5 views

Michael Weiss

Apr 2, 1999, 3:00:00 AM4/2/99
to
When last we met, we were looking at a vector space of
real-valued solutions to the classical wave equation on the line:

u_xx = u_tt

(where subscripts stand for partial derivatives, so this says that
d^2 u / dt^2 = d^2 u / dx^2 --- if you squint your eyes so the d's
look curly.)

The father and mother of all solutions are the travelling waves:

u(t,x) = cos(kx-wt), w = |k|
u(t,x) = sin(kx-wt), w = |k|

indeed, these form a "basis" --- where the scare-quotes mean, "we're
really doing Fourier transforms, but the formulas look a lot like the
usual formulas from intro linear algebra, except that the summation
signs are replaced with integrals."

The master plan is to turn this into a complex Hilbert space, called
the "1-foton space", and then form the Fock space of *that*; the Fock
space is also called the "multi-foton space".

[If you're listening, Oz, we're just following the outline I gave in
my last post. I talked about stuff like chi, and chi\$psi, and and a_0
+ a_1 chi + a_2 chi\$chi + ..., where the a_n are complex numbers.
There chi and psi were complex-valued functions defined over 3d space.

Now we're doing this in more detail, so we simplify the setting a bit.
(Also we'll use 'u' and 'v' instead of 'chi' and 'psi', and we call
our particles 'fotons' instead of 'photons'. See? *Much* simpler.)
First, we'll do functions on a line instead of functions on 3d space.
Think violin string. Think *infinite* violin string (great vibrato!)

Next, we put the 'wave' in 'wavefunction' by letting our functions
varying in time, and making them obey a wave-equation. (We *force*
them to obey the wave-equation. Any waves that would rather misbehave
like ozons get annihilated!) We picked the classical wave-equation,
which means "speed 1, all the time". Think light.

Finally, we'll do real-valued waves instead of complex-valued waves.
This means that fotons are neutral--- complex-valued waves correspond
to *charged* particles. No, I don't know why! But if we march
forward long enough, we'll find out.

Oops, the professor wants to know who's whispering in the back row!
Time to pay attention.]

So to get a complex Hilbert space we need to define two things, for
any vectors u(t,x) and v(t,x) in our space:

(a) The complex structure: what (a+bi) times u(t,x) means.
(b) The inner product: what <u(t,x), v(t,x)> is.

[What's that, Oz, you want to *why* we want to do this? Because the
professor said so!

Seriously, though, it's a puzzlement to me too. In plain old
non-relativistic QM, you've got a complex-valued function psi ---
let's say psi(x), defined on the line. The probability of finding
your particle in a given interval is proportional to the integral of
|psi|^2 over that interval.

That's the famous Born rule (one form of it, at least), and it makes
the psi function *really concrete*. An old familiar friend, nice and
easy to visualize. Indeed, when Schroedinger introduced his psi, and
physicists could deal with *that* instead of the
Heisenberg-Born-Jordan "shut up and calculate" matrices, well, they
all breathed a huge sigh of relief.

It would be nice to report that the answers to (a) and (b) will tell
us how to calculate the probability of finding a foton in a given
interval. (Note that |psi|^2 has a lot to do with the inner-product,
for the non-rel QM case.) Alas, they DON'T. Or if they do, HOW they
do is a closely guarded secret. The professor isn't telling.

Just between you and me and the internet, this particular puzzlement
is a real hang-up for me. It's the main conceptual reason (on my
end) the Schmoton thread went into hibernation so long.

The professor thinks we should just march forward, doing exercises and
building simple wooden boxes, and eventually the fog will lift and
we'll all say, "Why that's obvious!"

Maybe he's right. Or maybe he *try really hard to come up with some
motivation*, even it just means gesturing expansively for one class,
before we buckle back down to work.

Oops, I don't think he likes pedagogical advice from the back row...]

enough to define i times cos(kx-wt) and i times sin(kx-wt). To avoid
confusion, we write J instead of i, and note that J had better be a
linear operator on our vector space, satisfying the equation J^2 = -1.

The definition of J on our "basis" vectors is ridiculously simple:

J( cos(kx-wt) ) = -sin(kx-wt)
J( sin(kx-wt) ) = cos(kx-wt)

Of course, there was an elaborate patter song leading up to this. The
gimmick is to think of (u, u_t) as a point (q,p) in phase space. The
basis vectors simply cycle round, making big lazy circles in phase
space. Or actually big lazy *ellipses*: if u = cos(kx-wt), then the
pair (u, u_t) is just (cos(kx-wt), w sin(kx-wt)), which traces out an
ellipse. But by rescaling the p-axis, we can turn the ellipse into a
circle. And then the linear operation "turn back a quarter period"
starts looks like a *very* good candidate for the operator J. This
line of thought leads to the equation:

J(q,p) = (-p/w, wq) (if q = cos(kx-wt), p = w sin(kx-wt)
or if q = sin(kx-wt), p = -w cos(kx-wt))

So much for (a). How about (b)?

Well, that's not so clear. Somewhere in this thread, Prof. Wiz imposed
these requirements on a "good" complex structure:

Im <u,v> = w(u,v)
Re <u,v> = w(u,Jv)

where, just to confuse things further, w (i.e., omega) is the
symplectic form on our space, and not the angular frequency appearing
in the combination kx-wt.

I don't think he ever gave a formula for the symplectic form,
though...

Well, I can guess as well as the next fellow. The symplectic form on
a symplectic vector space eats two vectors and spits out a number. On
plain old 2D Euclidean space, w((q1,p1), (q2,p2)) = q1 p2 - q2 p1.
(You'll find a more extensive discussion somewhere on the schmotons

w(u,v) = integral u v_t - u_t v d(something)

What should we integrate over? And what should we use for
d(something)?

Well, let's integrate over the entire line! After all, u and v are
functions from -infinity to +infinity. (That's because we decided to
do the infinite violin string.)

As for d(something), after deep thought, mental chewing, extensive
cogitation, much fumbling and mumbling and mulling ---- I haven't the
foggiest idea. *Somehow* the choice of d(something) has to lead to a
relativistically invariant result.

We could Fourier transform u and v into U and V:

U(w,k) = (constant) integral exp(-ikx+wt) u(t,x) d(something)
V(w,k) = (constant) integral exp(-ikx+wt) v(t,x) d(something)

Maybe the symplectic form should be defined in terms of U and V?

john baez

Apr 4, 1999, 4:00:00 AM4/4/99
to
In article <7e11gn\$5b6\$1...@autumn.news.rcn.net>,

Michael Weiss <spam...@spamfree.net> wrote:
>When last we met, we were looking at a vector space of
>real-valued solutions to the classical wave equation on the line:
>
>u_xx = u_tt
>
>(where subscripts stand for partial derivatives, so this says that
>d^2 u / dt^2 = d^2 u / dx^2 --- if you squint your eyes so the d's
>look curly.)

When I squint they just look blurry. But frankly, anyone who needs
a curly d to remind them that the derivative is a partial one shouldn't
be playing this game. And for some reason that reminds me of an old
friend of mine who felt ripped-off in one of his math classes because
they only discussed "partial differential equations" - he wanted to

>The father and mother of all solutions are the travelling waves:
>
>u(t,x) = cos(kx-wt), w = |k|
>u(t,x) = sin(kx-wt), w = |k|
>
>indeed, these form a "basis" --- where the scare-quotes mean, "we're
>really doing Fourier transforms, but the formulas look a lot like the
>usual formulas from intro linear algebra, except that the summation
>signs are replaced with integrals."

Yes, and you'll know you're truly a physicist when you stop putting
quotes around the word "basis" like that. Only mathematician wimps
worry about whether their basis vectors actually like in the vector
space in question!

>The master plan is to turn this into a complex Hilbert space, called
>the "1-foton space", and then form the Fock space of *that*; the Fock
>space is also called the "multi-foton space".

Sounds good to me.

>So to get a complex Hilbert space we need to define two things, for
>any vectors u(t,x) and v(t,x) in our space:
>
>(a) The complex structure: what (a+bi) times u(t,x) means.
>(b) The inner product: what <u(t,x), v(t,x)> is.

Sounds good me.

>[What's that, Oz, you want to *why* we want to do this? Because the
>professor said so!

And to quote Firesign Theater: "I'm always right and I never lie."

Seriously, we can do a little better than argument by authority here.
We *need* a complex Hilbert space to do quantum theory! We want to
have self-adjoint operators to measure things like the energy (the
"Hamiltonian"), and we want to be able to calculate inner products
to figure out transition amplitudes. Without a complex Hilbert space,
we're sunk!

But you know all that, surely. You're not actually wondering why
we need a complex Hilbert space of foton states; you're just wondering
about the seemingly arcane procedure we're using to get our hands
on this complex Hilbert space. Would you feel better if I told
you there is a *unique* way to make the space of solutions of the
wave equation into a complex Hilbert space which is 1) invariant
under translations, rotations, reflections and Lorentz transformations
and 2) makes the Hamiltonian nonnegative? No? It certainly makes *me*
happy!

>Seriously, though, it's a puzzlement to me too. In plain old
>non-relativistic QM, you've got a complex-valued function psi ---
>let's say psi(x), defined on the line. The probability of finding
>your particle in a given interval is proportional to the integral of
>|psi|^2 over that interval.
>
>That's the famous Born rule (one form of it, at least), and it makes
>the psi function *really concrete*. An old familiar friend, nice and
>easy to visualize.

Heh. You teach 'em quantum mechanics and they whine about the
wavefunction psi. "Where did our point particles go?" they cry.
"This wavefunction is so ABSTRACT!" And the next thing you know,
you're teaching them quantum field theory and they say "where did
the wavefunction go? It was so CONCRETE!" You just can't win....

>It would be nice to report that the answers to (a) and (b) will tell
>us how to calculate the probability of finding a foton in a given
>interval. (Note that |psi|^2 has a lot to do with the inner-product,
>for the non-rel QM case.) Alas, they DON'T. Or if they do, HOW they
>do is a closely guarded secret. The professor isn't telling.

Actually, I don't think the concept of "finding a photon in a given
interval" makes much sense. The problem is that photons are massless
particles. Conventional wisdom is that there's no good "position
operator" for massless particles. If you'd like to switch gears
slightly and do a *massive* particle on the line, I'd be happy to
explain how what we're doing now is related to the nonrelativistic
Schrodinger equation you know and love.

Alternatively, we could talk about why there's no good position
operator for photons. I'm afraid I'd have to really learn a bit
That would be fun for me but perhaps less so for you.

Even without thinking too hard perhaps we can get a glimpse of
why it's tough to relate photoons to the nonrelativistic Schrodinger
equation and that psi you now claim is so "concrete". Photons move
at the speed of light! There ain't no nonrelativistic photons!
Also, there's a factor of 1/m in Schrodinger's equation, and photons
have zero mass!

>Just between you and me and the internet, this particular puzzlement
>is a real hang-up for me. It's the main conceptual reason (on my
>end) the Schmoton thread went into hibernation so long.

Really? And here I thought it was because you had a life.....

operator? I'll be glad to tell you the probability that a photon has
a given *momentum*! Do you really need to know about *position*???

>The professor thinks we should just march forward, doing exercises and
>building simple wooden boxes, and eventually the fog will lift and
>we'll all say, "Why that's obvious!"

Actually I do think it's good to understand what's going on. I really
didn't realize you were so desperately yearning for your good old
probability clouds in position space. I guess I'll either have to
get you to accept probability clouds in momentum space as an acceptable
substitute, or I'll have to come up with a clear explanation of why
you're not gonna get that position-space picture you love so much.

go make dinner. (Yes, behind every great wizard there is a great
witch, telling him he's toast if he doesn't hurry up and make
dinner....)

john baez

Apr 4, 1999, 4:00:00 AM4/4/99
to
In article <7e11gn\$5b6\$1...@autumn.news.rcn.net>,
Michael Weiss <spam...@spamfree.net> wrote:
>When last we met, we were looking at a vector space of
>real-valued solutions to the classical wave equation on the line:
>
>u_xx = u_tt
>
>(where subscripts stand for partial derivatives, so this says that
>d^2 u / dt^2 = d^2 u / dx^2 --- if you squint your eyes so the d's
>look curly.)

When I squint they just look blurry. But frankly, anyone who needs

a curly d to remind them that the derivative is a partial one shouldn't
be playing this game. And for some reason that reminds me of an old
friend of mine who felt ripped-off in one of his math classes because
they only discussed "partial differential equations" - he wanted to

>The father and mother of all solutions are the travelling waves:

>
>u(t,x) = cos(kx-wt), w = |k|
>u(t,x) = sin(kx-wt), w = |k|
>
>indeed, these form a "basis" --- where the scare-quotes mean, "we're
>really doing Fourier transforms, but the formulas look a lot like the
>usual formulas from intro linear algebra, except that the summation
>signs are replaced with integrals."

Yes, and you'll know you're truly a physicist when you stop putting

quotes around the word "basis" like that. Only mathematician wimps

worry about whether their basis vectors actually lie in the vector
space in question!

>The master plan is to turn this into a complex Hilbert space, called

>the "1-foton space", and then form the Fock space of *that*; the Fock
>space is also called the "multi-foton space".

Sounds good to me.

>So to get a complex Hilbert space we need to define two things, for
>any vectors u(t,x) and v(t,x) in our space:
>
>(a) The complex structure: what (a+bi) times u(t,x) means.
>(b) The inner product: what <u(t,x), v(t,x)> is.

Sounds good to me.

>[What's that, Oz, you want to *why* we want to do this? Because the
>professor said so!

And to quote Firesign Theater: "I'm always right and I never lie."

Seriously, we can do a little better than argument by authority here.
We *need* a complex Hilbert space to do quantum theory! We want to
have self-adjoint operators to measure things like the energy (the
"Hamiltonian"), and we want to be able to calculate inner products
to figure out transition amplitudes. Without a complex Hilbert space,
we're sunk!

But you know all that, surely. You're not actually wondering why
we need a complex Hilbert space of foton states; you're just wondering
about the seemingly arcane procedure we're using to get our hands
on this complex Hilbert space. Would you feel better if I told
you there is a *unique* way to make the space of solutions of the
wave equation into a complex Hilbert space which is 1) invariant
under translations, rotations, reflections and Lorentz transformations
and 2) makes the Hamiltonian nonnegative? No? It certainly makes *me*
happy!

>Seriously, though, it's a puzzlement to me too. In plain old

>non-relativistic QM, you've got a complex-valued function psi ---
>let's say psi(x), defined on the line. The probability of finding
>your particle in a given interval is proportional to the integral of
>|psi|^2 over that interval.
>
>That's the famous Born rule (one form of it, at least), and it makes
>the psi function *really concrete*. An old familiar friend, nice and
>easy to visualize.

Heh. You teach 'em quantum mechanics and they whine about the

wavefunction psi. "Where did our point particles go?" they cry.
"This wavefunction is so ABSTRACT!" And the next thing you know,
you're teaching them quantum field theory and they say "where did
the wavefunction go? It was so CONCRETE!" You just can't win....

>It would be nice to report that the answers to (a) and (b) will tell

>us how to calculate the probability of finding a foton in a given
>interval. (Note that |psi|^2 has a lot to do with the inner-product,
>for the non-rel QM case.) Alas, they DON'T. Or if they do, HOW they
>do is a closely guarded secret. The professor isn't telling.

Actually, I don't think the concept of "finding a photon in a given

interval" makes much sense. The problem is that photons are massless
particles. Conventional wisdom is that there's no good "position
operator" for massless particles. If you'd like to switch gears
slightly and do a *massive* particle on the line, I'd be happy to
explain how what we're doing now is related to the nonrelativistic
Schrodinger equation you know and love.

Alternatively, we could talk about why there's no good position
operator for photons. I'm afraid I'd have to really learn a bit
That would be fun for me but perhaps less so for you.

Even without thinking too hard perhaps we can get a glimpse of
why it's tough to relate photoons to the nonrelativistic Schrodinger
equation and that psi you now claim is so "concrete". Photons move
at the speed of light! There ain't no nonrelativistic photons!
Also, there's a factor of 1/m in Schrodinger's equation, and photons
have zero mass!

>Just between you and me and the internet, this particular puzzlement

>is a real hang-up for me. It's the main conceptual reason (on my
>end) the Schmoton thread went into hibernation so long.

Really? And here I thought it was because you, like, had a life.....

operator? I'll be glad to tell you the probability that a photon has
a given *momentum*! Do you really need to know about *position*???

>The professor thinks we should just march forward, doing exercises and

>building simple wooden boxes, and eventually the fog will lift and
>we'll all say, "Why that's obvious!"

Actually I do think it's good to understand what's going on. I really

john baez

Apr 6, 1999, 3:00:00 AM4/6/99
to
In article <7ed31k\$j...@edrn.newsguy.com>,
Daryl McCullough <da...@cogentex.com> wrote:

>I think what is confusing to some people (it was to me) in doing
>quantum field theory is that we're expecting there to be a wavefunction,
>and we think that the field amplitude u(x,t) is it. But it's not.
>
>Field amplitudes often don't have anything like a positive probability
>associated with them, for one thing.

Well, |u(x,t)|^2 is not a probability density, but you *can* compute
a probability density from u(x,t), at least in the case of the Klein-
Gordon equation.

>The field u(x,t) really is to be treated like a dynamical variable,
>like X(t) and P(t).

I think the word "really" is a bit too strong....

In my discussions with Michael I've been trying to emphasize the
2-faced nature of the space of solutions of the wave equation:

1) In the classical theory, the space H of solutions of the
wave equation plays the role of the classical phase space.

2) When equipped with a suitable complex structure, H becomes the
Hilbert spaces of 1-foton states in the quantum theory.

This is the origin of the term "second quantization": we can think
of vectors in H as states of a *classical* system, which we then
go ahead and quantize, or we can think of unit vectors in H as
states of a *quantum* system, which then go ahead and "second quantize"
by forming the Fock space on it.

You seem to be emphasizing side 1) above, which is fine, but I
think it's good to also remember side 2).

The simplest example is the harmonic oscillator, which Michael
and I have been discussing for a long time. But this is almost
too simple to understand, since the Hilbert space H in this case
is 1-dimensional. Michael used the word "notons" to describe
this quantum system, since it has no degrees of freedom. Perhaps
another good word would be "quanta"! An eigenstate of the harmonic
oscillator Hamiltonian can be regarded as composed of n such "quanta".

I think Michael understand this pretty well. My guess is that he's
worrying about the Hilbert space structure of H in the more complicated
case of "fotons". Part of the problem is that the familiar Schrodinger
equation is first-order, while the wave equation and Klein-Gordon
equation are second-order. This means that the probability density
for the Schrodinger equation is simply |psi|^2, while for the Klein-
Gordon equation it's a more complicated expression involving both
u and its first time derivative. I'll break down and tell Michael
what this expression is, and how to derive it, as soon as he gives
me permission to switch from the wave equation to the Klein-Gordon
equation - to avoid the added complications of massless particles.

james dolan

Apr 7, 1999, 3:00:00 AM4/7/99
to
daryl mccullough wrote:

-Since, to get something nonzero, each a must
-pair up with a corresponding a*,
-
- <0|(a + a*)^n |0>
-
-= the number of ways that the a's can pair
-up with the a*'s.

after talking it over with john baez the other day, i think i
understand most of what daryl said except for the sentence above,
which seems to me to involve a somewhat creative leap of logic
(perhaps in the spirit of the discussion so far).

let me try to act out an example here to a certain level of detail to
see if i can at least fake the mechanics of it.

i'll try the case n=4 again. (a+a*)^4 =

aaaa + a*aaa + aa*aa + a*a*aa +
aaa*a + a*aa*a + aa*a*a + a*a*a*a +
aaaa* + a*aaa* + aa*aa* + a*a*aa* +
aaa*a* + a*aa*a* + aa*a*a* + a*a*a*a*

in expanded form. then sufficient contemplation of daryl's facts #1-3
allegedly shows that each of the 16 terms above, when sandwiched
between "<0|" and "|0>", yields the number of feynman diagrams with
"chronological vertex profile" given by that term. that is,
interpreting "a" as standing for an annihilation vertex and "a*" for a
creation vertex, read the term from right to left (as daryl's
conventions seem to require) to obtain a sequence of four vertex
types, and then see how many feynman diagrams you can make from that
sequence. for example from the term "aaaa" in the upper left-hand
corner i get the sequence (annihilation, annihilation, annihilation,
annihilation), from which i can build a total of zero feynman
diagrams. in fact for all but two of the 16 terms, the sequence of
vertex types obtained from it admits no feynman diagrams. the two
feynman diagrams:

creation----------------------------------------annihilation

creation--------annihilation

and:

creation------------------------annihilation

creation------------------------annihilation

and "aa*aa*" which admits one feynman diagram:

creation--------annihilation creation--------annihilation

thus yielding a total of three feynman diagrams for all 16 terms put
together, which agrees with our previous calculations.

(in general, as each feynman diagram of the sort under consideration
has its unique sequence of vertex types, the number of feynman
diagrams you get from all the terms put together is equal to the total
number of feynman diagrams of that sort.)

that's about as much detail as i understand so far. it's not as close
yet as i'd like it to be to the aesthetic ideal of the attitude toward
combinatorics which prizes the giving of explicit and "robust"
bijections as explanations of equations between positive integers.

one more thing. while playing around i noticed that the equation:

integral("x^[2*n]*exp(-x^2/2)")/integral("exp(-x^2/2)")
= 1*3*...*(2*n-1)

is actually just a special case of the more general fact:

integral("x^[2*m*n]*exp(-x^[2*m]/(2*m))")/integral("exp(-x^[2*m]/(2*m))")
= 1*(2*m+1)*...*(2*m*(n-1)+1)

some even more general one) along the lines of what daryl and john and
others here have said about its special case? perhaps involving
combinatorics or feynman diagrams or quantum mechanics or integration
by parts or the gamma function or whatever?

john baez

Apr 8, 1999, 3:00:00 AM4/8/99
to
Okay, back to the technical task of quantizing the wave equation in
1+1 dimensions. We're taking the space of *real* solutions of the
wave equation and making it into a *complex* Hilbert space, so that
we can interpret unit vectors in this Hilbert space as states of a
massless neutral spin-0 particle - what Michael dubbed a "foton".

In article <7e11gn\$5b6\$1...@autumn.news.rcn.net>,
Michael Weiss <spam...@spamfree.net> wrote:

>So to get a complex Hilbert space we need to define two things, for
>any vectors u(t,x) and v(t,x) in our space:
>
>(a) The complex structure: what (a+bi) times u(t,x) means.
>(b) The inner product: what <u(t,x), v(t,x)> is.

Right.

If you don't mind, let me nitpick a little about basic mathematical
etiquette:

I wish you'd write u and v here instead of u(t,x) and v(t,x).
Let's save u(t,x) for the value of the function u the point (t,x),
and write just u when we are talking about the function u itself!

You may think I'm just being a grumpy old mathematician, and of course
I am, but I really don't want anyone to be fooled into thinking that
we're multiplying the *number* u(t,x) by (a+bi), or taking the inner
product of the *numbers* u(t,x) and v(t,x). We are really dealing
with the *functions* u and v.

Now I know you know this, Michael - you are probably writing things
this way to show off how you can mangle mathematics just like a real
physicist - but remember, there are impressionable young kids reading
this newsgroup, and we should set a good example.

>It would be nice to report that the answers to (a) and (b) will tell
>us how to calculate the probability of finding a foton in a given
>interval.

>Alas, they DON'T. Or if they do, HOW they do is a closely guarded

>secret. The professor isn't telling.

Well, I can promise you this: once we know the answers to (a) and
(b), we will know which operators on our Hilbert space are self-adjoint.
The self-adjoint operators are the ones that correspond to observables,
so we will be in the position to DETERMINE WHETHER THERE IS A DECENT
POSITION OPERATOR FOR FOTONS.

I think you'll agree that any decent position operator will be a
self-adjoint operator that transforms in the right way under translations:
translate your foton 2 meters left, the position operator increases by
2.

Once we get our complex Hilbert space, we can ask: does such an operator
exist? Is it unique? And I bet we can figure out the answers to
these questions! So we won't have to take anyone's word on these issues....

>Just between you and me and the internet, this particular puzzlement
>is a real hang-up for me.

Well, let's get rid of that hang-up.

>enough to define i times cos(kx-wt) and i times sin(kx-wt). To avoid
>confusion, we write J instead of i, and note that J had better be a
>linear operator on our vector space, satisfying the equation J^2 = -1.

Righto.

>The definition of J on our "basis" vectors is ridiculously simple:
>
>J( cos(kx-wt) ) = -sin(kx-wt)
>J( sin(kx-wt) ) = cos(kx-wt)

Yup.

>Of course, there was an elaborate patter song leading up to this.

Thanks, Michael - it's always nice to hear ones exposition described
as an "elaborate patter". Maybe I should rent a tux and learn to
tapdance while I explain this stuff.

>So much for (a). How about (b)?

>Well, that's not so clear. Somewhere in this thread, Prof. Wiz imposed
>these requirements on a "good" complex structure:
>
> Im <u,v> = w(u,v)
> Re <u,v> = w(u,Jv)
>
>where, just to confuse things further, w (i.e., omega) is the
>symplectic form on our space, and not the angular frequency appearing
>in the combination kx-wt.

Right. If you want to use a different letter for the symplectic
form, go ahead.... Anyway, the point is that we're taking a classical
phase space (in this case, a symplectic vector space) and equipping it
with some EXTRA STRUCTURE (in this case, a complex structure) in order
to quantize it. One typically needs a little EXTRA STRUCTURE to climb
from classical physics up to quantum physics, because the reverse
process - going from quantum to classical - forgets some structure.

>I don't think he ever gave a formula for the symplectic form,
>though...

Oh, but that's easy, once you understand the "elaborate patter".

>Well, I can guess as well as the next fellow. The symplectic form on
>a symplectic vector space eats two vectors and spits out a number. On
>plain old 2D Euclidean space, w((q1,p1), (q2,p2)) = q1 p2 - q2 p1.

Right, this is the standard symplectic structure for a classical
system with one degree of freedom. And we have seen how to decompose
the wave equation into a "direct integral" of simpler systems with
one degree of of freedom each. Each "mode" of the wave equation is
completely uncoupled to all the rest and acts like a 1-dimensional
harmonic oscillator. Glomming together the symplectic structures
for all these modes, we get the symplectic structure for the wave
equation!

>
> w(u,v) = integral u v_t - u_t v d(something)

An inspired guess!

>What should we integrate over? And what should we use for
>d(something)?

Well, maybe you should try out the simplest possible thing and
check it out on our "basis" of modes:

cos(kx-wt), sin(kx-wt) w = |k|

If you stick

u(t,x) = cos(kx-wt)

and

v(t,x) = sin(k'x-w't)

into your formula for the symplectic structure, you should get something
times a delta function delta(k-k'). Right? Do you see why this is what
we expect when we form a big symplectic vector space by glomming together
a bunch of little ones?

>Well, let's integrate over the entire line! After all, u and v are
>functions from -infinity to +infinity. (That's because we decided to
>do the infinite violin string.)

Sure, sounds good.

>As for d(something), after deep thought, mental chewing, extensive
>cogitation, much fumbling and mumbling and mulling ---- I haven't the
>foggiest idea. *Somehow* the choice of d(something) has to lead to a
>relativistically invariant result.

Well, dx is the only translation-invariant measure on the line, up
to a constant. You want relativistic invariance, so you certainly
want translation invariance, so if you're gonna integrate over the
line, you gotta integrate using the measure dx - or some number times
that.

So why don't you try that and see what you get for the particular
"basis" functions described above?

>We could Fourier transform u and v into U and V:
>
>U(w,k) = (constant) integral exp(-ikx+wt) u(t,x) d(something)
>V(w,k) = (constant) integral exp(-ikx+wt) v(t,x) d(something)
>
>Maybe the symplectic form should be defined in terms of U and V?

Well, we can do it this way too - both ways are easy and enlightening -
but why not try the other way first?

Michael Weiss

Apr 8, 1999, 3:00:00 AM4/8/99
to
Professor Wizard wants to talk about massive particles, where
Newton-Wigner localization provides an approximate sort of position
operator. I'm tempted--- I've had my sights on Newton-Wigner
localization for a while. But not just yet.

He's also referred to the massive literature on massless position
operators. Does it harbor the clues to the riddles that befuddle me?
Perhaps, but I have my doubts...

Let me begin with some less refined fare. I'll begin with a famous
old experiment, nearly bereft of mathematics, but elegant in its sheer
simplicity. Then I'll shift gears abruptly to mathematics--- rather
naive mathematics, but (I'm convinced) central the unravelling of my
confusions.

In 1910 or thereabouts, Geoffrey Taylor placed a very weak light
source in a box on one side of a two-pinhole screen. He placed a
photographic film on the other side, then he went sailing for a few
months. He calculated that there was (with overwhelming probability)
never more than one photon at a time in the box. Nonetheless, when he
returned from his topsails and spinnakers to develop the film, he
found the familiar two-point-source diffraction pattern--- the very
same diffraction pattern that Thomas Young observed over a hundred
years before.

Thomas Young's results we explain using field amplitudes. We square
the electric field, and use that as a measure of light intensity. Two
months before the mast, and the intensity has transformed itself into
a photon density.

This is old hat for Photon, Schmoton fans. The whole schmoton thread
got its start with the question: what is the relation between the
average photon density and the field amplitude? After puttering
around with some easy dimensional analysis, we launched into coherent
states and concluded that if a field has amplitude c, then the
matching coherent state has a Poisson distribution of quanta with
expectation value c^2/2. (This was a zero-dimensional field, i.e., a
field defined just at one point.)

All well and good--- but for a diffraction pattern, we'll need at a
two-dimensional field, at least! And for Taylor's experiment, we'd
like to say that the *probability* of detecting a photon at a given
spot on the film is *proportional* to the intensity you'd calculate
using Young's familiar field mathematics.

And Taylor's experiment is but a prototype. Leaf through the
literature of classical wave optics. Pick a page, and take the
low-intensity limit. Does the result survive, after replacing all
squared amplitudes (intensities) with photon detection probabilities?
I certainly hope so! (Answer no, and I'm ready to give ozons a second
look!)

But this means that the probability of detecting a photon *at a given
position* has to have *some* meaning in the brave new world of quantum
optics --- massless or no.

Do I need a photon position operator? Well, maybe. Or maybe I need a
quick course in scattering theory. After all, the schmoton thread has
resolutely stuck to *non-interacting* schmotons. But all those
non-interacting photons just slipped right through Taylor's box with
nary a trace....

I do see one out, of a sort. Taylor's experiment involved a
quasi-static quantum state of the electromagnetic field. After all,
no need to go sailing for months if your weak light source burns out
in two minutes! Perhaps for this special case, a spatial photon
probability density makes sense.

'Nuff physics, let's do some math. Last time ("The Lump in the
Carpet") I asked, with an aggrieved sense of injustice, why no one
bats an eyelash at a photon *momentum* operator. Surely we can
Fourier transform our way from k to x and back again to our heart's
content? I did have a bad feeling about x's without t's--- I've been
chided before for leaving out the 'c' in 'Schmoton' .
Still, how come k's without w's raise no eyebrows?

I now have a glimmer.

Fourier transforming u(x) into U(k) is a fine and dandy activity for a
lazy *non-relativistic* afternoon. Put that way, the solution is
obvious. Let u flood the spacetime plane, and Fourier transform
u(t,x) into U(w,k)!

U(w,k) = (const) integral exp i (kx-wt) u(t,x) dt dx
u(t,x) = (const) integral exp -i (kx-wt) U(w,k) dw dk

integrating over all spacetime, or over all of (w,k) space.

Now the great thing about dt dx is that it's relativistically
invariant. Ditto dw dk. I despair of rendering this in ascii-art,
but if you can't picture a nice square piece of dx dt stretching
itself area-preservingly into a rhombus, like one of those novelty
finger handcuffs, ---well, you need a good dose of Taylor and
Wheeler's Spacetime Physics.

Our u(t,x) does indeed flood the plain--- I can picture it shimmering
out in green and silver ripples as far as the eye can see. U(w,k) is
another school of fish entirely. We demanded that

u_tt = u_xx

fourmula above,

w^2 U(w,k) = k^2 U(w,k)

or w^2 = k^2. (Not a rigorous proof, but who cares?) (I think the
official lingo says that U is "supported on the light-cone".)

In other words, u(t,x) is made up of various sized dollops of waves
travelling left and right, but all with speed 1. (For more about the
plusses and minusses of +-k, +-w, see the "Seeing Double" episode in
http://math.ucr.edu/home/baez/schmoton, part II (gif or postscript).)

In the "Lump in the Carpet" post, I wondered what the formula for the
inner product of u and v could be, for u and v in our vector space of
foton wavefunctions. In non-relativistic QM, once you have an
inner-product, you're half-way to a probability measure.

I guessed:

integral (something with u(x)'s and v(x)'s) d(something)

but now a better bet looks like:

integral u(t,x)* v(t,x) dx dt (over all t,x)

or maybe

integral U(w,k)* V(w,k) dw dk (over all k^2 = w^2)

I guess I should pick one of these, Fourier transform it, and see what
I get...

But not now! (After all, what I *really* should be doing is my
taxes.)

But before signing off, let me pick some low-hanging from the previous
equations.

It's easy to see how we could scrounge a probability measure for k
from the second formula. After all, w^2=k^2, so w is virtually a
function of k! If we assume u is real (standard so far for fotons),
then U is completely determined by its values on the upper half (w>=0)
of the light cone. (U(-w,-k) = U(w,k)*.) We can restrict attention
to non-negative w, and set w=|k|.

So it would be natural to rewrite integral U(w,k)* V(w,k) dw dk as an
integral just over k. U(w,k) = U(|k|,k) = U(k), say. The integrand
becomes |U(k)|^2. The probability that a foton with momentum
wavefunction U(k) has momentum between A and B is proportional to the
integral of |U(k)|^2 dw dk, restricting attention to k's between A and
B. (You'll notice I'm being a bit coy about w. More on that in a
moment.)

For the *position* wavefunction u(t,x), we have no such reduction-- no
obvious way to get rid of t. It sticks to x like gum on its shoe.
Though I suppose in the static case (u(t,x) independent of t), we luck
out...

To finish the reduction to k alone--- to obliterate all traces of w---
we need to replace dw dk by a differential involving k alone.

Perplexing! After all, the light cone is a *one-dimensional* locus!
Its two-dimensional (dw dk) measure is zero. Talk about no visible
means of support! Of course this is one of those Dirac delta
thingies, but how to handle it?

After mulling and chewing and cogitating, I came up with this:

integral U(k)* V(k) dk/|k|

There's some geometric intuition behind this, but my post is long
enough already, and besides, Schedule B awaits...

Perhaps Professor Wiz would like to deduce something before *he's* up
to his neck in deductions!

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

 Nonetheless, Professor Wiz appears to have overstated his case,
Yiddish-wise.

Michael Weiss

Apr 12, 1999, 3:00:00 AM4/12/99
to
john baez wrote

>Would you feel better if I told
>you there is a *unique* way to make the space of solutions of the
>wave equation into a complex Hilbert space which is 1) invariant
>under translations, rotations, reflections and Lorentz transformations
>and 2) makes the Hamiltonian nonnegative

It does make me feel better. As I recall, *finding* that unique
Hilbertification was the last assignment you gave before the thread went on
summer hiatus.

>Heh. You teach 'em quantum mechanics and they whine about the
>wavefunction psi. "Where did our point particles go?" they cry.
>"This wavefunction is so ABSTRACT!" And the next thing you know,
>you're teaching them quantum field theory and they say "where did
>the wavefunction go? It was so CONCRETE!" You just can't win....

Not on Mondays, Wednesdays, or Fridays. And not on Tuesdays, Thursdays,
and Saturdays! Maybe on Sundays...

(Prof Wiz of course knows the old chestnut: that during the heyday of the
old quantum theory (1913-1925), when wave-particle duality was more than a
*philosophical* headache, physicists would say that the photon was a
particle on Mondays, Wednesdays, and Fridays, a wave on Tuesdays,
Thursdays, and Saturdays, and on Sundays it was a Divine Mystery.)

I wrote:

>>Just between you and me and the internet, this particular puzzlement
>>is a real hang-up for me. It's the main conceptual reason (on my
>>end) the Schmoton thread went into hibernation so long.

The professor replies:

>Really? And here I thought it was because you had a life.....

I said the main *conceptual* reason.

>Actually, I don't think the concept of "finding a photon in a given
>interval" makes much sense. The problem is that photons are massless
>particles. Conventional wisdom is that there's no good "position

>operator" for massless particles. ...I really

>didn't realize you were so desperately yearning for your good old
>probability clouds in position space. I guess I'll either have to
>get you to accept probability clouds in momentum space as an acceptable
>substitute, or I'll have to come up with a clear explanation of why
>you're not gonna get that position-space picture you love so much.

If the professor will permit a little class-hijacking, I'd like to spend
some time exploring this. But not in full wizardly detail! I can hear the
reproaches already: "...and so if you compute the semi-excellent topoidal
adjoint of the Wick rotation of the Froobish-Smedley photon pseudo-position
quasi-operator in the braided representation, you'll see that it fails to
satisfy the Baskerville equation. What's that? You tried to compute it,
and you got stuck because you needed the formula for the unique
Lorentz-invariant inner-product on foton-space that makes the Hamiltonian
non-negative... the homework I ASKED YOU TO DO A YEAR AGO!!!???"

I have in mind something a little more hand-wavy. You know what I mean:
"if we *did* try to define a position operator this way, we'd get stuck
because m=0. But Froobish suggested this work-around (I won't give all the
technical details)--- that's called a quasi-operator. However, then the
lump in the carpet shows up over here! ..." The sort of discussion you
give all the time on TWF, for the stuff that's new enough to you (and often
everyone) so it still looks complicated, and not as easy and familiar as
plain old quantum field theory!

Let me see how far I can get with this discussion in the rest of this post.
Like that kid in the old Platonic dialog, I sometimes wonder if I didn't
understand all this stuff in a previous existence and have just forgotten
it....

For starters, what's with this position operator jazz? The nice concrete
(and yes, *anschaulich* wavefunction psi--- as I remarked recently, the
dirty little secret of "collapse" is that it is *really easy to
visualize* --- pace on the philosophical migraines)--- the wavefunction
psi tells us the probability of finding our non-relativistic particle in
the interval [A,B]:

Prob(particle in [A,B]) = integral on [A,B] of psi(x)* psi(x) dx /
||psi||^2

where ||psi|| is the norm of psi:

||psi||^2 = integral over entire line of psi(x)* psi(x) dx

Now the position operator is a bit different. That helps us compute the
expectation value for the q observable for psi:

<psi, Q psi>/||psi||^2 = integral of psi(x)* x psi(x) dx / ||psi||^2, over
the whole line

so Q is the operator taking psi(x) to x psi(x). Using the grungy
frequentist interpretation of probability theory, that's the average value
I'd probably get if I measured the position of the particle a
zillion-jillion times. (Another pace for frequentist/bayesian debate. If
you prefer, that's the average value I'd bet schmotons on getting.)

Let's *pretend* I'd done my homework, and I knew what formula to use for
<u, v>, where now u and v are functions belonging to foton-space. So I
compute:

<u, Qu> / <u,u>

Oops, now I have to define Q! And that's not gonna be easy, or so we've
been warned. So that's *one* place the lump can show up.

But what about the *other way* to give some concreteness to psi: the
integral of psi(x)* psi(x) over the interval [A,B]?

Why can't I just say that, if:

<u,v> = integral (some expression with u and v) d(something) over the whole
line

then:

Prob(foton is in [A,B]) = integral (same stuff) over the interval [A,B]
??

Hmmmm... maybe a little Aristotelian logic can help. The "if X, then Y" I
just stated looks fine to me. Since we're told that Y doesn't cut it---
that there *is* no way to give meaning to the phrase "Prob(foton is in
[A,B])" --- then the antecedent must be to blame. (Gotta point the
finger-of-blame *somewhere*.) So maybe the formula for <u,v> *isn't* of
the form:

<u,v> = integral (some expression with u and v) d(something) over the whole
line

On the other hand, the professor said that:

>I'll be glad to tell you the probability that a photon has
>a given *momentum*!

Fourier-transforming u(x) and v(x) into U(k) and V(k), I guess this means
that the formula for <U,V> *does* have the form:

<U,V> = integral of (some expression with U and V) d(something) over the
whole line

and U(k), I should be writing u(t,x) and U(w,k). And relativity certainly
has some disapproving things to say about x's without t's and k's without
w's! But for *some* reason, in *this* case, x's without t's get read the
riot act, while k's without w's get a free pass and a ticket to ride!

Let's look at this a bit more physically. People certainly *do* do
experiments where they, "Hey, guys and gals, I just detected a photon! At
coordinates (3.256, 5.901, 4.443, 2.718), accurate to three decimal
places!" That's what all those diffraction pattern experiments are all
about! (Not to mention the EPR experiments, and Hanbury-Twiss, and
anti-bunchings, and dozens of others I'm sure I don't know about.)

How about an operational definition? Do a zillion-jillion experiments with
identically prepared fotons, all with the same foton-function u. Let's
assume we have *some* kind of foton detector that gives us *some*
approximate measurement of where the heck the foton popped up. The
handy-dandy Froobish film detector, accurate to 3 decimal places.

Let's say that our common foton-function u is chosen so that there's almost
no chance of detecting a foton outside the interval [-1000, +1000]. So all
but six or seven (let's say) of our foton-sighting are in this interval.
(There could be an ambiguous case or two, since Froobish film is only
accurate to 3 decimal places.) Surely we can take the ratio

fotons detected between 2 and 4 / total fotons detected

as an estimate for the probability of detecting a foton in the interval
[2,4]. (Again, there could be an ambiguous case or two, but we're
detecting several jillion fotons!)

I've emphasized the limited accuracy of the detector, because I recall that
you get into trouble if you go for *infinite* accuracy. Infinite accuracy
means infinite energy, which means fotons popping up all over the wazoo,
making messes on the furniture and cleaning out the refrigerator. (Think
tribbles. Think wizards and Mickey Mouse and brooms...) Also there was
some spicy math involved, something about densely-defined linear
operators....

This experimental arrangement has to have *some* representation in the
formalism. You can only sweep so much dirt under the carpet!

>I have to
>go make dinner.

Got it. Insomnia will only carry you so far. Besides, the taxes await.

Toby Bartels

Apr 16, 1999, 3:00:00 AM4/16/99
to
Hello again!

Michael Weiss wrote at last:

>After mulling and chewing and cogitating, I came up with this:

> integral U(k)* V(k) dk/|k|

>There's some geometric intuition behind this, but my post is long
>enough already, and besides, Schedule B awaits...

Having finished my taxes already, I can speak a bit on this matter.
I'm not about to confirm the integral_k U(k)* V(k) bit,
but dk/|k| is definitely the correct measure,
because it alone is Lorentz invariant.
This may be what you mean by "geometric intuition",
but I think Lorentz invariance is more than intuitive --
it's rigorously necessary.

-- Toby Bartels
to...@ugcs.caltech.edu

john baez

Apr 17, 1999, 3:00:00 AM4/17/99
to
In article <7etvac\$ls2\$1...@pravda.ucr.edu>,
Michael Weiss <spam...@spamfree.net> wrote:

[Geoffrey Taylor's marvelous experiment....]

>But this means that the probability of detecting a photon *at a given
>position* has to have *some* meaning in the brave new world of quantum
>optics --- massless or no.
>
>Do I need a photon position operator? Well, maybe.

Hmm. Maybe what you need is some photon number operators! Remember,
long ago we began this thread by tackling your question "what's the
expected number of photons per unit volume in laser light of a given
amplitude and frequency?"

It's easy to figure this out by a quick back-of-the-envelope calculation,
so we did that. Then we realized that this problem really involved just
a single vibrational mode of the electromagnetic field. Since each mode
acts just like a harmonic oscillator, we turned to discuss the harmonic
oscillator and related issues: coherent states, annihilation and creation
operators a and a*, and the number operator a*a. We figured out that in a
coherent state there was not a definite number of quanta, but instead
a Poisson distribution of various numbers of quanta.

Okay? We were on the right track and it wouldn't hurt much to stay
on it! If we want to understand quantum optics, diffraction patterns,
and all that jazz, we need some observables that tell us how many
photons there are in a given state, or set of states. Simplifying
a bit: take one molecule of silver iodide in Geoffrey Taylor's
photographic plate. Photons in a certain state will cause this
molecule to undergo a chemical reaction, creating a speck on the
plate. To figure out the probability that this will happen in a
certain amount of time, we need to know the expected number of photons
in this state.

Look, Ma: no photon position operators!

A photon position operator would be needed if we had a single photon
and were eager to know exactly where it was. But this is rarely, and
arguably never, what we really need to know. And if you look for photon
position operators in the textbooks, you don't find them. Instead, you
find lots of photon number operators: operators describing how many
photons there are in a given state or states.

Now, knowing you, you'll probably try to argue that if we can count
the number of photons in any given region of space, we can use this
to *locate* a photon if we happen to be in a state with just a single
photon.

This is approximately correct, but some very funny things happen.
For example, if we try to define an operator to count exactly how
many photons there are in a region with perfectly sharp boundaries,
we get a divergence: it will always say there are INFINITELY MANY
photons in this region! Crudely speaking, the reason is that the
very process of trying to determine whether a photon is in or out
of this region will, by virtue of the insanely perfect precision we
are demanding, create infinitely many photons. But if we "smear"
the boundaries of the region a bit we can get a finite answer -
which, however, depends on how we smear it out.

It's not so hard to do the calculation that shows number operators
have this funny property. I'm dying to lead you through it. But
first we need a few preliminary facts. Most importantly, we need
to know the Hilbert space for a single photon. Or, if we go down
to 2-dimensional spacetime and massless spin-0 particles for
simplicity, the Hilbert space for a single "foton".

So - you have been soldiering away trying to determine the inner
product in this Hilbert space, and you've done very well, so now
I should tell you what the answer is, or more precisely, what
the answers are. We have two answers, one if we work in momentum
space, the other in position space. As usual when dealing with
wave equations, everything looks simpler in momentum space.

As you note, if we take a real solution u of the wave equation in
2 dimensions, and take its Fourier transform, we get a distribution
U supported on the lightcone in momentum space, because the wave
equation

u_tt = u_xx

becomes the equation

w^2 U(w,k) = k^2 U(w,k)

which says that U vanishes except where w^2 = k^2.

>(Not a rigorous proof, but who cares?)

Oh, it's perfectly rigorous once you pad it with suitable verbiage.
But let's make like physicists and skip that crud.

Now, what distributions on the lightcone do we get this way?
Well, ignoring mathematical niceties, we get all distributions
with the property that

U(-w,-k) = U(w,k)*

This is the Fourier transform of the condition that u is real-valued!
This equation lets us figure out U on the backwards lightcone (w < 0)
if we know U on the forwards lightcone (w > 0).

Now since everything is simple in momentum space, we expect that
the inner product will be the simplest thing we can think of.

So, what is it? What's the inner product of a solution u and
a solution v? What's <u,v>, my friend?

>After mulling and chewing and cogitating, I came up with this:
>
> integral U(k)* V(k) dk/|k|

Right! You just Fourier transform u and v to get U and V,
and integrate one times the complex conjugate of the other
over the lightcone. Easy as pie! The only tricky bit is:
what measure should we use on the lightcone?

As always, when in doubt, we appeal to *symmetry* to narrow
down the field of choices. Obviously we want a Lorentz-invariant
measure on the lightcone, since our inner product should be
Lorentz-invariant. And the gods of symmetry smile on us here:
in any dimension, there's a unique Lorentz-invariant measure
on the lightcone, at least up to a constant factor. And it's
particularly simple in the example we're looking at, 2d spacetime:
it's just

dk/|k|

The reason, as you subtly hinted, is the way Lorentz transforms
act here: they stretch in one lightlike direction while squashing
by the same factor in the other lightlike direction, keeping the
area constant. I.e., if we use "lightcone coordinates" in momentum
space:

r = k + w
s = k - w

a Lorentz transform acts by multiplying r by some factor while
dividing s by the same factor.

So to get our Lorentz-invariant measure on the lightcone, we just
need a measure on the halfline [0,infinity) that is invariant
under dilations. And up to a constant factor there's just one:

dk/|k|

which works because

d(ak)/|ak| = dk/|k|

when we dilate by some factor a > 0.

So you've definitely got the right formula. Great!!!!!!!!

Now what about position space? Well, this post is too long
already, but I'll tell you this. The imaginary part of the
inner product <u,v> is given by the formula you guessed, or
almost guessed:

Im<u,v> = integral (u vdot - v udot) dx

where we integrate over the line {t = 0}, or for that matter
any line {t = constant}.

[By the way, I only guarantee that this formula agrees with our
other one - the momentum-space formula for <u,v> - up to a
constant factor. Personally I would choose the conventions
to make sure that

Im<u,v> = integral (u vdot - v udot) dx

because I regard this formula as sacred.]

Now it's not obvious that this formula for Im<u,v> is
Lorentz-invariant, but there's an easy way to show it is:
just use the wave equation!

What about a position-space formula for Re<u,v>? Ah, here things
get a bit more tricky....

john baez

Apr 17, 1999, 3:00:00 AM4/17/99
to
In article <7etv1b\$8j4\$3...@pravda.ucr.edu>,
Michael Weiss <spam...@spamfree.net> wrote regarding photon position
operators....

>If the professor will permit a little class-hijacking, I'd like to spend
>some time exploring this. But not in full wizardly detail! I can hear the
>reproaches already: "...and so if you compute the semi-excellent topoidal
>adjoint of the Wick rotation of the Froobish-Smedley photon pseudo-position
>quasi-operator in the braided representation, you'll see that it fails to
>satisfy the Baskerville equation. What's that? You tried to compute it,
>and you got stuck because you needed the formula for the unique
>Lorentz-invariant inner-product on foton-space that makes the Hamiltonian
>non-negative... the homework I ASKED YOU TO DO A YEAR AGO!!!???"

>I have in mind something a little more hand-wavy. You know what I mean:
>"if we *did* try to define a position operator this way, we'd get stuck
>because m=0. But Froobish suggested this work-around (I won't give all the
>technical details)--- that's called a quasi-operator. However, then the
>lump in the carpet shows up over here! ..." The sort of discussion you
>give all the time on TWF, for the stuff that's new enough to you (and often
>everyone) so it still looks complicated, and not as easy and familiar as
>plain old quantum field theory!

Umm, err, that sounds great, but the problem is, even though this is
"easy familiar plain old quantum field theory", I'm not very familiar
with the literature on position operators for photons. So I can't give
you a nice easy painless guided tour. We can go on a safari if you like.
But we may have to hack away at some thickets here and there.

Now you may wonder *why* I'm not familiar with the literature on
position operators for photons. After all, isn't it important to
know where photons are? NO! Practical work in quantum optics and
quantum field theory seems to get along perfectly fine without a
position operator for photons. Strangely, position operators are
not really so important in quantum field theory! It seems the only
people who worry about them are people who have nothing better to do.

Of course that wasn't always true: luminaries such as Newton and
Wigner have written about position operators in relativistic quantum
mechanics.  But now, it seems that only an intrepid and perhaps
foolhardy few dare ponder position operators for photons.

I regularly see papers with abstracts that say things like "We solve
the problem raised by Baskerville by defining a position quasioperator
in lightcone coordinates", "Froobish's attempt to solve the Baskerville
paradox fails because his quasioperator is not Hermitian", and "Glompf's
claim that our position quasioperator is not Hermitian is incorrect; he is
using a definition of Hermitian that only applies to ordinary operators".
And so on... a slow, endless trickle of papers, oozing out from some
huge swamp of confusion somewhere in the middle of darkest Academia....

Now I admit that this seems strange. But before journeying into the
trackless wastes of literature on the subject, why don't we just *think*
a bit first? This is, after all, easy familiar plain old quantum field
theory! We should be able to decide what properties a "position operator"
should have, and tell for ourselves if there is one or not. I'll do the
mathematical heavy lifting if we run into trouble... but it looks
like you've already started the work! Good!

>For starters, what's with this position operator jazz? The nice concrete
>(and yes, *anschaulich* wavefunction psi--- as I remarked recently, the
>dirty little secret of "collapse" is that it is *really easy to
>visualize* --- pace on the philosophical migraines)--- the wavefunction
>psi tells us the probability of finding our non-relativistic particle in
>the interval [A,B]:
>
>Prob(particle in [A,B]) = integral on [A,B] of psi(x)* psi(x) dx /
>||psi||^2
>
>where ||psi|| is the norm of psi:
>
>||psi||^2 = integral over entire line of psi(x)* psi(x) dx

Right: the reason why |psi|^2 represents a probability density in
nonrelativistic quantum mechanics is that the Hilbert space is L^2(R).

>Now the position operator is a bit different. That helps us compute the
>expectation value for the q observable for psi:
>
><psi, Q psi>/||psi||^2 = integral of psi(x)* x psi(x) dx / ||psi||^2, over
>the whole line
>
>so Q is the operator taking psi(x) to x psi(x).

Right: but the real reason why Q deserves the name "position operator" is
that spatial translations act on L^2(R) in such a way that when we
apply spatial translation by the distance s to the wavefunction psi,
the quantity <psi, Q psi>/||psi||^2 increases by s. That's
the property "position" is supposed to have, right? When we push a
rock two meters to the left, its "position" increases by 2.

More precisely, we want not just the *mean* value of Q, but the whole
*probability distribution* of values, to get shifted by s when we apply
a translation of distance s.

Now what about fotons? From our discussion above, it seems clear that
we need to follow these steps to see if there is a "position operator"
for a foton:

1) Determine the Hilbert space H for a foton.
2) Determine the action of spatial translations on this Hilbert space.
Translation by the distance s should correspond to a unitary operator U(s).
3) See if there is a self-adjoint operator Q on H such that

U(-s)QU(s) = Q + sI

Also see if Q is unique. If so, we say Q is "the position operator".

See what I've done? I've taken your remarks and axiomatized them, making
sure to assume nothing ahead of time about what the Hilbert space H, the
operator U(s), or the operator Q look like. For example, it would be
disastrous to assume the position operator looks like multiplication by
x, without some good reason to assume this.

Now the 3-step program above looks pretty easy to me. I haven't done
it yet, but this is the kind of stuff I'm paid to do and it seems routine.
So I suggest that we go ahead and do it! Of course (heh-heh) the first
step just *happens* to be the homework problem I've been trying to get
you to do for year! But in other posts you show distinct signs of doing
it, so this should not be an obstacle. Nor should step 2: we all know how
to translate solutions of the wave equation. The only real work comes in
step 3. This is where I'll need to think a bit. But let's get the first
2 steps nailed down first, okay?

>Let's *pretend* I'd done my homework, and I knew what formula to use for
><u, v>, where now u and v are functions belonging to foton-space. So I
>compute:
>
><u, Qu> / <u,u>
>
>Oops, now I have to define Q! And that's not gonna be easy, or so we've
>been warned. So that's *one* place the lump can show up.

Right. We should not assume it has any particular form ahead of time;
we should characterize it axiomatically and either prove that it exists
or that it doesn't. (By the way, I really have no opinion as to which
way it will turn out! Sometimes conventional wisdom is wrong. Also,
we're talking about "fotons" here, not photons. I don't know if they
work the same way.)

>Why can't I just say that, if:
>
><u,v> = integral (some expression with u and v) d(something) over the whole
>line
>
>then:
>
>Prob(foton is in [A,B]) = integral (same stuff) over the interval [A,B]
>??
>
>Hmmmm... maybe a little Aristotelian logic can help.

[logic deleted - this is a physics newsgroup]

>So maybe the formula for <u,v> *isn't* of the form:
>
><u,v> = integral (some expression with u and v) d(something) over the whole
>line

I think there *is* such an expression, but I'm pretty sure that if
it exists, it is a *nonlocal* function of u and v. This may or may
not be a real problem, but it's certainly enough to make things
a bit scary....

However, I think we should proceed a bit more systematically, as I've
outlined. In a while I'll reply to your other post, where you actually
start working out that foton inner product. Without that, we can't
really do anything - quantum mechanics demands a Hilbert space!

------------------------------------------------------------------------
 Yes, it's a little-known fact that when not working at
the mint or inhaling mercury vapor, Newton spent his later years
developing relativistic quantum mechanics - but he encrypted this
work and embedded it in an unreadable tome on the Book of Revelations,
which was only recently discovered at the Bodleian library.

john baez

Apr 17, 1999, 3:00:00 AM4/17/99
to
In article <7f89n1\$eh5\$1...@pravda.ucr.edu>,
Michael Weiss <spam...@spamfree.net> wrote:

>Hmm, back in my mathematical logic days, I would have written (lambda
>t)(lambda x) u(t,x). That gets tiresome when lambda is 6 characters long!

Yes, but the point of mathematical logic is largely to make mathematics
as unintuitive as possible, so that even the most obvious theorems
seem worthy of enormous tomes. (Just kidding.) Here we want the
mathematical machinery to grind away unobtrusively in the background
while we ponder cosmic issues like whether you can ever tell where a
photon really is. We need precision but not at too great a cost of
convenience.

>But you touch on an important point. Even in sloppy physics-land, one
>needs to distinguish between u(t,x) at a point, u(t,x) as a function of t
>and x (what the physicists call a scalar field), and u(t,x) as a function
>of x alone for some fixed value of t (what they call a scalar field on a
>spacelike-slice of spacetime).

Right. All 3 are important, and significantly different.

>Indeed, the difference between those last two notions *was* to blame for
>some of my mental murk.

Aha! Well, then I feel less guilty for making such pedantic point.
I only bothered making it at all because you are normally such an
avatar of precision - a real role model for the kids.

>As for notation--- well, how about x |-> u(t,x) for (lambda x) u(t,x), and
>(t,x) |-> u(t,x) for (lambda t)(lambda x) u(t,x)? And u by its lonesome
>will stand for (t,x) |-> u(t,x) --- unless, of course, context says it
>means something else.

Here's a commonly used notation which has the advantage of using fewer
characters: u for the function of (t,x), u(t,.) for the function of x
with t fixed, and u(t,x) for the number with t and x fixed. The idea is
to use a dot for a variable that's still busy varying, if you know what
I mean.

Ralph Frost

Apr 19, 1999, 3:00:00 AM4/19/99
to
bj flanagan wrote:

> bj: Well, perhaps we need some different mathematics.

Has anyone ever tried analog math? It doesn't seems to be simple or
perfect, by a far cry, but it sure is DIFFERENT!

I read recently, maybe in Ian Stewart's "Magical Mathematical Mazes",
something like when a problem is unassailable by one method, it may turn
out to be almost trivial when approached from a different direction.

Analog is different.

Don't try to convince me it is not.

JAT

Best regards,
Ralph Frost
ref...@dcwi.com

Toby Bartels

Apr 20, 1999, 3:00:00 AM4/20/99
to
John Baez <ba...@galaxy.ucr.edu> wrote at last:

>Here's a commonly used notation which has the advantage of using fewer
>characters: u for the function of (t,x), u(t,.) for the function of x
>with t fixed, and u(t,x) for the number with t and x fixed. The idea is
>to use a dot for a variable that's still busy varying, if you know what
>I mean.

Thus you can say "u(.,.)" instead of "u", but that would be silly.
Anyway, the real reason I wrote in is to say that,
if you write this notation up by hand or in TeX or something,
the dot should be centred, not low like a period.

-- Toby
to...@ugcs.caltech.edu

Ralph Frost

Apr 20, 1999, 3:00:00 AM4/20/99
to
john baez wrote:
------------------------------------------------------------------------
>  Yes, it's a little-known fact that when not working at
> the mint or inhaling mercury vapor, Newton spent his later years
> developing relativistic quantum mechanics - but he encrypted this
> work and embedded it in an unreadable tome on the Book of Revelations,
> which was only recently discovered at the Bodleian library.

Pardon my surly post-traumatic-stress induced paranoia, but, is this
statement about some recent discovery related to Newton's writing in
the Bodleian library, true or just a high level academic troll?

Where might the Bodleian library be? Any reference to the thing you're

Thanks much.

bj flanagan

Apr 23, 1999, 3:00:00 AM4/23/99
to
In article <371B9792...@dcwi.com>,

Ralph Frost <ref...@dcwi.com> wrote:
> bj flanagan wrote:
>
> > bj: Well, perhaps we need some different mathematics.
>
> Has anyone ever tried analog math? It doesn't seems to be simple or
> perfect, by a far cry, but it sure is DIFFERENT!

bj: What do you mean by "analog" math?

-----------== Posted via Deja News, The Discussion Network ==----------

Ralph Frost

Apr 23, 1999, 3:00:00 AM4/23/99
to
john baez wrote:
>
> In article <371B9C6A...@dcwi.com>, Ralph Frost <ref...@dcwi.com> wrote:

> >john baez wrote:
>
> >>  Yes, it's a little-known fact that when not working at
> >> the mint or inhaling mercury vapor, Newton spent his later years
> >> developing relativistic quantum mechanics - but he encrypted this
> >> work and embedded it in an unreadable tome on the Book of Revelations,
> >> which was only recently discovered at the Bodleian library.
>
> >Pardon my surly post-traumatic-stress induced paranoia, but, is this
> >statement about some recent discovery related to Newton's writing in
> >the Bodleian library, true or just a high level academic troll?
>
> It was a joke:

Ha ha ha. I get it now?

> ...it
> *has* been suggested that his eccentric behavior might be the
> result of inhaling mercury fumes in his work on alchemy.

But, on second thought, *Which eccentric behavior*?? I'm guilty, too,
of suggesting his alchemical meditations influenced his behavior after
reading (Dobbs, Betty Jo Teeter, "The foundations of Newton's Alchemy"
Oxford, 1979). However, the same selective criticism and isolated
inference is applicable to just about everyone in one form or other?

That is, transcendence, discovery, life, etc., is what happens when you
are busy doing other things.

..
> thereafter, because it was heretical - he was a professor at Trinity

It's good we are beyond that heresy stuff, and can settle into to doing
the straightforward science, huh?

>
> The classic reference for all this stuff is Richard Westfall's "Never
> at Rest: A Biography of Isaac Newton".

Thanks. Dobbs paints another good picture of some of these shadowy
undercurrents. Recommended for those who like to ponder "WHY?"
questions...

Oops, obilgatory physics content related to photon thread: I mean,
"Dobbs shed _LIGHT_ on another aspect of Newton's marvelous life."

(Greg Weeks)

Apr 23, 1999, 3:00:00 AM4/23/99
to
Michael Weiss (spam...@spamfree.net) wrote:
: 'Nuff physics, let's do some math. Last time ("The Lump in the

: Carpet") I asked, with an aggrieved sense of injustice, why no one
: bats an eyelash at a photon *momentum* operator. Surely we can
: Fourier transform our way from k to x and back again to our heart's
: content?

Nope! The momentum operator gives you the momentum-space wave function
only up to an undefined phase at each momentum. So you don't have a
well-defined function to apply your Fourier transform to.

Greg

Minhyong Kim

Apr 26, 1999, 3:00:00 AM4/26/99
to
I understood at the level of mathematical formalism
why there wasn't a position representation for photons,
but this thread convinced me that I don't understand
anything. So let me contribute a naive question:

If someone says something like `It took 200 years for
the light from this supernova to reach the earth'
what does this mean?
I suppose roughly speaking there's some state vector
with a lot of photons that have just been created
by the supernova and which under time evolution (for 200 years)
turns into
a state where some of the photons are interacting with
detectors on earth. But it would be nice to have
a more technical explanation of why it isn't natural
to have a position operator, given that we do
use sentences like above with reasonable clarity.

john baez

Apr 26, 1999, 3:00:00 AM4/26/99
to
now, and maybe more later....

Daryl McCullough <da...@cogentex.com> wrote:

>ba...@galaxy.ucr.edu (John Baez) says...

>>Michael Weiss <spam...@spamfree.net> wrote:

>>>After mulling and chewing and cogitating, I came up with this:
>>> integral U(k)* V(k) dk/|k|

>>Right!
>
>I'm a little puzzled as to why this is the case. Way back when
>we first did the simple, 1-D harmonic oscillator, we used the
>inner product:
>
>1. (x_1,p_1) . (x_2,p_2)
> = (m w) x_1 x_2 + p_1 p_2/(m w)
> + i (x_1 p_2 - x_2 p_1)

Right. While it may not be obvious, Michael's formula above
reduces to this formula (with m = 1) if we consider a single
mode of the wave equation corresponding to waves of frequency
w. This is something we need to talk about - Michael and I,
that is.

>Now, if we let U(k,t) = integral u(x,t) exp(-ikx) dx, [...]

One thing to note: your U(k,t) is different from Michael's
U(k,w). His U(k,w) was obtained by taking a solution of
the wave equation u(x,t) and taking its Fourier transform
in both the space and time variables. Since his U(k,w)
vanishes unless k = w, he then abbreviated it to U(k).

Your U(k,t) is obtained by fixing t and taking the Fourier
transform of u(x,t) in only the space variable. This is also
very fun to do. But you will get a very different-looking
formula for the inner product.

It's also good to express the inner product directly in
terms of the original function u(x,t). Again this looks
quite different. But all these formulas are just different
views of the same elephant.

john baez

Apr 30, 1999, 3:00:00 AM4/30/99
to
In article <7fub2o\$5da\$1...@news.math.arizona.edu>,
Minhyong Kim <k...@math.arizona.edu> wrote:

>I had thought I understood at the level of mathematical
>formalism why there wasn't a position representation for photons,
>but this thread convinced me that I don't understand anything.

Well, this game is indeed pretty subtle, but could you tell
me what you *thought* you understood? I too once thought I
knew why there wasn't a position representation for massless
particles - but now I'm in great danger of constructing one!

Actually I posed this as a problem for Michael and I to ponder.
But let me destroy the suspense by saying where I've gotten
so far. To keep the math simple, I'm considering massless spin-0
particles in 1+1 dimensions: mere "fotons" instead of full-fledged
photons. But first let's consider massive particles, and then
let's see what happens when we set the mass m equal to zero.

States of a massive spin-0 particle in 1+1 dimensions are just
normalizable real-valued solutions of the Klein-Gordon equation

((d/dt)^2 - Laplacian + m^2) phi = 0

And these are in one-to-one correspondence with normalizable *complex*
solutions of Schrodinger's equation

i (d/dt) psi = H psi

where the Hamiltonian

H = sqrt(m^2 - Laplacian)

is defined using the Fourier transform.

The correspondence works as follows: the real and imaginary parts of psi
are cooked up from phi and its first time derivative phidot, respectively.
The real part of psi is given by

(m^2 - Laplacian)^{1/4} phi

and its imaginary part is given by

(m^2 - Laplacian)^{-1/4} phidot

Again, these funny-looking operators are defined via Fourier transform.
By the way, I got these formulas wrong in a previous article. I think
I got them right this time. 

Now I happen to know that this correspondence is actually a *unitary*
operator mapping the Hilbert space of a massive spin-0 particle to
L^2(R). This lets us define observables for massive spin-0 particles
by giving self-adjoint operators on L^2(R).

In particular, I can define a position operator for these particle in
the usual way: just multiplication by x. The trick, of course, is
that we don't multiply phi by x! We multiply psi by x! One can
check that this satisfies everything a "position operator" should.

(I listed some criteria for a "position operator" in a previous post:
we want it to be self-adjoint, and we want it to transform under
translations in the correct way.)

Now what happens when we set m = 0? Well, the above formulas said

psi = (m^2 - Laplacian)^{1/4} phi + i (m^2 - Laplacian)^{-1/4} phidot

so when we set m = 0 we get

psi = (- Laplacian)^{1/4} phi + i (- Laplacian)^{-1/4} phidot

The question is, does this really define a unitary operator from
the Hilbert space of a massless spin-0 particle to L^2(R)? If so,
we can get a position operator for massless spin-0 particles in
the same way as for massive ones. The subtlety lies in some nuances
regarding Sobolev spaces. I could say more but I'm afraid most
people wouldn't enjoy it, so I'll relegate it to a footnote. 

>If someone says something like `It took 200 years for
>the light from this supernova to reach the earth'
>what does this mean?

In practice, it means some astronomer has studied enough
"standard candles" - stars of presumably known brightness -
to be willing to guess that a certain supernova is 200
light-years away! :-)

By the way, a supernova that close would be pretty damn bright.

But I'm just joking around; I know what you're trying to get at,
and I agree that it's a trickier problem than one might at first
think! What I want to eventually do on this thread is explain
why *approximate* position operators are not so hard to come by.
It's only when we try to locate particles too precisely that the
trouble starts. Locating a photon of visible light to within
a few light-years is really not a problem. But I want to work
my way to this issue slowly and systematically, at least if any
of the "students" like Michael are able to stick with the "course".

...................................................................

 The trick is to remember that for normalizable solutions of
the Klein-Gordon equation, phi lies in the Sobolev space H^{1/2},
while phidot lies in H^{-1/2}. So roughly, we have to take 1/2 a
derivative of phi to get something in L^2, and we have to take -1/2
a derivative of phidot to get something in L^2. The operator
(m^2 - Laplacian)^{1/4} roughly has the effect of taking 1/2 a
derivative.

 For the experts: the Hilbert space of a massive spin-0
particle is H^{1/2} + H^{-1/2}, and the operator (- Laplacian)^{1/4}
is unitary from H^{1/2} to L^2, while (- Laplacian)^{-1/4} is
unitary from H^{-1/2} to L^2, at least if we give these Sobolev
spaces the right inner products. But *all* of these facts break
down when m = 0. The question is whether they break down in a
way that makes the position operator still well-defined and