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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.

Pricklebush Tickletush
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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: http://physics.stackexchange.com/q/66977/ –  Jun 15 '13 at 21:32
  • I think all of this discussion is hanging on the fact that "normal Heisenberg uncertainty principle" is relating to position and momentum. The principle applies to photon (two slit experiment for instance, or polarization etc.), not in the normal sens according to Wikipedia... – Martigan Jul 09 '14 at 11:18

4 Answers4

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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.

Ron Maimon
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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 $\Delta p$. – Pricklebush Tickletush Aug 26 '12 at 05:40
  • @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 05:47
  • 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. – Pricklebush Tickletush Aug 26 '12 at 08:43
  • @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 09:49
  • The language you are using is a little tough to parse, but it more-or-less agrees with some of the literature I've seen on the topic. – Pricklebush Tickletush Aug 26 '12 at 19:42
  • @AlecS: I'm afraid that the language is simply something you will have to get to grips with; the technical content of Ron's answer is precise (as usual) if a bit abstract. If you are really invested in this topic, I would recommend a good quantum optics book, which will discuss the necessary physics without necessarily invoking deep QFT arguments (e.g. Mandel and Wolf). – genneth Aug 26 '12 at 23:33
  • Perhaps the following will help you to sharpen the intuition and at least make plausible what is happening: accept that there exist ladder operators $a_k$ and $a_k^\dagger$ which obey the right commutation relations (suppressing the polarisation index). Naively one would expect to be able to construct a position operator $V(r)$ which could be used to build a localised state. However, note that even classically, due to gauge symmetry there is a constraint on what the potential must look like ($\nabla \cdot E = 0$)... – genneth Aug 26 '12 at 23:48
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    This means that in trying to construct an arbitrary potential by piecing together "point masses" actually creates a much less localised potential than is possible in the free matter case. In particular, the operators $V(r)$ at different $r$'s do not properly commute with each other --- so that precludes the possibility of associating a wavefunction with each photon, which depends on the position operators forming a completely commuting set. – genneth Aug 26 '12 at 23:51
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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! – Pricklebush Tickletush Mar 16 '13 at 20:02
  • I see that the thread is quite old, but I have a question: does it mean that - strictly speaking - the Young two-slit experiment with photons is not illustration of Heisenberg uncertainty principle, but the same experiment with electrons (for example) is? – Leos Ondra Feb 14 '18 at 07:09
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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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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...

bla
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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... – bla Aug 26 '12 at 07:23
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    Wonderful. But why not the Dirac equation, then? – Pricklebush Tickletush Aug 26 '12 at 07:34
  • 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... – bla Aug 26 '12 at 07:59
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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.

mwengler
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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.

     – Pricklebush Tickletush Aug 26 '12 at 19:15
  • Sorry, I'm withdrawing my "answered" check for now. According to several sources (http://pra.aps.org/pdf/PRA/v79/i3/e032112) (https://docs.google.com/viewer?a=v&q=cache:djf1u49p9WcJ:bayes.wustl.edu/etj/articles/review.extended.prob.pdf+&hl=en&gl=us&pid=bl&srcid=ADGEESg4snrOkd4hCQkvjZxcHcZFR-O5x7nZ90plCgW6lKNsP7BlAocqZBE5pso4GW6sxNTklwQv5h1aKScNbAcrSmH6SV3rSf1STPn8Jht5NlmRXmw4wWPSgtaQ70TCJ_3NhJJfxFXg&sig=AHIEtbT2CFFI2L0yuWk5FM-NAn07vZEwaw) (see just below 3.49) photons cannot be sharply localized at ALL -- even with an arbitrarily high uncertainty in the momentum. – Pricklebush Tickletush 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 00:01
  • See my comments to Ron's answer below. The problem is the gauge invariance, which is really a simple-looking (but actually complex) way to build in a non-locality in the EM field. Attempt to build a position operator from the ladder/momentum operators, and you will find that they don't commute. – genneth Aug 27 '12 at 00:10
  • For further details, see section 12.11 in Mandel and Wolf, where the calculations for a physical detector is shown. In the limit where the volume of the detector is much larger than the wavelength the intensity operator pretty much counts the number of photons in that volume. For a photon "localised" at the origin, the probability of detection at distance $r$ falls off as $r^7$. – genneth Aug 27 '12 at 00:15
  • @genneth experimentally it seems possible to define notion of position for photons. e.g. imagine a very tiny source of light; then you know whatever photon that is coming to you is coming right from that source and hence that it once had a quite well defined position. – user10001 Aug 27 '12 at 04:53
  • @dushya: this breaks down badly once the source is much smaller than the relevant wavelengths. But at all sizes there are "leaks" which prevent the mathematics from working out correctly. The lack of theoretical localisability is well-known and not mysterious; it is also measurable and quite crucial to several aspects of quantum optics. – genneth Aug 27 '12 at 14:19
  • @genneth There is no difference between light and electrons in anything you are saying, except it is very rare we see electrons coming from a source smaller than electron wavelength because electron rest mass makes its wavelength very small. A pulse of light 1 ns long is 30 cm long and is completely localised throughout its entire time of flight, you can put things in its path before and after it passes without stopping it, but put something in its path while it is passing and you stop the beam. – mwengler Aug 27 '12 at 17:28
  • @mwengler genneth is right. though intuitively its hard to believe but you can find exact mathematical results in this very old paper and references in it. – user10001 Aug 27 '12 at 18:10
  • @dushya Consider an electric field in the $z$-direction at time 0 $E_z(x,y,z,0) = $\delta(x)\delta(y)\delta(z)$. It is a straightforward matter to calculate "classically" (and result is correct for quantum) the $E_z(x,y,z,t)$. You will never detect a photon at $r>ct$, the photons are localized inside the appropriate light-sphere. Please describe what is meant by "photons cannot be localized" in light of this simple experimental fact. – mwengler Aug 27 '12 at 19:56
  • @mwengler: that field violates the Gauss constraint of $\nabla \cdot E = 0$ in free space. – genneth Aug 30 '12 at 11:54
  • @genneth ∇⋅E=0 is a static constraint, I am not talking about static fields. I am talking about an initial condition at time $t=0$ which will not be static at t>0. Every light beam and radio wave we know of violates ∇⋅E=0 in free space. – mwengler Aug 30 '12 at 14:13
  • @mwengler: I don't even know where to begin with that. The divergence-free-ness of $E$ is not a static constraint. It is never violated without charges. If you think so, you need to re-learn electrodynamics. – genneth Sep 01 '12 at 16:33
  • @genneth of course you are right, I was wrong about why you are allowed to "violate" ∇⋅E=0, which is of course because $\nabla E=\rho/\epsilon_0$. So at time t=0 some charge shows up at the origin creating a delta function of electric field at the origin, and then run time propagation from there. – mwengler Sep 03 '12 at 00:18
  • @genneth Sorry for the necro, but this is too relevant. Once I was going to make a presentation on this topic, and I was making the $\nabla\cdot E=0$ vs. $E_z=\delta(x,y,z)$ argument. My professor then looked at me weird and added a single stroke: $E_z=\delta'(x,y,z)$. I have yet to find any reason why this would be wrong. Can you? – Pato Raimundo Oct 25 '21 at 06:27
  • Since we got a revival of this thread, If photons could not be localized, you could not take a picture. You could not use your eyes to tell what is out in the world by localizing on your retina the photons coming from the various objects out there. You could not burn a hole in a sheet of paper using a magnifying glass to image the sun onto a piece of paper. The math is the tail, the overwhelming obviousness of localized photons in real life is the dog. If you can't figure out the math you are not doing it right because photons ARE localized, you know this if you understand imaging optics. – mwengler Mar 25 '22 at 16:38