Is the canonical commutation relationship really just a low-mass approximation?
Jay R. Yablon
When apply we the canonical :
[x^i,p_j] = ih delta^i_j, i=1,2,3 (1)
is this really just an approximate relationship in the limiting case
where a mass m-->0?
That is, is (1) really of the form:
[x^i,p_j] = ih delta^i_j + F(m)^i_j (1)
where F(m)^i_j --> 0 as m --> 0?
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
Hendrik van Hees
momentum components is exact, supposed you can define self-adjoint
position and momentum operators at all. Its origin is however not some
hand-waving argument called "canonical quantization" (which by the way
doesn't work in almost all other cases than cartesian coordinates ;-))
but originates from the definition of momentum as the generator of
spatial translations in the sense of representation theory of Lie
algebras and Lie groups on Hilbert space.
In relativistic quantum theory (i.e., relativistic QFT) it's not easy to
define a position operator with the appropriate properties. E.g., it's
far from obvious (and not clear to me) whether one can define a
position operator for massless particles with spin, e.g., for photons.
Fortunately in practice this problem is of academic interest only since
one can define observable quantities like cross sections and the like
without reference to a position operator as is shown in any textbook of
qft. In my opinion the best textbook reference is Weinberg's Quantum
Theory of Fields.
Jay R. Yablon wrote:
> I seem to recall having this discussion before, but need to refresh.
>
> When apply we the canonical :
>
> [x^i,p_j] = ih delta^i_j, i=1,2,3 (1)
>
> is this really just an approximate relationship in the limiting case
> where a mass m-->0?
>
> That is, is (1) really of the form:
>
> [x^i,p_j] = ih delta^i_j + F(m)^i_j (1)
>
> where F(m)^i_j --> 0 as m --> 0?
>
> Thanks,
>
> Jay
--
Hendrik van Hees Institut f�r Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universit�t Gie�en
Fax: +49 641 99-33309 D-35392 Gie�en
http://theory.gsi.de/~vanhees/faq/
Arnold Neumaier
>
> When apply we the canonical :
>
> [x^i,p_j] = ih delta^i_j, i=1,2,3 (1)
>
> is this really just an approximate relationship in the limiting case
> where a mass m-->0?
>
> That is, is (1) really of the form:
>
> [x^i,p_j] = ih delta^i_j + F(m)^i_j (1)
>
> where F(m)^i_j --> 0 as m --> 0?
No. it is an exact relationship that is at the basis of quantum
mechanics, valid for total position and momentum of any physical
system with nonzero total mass, even in the relativistic case and
in quantum field theory.
For systems of mass 0 and spin >1/2 , the position operator
is nonexistent, and therefore no associated CCR exists.
See Section S2g ''Particle positions and the position operator''
of my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
Arnold Neumaier
juanR...@canonicalscience.com
Hendrik van Hees wrote on Sun, 10 May 2009 19:00:59 +0000:
(...)
> In relativistic quantum theory (i.e., relativistic QFT) it's not easy t=
o
> define a position operator with the appropriate properties.
x in (1) is the particle position operator. And as explained in textbooks
position is downgraded to parameter in relativistic QFT. This is that
Mandl and Shaw texbook (revised version) in its page 10 defines as:
an important difference between a quantized field theory and
non-relativistic quantum mechanics.
> E.g., it's
> far from obvious (and not clear to me) whether one can define a positio=
n
> operator for massless particles with spin, e.g., for photons.
There is not position operator for particles (massless or not) in RQFT.
> Fortunately in practice this problem is of academic interest only since
> one can define observable quantities like cross sections and the like
> without reference to a position operator as is shown in any textbook of
> qft. In my opinion the best textbook reference is Weinberg's Quantum
> Theory of Fields.
Fortunately for particle physicists who, as well explained in the chapter
3 of Weinberg (age 107), are only interested in scattering amplitudes for
atemporal processes over an infinite volume involving one or two particle=
s
(Weinberg considers the three particle case exceptional for him!).
Beyond particle accelerators, cross sections and all the S-matrix
formalism are often unuseful. E.g., take electron trasfer reactions in a
vessel; you just cannot assume that the electron was initially infinitely
separated from others electrons collide and again is free at infinite
distance :-D
(...)
> Jay R. Yablon wrote:
>
>> I seem to recall having this discussion before, but need to refresh.
>>
>> When apply we the canonical :
>>
>> [x^i,p_j] = ih delta^i_j, i=1,2,3 (1)
--
http://www.canonicalscience.org/
Usenet Guidelines:
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Arnold Neumaier
> Hendrik van Hees wrote on Sun, 10 May 2009 19:00:59 +0000:
>
> Hendrik van Hees wrote on Sun, 10 May 2009 19:00:59 +0000:
>
> (...)
>
>> In relativistic quantum theory (i.e., relativistic QFT) it's not easy t=
> o
>> define a position operator with the appropriate properties.
>
> x in (1) is the particle position operator. And as explained in textbooks
> position is downgraded to parameter in relativistic QFT. This is that
> Mandl and Shaw texbook (revised version) in its page 10 defines as:
>
> an important difference between a quantized field theory and
> non-relativistic quantum mechanics.
>
>> E.g., it's
>> far from obvious (and not clear to me) whether one can define a positio=
> n
>> operator for massless particles with spin, e.g., for photons.
>
> There is not position operator for particles (massless or not) in RQFT.
This is not true. Even though there is a parameter called x and referred
to as 4-dimensional position, there is also an vector defining a
3-dimensional position operator, if a relativistic system is not massless.
For any relativistic theory possesses the Poincare group as a symmetry
group, whose infinitesimal generators satisfy the standard commutation
rules of the Poincare algebra. But given these, the standard
construction by Newton and Wigner gives (in each Lorentz frame)
a 3-dimensional position operator with commuting components, and
the associated conjugate momentum operators. These play exactly the
same role as the position and momentum operators in nonrelativistic
quantum mechanics.
Juan R.
> juanR...@canonicalscience.com schrieb:
>> Hendrik van Hees wrote on Sun, 10 May 2009 19:00:59 +0000:
(...)
>> There is not position operator for particles (massless or not) in RQFT.
>
> This is not true. Even though there is a parameter called x and referred
> to as 4-dimensional position, there is also an vector defining a
> 3-dimensional position operator, if a relativistic system is not
> massless.
First, that four-dimensional parameter x^b = (t,x) has nothing to see with
position or time [1]:
"Every physicist would easily convince himself that all quantum
calculations are made in the energy-momentum space and that the
Minkowski x^b are just dummy variables without physical meaning
(although almost all textbooks insist on the fact that these variables
are not related with position, they use them to express locality of
interactions!)"
> For any relativistic theory possesses the Poincare group as a symmetry
> group, whose infinitesimal generators satisfy the standard commutation
> rules of the Poincare algebra. But given these, the standard
> construction by Newton and Wigner gives (in each Lorentz frame) a
> 3-dimensional position operator with commuting components, and the
> associated conjugate momentum operators. These play exactly the same
> role as the position and momentum operators in nonrelativistic quantum
> mechanics.
>
> See Section S2g ''Particle positions and the position operator'' of my
> theoretical physics FAQ at
> http://www.mat.univie.ac.at/~neum/physics-faq.txt
My original claim was about the lack or position operator x for particles
in relativistic quantum field theory (RQFT).
In your FAQ you have a section S2i. "Position operators in relativistic
quantum field theory" where you write:
"In relativistic quantum field theory in its usually given form,
position is promoted to the same status as time, and hence becomes a
parameter in the qwuantum field, while in quantum mechanics it is an
operator vector."
With the typo in the original. This part of your FAQ is right.
Your section S2g ''Particle positions and the position operator'', is
*not* about RQFT but about RQM.
This section is incorrect. In RQM there is an position operator x, but it
cannot be consistently interpreted. The failure to obtain a correct
position operator x was the reason which the founders of RQFT downgraded x
to *parameter*.
You cite several RQM papers about the Newton-Wigner consistent position
operator. Take for instance
L.L. Foldy and S.A. Wouthuysen,
On the Dirac Theory of Spin 1/2 Particles and Its Non-Relativistic
Limit, Phys. Rev. 78 (1950), 29-36.
They start by noticing some difficulties of the position operator x. Then
they introduce an operator X in (23), and refer to it as the new
"position" operator (the "" are in the original and the reason which this
is not the position operator but a "position" operator is given below).
In the same page they do clear that the position operator x can be
splinted as
x = X + (X - x)
They name X the "/mean-position/ operator". Evidently (X - x) measures the
difference with the particle position operator. They also write in a
footnote that X is that Newton and Wigner found.
They also give the "mean velocity operator" associated to X.
Contrary to your claims, X is not the position operator associated to the
particle but only an averaged mean-position operator. The operator
continues being x with all the known deficiencies.
Moreover, the operator X is useless in RQFT; the arguments for quantum
fields continue being the parameters x and t, the scalar interaction in
QED continues being given by an expression (e^2 / |x-y|), the S-matrix
integrations are done over d^3x and dt and so on...
Regards
[1]
The notions of localizability and space: from Eugene Wigner to Alain
Connes 1989: Nucl. Phys. Proc. Suppl. 6, 222. Bacry, H.
Jay R. Yablon
"Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
news:4A07225B...@univie.ac.at...
Does this also remain exact for *interacting* fermions, where there is a
potential A^v employed in Dirac's equation and p^v-->p^v+eA^v?
Thanks,
Jay
Arnold Neumaier
>
> "Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
> news:4A07225B...@univie.ac.at...
>> Jay R. Yablon schrieb:
>>>
>>>
>> mechanics, valid for total position and momentum of any physical
>> system with nonzero total mass, even in the relativistic case and
>> in quantum field theory.
>>
>> For systems of mass 0 and spin >1/2 , the position operator
>> is nonexistent, and therefore no associated CCR exists.
>> See Section S2g ''Particle positions and the position operator''
>> of my theoretical physics FAQ at
>> http://www.mat.univie.ac.at/~neum/physics-faq.txt
>
> potential A^v employed in Dirac's equation and p^v-->p^v+eA^v?
It holds for any Poincare-invariant quantum theory, as I explained
in my mail on this topic that appeared on s.p.r on 2009-05-14.
In this equation, p must be the vector whose components are
the generators of the space translations. In the standard
representation (the instant form in the terminology of Dirac,
see Section S2d of my FAQ), these generators are explicitly
computable for all field theories, as generators of corresponding
symmetries of the action.
For gauge theories, I don't remember whether these are the sum over
the single-particle p^v or the sum over the single-particle p^v+eA^v
(and have currently no time to check).
For particles in a nontrivial external field, however, the field
breaks translation invariance, and the symmetry argument no longer
applies. I can't immediately interpret this situation.
Arnold Neumaier
Arnold Neumaier
> Arnold Neumaier wrote on Thu, 14 May 2009 21:23:51 +0200:
>
>> juanR...@canonicalscience.com schrieb:
>>> Hendrik van Hees wrote on Sun, 10 May 2009 19:00:59 +0000:
>
> (...)
>
>>> There is not position operator for particles (massless or not) in RQFT.
>> This is not true. Even though there is a parameter called x and referred
>> to as 4-dimensional position, there is also an vector defining a
>> 3-dimensional position operator, if a relativistic system is not
>> massless.
>
> First, that four-dimensional parameter x^b = (t,x) has nothing to see with
> position or time [1]:
>
> "Every physicist would easily convince himself that all quantum
> calculations are made in the energy-momentum space and that the
> Minkowski x^b are just dummy variables without physical meaning
> (although almost all textbooks insist on the fact that these variables
> are not related with position, they use them to express locality of
> interactions!)"
My statement does not contradict this observation of Wigner.
>
>> For any relativistic theory possesses the Poincare group as a symmetry
>> group, whose infinitesimal generators satisfy the standard commutation
>> rules of the Poincare algebra. But given these, the standard
>> construction by Newton and Wigner gives (in each Lorentz frame) a
>> 3-dimensional position operator with commuting components, and the
>> associated conjugate momentum operators. These play exactly the same
>> role as the position and momentum operators in nonrelativistic quantum
>> mechanics.
>>
>> See Section S2g ''Particle positions and the position operator'' of my
>> theoretical physics FAQ at
>> http://www.mat.univie.ac.at/~neum/physics-faq.txt
>
> My original claim was about the lack or position operator x for particles
> in relativistic quantum field theory (RQFT).
Any relativistic quantum theory, whether a field theory or not,
must contain the generators of the Poincare group and hence has a
position operator, unless one restricts onself to the zero mass
sector of the theory.
> In your FAQ you have a section S2i. "Position operators in relativistic
> quantum field theory" where you write:
>
> "In relativistic quantum field theory in its usually given form,
> position is promoted to the same status as time, and hence becomes a
> parameter in the qwuantum field, while in quantum mechanics it is an
> operator vector."
>
> With the typo in the original. This part of your FAQ is right.
Thanks for spotting the misprint.
> Your section S2g ''Particle positions and the position operator'', is
> *not* about RQFT but about RQM.
It is about the Schroedinger picture, which, as shown there, must exists
whenever there is a consistent probability interpretation for spatial
detection.
''Therefore, if we have a quantum system defined in an arbitrary
Hilbert space in which a momentum operator is defined, the necessary
and sufficient condition for the existence of a spatial probability
interpretation of the system is the existence of a position operator
with commuting components which satisfy standard commutation
relations with the components of the momentum operator and the
angular momentum operator.''
This is valid independent of particle or field theory, and independent
of relativistic or not.
The second half of this section assumes an irreducible representation
of the Poincare group, and hence applies to all relativistic theories.
Of course, in a multiparticle theory or a field theory, the full
theory is in a reducible representation of this group, but the
center of mass of the whole system is in an inrreducible representation.
Thus the whole argument applies.
> This section is incorrect. In RQM there is an position operator x, but it
> cannot be consistently interpreted.
This claim is simply false.
I showed in section S2g that the interpretation of the Newton-Wigner
position operator is exactly the same as the position operator in
the nonrelativistic quantum mechanics of a single particle.
I never heard anyone say that this interpretation si not consistent.
> L.L. Foldy and S.A. Wouthuysen,
> On the Dirac Theory of Spin 1/2 Particles and Its Non-Relativistic
> Limit, Phys. Rev. 78 (1950), 29-36.
>
> They start by noticing some difficulties of the position operator x. Then
> they introduce an operator X in (23), and refer to it as the new
> "position" operator [...] They also write in a
> footnote that X is that Newton and Wigner found.
> Contrary to your claims, X is not the position operator associated to the
> particle but only an averaged mean-position operator.
This is their name for it. People can name things anything they like.
it doesn't alter the provable properties of the named object.
Their analysis is not the final arbiter on the question.
My analysis is completely different and unassailable.
> Moreover, the operator X is useless in RQFT;
No, unless you find the probability intepretation of quantum mechanics
useless. Most people don't.
The existence of the Newton-Wigner operator is necessary and
sufficient for a probability interpretation of being localized
at a point in 3-space.
Arnold Neumaier
ultram...@hotmail.com
Neumaier) What is the reason why position or momentum operators cannot
be defined for massless particles. Is the reason definitive? Can these
always be defined for massive particles with spin? In other words, I
want to understand the exact obstructions.
For Van Hees, I want to ask exactly what are the observables he claims
that dont depend on position, and WHY it is that only these are
important from a physical point of view, and separately I want to ask
whether we can do away with position altogeteher or do we need it, but
only within a boundary of error-- After all we care that the particles
that we'll do the experiment with ahve a large probability of being
inside the accelerator apparatus, no?
Arnold Neumaier
> Hi I just saw this post and would like to ask some questions: (esp to
> Neumaier) What is the reason why position or momentum operators cannot
> be defined for massless particles. Is the reason definitive?
Yes, except for spin 0 or 1/2. See Section S2g (Particle positions
and the position operator) of my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
> Can these always be defined for massive particles with spin?
Yes. See again the above.
> In other words, I want to understand the exact obstructions.
The main reason is that, in the massive case, the algebra of
expressions in p, q and the spin generators is isomorphic to the
algebra of expressions in the Poincare generators and the spin
generators on the corresponding Hilbet space, while in the
massless case, the latter space is too small when only one
irreducible representation is present.
> For Van Hees, I want to ask exactly what are the observables he claims
> that dont depend on position,
Scattering cross sections do not depend on position but only on
angles.
> and WHY it is that only these are
> important from a physical point of view, and separately I want to ask
> whether we can do away with position altogteher or do we need it,
The position operator exists for any massive system separated enough
to be approximated well by an irreducible representation of the
Poincare group. Thus the question whether or not it is needed
is practically irrelevant.
But it is needed to give the concept of ''probability of being
in an arbitrary circular region of space'' a logically coherent
meaning. For a photon, one can attach to this phrase no clear
meaning since the photon is massless.
> but only within a boundary of error-- After all we care that the particles
> that we'll do the experiment with have a large probability of being
> inside the accelerator apparatus, no?
True, but these particles are all massive, so there is no problem.
Arnold Neumaier.
Hendrik van Hees
> But it is needed to give the concept of ''probability of being
> in an arbitrary circular region of space'' a logically coherent
> meaning. For a photon, one can attach to this phrase no clear
> meaning since the photon is massless.
From a mathematical point of view you may be right, but everyday
experience tells us that we are able to use light for very precise
measurements of position. Every time, you point with a laser pointer
during a talk you are giving, you disprove what you are saying. You
can point pretty precisely where you want to point.
The reason is simply that the wall you point to is massive, and thus
the interaction region of the em. wave with the massive wall, leading
to reflection (and of course also some absorption) of the wave, so
that you can see the spot on the wall. Again, in your eye the light
wave affects some small well-defined region on your retina leading to
the picture of the spot after some signal processing in your brain.
That's why I said previously that we don't need a position operator
for photons to understand its measurable properties: The location of
a photon within the uncertainty given by the detector's resolution at
the moment of measurement is simply defined by the position of the
detector where it is registered.
It's clear that we cannot measure a photon's position without a
detector of any kind. Thus in practice the lack of a position
operator for photons is quite irrelevant.
Arnold Neumaier
> Arnold Neumaier wrote:
>
>> But it is needed to give the concept of ''probability of being
>> in an arbitrary circular region of space'' a logically coherent
>> meaning. For a photon, one can attach to this phrase no clear
>> meaning since the photon is massless.
>
> From a mathematical point of view you may be right, but everyday
> experience tells us that we are able to use light for very precise
> measurements of position. Every time, you point with a laser pointer
> during a talk you are giving, you disprove what you are saying. You
> can point pretty precisely where you want to point.
Nobody knowws where a photon in flight is. One at best knows where
a photon had been in the moment it died. A photon in a monochromatic
beam is smeared along the whole beam (or at least a very extended
portion of it) and has no definite position. This is obvious from
the fact that its eigentime is the same everywhere along the beam -
it at all these positions simultaneously.
> The reason is simply that the wall you point to is massive, and thus
> the interaction region of the em. wave with the massive wall, leading
> to reflection (and of course also some absorption) of the wave, so
> that you can see the spot on the wall. Again, in your eye the light
> wave affects some small well-defined region on your retina leading to
> the picture of the spot after some signal processing in your brain.
The position where a photon is detected is indeed fixed by where
the beam ends, by interacting with a piece of massive matter that
is absorbing enough.
But it is the position of the absorbing material and not the
position of the photon that is observed.
This is fully consistent with the fact that there is no position
operator for a photon, but that there is one for massive objects.
> That's why I said previously that we don't need a position operator
> for photons to understand its measurable properties: The location of
> a photon within the uncertainty given by the detector's resolution at
> the moment of measurement is simply defined by the position of the
> detector where it is registered.
>
> It's clear that we cannot measure a photon's position without a
> detector of any kind. Thus in practice the lack of a position
> operator for photons is quite irrelevant.
The lack of a position operator gives just the right insight to
interpret everything consistently.
Arnold Neumaier
Hendrik van Hees
Arnold Neumaier wrote:
> Nobody knowws where a photon in flight is. One at best knows where
> a photon had been in the moment it died. A photon in a monochromatic
> beam is smeared along the whole beam (or at least a very extended
> portion of it) and has no definite position. This is obvious from
> the fact that its eigentime is the same everywhere along the beam -
> it at all these positions simultaneously.
Sure, but since a photon is not observable without interaction with a
detector, it doesn't even make sense to ask where it is without
detecting it.
> But it is the position of the absorbing material and not the
> position of the photon that is observed.
Exactly: Since there's no position operator for a photon, one cannot
define the observable position within QED. Thus within QED the very
notion of a photon's position is meaningless.
>
> This is fully consistent with the fact that there is no position
> operator for a photon, but that there is one for massive objects.
Precisely!
Marc Nardmann
> What is the reason why position or momentum operators cannot
> be defined for massless particles. Is the reason definitive?
Arnold Neumaier replied:
> Yes, except for spin 0 or 1/2. See Section S2g (Particle positions
> and the position operator) of my theoretical physics FAQ at
> http://www.mat.univie.ac.at/~neum/physics-faq.txt
A position *observable* for photons can be defined as a positive
operator-valued measure (POVM); this observable has all desired
properties. POVM measures are the today (quite) generally accepted
mathematical formalisation of the notion "observable" in quantum theory
(cf. e.g. "POVM" in Wikipedia). POVM observables are more general than
"operator observables" (which can be described equivalently as
projection-valued measures). Therefore Wightman's no-go theorem does not
apply to them.
This should be mentioned in the FAQ, I guess. I can't give a precise
reference right now, because I don't have access to the literature. If I
remember correctly, the construction of the photon position observable
goes back to Karl Kraus (~1980). It should be mentioned in any decent
book which discusses POVM observables.
-- Marc Nardmann
Hendrik van Hees
along the lines you suggested.
I guess the idea is precisely to formalize what I tried to say in the
handwaving physicist's way: The photon position is defined by
interactions with the measurement device (like a photo plate or more
modern versions thereof like CCDs). Isn't this the physics starting
point to introduce POVMs as generalized state-preparation descriptions?
One considers the measurement as coupling of the object (or observable)
to measure with the measurement device, described as another quantum
system (of "macroscopic size") and then to trace out the measurement
device again, leading to a mixed state of the measured object, which in
general is not decomposed into mutually orthogonal projectors but more
generally into non-orthogonal ones.
I think, a nice book to learn about this formalism is
A. Peres, Quantum Mechanics Concepts and Methods, Kluwer
I'm not sure whether the definition (or better formalization) of
photon-position measurements in terms of POVMs can be found in this
book, since I don't have it available at the moment.
Marc Nardmann wrote:
> A position *observable* for photons can be defined as a positive
> operator-valued measure (POVM); this observable has all desired
> properties. POVM measures are the today (quite) generally accepted
> mathematical formalisation of the notion "observable" in quantum
> theory (cf. e.g. "POVM" in Wikipedia). POVM observables are more
> general than "operator observables" (which can be described
> equivalently as projection-valued measures). Therefore Wightman's
> no-go theorem does not apply to them.
>
> This should be mentioned in the FAQ, I guess. I can't give a precise
> reference right now, because I don't have access to the literature. If
> I remember correctly, the construction of the photon position
> observable goes back to Karl Kraus (~1980). It should be mentioned in
> any decent book which discusses POVM observables.
--
Hendrik van Hees Institut fï¿œr Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universitï¿œt Gieï¿œen
Fax: +49 641 99-33309 D-35392 Gieï¿œen
http://theory.gsi.de/~vanhees/faq/
Marc Nardmann
> That sounds interesting. I cannot find anything with Google Scholar
> along the lines you suggested.
E.g.: Busch, Grabowski, Lahti: Operational quantum physics. The pages
92-94 (which can be seen on Google Books) contain a short introduction
and references (but one would need a hardcopy of the book to read the
bibliography and find out what their references 3.27 and 3.28 actually are):
The authors give one definition of a photon position observable. They
say that there are many nonequivalent possible definitions.
-- Marc Nardmann
Juan R.
> Juan R. Gonz�lez-�lvarez schrieb:
>> Arnold Neumaier wrote on Thu, 14 May 2009 21:23:51 +0200:
>>
>>> juanR...@canonicalscience.com schrieb:
>>>> Hendrik van Hees wrote on Sun, 10 May 2009 19:00:59 +0000:
(...)
> It is about the Schroedinger picture, which, as shown there, must exists
> whenever there is a consistent probability interpretation for spatial
> detection.
> ''Therefore, if we have a quantum system defined in an arbitrary Hilbert
> space in which a momentum operator is defined, the necessary and
> sufficient condition for the existence of a spatial probability
> interpretation of the system is the existence of a position operator
> with commuting components which satisfy standard commutation relations
> with the components of the momentum operator and the angular momentum
> operator.''
> This is valid independent of particle or field theory, and independent
> of relativistic or not.
The conmutation relations in non-relativistic theory give observables x
and p. In relativistic theory the same relations forbid existence of an
observable x due to light cone causality restrictions.
The imposibility to use x as observable was finally the reason which x was
downgraded to parameter in RQFT.
And it is also the reason which one-particle states in RQFT are given by
spin-helicity, charge, mass and four-momenta p^b (See Weinberg cited
below), but not by *x*.
As explained in the page 10 of Mandl and Shaw well-known texbook (revised
edition) in quantum field theory, *x* is an operator in quantum mechanics
but is *not* an operator in quantum field theory, being this,
"an important difference between a quantized field theory and
non-relativistic quantum mechanics."
http://www.amazon.com/Quantum-Field-Theory-F-Mandl/dp/0471105090
(...)
>> This section is incorrect. In RQM there is an position operator x, but
>> it cannot be consistently interpreted.
>
> This claim is simply false.
>
> I showed in section S2g that the interpretation of the Newton-Wigner
> position operator is exactly the same as the position operator in the
> nonrelativistic quantum mechanics of a single particle.
This is *not* I said. I wrote about *x* not about the NW operator *X*.
Moreover, the same references you cite in your FAQ
http://www.mat.univie.ac.at/~neum/physics-faq.txt
in "S2g. Particle positions and the position operator" clearly state that
*X* is not the position operator for a particle.
E.g. this is done clear in
L.L. Foldy and S.A. Wouthuysen,
On the Dirac Theory of Spin 1/2 Particles and Its Non-Relativistic
Limit, Phys. Rev. 78 (1950), 29-36.
explaining that *x* (in their notation) is the position operator in
relativistic quantum mechanics and *X* (again in their notation) is the
mean-position operator. *X* and *x* are different objects.
Foldy and Wouthuysen name x and x' (old and new representation
respectively) the *position operators* and X and X' the mean-position
operators.
(...)
>> Contrary to your claims, X is not the position operator associated to
>> the particle but only an averaged mean-position operator.
>
> This is their name for it. People can name things anything they like. it
> doesn't alter the provable properties of the named object.
(i)
Names "average-position" or "mean-position" operator are rather standard
in the literature in the topic of localization.
(ii)
Choice of name was not arbitrary because *X*, the mean-position operator,
verifies
x = X + deviation
as explained in above Foldy and Wouthuysen reference.
If you *average* the operator x over a region of spacetime of order of the
Zbw you will find that deviation term cancel out and x_avg = X.
(iii)
The fact that *X* coincides with an *averaged* x is the reason which *X*
is named the "averaged" or "mean-position" operator.
*X* would not be confused with *x*, the true position operator for a
particle.
>> Moreover, the operator X is useless in RQFT;
>
> No, unless you find the probability intepretation of quantum mechanics
> useless. Most people don't.
The well-known imposibility to maintain a consistent probability
interpretation for relativistic localization was the reason which the
position operator x of QM was downgraded to *parameter* in RQFT.
RQFT is formulated over spacetime parameters x and t. You even recognize
this in another part of your FAQ
"S2i. Position operators in relativistic quantum field theory"
where you write:
"In relativistic quantum field theory in its usually given form,
position is promoted to the same status as time, and hence becomes a
parameter in the quantum field, while in quantum mechanics it is an
operator vector.
This poses the question of whether there is a position operator in
relativistic quantum field theory. Many people think that there is none.
But even though there is a parameter called x and referred to as
4-dimensional position, there is also an vector defining a 3-dimensional
position operator, provided the relativistic system under consideration
is not massless."
First you just talk about RQFT, where as I just said in this thread there
is not position operator but position parameter.
Then you seem to think this choice was fortitous or that many people think
so, but x is an parameter because during development of quantum field
theory was showed that no consistent position operator x exist.
Finally you confound the quantum mechanics position operator x with the
Newton-Wigner *mean-position* operator, which do not describe the position
of a relativistic particle but only in an average sense.
The references you cite in your FAQ are about relativistic quantum
mechanics, where the Dirac and Klein gordon equations are wave equations
and x is an operator.
In relativistic quanutm field theory the Dirac and Klein gordon equations
are *not* wave equations and x is a parameter.
Lacking one x operator, the NW decomposition
x = X + deviation
makes no sense because x is not more an operator in quantum field theory
and the NW operator X play no role in RQFT. I am still to find a textboook
in quantum field theory using the Newton Wigner operator *X* for
something, less still in some fundational issue as your above claim that X
is needed for "the probability intepretation of quantum mechanics".
Quantum field theory is formulated without a position operator x, the
probabilistic interpretation holds in the dynamical variables for quantum
fields and the Newton Wigner operator X plays no fundamental role in any
treatment of quantum field theory I know
http://www.amazon.com/Quantum-Field-Theory-F-Mandl/dp/0471105090
http://www.amazon.com/Quantum-Field-Theory-Modern-Introduction/dp/
0195076524/ref=sip_rech_dp_8/182-8162622-0161351
http://www.amazon.com/Quantum-Theory-Fields-Foundations/dp/0521670535/
ref=sr_1_1?ie=UTF8&s=books&qid=1243674490&sr=1-1
Etc.
The fact is that particles just cannot be consistently localized in RQFT,
reason which quantum field theory is formulated over parameters x which
are not related to physical measurement of position, as remarked in the
next reference:
"Every physicist would easily convince himself that all quantum
calculations are made in the energy-momentum space and that the
Minkowski x^b are just dummy variables without physical meaning
(although almost all textbooks insist on the fact that these variables
are not related with position, they use them to express locality of
interactions!)"
The notions of localizability and space: from Eugene Wigner to Alain
Arnold Neumaier
Marc Nardmann schrieb:
I added the following to Section 2g of my FAQ:
If the concept of an observable is not tied to that of a Hermitian
operator but rather to that of a POVM (positive operator-valued
measure), there is more flexibility, and covariant POVMs for positon
measurements can be meaningfully defined, even for photons. See, e.g.,
A. Peres and D.R. Terno,
Quantum Information and Relativity Theory,
Rev. Mod. Phys. 76 (2004), 93.
[see, in particular, (52)]
K. Kraus, Position observable of the photon, in:
The Uncertainty Principle and Foundations of Quantum Mechanics,
Eds. W. C. Price and S. S. Chissick,
John Wiley & Sons, New York, pp. 293-320, 1976.
M. Toller,
Localization of events in space-time,
Phys. Rev. A 59, 960 (1999).
P. Busch, M. Grabowski, P. J. Lahti,
Operational Quantum Physics,
Springer-Verlag, Berlin Heidelberg 1995, pp.92-94.
Note that a POVM describes the statistics of a measurement process
rather than some underlying reality. This is reflected in the fact
that there are many possible nonequivalent possible definitions of
POVMs, all pertaining to possible different ways to get a measured
position.
Therefore, the POVM does not allow to talk about the position of a
photon - which could exist only if the corresponding operator existed -,
but only about th measured position: The photon is somewhere near the
value obtained by the measurement, without any more definite statement
being possible.
Arnold Neumaier
student
wrote:
>
> Note that a POVM describes the statistics of a measurement process
> rather than some underlying reality. This is reflected in the fact
> that there are many possible nonequivalent possible definitions of
> POVMs, all pertaining to possible different ways to get a measured
> position.
I think you are bringing in your interpretational preferences here.
POVMs include PVMs (i.e., 'Hermitian operator' observables). Like PVMs,
they correspond to a number of possible implementations (there is more
than one way to measure any given observable). Like PVMs, they give the
answer to ANY statistical question about the measurement outcomes. Like
PVMs, there is a theory describing the state after measurement, based on
details of the measuring interaction. Operationally speaking, they have
precisely as much 'reality' as PVMs do.
Further, they are necessary for describing the maximal information that
can be extracted from quantum communication channels, as well as for
describing the optimal measurement for discriminating between members of
a set of (more than two) non-orthogonal states. So how much 'reality'
are you prepared to give up?! Note also that POVMs can always in
principle be implemented via unitary interaction with a probe followed
by measurement of a PVM ('Hermitian operator') on the probe. In this
sense they are a necessary (and full) completion of the naive concept of
observable.
> Therefore, the POVM does not allow to talk about the position of a
> photon - which could exist only if the corresponding operator existed -,
> but only about th measured position: The photon is somewhere near the
> value obtained by the measurement, without any more definite statement
> being possible.
I guess this makes your interpretational bias clearer, but it doesn't
justify it. You distinguish 'the position' from 'the measured position'.
However, you might recall that Bohr (and Wheeler) made no such
distinction ('no phenomenon is a phenomenon until it is an observed
phenomenon'). Hence you appear to be trying to add an extra element to
the basic formalism.
Actually, by your own argument, nothing has a 'real' position. In
particular, consider measuring the position of an electron, and getting
the result X=a. So, where is the electron? One thing is for sure: it
is not described by the eigenket |a> (which is not a member of the
Hilbert space, and would have infinite energy). So, we only know that
the electron is 'somewhere near the value obtained by measurement' -
nothing to do with PVMs vs POVMs!
Finally, it is worth recalling von Neumann's famous "Grundlagen der
Quantenmechanik (1932?), published in English as "Foundations of quantum
mechanics". In the last section he describes a position measurement -
via the displacement of a probe (i.e, 'meter' or 'measuring apparatus'),
that is coupled to the particle. The statistics of von Neumann's
measurement are NOT described by a Hermitian operator on the Hilbert
space of the system. They are described by a more general POVM. Case
closed.
Arnold Neumaier
[on june 28 in the thread ``Photon position observable'';
sorry for my late reply - I was busy with other work]
> On Jun 26, 4:41 pm, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
>> Note that a POVM describes the statistics of a measurement process
>> rather than some underlying reality. This is reflected in the fact
>> that there are many possible nonequivalent possible definitions of
>> POVMs, all pertaining to possible different ways to get a measured
>> position.
>
> I think you are bringing in your interpretational preferences here.
Yes. I do this always - whenever I write a mail, or a piece in the FAQ.
Without preferences no quality. I try to present the _best_ view,
and ``best'' always implies interpretational preferences.
> Further, they are necessary for describing the maximal information that
> can be extracted from quantum communication channels, as well as for
> describing the optimal measurement for discriminating between members of
> a set of (more than two) non-orthogonal states. So how much 'reality'
> are you prepared to give up?! Note also that POVMs can always in
> principle be implemented via unitary interaction with a probe followed
> by measurement of a PVM ('Hermitian operator') on the probe. In this
> sense they are a necessary (and full) completion of the naive concept of
> observable.
For my - less naive than you may think - view of observables (including
POVMs), see Chapters 7.4 and 7.5 of my book
A. Neumaier and D. Westra,
Classical and Quantum Mechanics via Lie algebras,
arXiv:0810.1019
I think this represents actual practice better than the idealized
pictures found in common textbooks.
>> Therefore, the POVM does not allow to talk about the position of a
>> photon - which could exist only if the corresponding operator existed -,
>> but only about the measured position: The photon is somewhere near the
>> value obtained by the measurement, without any more definite statement
>> being possible.
>
> I guess this makes your interpretational bias clearer, but it doesn't
> justify it. You distinguish 'the position' from 'the measured position'.
> However, you might recall that Bohr (and Wheeler) made no such
> distinction ('no phenomenon is a phenomenon until it is an observed
> phenomenon'). Hence you appear to be trying to add an extra element to
> the basic formalism.
The distinguishing point (that is independent of any nonstandard
interpretation) is that for a PVM, it is possible to have (generalized)
states that give a particular response with certainty, while for a POVM,
this is generally not the case.
> Actually, by your own argument, nothing has a 'real' position. In
> particular, consider measuring the position of an electron, and getting
> the result X=a. So, where is the electron? One thing is for sure: it
> is not described by the eigenket |a> (which is not a member of the
> Hilbert space, and would have infinite energy).
This is because an idealized von-Neumann measurement that would allow
to infer |a> is impossible for continuous observables. But it is
theoretically possible to make an ideal measurement that establishes
position in a particular box, and then it is known that the state of
the electron is a superpostions of eigenkets |a> with a in this box.
In this sense, the position of an electron is more real than the
position of a photon.
> So, we only know that
> the electron is 'somewhere near the value obtained by measurement' -
> nothing to do with PVMs vs POVMs!
With the usual POVM's for photon position, you cannot establish the
existence of exactly one particle in a particular box, while with a
PVM for electron position, you can.
But in the light of your comments, I modified the passage in Section
S2g. Particle positions and the position operator of my theoreetical
physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
which now reads as follows:
<begin quote>
Note that a POVM describes the statistics of a particular measurement
process rather than some underlying reality. This is reflected in the
fact that there are many possible nonequivalent possible definitions
of POVMs, pertaining to the possible different ways to get a measured
position.
Therefore, the concept of a photon position is necessarily subjective,
since it depends on the POVM used, hence on the way the
measurement is performed. It does not describe something objective.
The POVM does not allow one to talk about the position of a photon
- which could exist only if the corresponding operator existed -,
but only about the measured position: The photon is somewhere near the
range of values established by the measurement, without any more
definite statement being possible. On the other hand, for observables
corresponding to Hermitian operators, there are states in which
a definite statement is (at least theoretically) possible that the
observable has a value in a given range.
<end quote>
Arnold Neumaier