corrections to omega(0). It turns out that they are
infinite. It is also impossible to solve the exact QFT equations
directly. What to do then?

It was recognized that the difficulty is connected with the self-
action, often wrongly spoken of as of a �too strong� interaction. So
the origin of the difficulty is in the badly guessed exact equations.
Also it was one day discovered that discarding these infinite
corrections led to good results. In order not to get lost in such a
discarding (corrections to the electron mass, to the photon mass, and
to the charge), a counter-term addenda were introduced into the
initial Lagrangian. Again, the role of counter-terms was to carry out
the subtractions of infinite corrections to the initial constants in a
consistent way. But with time some people started to insist that it is
the initial (�bare�) parameters that are infinite (?) and the
perturbative corrections MUST be infinite to give a finite sum.

Of course, it was not even a stretch but pure rubbish: the corrections
were divergent with the _finite_ original parameters, which are equal
to unity in the appropriate units m = 1, e = 1.

The most funny thing is that after all renormalizations the solution
remains A*sin(omega*t) with the same numerical value of omega (the
same electron mass and photon frequency, if you like). Much effort was
spent in order to �justify� the renormalizations. Now I see people say
that free particles are not observable, omega is not observable, etc.
Counter-terms are also non observable. So the whole Lagrangian is made
of non-observable stuff and serves now as a fitting curve instead of
being a physical theory.

Bob_for_short

unread,
Apr 2, 2010, 6:34:30 PM4/2/10
to
On Apr 2, 7:23 pm, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
wrote:

> On 1 avr, 18:44, Arnold Neumaier <Arnold.Neuma...@univie.ac.at> wrote:
>
> > Omega has no physical meaning, except if the system under study is
> > already harmonic. ...
> ...
> > Thus one tries to choose the
> > frequency in a way that the errors remain reasonable.
> ...
As to omega, in QED the operators a and a^* are labeled with it, the
total Hamiltonian is a sum of "elementary" oscillator Hamiltonians
over different wave vectors |k| = omega/c. Each "elementary"
Hamiltonian contains kinetic and potential energies of a particular
oscillator with a particular frequency omega(k). In course of charge
scattering, apart from change in the charge momenta, there is change
in populations of these quantum oscillators (absorption/emission of
real photons) and nothing else. So everything is physical and
observable.

If you want to work out a simple QM problem with an arbitrary
potential U(x) that has a local quadratic minimum U0(x), you may
safely choose the corresponding oscillator omega. The total potential
U will be a non linear function of the creation/annihilation operators
and any stationary state in U |E_i> will be represented as a
superposition of oscillator wave functions with different populations:
|E_i> = sum_n A_i_n|n>, with coefficients A_i_n = <n|E_i>. Of course,
you can choose any other frequency but the sum coefficient may become
not decreasing with n. Decreasing coefficients provide rapid
convergence of the spectral sum. Let us distinguish the spectral sum
and the perturbation series.

Apparently Arnold was speaking of "optimization" of omega for a given
(finite) number N of spectral addenda (it is the spectral sum which is
truncated, not the power series). The exact solution is not sensitive
to the omega choice but we never deal with the exact solutions - we
limit ourselves with several spectral addenda A_i_n|n>.

It is the coefficients A_i_n that are expanded in the perturbation
series: A_i_n= A_i_n^(0) + epsilon*A_i_n^(1) + ... . So truncated
spectral sum and perturbation series are frequency-dependent and may
be "optimized" with a variational principle, I guess.

In my opinion, if "anharmonicity" (U - U0) contributions are
relatively "small", one can take the natural frequency for defining
operators. If "anharmonicity" is too "strong", I am not sure at all if
we should represent the solution as a superposition of oscillator
solutions of a "better" omega. Maybe another basis (different not only
with omega) will be more practical (better convergent).

Igor Khavkine

unread,
Apr 3, 2010, 10:21:38 AM4/3/10
to
On Apr 2, 7:23 pm, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
wrote:

> [...] Much effort was


> spent in order to justify the renormalizations.

No effort is needed if QFT is constructed without divergent
intermediate quantities, as with using the method of Epstein and
Glaser described in Scharf's QED book (though there is no problem with
justification even if divergent intermediate quantities do appear).
You spend too much time arguing against straw men. If you want your
objections to be taken seriously, point to an unacknowledge
mathematical flaw in a modern treatment of QFT, as in Scharf's book or
to a quantitative disagreement of QED with experimental measurements
(where non-QED physics is not an issue). You've been challenged to
provide this sort of evidence more than once, yet it has yet to
materialize.

> Now I see people say
> that free particles are not observable, omega is not observable, etc.
> Counter-terms are also non observable. So the whole Lagrangian is made
> of non-observable stuff and serves now as a fitting curve instead of
> being a physical theory.

What cannot be put into correspondence with a quantitative reading
from an experiment is not observable, there is no mystery in that
statement. All of physics is a giant exercise in curve fitting. If you
disagree, then you are in aesthetic argument with philosophers of
science, which should be voiced in another forum.

Igor

Bob_for_short

unread,
Apr 3, 2010, 1:18:33 PM4/3/10
to
On 28 mar, 11:15, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
> And, he pointed out
> that in the series expansion that for a potential with a minimum at
> zero, the zero and first order terms drop out and the coefficient of the
> leading term is (1/2!) d^2V/dx^2 |_x=0, which means that the harmonic
> oscillator will be the leading order for most or all physical potentials
> of interest.  (Is this also how "perturbation theory" starts out?)  

Generally the perturbation theory stars from some exactly solvable
potential U0 or Hamiltonian H0, not obligatory from a quadratic or
oscillator one. The difference between the exact (but impossible to
solve) Hamiltonian H and the exactly solvable (but approximate)
Hamiltonian H0 is the perturbation to take into account.

If the series in "powers of the perturbation" converges quickly, the
truncated series may directly serve for approximating the sought
eigenvalues and eigenstates.

If the series in "powers of the perturbation" converges badly for one
or another reason, one may employ some non-linear transformation to
the truncated series to approximate the sought eigenvalues and
eigenstates, for example, Padé approximations. For example, exp(-x) =
1 - x + x^2/2 is only "good" for sufficiently small x. The bad thing
about this truncated series is that it grows as x^2/2 at large x
instead of decaying. Using a non-linear approximation exp(-x) = 1/(1 +
x + x^2/2) extends the region of "good" accuracy to larger x.

There are in fact many methods of constructing approximations for
different cases of bad convergence of a truncated series.

Bob_for_short

unread,
Apr 4, 2010, 4:53:56 PM4/4/10
to
On 3 avr, 16:21, Igor Khavkine <igor...@gmail.com> wrote:

> All of physics is a giant exercise in curve fitting. ...

I remember Max Plank and his formula that first was written as a data
fitting curve. Fortunately it fitted too perfectly to be just a good
approximation: it was exact!

It is difficult to imagine what it would have been with the idea of
light quanta if the exact curve had been more complicated and had not
been guessed so luckily.


Jay R. Yablon

unread,
Apr 5, 2010, 9:09:36 PM4/5/10
to
"Igor Khavkine" <igo...@gmail.com> wrote in message
news:8da83a1a-cca3-4f59...@z3g2000yqz.googlegroups.com...

> On Apr 2, 7:23 pm, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
> wrote:
>
. . .

> All of physics is a giant exercise in curve fitting. If you
> disagree, then you are in aesthetic argument with philosophers of
> science, which should be voiced in another forum.
>
> Igor

That is only half right, and presents a crimped view of what physics is
all about.

If physics was simply about curve fitting, then there would have been no
need for Maxwell to do anything, because Faraday, Ampere and Gauss
already had completed successful exercises to fit the curves of what was
observed for magnetism, current flow, and static charge. Not only is
one fitting curves, but one is always seeking to enlarge the curves, so
that a single curve can fit all of the data that was hitherto
represented in several separate curves.

One cannot not and ought not deny the strong impulse toward unifying
diverse phenomena and developing aesthetically pleasing understandings
of what is observed, always with the caveat that in the end,
experimental observation or lack thereof trumps aesthetics. That is,
physics involves finding the best fitting curves in the most
aesthetically pleasing fashion, with the least number of underlying
assumptions and with as many curves as possible combined into one.

To be fair, perhaps your clever characterization of physics as a "giant"
exercise in curve fitting was intended to encompass the unification
inherent in fitting the results of "less giant" curves. But if so, it
might behoove you to talk more explicitly about the relation between
curve fitting and aesthetics and the drive to simplicity that is in the
back of the minds of many who are working to do physics. For someone to
state, essentially, "those infinities in renormalization are ugly even
if the results do fit the observational curve" is a valid observation to
make about the physics, and not merely a matter of science philosophy
unsuited to physics discussion. Many who do physics are trying to
understand nature. They are indeed fitting curves, but they are doing
more than just fitting curves. There are other considerations at work,
and one ought not so crimp the discussion as to pretend otherwise.

Jay

Igor Khavkine

unread,
Apr 5, 2010, 9:10:01 PM4/5/10
to
On Apr 4, 10:53 pm, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
wrote:

> On 3 avr, 16:21, Igor Khavkine <igor...@gmail.com> wrote:
>
> > All of physics is a giant exercise in curve fitting. ...
>
> I remember Max Plank and his formula that first was written as a data
> fitting curve. Fortunately it fitted too perfectly to be just a good
> approximation: it was exact!

I hope you realize that in physics nothing is exact. That includes
Planck's formula, which despite not being exact is at times such a
good aproximation that it can be colloquially referred to as such.

> It is difficult to imagine what it would have been with the idea of
> light quanta if the exact curve had been more complicated and had not
> been guessed so luckily.

I find it difficult to imagine what the implicit point you are trying
to make is. I cannot see the value of agreeing or disagreeing with
such a vague statement. Luckily, the progress of science does not
depend on the limited imagination of any one person, be it either one
of us.

Igor

Richard D. Saam

unread,
Apr 7, 2010, 4:20:15 AM4/7/10
to
If all physics is curve fitting, here is a challenge in that regard.

Look at the log log plot of
Extragalactic Gamma-Ray Diffuse Background(EGDB):
http://arxiv.org/abs/1002.4415v1 (figure 4)
EGDB is apparently not generated by known point sources.

There are two features potentially worth noting
in this rather simple log log plot.

1. The 56 Mev EGRET cutoff or peak feature
(yet to be confirmed by FERMI).
2. The EGRET/FERMI log linear (-1/3 power) fall off
of E(Mev)/area/time/steradian with E(Mev).

By curve fitting this data in a mode fostered by Max Planck,
are we observing something physically very fundamental linking 1 and 2
into a continuous spectrum,
something akin to the ubiquitous CMBR spectrum?
What would the source of such an ubiquitous EGDB spectrum be?

There does not appear to be any known particle energetics that can
produce such a spectrum.

Richard D. Saam

Bob_for_short

unread,
Apr 7, 2010, 4:20:16 AM4/7/10
to
On 6 avr, 03:10, Igor Khavkine <igor...@gmail.com> wrote:

> I hope you realize that in physics nothing is exact. That includes

> Planck's formula.

I did not know that the black body radiation law (Plank formula) was
inexact. At Plank's time researchers could model a black body well and
obtained a lot of data. Plank formula contains only two parameters (at
a given frequency): temperature and h-bar. Factually h-bar was the
only fitting parameter in the given analytical expression. Plank was
really surprised to see that once chosen, the parameter was unique for
all available curves. This made him search for a physical reason
behind this lucky coincidence. Thus a hypothesis of light quanta was
introduced. It was not an easy step and Plank himself resisted for a
long time.


>
> > It is difficult to imagine what it would have been with the idea of
> > light quanta if the exact curve had been more complicated and had not
> > been guessed so luckily.
>
> I find it difficult to imagine what the implicit point you are trying
> to make is.

It is my drawback not to be able to express myself clearly. I wanted
to say that without such a luck Plank would not have invented the
light quanta.

Concerning the bare particles and counter-terms (that do not exist in
nature but are the only entities involved in the Lagrangian), I think
it is not the brightest physical "insight" ever proposed to explain
the luck of renormalizations. In particular, Scharf is shy of that
"pearl". I quite understand his desire to find another justification
of renormalizations.

Igor Khavkine

unread,
Apr 7, 2010, 3:33:37 PM4/7/10
to
On Apr 7, 10:20�am, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
wrote:

> On 6 avr, 03:10, Igor Khavkine <igor...@gmail.com> wrote:
>
> > I hope you realize that in physics nothing is exact. That includes
> > Planck's formula.
>
> I did not know that the black body radiation law (Plank formula) was
> inexact. [...]

Then you must not be aware of error bars and/or possible deviations of
any physical system from its idealized theoretical description.

> Concerning the bare particles and counter-terms (that do not exist in
> nature but are the only entities involved in the Lagrangian),

Never a truer thing has been said.

> I think
> it is not the brightest physical "insight" ever proposed to explain
> the luck of renormalizations.

Followed by a non-sequitur. No luck is involved, as has been
repeatedly explained.

Igor

Bob_for_short

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Apr 8, 2010, 12:31:19 PM4/8/10
to
On Apr 7, 9:33 pm, Igor Khavkine <igor...@gmail.com> wrote:
> On Apr 7, 10:20 am, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
> wrote:
>
> > On 6 avr, 03:10, Igor Khavkine <igor...@gmail.com> wrote:
>
> > > I hope you realize that in physics nothing is exact. That includes
> > > Planck's formula.
>
> > I did not know that the black body radiation law (Plank formula) was
> > inexact. [...]
>
> Then you must not be aware of error bars and/or possible deviations of
> any physical system from its idealized theoretical description.
>

Yes, I am well aware of them. Any physical body radiation is different
from the black body radiation and even the black body radiation is an
average curve with its statistical dispersion. Nevertheless the black
body modeled as a hole in a thermostat has a unique feature -
independence of the average curve of the thermostat material. So it is
a characteristic of the electromagnetic radiation subsystem solely.
This particular feature was noticed and used by experimentalists in
their studies.

I myself used to work with radiating finite volume plasmas and thin
films whose radiated spectral power is quite different from the Plank
law.

> No luck is involved, as has been repeatedly explained.

The words about "miraculously working" renormalization prescription
were written well before me. I just join this opinion because I found
my own explanation why this is a luck rather than a triumph.

Igor Khavkine

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Apr 9, 2010, 4:21:00 PM4/9/10
to
On Apr 8, 6:31 pm, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
wrote:

> On Apr 7, 9:33 pm, Igor Khavkine <igor...@gmail.com> wrote:
> > On Apr 7, 10:20 am, Bob_for_short <vladimir.kalitvian...@wanadoo.fr> wrote:
> > > On 6 avr, 03:10, Igor Khavkine <igor...@gmail.com> wrote:
>
> > > > I hope you realize that in physics nothing is exact. That includes
> > > > Planck's formula.
>
> > > I did not know that the black body radiation law (Plank formula) was
> > > inexact. [...]
>
> > Then you must not be aware of error bars and/or possible deviations of
> > any physical system from its idealized theoretical description.
>
> Yes, I am well aware of them. Any physical body radiation is different
> from the black body radiation and even the black body radiation is an
> average curve with its statistical dispersion.

Excellent. Now is a good time to retract your previous statement about
exactness then.

> Nevertheless the black
> body modeled as a hole in a thermostat has a unique feature [...]

Nothing to do with exactness.

Igor

Bob_for_short

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Apr 10, 2010, 4:07:26 AM4/10/10
to
On Apr 9, 10:21 pm, Igor Khavkine <igor...@gmail.com> wrote:
>
> > > > > I hope you realize that in physics nothing is exact. That includes
> > > > > Planck's formula.
>
> > > > I did not know that the black body radiation law (Plank formula) was
> > > > inexact. [...]
>
> > > Then you must not be aware of error bars and/or possible deviations of
> > > any physical system from its idealized theoretical description.
>
> > Yes, I am well aware of them. Any physical body radiation is different
> > from the black body radiation and even the black body radiation is an
> > average curve with its statistical dispersion.
>
> Excellent. Now is a good time to retract your previous statement about
> exactness then.

I disagree. The average (mean) value is quite certain. It exactly
coincides with the Planck formula. The latter is also derived as a
statistical average (mean). This excellent coincidence obtained for
different T and different experiments made M. Planck seriously think
that it could be derived from some fundamental principle.

Vladimir.

Igor Khavkine

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Apr 11, 2010, 11:38:43 PM4/11/10
to
On Apr 10, 10:07 am, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
wrote:

> On Apr 9, 10:21 pm, Igor Khavkine <igor...@gmail.com> wrote:

> > Excellent. Now is a good time to retract your previous statement about
> > exactness then.
>
> I disagree. The average (mean) value is quite certain. It exactly
> coincides with the Planck formula.

You are aware that the "average (mean) values" themselves, like all
practical measurements, come with error bars, Yes or No?

[snip of text irrelevant to the point of discussion]

Igor

Bob_for_short

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Apr 12, 2010, 10:01:33 AM4/12/10
to
On 12 avr, 05:38, Igor Khavkine <igor...@gmail.com> wrote:
> On Apr 10, 10:07 am, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
> wrote:
>
> > On Apr 9, 10:21 pm, Igor Khavkine <igor...@gmail.com> wrote:
> > > Excellent. Now is a good time to retract your previous statement about
> > > exactness then.
>
> > I disagree. The average (mean) value is quite certain. It exactly
> > coincides with the Planck formula.
>
> You are aware that the "average (mean) values" themselves, like all
> practical measurements, come with error bars, Yes or No?
>

Let us consider a certain frequency omega. The experimental data
(points) for it come with error bars but after averaging one obtains a
quite certain number. Ensemble of such averaged data points for
different frequencies represents a curve F(omega) which is compared to
the theory.

The point I would like to underline is not in error bars but in how it
is difficult to develop new physical ideas.

Arnold Neumaier

unread,
Apr 12, 2010, 4:52:29 PM4/12/10
to
Bob_for_short wrote:
> On 12 avr, 05:38, Igor Khavkine <igor...@gmail.com> wrote:
>> On Apr 10, 10:07 am, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
>> wrote:
>>
>>> On Apr 9, 10:21 pm, Igor Khavkine <igor...@gmail.com> wrote:
>>>> Excellent. Now is a good time to retract your previous statement about
>>>> exactness then.
>>> I disagree. The average (mean) value is quite certain. It exactly
>>> coincides with the Planck formula.
>> You are aware that the "average (mean) values" themselves, like all
>> practical measurements, come with error bars, Yes or No?
>>
>
> Let us consider a certain frequency omega. The experimental data
> (points) for it come with error bars but after averaging one obtains a
> quite certain number.

Even a quite certain number is still quite uncertain. If the initial
certainty is sigma then after N independent repetitons, the average
still has an uncertainty of sigma/sqrt(N). No maount of averaging makes
the error disappear.

Bob_for_short

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Apr 15, 2010, 7:12:34 AM4/15/10
to
On 12 avr, 22:52, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
wrote:

>
> > Let us consider a certain frequency omega. The experimental data
> > (points) for it come with error bars but after averaging one obtains a
> > quite certain number.
>
> Even a quite certain number is still quite uncertain. If the initial
> certainty is sigma then after N independent repetitions, the average
> still has an uncertainty of sigma/sqrt(N). No amount of averaging makes
> the error disappear.

Let us not overcomplicate a simple thing. If we speak of photon number
fluctuations, it is rather small due too big N. There may be
uncertainties in measuring omega, temperature, and the measured power.
A grey body has also a spectral power different form the Plank law, I
agree. The fact is that even at that time the experimentalists and
theorists recognised the black body radiation as a fundamental subject
(radiation subsystem of the thermostat) and made good measurements.
Plank first wrote this formula as an approximation that joins two
asymptotics in one formula. But too good agreement with different
experiments stroke him. Then he started deriving it with help of
Boltzmann method. Doing so, he had to admit a bizarre idea of energy
quantization for a given radiation mode. He writes himself about that.
It is a part of history now and a too good coincidence of the
initially simple analytical approximation (interpolation between two
asymptotics) with the available data played a crucial role in his
later derivations.

Igor Khavkine

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Apr 15, 2010, 12:31:21 PM4/15/10
to
On Apr 15, 1:12 pm, Bob_for_short <vladimir.kalitvian...@wanadoo.fr>
wrote:

> On 12 avr, 22:52, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
>
> > > Let us consider a certain frequency omega. The experimental data
> > > (points) for it come with error bars but after averaging one obtains a
> > > quite certain number.
>
> > Even a quite certain number is still quite uncertain. If the initial
> > certainty is sigma then after N independent repetitions, the average
> > still has an uncertainty of sigma/sqrt(N). No amount of averaging makes
> > the error disappear.
>
> Let us not overcomplicate a simple thing. If we speak of photon number
> fluctuations, it is rather small due too big N. There may be
> uncertainties in measuring omega, temperature, and the measured power.
> A grey body has also a spectral power different form the Plank law, I
> agree.

Good. I'm glad to see that you've retracted your previous statement
about Planck's law being in exact agreement with experimental data,
even if only in the form of an implicit admission. Lets hope this
spirit of intellectual honesty pervades parallel conversations as
well.

[...snip discussion irrelevant to the point I originally raised...]

Yes, the history of science is full of serendipity.

Igor

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