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Wikipedia claims the following:

More generally, the normal concept of a Schrödinger probability wave function cannot be applied to photons. Being massless, they cannot be localized without being destroyed; technically, photons cannot have a position eigenstate and, thus, the normal Heisenberg uncertainty principle does not pertain to photons.

Edit:

We can localize electrons to arbitrarily high precision, but can we do the same for photons? Several sources say "no." See eq. 3.49 for an argument that says, in so many words, that if we could localize photons then we could define a current density which doesn't exist. (Or something like that, I'll admit I don't fully understand.)

It's the above question that I'd like clarification on.

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Here it should be mentioned that in experiments like single slit diffraction photons are observed to follow uncertainty principle. –  user10001 Aug 26 '12 at 11:01
    
related: physics.stackexchange.com/questions/66977/… –  Ben Crowell Jun 15 '13 at 21:32
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The relation $p={h\over \lambda}$ applies to photons, it has nothing to do with the uncertainty principle. The issue is localizing the photons, finding out where the are at any given time.

The position operator for a photon is not well defined in any usual sense, because the photon position does not evolve causally, the photon can go back in time. The same issue occurs with any relativistic particle when you try to localize it in a region smaller than its Compton wavelength. The Schrodinger position representation is only valid for nonrelativistic massive particles.

There are two resolutions to this, which are complementary. The standard way out it to talk about quantum fields, and deal with photons as excitations of the quantum field. Then you never talk about localizing photons in space.

The second method is to redefine the position of a photon in space-time rather than in space at one time, and to define the photon trajectory as a sum over forward and backward in time paths. This definition is fine in perturbation theory, where it is an interpretation of Feynman's diagrams, but it is not clear that it is completely correct outside of perturbation theory. I tend to think it is fine outside of perturbation theory too, but others disagree, and the precise nonperturbative particle formalism is not completely worked out anywhere, and it is not certain that it is fully consistent (but I believe it is).

In the perturbative formalism, to create a space-time localized photon with polarization $\epsilon$, you apply the free photon field operator $\epsilon\cdot A$ at a given space time point. The propagator is then the sum over all space-time paths of a particle action. The coincidence between two point functions and particle-paths This is the Schwinger representation of Feynman's propagator, and it is also implicit in Feynman's original work. This point of view is downplayed in quantum field theory books, which tend to emphasize the field point of view.

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I'll have to give this a thorough looking. There's a lot of unfamiliar stuff in here. As far as the De Broglie relation, I meant its use in finding $\Delta p$. –  santa claus Aug 26 '12 at 5:40
    
@AlecS: You can find $\Delta p$, but there is no $\Delta X$ because photons don't have a real position operator defined at one time-slice. The photon concept, like all relativistic particle concepts, is a quantum field picture which is justified in perturbation theory, but hard to justify outside it (but it should be possible). –  Ron Maimon Aug 26 '12 at 5:47
    
I wish I could give this comment more of my time because it looks like you put a lot of thought into it. But on the surface it doesn't seem to answer my question. There are a lot of new words here, but not a lot of new explanation. –  santa claus Aug 26 '12 at 8:43
    
@AlecS: ok here's a rephrasing: photons don't have a position wavefunction, so the uncertainty principle can't be formulated. The photon still has a qualitative uncertainty principle, because it has a space-time localization, just not a 3-dimensional uncertainty principle like a nonrelativistic particle. –  Ron Maimon Aug 26 '12 at 9:49
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@RonMaimon Reading this again more than half a year later, it actually makes a lot more sense to me. I just needed a little more background. I'm marking this answer as correct. Thanks! –  santa claus Mar 16 '13 at 20:02
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There are no clearcut yes/no answers to these questions.

We can localize electrons to arbitrarily high precision[...]

This is not quite right. A simple conceptual argument is the following. If you try to localize electrons to a region that is small compared to the Compton wavelength, the uncertainty principle says that the localized state has to be built out of a range of energies that is big compared to $mc^2$. Therefore it has to include negative-energy states, the interpretation being that any attempt to measure the position of an electron to such high precision ends up creating electron-positron pairs. This means that it's not an eigenstate of particle number, and we no longer have any meaningful notion of measuring the position of "the" electron.

but can we do the same for photons? Several sources say "no."

Again, this is not quite right. Photons, just like electrons, can be localized to some extent, just not to an unlimited extent. It used to be believed that they couldn't be localized so that their energy density fell off faster than $\sim r^{-7}$, but it turns out that they can be localized like $e^{-r/l}$, where $l$ can be as small as desired (Birula 2009).

the normal concept of a Schrödinger probability wave function cannot be applied to photons

Not necessarily true. See Birula 2005. A more accurate statement would be that you have to give up some of the usual ideas about how God intended certain pieces of quantum-mechanical machinery, e.g., inner products, to work.

Being massless, they cannot be localized without being destroyed

A more accurate statement would be that they can't be localized perfectly (i.e., like a delta function).

technically, photons cannot have a position eigenstate and, thus, the normal Heisenberg uncertainty principle does not pertain to photons.

This is a non sequitur. The HUP has been reinvented multiple times. Heisenberg's 1927 paper discusses it in terms of limitations on measurement. Later it was reimagined as an intrinsic limit on what there was to know. It has also been formalized mathematically in a certain way, and then proved mathematically within this formalism. What the WP author probably had in mind was that these proofs are written on the assumption that there is a position operator and that there are position eigenstates that act like delta functions. Just because those particular proofs of a certain version of the HUP fail for photons, that doesn't mean there is no HUP for photons. You can confine a photon in an optical cavity, and a version of the HUP then follows immediately from applying the de Broglie relation to the two traveling waves that make up the standing wave.

The interpretation of this kind of thing is not at all simple. A couple of papers with good physics discussions are De Bievre 2006 and Halvorson 2001.

I. Bialynicki-Birula, "Photon wave function," 2005, http://arxiv.org/abs/quant-ph/0508202

I. Bialynicki-Birula and Z. Bialynicki-Birula, "Why photons cannot be sharply localized," Phys Rev A27 (2009) 032112. A freely available paper describing similar results is Saari, http://arxiv.org/abs/quant-ph/0409034

De Bievre, "Where's that quantum?," 2006, http://arxiv.org/abs/math-ph/0607044

Halvorson and Clifton, "No place for particles in relativistic quantum theories?," 2001, http://philsci-archive.pitt.edu/195/

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Absolutely yes, the uncertainty principle applies to photons nearly identically to how it applies to electrons. To see a great example of a localized traveling wave function which could apply to either a photon or an electron, see the wikipedia article on Wave Packets.

The original wikipedia quote is nonsense, and I have modified the original wikipedia article to remove it.

The energy eigenstates of a photon in free space are also eigenstates of momentum and are monochromatic. So at frequency $f$ the energy is $E=hf$ and the momentum is $p=E/c=hf/c$. The correct statement is "a photon in a momentum eigenstate can't be localized." Guess what, neither can an electron in free space in a momentum eigenstate be localized. If momentum is certain, uncertainty in position is infinite, i.e. can't be localized. As with electrons, so with photons. And electrons have a finite rest mass and therefore finite eigenstates.

So how do I localize a photon? Experimentally, I have a light source with a shutter. I can open the shutter for 1 ns, otherwise it is closed. You can be sure when I do that I have a burst of electromagnetic energy of about 30 cm physical extent along the direction of travel. That burst of energy is traveling at 30 cm/ns. So every photon that made it through that open shutter has now got finite position uncertainty, even though it's expected position is a function of time, just like a car driving down the road at 100 kph has finite position uncertainty even as its position changes with time.

Theoretically, I create a wave packet which Wikipedia describes beautifully. A localized photon, just a like a localized anything, is no longer monochromatic, is no longer an eigenstate of momentum and energy. No difference here between a photon and an electron.

I am shocked the wikipedia article on the photon has such nonsense in it. I went to wikipedia and removed that paragraph from the article and put a comment in the talk section to describe why.

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> The correct statement is "a photon in a momentum eigenstate can't be localized." Of course -- this is first-year quantum mechanics. Thanks for your explanation, I'm curious to hear what comes up in the talk section of wikipedia. –  santa claus Aug 26 '12 at 19:15
    
Sorry, I'm withdrawing my "answered" check for now. According to several sources (pra.aps.org/pdf/PRA/v79/i3/e032112) (docs.google.com/…) (see just below 3.49) photons cannot be sharply localized at ALL -- even with an arbitrarily high uncertainty in the momentum. –  santa claus Aug 26 '12 at 19:40
    
@AlecS I can't read the PRA ref, don't have login. In the google docs ref, what you cite is an unpublished one line quote from a referee of the paper. I call B.S. on the referee, and challenge you or anyone else to find a sensible refutation to what I say about localization. –  mwengler Aug 26 '12 at 21:11
    
Sorry mwengler, but you are simply incorrect. As Ron mentions below, there is no Schrodinger (that is, position operator) basis for photons. Furthermore, I would advise you to revert your wikipedia edit, because that sentence was in fact completely correct, if obtuse. –  genneth Aug 26 '12 at 23:30
    
@genneth can you give me a reference that goes beyond a single sentence saying it can't be done? I'd love to understand how a photon represented by a wave packet is any different from an electron represented by a wave packet. –  mwengler Aug 27 '12 at 0:01
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In addition to what was discussed already, and besides the fact the Schrödinger formalism is not relevant for photons, a good place to start in my view is in Roy Glauber's work (or some other introductory text to quantum optics). There, you'd see different uncertainties arising, such as between the photon number and phase, etc...

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Frankly, I don't think that by explaining something by a bunch of other names (i.e. feynamnn's propagator etc) an understanding can arise. Maybe this bud of explanation would work: If you understand why photons are relativistic (the do travel fast...) then you would expect a QM description that would also be Lorentz invariant (familiar special relativity?). This means that the equation used will be symmetric to translations (time&space) and rotations. Alas, Schrödinger's eq isn't, it has a 1st derivative in time, and 2nd derivatives in space. Hence it cannot describe relativistic particles... –  natan Aug 26 '12 at 7:23
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Wonderful. But why not the Dirac equation, then? –  santa claus Aug 26 '12 at 7:34
    
Why Dirac? why not Klein Gordon to begin with? (then go to dirac...), and then you'll need to describe an eq that captures the QM nature of the electromagnetic field. This time, a gauge symmetry is needed, specifically, the Abelian U(1) symmetry of a complex number, which reflects the ability to vary the phase of a complex number without affecting observables or real valued functions made from it (such as the energy or the Lagrangian). But with these my friend Alec, we already long left the realms of Schrödinger. I think you'd enjoy the graduate courses in physics if you want to dig more... –  natan Aug 26 '12 at 7:59
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