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EM wave function & photon wavefunction

+ 7 like - 0 dislike
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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?

This post imported from StackExchange Physics at 2015-04-22 11:16 (UTC), posted by SE-user Jia Yiyang
asked May 19, 2012 in Theoretical Physics by Jia Yiyang (2,465 points) [ no revision ]
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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.

This post imported from StackExchange Physics at 2015-04-22 11:16 (UTC), posted by SE-user Ron Maimon
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.

This post imported from StackExchange Physics at 2015-04-22 11:16 (UTC), posted by SE-user Jia Yiyang
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.

This post imported from StackExchange Physics at 2015-04-22 11:16 (UTC), posted by SE-user Ron Maimon
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)

This post imported from StackExchange Physics at 2015-04-22 11:16 (UTC), posted by SE-user Jia Yiyang
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 above 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?

This post imported from StackExchange Physics at 2015-04-22 11:16 (UTC), posted by SE-user Luboš Motl
@JiaYiyang: It is not hard to explain the success of the Dirac equation in an external potential--- this is a case where the eigenstates of the equation have a particle and field interpretation simultaneously. I am writing a long answer, because this is never properly explained in the literature, at least not in one place.

This post imported from StackExchange Physics at 2015-04-22 11:16 (UTC), posted by SE-user Ron Maimon
@RonMaimon: I'd be very interested to read such a thing if you have the time to put something together (even if brief)

This post imported from StackExchange Physics at 2015-04-22 11:16 (UTC), posted by SE-user twistor59

2 Answers

+ 5 like - 0 dislike

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.

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Arnold Neumaier
answered May 21, 2012 by Arnold Neumaier (12,385 points) [ no revision ]
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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.

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Jia Yiyang
@WetSavannaAnimalakaRodVance: I don't agree that you can measure one-photon state using the same apparatus and set ups measuring classical EM fields. What we measured classically is the expectation value $\langle \alpha|\mathbf{E}|\alpha\rangle$, and if $|\alpha\rangle$ is any 1-photon state, the expectation is 0 every where in space. In this post(me, arnold, lubos) the conclusion reached is that given a (positive-frequency) solution to Maxwell's equation, without further clarification, it can be a classical em wave, or the wavefunction to 1-photon state, not both.

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Jia Yiyang
@JiaYiyang your last sentence is what I mean - there is an isomorphism between what you're probing with classical apparatus and the corresponding one-photon state. Of course they're not the same thing, indeed photons are almost impossible to detect without destroying them, let alone glean their detailed state - that's what I meant when I said "we are probing exact models of one-photon states". But the model is in all other respects exact, so I still think the idea is pretty neat.

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance
@JiaYiyang Also, for the purposes of the link I gave, I argue that one can in principle copy a boson repeatedly and thus amplify its state to be classically probed: in practice the no-cloning theorem will stop you doing this for an arbitrary state, but if you can come up with a preparation procedure to realise a particular one photon state repeatedly, then you can do it in practice too - e.g. in a laser. So therefore the phase of the one photon wavefunction is meaningful and not amenable to being "gauged to something else" as the electron's wavefunction's phase is.

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance
@WetSavannaAnimalakaRodVance: Ok, I see what you mean. I read the first part(will read the rest later) your linked post. It indeed seems to be some content in this isomorphism, but I'm not convinced it's strong enough to differentiate the physical significance of photon and electron wave function in the way you suggested, even for electron $|\psi|^2$ is not the only thing we can measure, clearly we can reveal more information about $\psi$ itself by measuring some more observables, that is, some $\psi$'s can have identical $|\psi|^2$, but not identical $\langle\psi|A|\psi\rangle$ for all A's.

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Jia Yiyang
Most recent comments show all comments
@JiaYiyang: The relevant formula is ⟨α|F(x)|α⟩=α(x), where F(x) is the operator version of the silberstein vector, α is a solution of the homogeneous Maxwell equations in Silberstein form and |α⟩ is the corresponding coherent state.

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Arnold Neumaier
@JiaYiyang: Useful interactions can be found in Weinberg's 1964 paper on massless fields.

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Arnold Neumaier
+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2015-04-22 11:16 (UTC), posted by SE-user Luboš Motl
answered May 20, 2012 by Luboš Motl (10,248 points) [ no revision ]
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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).

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Jia Yiyang
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?

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Luboš Motl
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)?

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Jia Yiyang
There is absolutely no difference in this respect between the Dirac and Maxwell field. I have already demonstrated it about three times.

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Luboš Motl
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)

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Jia Yiyang
Most recent comments show all comments
It is not true you won't find a textbook that discusses $\langle 0|E(x)|p\rangle$ and even if it were true, I don't understand why it would be relevant. It is not true that all things that may be discussed are inevitably written in some textbooks. One may ask thousands of questions that are not discussed in the textbooks; this doesn't imply any contradiction. On the other hand, there's a simple reason why the expectation value of a Dirac field is the same state isn't discussed and I have already explained it: it is identically zero because of the Grassmann grading.

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Luboš Motl
Two answers above, "morally wrong" should have been "morally right".

This post imported from StackExchange Physics at 2015-04-22 11:17 (UTC), posted by SE-user Luboš Motl

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