What, then, is a photon's wave-function?
Michael Weiss
The photon is simply a wavefunction - or more precisely, a solution
of Maxwell's equations.
and later:
Please, folks, don't start talking again about the lack of a Newton-
Wigner localization for massless particles; I'm using the other localization,
in which I treat the photon's wavefunction as simply a solution of Maxwell's
equations. I'm just trying to get across the idea of wave-particle duality,
without worrying about these nuances.
Now John, much as I'd like to, I can't let you off the hook that easily.
Didn't you spend endless hours pounding the significance
of coherent states into my brain, until finally a small flash
radiated enlightenment through my mental aether---
that a solution of Maxwell's equations is *not* the same thing
as a single-photon state? For one thing, there is field-strength/
photon number duality, so beautifully illustrated by those
coherent states.
Ooops, storm clouds are gathering above the wizard's brows!
Better put up a lightening rod! True enough, you didn't say that
the *photon* is the same as a solution of Maxwell's equations---
you said that about the photon's *wavefunction*!
But alas, now my confusions have just begun! What the heck
*is* a wavefunction? What does a function have to do to
deserve that honorific? Just satisfy a wave-equation?
I suppose that answer might satisfy a pure mathematician, at
least one fond of PDEs....
For the ordinary non-relativistic wavefunction psi(x), we have the
good old Born formulas for probabilities and expectation values
in terms of psi. Not for nothing did Born win his Nobel. Thanks
to him, when we "shut up and calculate", at least we know what
we are calculating: probabilities! Of course, we still may not
know what a probability really *is*, but you can't have everything....
Now, a classical solution to Maxwell's equations sure as heck
*looks* like it has something to say about the probability of detecting
a photon. Where do the silver nitrate grains turn dark, where does
the CCD chatter away, where, in short, do the photons flock,
except at spots where E^2 + B^2 is large.
Coincidence? I think not.
But alas, poking around in a few QFT books tells me a rather different
story. Mind you, I'm quite prepared to believe that the tale I read is
*not* the tale they mean to tell. But therein lies my confusion.
In Haag, _Local Quantum Physics_, I get this formula for the inner-product
of psi and chi:
integral (psi* (d chi/dt) - (d psi*/dt) chi) dx
For starters, if psi is real, then this formula yields zero for the
inner product of psi with itself. OK, so the wavefunction can't be real.
But then what kind of wavefunction is a classical (real-valued) solution
to Maxwell's equations?
For another thing, we have that irritating fact you mentioned, about
NW localization. As Haag puts it, "For massless particles the concept
of localization is not appropriate". Can't you cook up
a solution to Maxwell's equations that has compact support--- that is
zero outside some bounded region, non-zero inside,
and that is smooth everywhere? And (assuming the answer is yes),
doesn't this "wavefunction" drop some rather heavy hints about
where a photon is *not*?
I do gather from Haag that probability amplitude in momentum space
are just hunky-dory, even for photons. This does make me a touch
more sanguine about those diffraction photographs. After all,
a diffraction grating ---nay, an ordinary lens--- ain't nothing but a
Fourier transform. The diffraction photograph tells us, not *where*
the photon was at time t_0, but in *which directions* the light was strong.
Yet surely also tells us *something* about position. It's a bit like a
campaign finance statement--- it tells you which plane waves contributed
heavily to the candidate's election effort, and certainly suggests just
*where* we can expect him to place his votes.
I'm just trying to get across the idea of wave-particle duality,
without worrying about these nuances.
Ah, but are they merely nuances? Consider two versions
of wave-particle duality that QFT denies:
- the waves guide the particles
- a particle is just a well-localized wave packet and nothing more.
So these are wrong, but what is right? To say, "a wave is somehow
associated with a photon", and nothing else, doesn't put food for
thought on the table.
john baez
Michael Weiss <we...@spamfree.net> wrote:
>Now John, much as I'd like to, I can't let you off the hook that easily.
If you really wanted to, you could. But you don't....
>Didn't you spend endless hours pounding the significance
>of coherent states into my brain, until finally a small flash
>radiated enlightenment through my mental aether ---
>that a solution of Maxwell's equations is *not* the same thing
>as a single-photon state?
I think you understand what's going on perfectly well. You just
want me repeat some mantras to reassure you that we're both still
sane. Okay:
1) FROM CLASSICAL TO QUANTUM:
Suppose we want to quantize a system whose classical phase
space is a symplectic vector space. First we turn the classical
phase space into a complex Hilbert space H. Unit vectors in
H will represent "single-particle states" in our quantum theory.
(As usual, two unit vectors differing only by a phase represent
the same state.) Then we form the Fock space on H - call it K.
This is the Hilbert space of our quantum theory - unit vectors
in K represent states with an arbitrary number of particles.
2) FROM QUANTUM BACK TO CLASSICAL:
Suppose we have a state of our classical system - what is its
quantum analog? A good answer to this question is provided by
the theory of coherent states. The state of our classical
system is a point p in H. Let p = tv where v is a unit vector
in H and t is a real number. Associated to v we have an operator
on K called the "creation operator" a*(v), which creates particles
of type x. Apply the operator exp(t a*(v)) to the vacuum vector
in K, and normalize the resulting vector. This gives us a unit
vector in K which we call the "coherent state" corresponding to
the classical state p. It's a good quantum approximation to the
classical state we started with. Call it Coh(p).
3) IN THE CASE OF ELECTROMAGNETISM:
Here we take H to be the space of solutions of the source-free
Maxwell equations (i.e., the equations with charge density and
current set to zero). More precisely, we take a space of smooth
solutions that go to zero rapidly at infinity, note that this
space becomes a complex vector space with an inner product in an
almost unique Poincare-invariant way, and complete it to get a
a complex Hilbert space Then:
a) a point p in H corresponds to a *classical* solution of the
source-free Maxwell equations.
b) a unit vector v in H describes the state of a single photon in
the *quantum* theory of electromagnetism,
and
c) the above recipe gives us a coherent state Coh(p) in K, which is
a state of the quantized electromagnetic field serving as a good
approximation to the classical state p.
Item b) is what I meant when I said "the state of a single photon
is a solution of the classical source-free Maxwell equations". I'm
sorry if the slight vagueness in this formulation got you scared -
quantum field theorists talk this way all the time, so it's good to
get used to it.
>But alas, now my confusions have just begun! What the heck
>*is* a wavefunction? What does a function have to do to
>deserve that honorific? Just satisfy a wave-equation?
Well, in modern quantum field theory, people feel free to call any
vector in any Hilbert space "a wavefunction", so long as they are
using it to describe the state of a quantum system! That's what
I'm doing when I call the state of a photon, or by extension any
solution of the classical source-free Maxwell equations, a
"wavefunction". Again this may be a bit confusing, but real
physicists do it all the time, so it's best to get used to it.
>I suppose that answer might satisfy a pure mathematician, at
>least one fond of PDEs....
Note that in the case at hand our "wavefunction", i.e. our solution
of Maxwell's equations, really does satisfy a wave equation. It's
a function and it waves - it's a wavefunction. :-)
More later about Newton-Wigner localization and all that jazz.
P.S. - I may have described coherent states a bit more abstractly
this time than I did before, by assigning creation operators to
*all unit vectors* in the Hilbert space H. I forget if I did that
before. Slicker still, we can assign creation operators to *all*
vectors in the Hilbert space by defining a*(tv) = ta*(v), i.e.,
by demanding linearity. Then we get
Coh(p) = exp(-||p||^2) exp(a*(p))|0>
where this time I've bothered to remember the normalizing factor
and stick it in front.
Michael Weiss
I hope Paul Kinsler is listening. Treat most of this
as a request for clarification, by adding an implicit
"is this right?" after all my sentences. I do work up
to some more specific semi-experimental questions
by the end.
John Baez writes:
the above recipe gives us a coherent state Coh(p) in K, which is
a state of the quantized electromagnetic field serving as a good
approximation to the classical state p.
Yes, that much I still remember. Now I'm starting to wonder about
some of those non-coherent states of light. In particular, a
one-photon state.
First we turn the classical
phase space into a complex Hilbert space H. ...
Then we form the Fock space on H - call it K.
This is the Hilbert space of our quantum theory - unit vectors
in K represent states with an arbitrary number of particles.
In particular, if v is a unit vector in H, then I'm wondering
about a*(v) |0>. Terminology digression: Do quantum opticians
use the phrase "one-photon state" interchangably
both for the unit vector v in H, and for the vector a*(v) |0> in K?
They can always crank up the precision when needed,
as I gather Paul Kinsler is doing when he writes:
You seem to be confusing the shape of the field mode
(which _is_ well defined at all points in space), with the E
(or B) field we might (in principle) measure as resulting
from that mode.
I.e., the field mode corresponds to v; we measure the electric field
E on some quantum state obtained *somehow* from v. We could
measure E on a*(v) |0>, or we could measure E on one of the
coherent states exp(t a*(v)) |0>. If t is large enough, then E will be
fairly well-defined. If t is small, then not so well-defined.
Measuring E on a*(v) |0> will give random results.
Incidentally, Greg Weeks said earlier that coherent states have
gazillions of photons, or at least jillions. Strictly speaking though
the state exp(t a*(v)) |0> is coherent even when t is small.
OK, enough set-up, now on to experiments. For starters, the classic
(but not *classical*) Taylor experiment.
The discussion so far seems to treat this as an example of one-photon
states, since the probability is extremely small that two photons were
ever in the apparatus at the same time.
That seems nice and intuitive, but to be picky and precise, our
waveforms--- our vectors v in H, our solutions to the source-free
Maxwell equations--- are really functions on *all of spacetime*.
Oh, I suppose you don't have to do *all* of spacetime, you can slice
out a patch and impose boundary conditions on the edges. Still,
in the Taylor experiment that patch seems to be several months long
and but a few feet wide. And the quantum light state inside that patch
*must* have been a multiphoton state, since after all Taylor got
an interference pattern, with scads of darkened silver nitrate grains
on the photograph.
It sounds like there is some kind of approximation principle at work,
which says that you can treat a weak light beam over a short enough
time period as if it were a single-photon state. And also that
you can treat what Taylor actually did as equivalent to a large
number of independent trials, each using these approximations
to one-photon states.
I suppose the second assumption is the standard statistical
hypothesis of all of experimental science. You run the same
experiment a million times on the Large Hadron Collider and
collect results, then you assume that you'd get the same statistics
if you built a million LHC's and ran the experiment once on
each of them. Better assume that, 'cause no one is funding the
second approach!
So (back to the first assumption), do weak coherent beams
ever look like one-photon states?
The probability distribution for photon number for exp(t a*) |0>
is a Poisson distribution with mean t^2/2, so if t=sqrt(2), then we have
a quantum state with an *average* of one-photon.
Is that somehow "close enough" to a true one-photon state
for experimental purposes?
Can we distinguish, in practice, a true one-photon state from
a coherent state with mean photon number equal to 1?
john baez
Michael Weiss <we...@spamfree.net> wrote:
>John Baez intoned:
> First we turn the classical
> phase space into a complex Hilbert space H. ...
> Then we form the Fock space on H - call it K.
> This is the Hilbert space of our quantum theory - unit vectors
> in K represent states with an arbitrary number of particles.
>
>In particular, if v is a unit vector in H, then I'm wondering
>about a*(v) |0>. Terminology digression: Do quantum opticians
>use the phrase "one-photon state" interchangably
>both for the unit vector v in H, and for the vector a*(v) |0> in K?
I wouldn't be surprised: quantum field theorists sure do. The Fock
space K is the symmetrized tensor algebra on H, suitably completed to
form a Hilbert space, so it's very natural to think of H as sitting
inside K.
>Incidentally, Greg Weeks said earlier that coherent states have
>gazillions of photons, or at least jillions. Strictly speaking though
>the state exp(t a*(v)) |0> is coherent even when t is small.
Right, and....
>It sounds like there is some kind of approximation principle at work,
>which says that you can treat a weak light beam over a short enough
>time period as if it were a single-photon state.
Right. A very weak beam of coherent light is a state of the form
exp(t a*(v)) |0> for t very small - suitably normalized, of course.
Expand this as a power series in t and behold: this state is a linear
combination of the vacuum state, the one-particle state a*(v) |0>,
and puny amounts of states with more particles!
So we can figure out what this state will do, pretty accurately, by
figuring out what the vacuum and the 1-particle state will do.
In particular when we do those sneaky experiments I mentioned
involving interference of very weak beams of light, we can figure
out what'll happen by thinking about 1-particle states. (We
need to think about the vacuum, too, but that's usually rather
dull.)
>Can we distinguish, in practice, a true one-photon state from
>a coherent state with mean photon number equal to 1?
Yes, because the latter will have a bunch of the vacuum state in
it - i.e., sometimes when you look you don't see a photon at all!
And to balance that out, sometimes you'll see more than one.
Mikko S Kiviranta
>antibunched light from laser diodes and feedback for years now,
>unfortunately I dont have any references to hand ... but if you
>look around on http://rsphy1.anu.edu.au/ you might find something.
>
>Actually this was a bit controversial, as it seemed a bit weird that
>you could make non-classical light by using classical feedback,
>but it can be done -- although I think there is a limit to how
>antibunched the output can be with this kind of scheme.
Hmmm ... I wonder how this relates with Wiseman & Milburn
(Phys Rev A, vol 49 no 2 p 1350). An excerpt from the abstract:
"However, we also show that feedback mediated by homodyne detection
(or any other extracavity measurement) cannot produce nonclassical
light unless the cavity dynamics can do so without feedback.
Furthermore, in systems which already exhibit squeezing, such
feedback can only degrade the squeezing in the output".
Best regards,
Mikko
john baez
Dr Paul Kinsler <ee...@eensgi4.leeds.ac.uk> wrote:
>Mikko S Kiviranta <m...@beta.hut.fi> wrote:
>> I guess 'coherent state' as described above is just a technical term
>> for a state which is an eigenstate of the annihilation operator: I don't
>> see what is so 'coherent' about that kind of a state.
>It's the closest quantum thing you can make to a classical
>state with well defined amplitude and phase.
That's a coherent state of the electromagnetic field. But if you're
just learning about this stuff, it's good to think about coherent
states of a quantum particle in 1 dimensional space. This is a state
that minimizes the product of uncertainties (Delta p) (Delta q)
while keeping Delta p and Delta q equal to each other. In a recent
post I showed you can get all these states by translating the ground
state of the harmonic oscillator Hamiltonian in both position space
and momentum space. These states are also eigenstates of the
annihilation operator!
One can also consider "squeezed" states, which minimize
(Delta p) (Delta q) but for which Delta p and Delta q need not
be equal. In that same recent post of mine, I showed all these
are obtained from coherent states by dilation.
For an introduction to coherent states, try this website:
http://math.ucr.edu/home/baez/schmoton
Dr Paul Kinsler
> >Actually this was a bit controversial, as it seemed a bit weird that
> >you could make non-classical light by using classical feedback,
> >but it can be done -- although I think there is a limit to how
> >antibunched the output can be with this kind of scheme.
> Hmmm ... I wonder how this relates with Wiseman & Milburn
> (Phys Rev A, vol 49 no 2 p 1350). An excerpt from the abstract:
> "However, we also show that feedback mediated by homodyne detection
> (or any other extracavity measurement) cannot produce nonclassical
> light unless the cavity dynamics can do so without feedback.
> Furthermore, in systems which already exhibit squeezing, such
> feedback can only degrade the squeezing in the output".
Hmm. I'd belive what they say, because they certainly know
more about the subject than I. However, there may be some
difference in the definitions of "non classical" used here.
Note that "squeezing" often refers to quadrature-squeezing,
which is not the same as antibunching.
--
------------------------------+------------------------------
Dr. Paul Kinsler
Institute of Microwaves and Photonics
University of Leeds (ph) +44-113-2332089
Leeds LS2 9JT (fax)+44-113-2332032
United Kingdom P.Ki...@ee.leeds.ac.uk
WEB: http://www.ee.leeds.ac.uk/staff/pk/P.Kinsler.html