Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

According to this review

Photon wave function. Iwo Bialynicki-Birula. Progress in Optics 36 V (1996), pp. 245-294. arXiv:quant-ph/0508202,

a classical EM plane wavefunction is a wavefunction (in Hilbert space) of a single photon with definite momentum (c.f section 1.4), although a naive probabilistic interpretation is not applicable. However, what I've learned in some other sources (e.g. Sakurai's Advanced QM, chap. 2) is that, the classical EM field is obtained by taking the expectation value of the field operator. Then according to Sakurai, the classical $E$ or $B$ field of a single photon state with definite momentum p is given by $\langle p|\hat{E}(or \hat{B})|p\rangle$, which is $0$ in the whole space. This seems to contradict the first view, but both views make equally good sense to me by their own reasonings, so how do I reconcile them?

share|improve this question
    
The photon doesn't have a non-relativistic wavefunction because it is never slowly moving. The vector potential can be interpreted in certain ways as a relativistic wavefunction for a four-dimensional photon, but this requires understanding four-dimensional propagators in a particle view. It is probably best to follow Sakurai and ignore the concept of the wavefunction of the photon until you study the path integral for particle paths. –  Ron Maimon May 19 '12 at 20:10
    
I did study some path integral, but I guess you mean something else by " path integral for particle paths"? Anyway I do want to get some clarifications on the issue now. Sakurai's view gives me some confusions too: spectrum of one photon state is determined by the classical Maxwell's equations, yet the one-photon state vectors do not correspond to the eigensolutions of Maxwell according to Sakurai's prescription. On the contrary, for Dirac particles it seems to be universally agreed that the one-particle states are the solutions of Dirac equation. –  Jia Yiyang May 20 '12 at 2:22
    
It's not much different for Dirac equation--- the antiparticles are there, so you either describe the particle zig-zagging in time (which is all I meant by particle-path path-integral, Feynman propagators), or you fail to describe single particle. The Dirac equation is wrong as a single particle equation, this is Klein's paradox. The fields never correspond directly to the particle wavefunction, only in a Feynman description, and then the particle wavefunction in four dimensions (zig-zagging in time) integrated over all internal proper times obeys the linearized field equation. –  Ron Maimon May 20 '12 at 6:00
    
This reply will be more or less the same as the one I replied Lubos: one takes solutions of Dirac equation as one-particle space and build up a Fock space based on it(c.f. Thaller.b "the Dirac equation"chap 10), or else it is hard to explain the success of applying Dirac's equation to hydrogen atom. The correspondence between state kets and solutions of Dirac equation is given by $\psi(x)=\langle 0|\hat{\psi}(x)|p\rangle$ (Sakurai, weinberg) –  Jia Yiyang May 20 '12 at 6:19
1  
@twistor59: I wrote a long answer yesterday, but I got confused on something stupid, and I am still editing it. The main issue is defining 4-d wavefunctions from the Feynman/Schwinger proper-time formulation. This is possible for sure, Feynman always thought this way, but I thought at first I could give a short-cut to making it precise using Parisi's stochastic quantization idea (which adds a fifth dimension like proper time), but this doesn't work quite, SQ is different. I need to edit it down to the things that actually work, I'll try to finish tommorrow. –  Ron Maimon May 21 '12 at 9:19
show 3 more comments

2 Answers

up vote 3 down vote accepted

As explained by Iwo Bialynicki-Birula in the paper quoted, the Maxwell equations are relativistic equations for a single photon, fully analogous to the Dirac equations for a single electron. By restricting to the positive energy solutions, one gets in both cases an irreducible unitary representation of the full Poincare group, and hence the space of modes of a photon or electron in quantum electrodynamics.

Classical fields are expectation values of quantum fields; but the classically relevant states are the coherent states. Indeed, for a photon, one can associate to each mode a coherent state, and in this state, the expectation value of the e/m field results in the value of the field given by the mode.

For more details, see my lectures
http://www.mat.univie.ac.at/~neum/ms/lightslides.pdf
http://www.mat.univie.ac.at/~neum/ms/optslides.pdf
and Chapter B2: Photons and Electrons of my theoretical physics FAQ.

share|improve this answer
    
I see, so you(and Luboš) mean if we just look at the c-number solutions of Maxwell equations without other specification, the interpretation is two-fold, they can either be classical EM fields or quantum wavefunctions, and we must specify which one it is. –  Jia Yiyang May 22 '12 at 11:42
    
They can be (a) classical fields, or (b) 1-Photon wave function, or (c) modes of a coherent state; the latter two only if the negative energy part is ignored or absent. –  Arnold Neumaier May 22 '12 at 14:01
    
@JiaYiyang you can think of Arnold's last statement as what I call the "one-photon correspondence principle", so that there is an isomorphism between every classical coherent field and one photon states (the isomorphism being "throw away negative frequency part" one way and "get back unique real valued field by adding complex conjugate" the other way. This will work for any Boson: we can realize any classical state by copying Boson states putting as many Bosons in the same state as we like together: .... –  WetSavannaAnimal aka Rod Vance Aug 19 '13 at 1:25
    
@JiaYiyang ....this is my interpretation for why people never talk of making Maxwell's equations $U(1)$ gauge invariant as they do with the Dirac equation, see physics.stackexchange.com/a/73330/26076 I find it quite thrilling to think that when one probes e.g. any laboratory setup with microwaves, we are probing exact models of one-photon states and we can do this with classical interferometers, vector voltmeters, hammers and pieces of string (the last two just :) to emphasise how one can get accurate pictures of one photon states with truly rudimentary, macroscopic measurements). –  WetSavannaAnimal aka Rod Vance Aug 19 '13 at 1:30
    
@JiaYiyang and Arnold: This may seem like a really daft question: in section 7 of the paper, Bialynicki-Birula proves the representation bit of your extremely elegant and pithy summary, but I don't see how he proves the representation is irreducible. Is this obvious from other physics that I am missing? –  WetSavannaAnimal aka Rod Vance Aug 19 '13 at 1:55
show 8 more comments

The expectation values $$ \langle p | \vec E(\vec x) | p\rangle $$ and similarly for $\vec B(\vec x)$ vanish for a simple reason: the state $|p\rangle$ is by definition translational symmetric (translation only changes the phase of the state, the overall normalization) so the expectation values of any field in this state has to be translationally symmetric, too (the phase cancels between the ket and the bra).

So if you expect to see classical waves in expectation values in such momentum eigenstates, you are unsurprisingly disappointed. Incidentally, the same thing holds for any other field including the Dirac field (in contrast with the OP's assertion). If you compute the expectation value of the Dirac field $\Psi(\vec x)$ in a one-particle momentum eigenstate with one electron, this expectation value also vanishes. In this Dirac case, it's much easier to prove so because the expectation values of all fermionic operators (to the first or another odd power) vanish because of the Grassmann grading.

The vanishing of the expectation values of fields (those that can have both signs, namely the linear functions of the "basic" fields connected with the given particle) would be true for any momentum eigenstates, even multiparticle states which are momentum eigenstates simply because the argument above holds universally. You may think that this vanishing is because the one-particle momentum eigenstate is some mixture of infinitesimal electromagnetic waves that are allowed to be in any "phase" and these phases therefore cancel.

However, the formal relationship between the classical fields and the one-particle states still holds if one is more careful. In particular, one may construct "coherent states" which are multiparticle states with an uncertain number of particles which are the closest approximations of a classical configuration. You may think of coherent states as the ground states of a harmonic oscillator (and a quantum field is an infinite-dimensional harmonic oscillator) which are shifted in the position directions and/or momentum directions, i.e. states $$ |a\rangle = C_\alpha \cdot \exp(\alpha\cdot a^\dagger) |0\rangle $$ This expression may be Taylor-expanded to see the components with individual numbers of excitations, $N=0,1,2,3,\dots$ The $C_\alpha$ coefficient is just a normalization factor that doesn't affect physics of a single coherent state.

With a good choice of $\alpha$ for each value of the classical field (there are many independent $a^\dagger(k,\lambda)$ operators for a quantum field and each of them has its $\alpha(k,\lambda)$), such a coherent state may be constructed for any classical configuration. The expectation values of the classical fields $\vec B,\vec E$ in these coherent states will be what you want.

Now, with the coherent state toolkit, you may get a more detailed understanding of why the momentum eigenstates which are also eigenstates of the number of particles have vanishing eigenvalues. The coherent state is something like the wave function $$ \exp(-(x-x_S)^2/2) $$ which is the Gaussian shifted to $x_S$ so $x_S$ is the expectation value of $x$ in it. Such a coherent state may be obtained by an exponential operator acting on the vacuum. The initial term in the Taylor-expansion is the vacuum itself; the next term is a one-particle state that knows about the structure of the coherent state – because the remaining terms in the Taylor expansions are just gotten from the same linear piece that acts many times, recall the $Y^k/k!$ form of the terms in the Taylor expansion of $\exp(Y)$: here, $Y$ is the only thing you need to know.

On the other hand, the expectation value of $x$ in the one-particle state is of course zero. It's because the wave function of a one-particle state is an odd function such as $$ x\cdot \exp(-x^2/2) $$ whose probability density is symmetric (even) in $x$ so of course that the expectation value has to be zero. If you look at the structure of the coherent state and you imagine that the $\alpha$ coefficients are very small so that multiparticle states may be neglected for the sake of simplicity, you will realize that the nonzero expectation value of $x$ in the shifted state (the coherent state) boils down to some interference between the vacuum state and the one-particle state; it is not a property of the one-particle state itself! More generally, the nonzero expectation values of fields at particular points of the spacetime prove some interference between components of the state that have different numbers of the particle excitations in them.

The latter statement should be unsurprising from another viewpoint. If you consider something like the matrix element $$ \langle n | a^\dagger | m \rangle $$ where the bra and ket vectors are eigenstates of a harmonic oscillator with some number of excitations, it's clear that it's nonzero only if $m=n\pm 1$. In particular, $m$ and $n$ cannot be equal. If you consider the expectation values of $a^\dagger$ in a particle-number eigenstate $|n\rangle$, it's obvious that the expectation value vanishes because $a$ and $a^\dagger$, and they're just a different way of writing linear combinations of $\vec B(\vec x)$ or $\vec E(\vec x)$, are operators that change the number of particle excitations by one or minus one (the same for all other fields including the Dirac fields).

So if you want to mimic a classical field or classical wave with nonzero expectation values of the fields, of course that you need to consider superpositions of states with different numbers of particle excitations! But it's still true that all these expectation values are already encoded in the one-particle states. Let me summarize it: the right states that mimic the classical configurations are $\exp(Y)|0\rangle$ where $Y$ is a linear combination of creation operators (you may add the annihilation ones but they won't make a difference, except for the overall normalization, because annihilation operators annihilate the vacuum). Such coherent exponential-shapes states have nonzero vevs of any classically allowed form that you may want. At the same moment, the exponential may be Taylor-expanded to $(1+Y+\dots)$ and the linear term $Y$ produces a one-particle state that is the ultimate "building block" of the classical configuration. But if you actually want to calculate the vevs of the fields, you can't drop the term $1$ or others, either: you need to include the contributions of the matrix elements between states with different numbers of the particle excitations.

share|improve this answer
    
Thanks for the reply, but I'm aware of this, and this is exactly why I'm confused because the first reference suggests one-photon wave function is a nontrivial solution of Maxwell's equation. As for Dirac particles the situation is different because the relation between a one-particle state vector and the solution of Dirac equation is given by(c.f.sakurai chap 3-10; weinberg chap 14.1): $\psi(x)=\langle 0|\hat{\psi}(x)|p\rangle$, or from a second quantization point of view, people seem to agree on taking solutions of Dirac equation as the one-particle space(c.f. Thaller.b chap 10). –  Jia Yiyang May 20 '12 at 5:58
    
Dear Jia, a one-photon wave function is a nontrivial solution of Maxwell's equations and and a one-electron wave function is a nontrivial solution of Dirac's equation. There is no difference here. Note that the classical field you calculated from Dirac wasn't the expectation value in the same state of momentum $p$: it was the matrix element between $p$ and the vacuum. When you do the same thing for the Maxwell field, it will work just as well. That's what my answer was about: it's about the mixtures from bras and kets with different occupation numbers. What's the problem? –  Luboš Motl May 20 '12 at 6:27
    
Let me put it this way: if we take the c-number solutions of the field equations as classical fields, why different rules of assigning kets to classical fields(for Dirac and EM field)? –  Jia Yiyang May 20 '12 at 6:35
    
There is absolutely no difference in this respect between the Dirac and Maxwell field. I have already demonstrated it about three times. –  Luboš Motl May 20 '12 at 7:45
    
Let me try to phrase my confusion clearer: I understand single-particle expectation values are 0 in both Dirac field and EM field, and I understand the matrix element \langle 0|field\ operator|p\rangle gives plane wavefunction in both cases. I guess my confusion comes from the following content of quite a few textbooks: (1)Relativistic wave equations are understood as field equations(EM,Dirac etc.) (2)c-number solutions of field equations are understood as classical fields(this is usually mentioned for EM field, but I presume this is also the case for Dirac since they are both field equations) –  Jia Yiyang May 21 '12 at 2:57
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.