20

In his ArXiv paper What is Quantum Field Theory, and What Did We Think It Is? Weinberg states on page 2:

In fact, it was quite soon after the Born–Heisenberg–Jordan paper of 1926 that the idea came along that in fact one could use quantum field theory for everything, not just for electromagnetism. This was the work of many theorists during the period 1928–1934, including Jordan, Wigner, Heisenberg, Pauli, Weisskopf, Furry, and Oppenheimer. Although this is often talked about as second quantization, I would like to urge that this description should be banned from physics, because a quantum field is not a quantized wave function. Certainly the Maxwell field is not the wave function of the photon, and for reasons that Dirac himself pointed out, the Klein–Gordon fields that we use for pions and Higgs bosons could not be the wave functions of the bosons. In its mature form, the idea of quantum field theory is that quantum fields are the basic ingredients of the universe, and particles are just bundles of energy and momentum of the fields.

Weinberg clearly says the Maxwell field is NOT a wavefunction of the photon (bolded line in quote). However, many on here seem to think it is. Who's right? Please explain. Thanks!

user253751
  • 1,496
sbphysics
  • 341
  • 6
    Who thinks its the wavefunction? I've never seen that statement (yours, not Weinberg's) anywhere. – Prahar Mar 18 '15 at 16:31
  • Answering "Who is right?" requires answering the primary question of "Is the Maxwell field not the wave function of the photon?" Thus a definite physics question has to be answered, with good reasoning. I think most are able to see that easily. – sbphysics Mar 18 '15 at 18:36
  • @ Prahar, I do have difficulty in the semantic use of fields and wavefunction. I had read that EM fields are wavefunctions for photon. Perhaps Maxwell fields are different than EM fields. I am still learning the usage. – sbphysics Mar 18 '15 at 19:10
  • 2
    There is, of course, a relationship between the classical electromagnetic field and photons, but it is much more subtle than the field being the wave function of a photon which is nonsensical for several reasons starting with Ján's post below. I generally assume that people who write things like that are either lying to children (if I think they actually know better) or parroting something they heard (if I have no reason to believe they no better). I presume that Weinberg means exactly what he says. – dmckee --- ex-moderator kitten Mar 18 '15 at 20:15
  • 1
    I won't make this an answer, because it's just a reference, but I've been reading my way through Scully and Zubairy's Quantum Optics and their first chapter actually deals extensively with the problem of talking about the "wavefunction of the photon." – zeldredge Mar 18 '15 at 22:10

3 Answers3

21

Weinberg is right.

The issue here is with the usual interpretation of the wavefunction as an amplitude density. This implies being able to localize the particle in an arbitrarily small region. However, it is not possible to localize photons (or any massless particles with spin, for that matter).

The reason for this is the careful definition of what localization means mathematically. It means that there must exist a projection operator with certain properties that intuitively correspond to the idea of measuring a particle at a given location. For a massive particle, one can show such a projection operator exists by looking at the little group (the subgroup of Lorentz transformations that leave the "rest frame" invariant). Because there is a rest frame for the massive particle, the little group is $SO(3)$, the group of three dimensional spatial rotations. If you "quotient out" the little group you're left only with boosts, and since the space of boosts is homeomorphic to $\mathbb{R}^3$, you can use them to define a position operator. A particle being "localized" then just means that you're not allowed to perform translations without changing the description of the physical state. In other words, localizing a particle breaks translational symmetry. So far, so good.

For a massless particle, there is no rest frame, so you must say the particle's momentum lies along a spatial direction and consider what transformations leave the momentum invariant. Then, the little group is $ISO(2)$, the group of translations and rotations in the plane orthogonal to the momentum. Already we start to see the problem: the little group is "intruding" on the possible characterizations of position states. This is no problem for a spinless particle -- intrude away -- but for a vector particle the translations correspond to gauge transformations, which means you can't project out states that break translational symmetry without also breaking gauge invariance -- a big no-no. So a photon can't be conventionally localized, with a Maxwell field for a "wavefunction" or anything else.

A more heuristic way of saying the same thing is to imagine multiplying the Maxwell field by a position operator. In the vacuum it's supposed to be divergenceless, but any scalar function that depends on the position breaks this condition. What Wightman has shown is that it's impossible to construct a position operator consistently, scalar or otherwise.

I have bastardized a long mathematical story so I encourage you to read the original references. If you're not familiar with the little group and classifications of particles under the Poincaré group I recommend you start with chapter 2 of Weinberg Vol. 1. Then read here for Newton and Wigner's original proof, here for the more general construction by Wightman, and this and this for weaker notions of photon localization that actually make sense but rule out interpreting the Maxwell field as a wavefunction.

PS: In case you're wondering, Weinberg's second remark that the Klein-Gordon field cannot be interpreted as a wavefunction is also correct. Here however we're led to the standard story about negative energy states and propagation outside the light cone that you can see in pretty much any QFT textbook.

Leandro M.
  • 1,694
  • The best I could gather from your answer is that no wavefunction can exist for a photon that allows one to localize the photon, imply that a wavefunction does exist but it is not the maxwell field. Is that what you meant? Thank you for the links, by the way. – sbphysics Mar 18 '15 at 22:32
  • 1
    That's the gist of it, but because of the theorem due to Wightman any "wavefunction" for the photon will necessarily have undesirable properties. It's a "pick your poison" kind of issue. – Leandro M. Mar 18 '15 at 23:13
  • 1
    This is a fantastically clear explanation of an argument that often descends into total gobbeltygook. – Zo the Relativist Mar 19 '15 at 02:40
6

Wave function for $N$ particles is a scalar function of $3N$ spatial coordinates.

Wave function for $N$ photons, if named so at all, should thus also be a scalar function of $3N$ spatial coordinates.

EM field, on the other hand, is a vector function of 3 spatial coordinates.

Any chosen wave function is non-unique; there are many different wave functions, related by gauge transformation that lead to same probabilities and it does not matter which one is used.

On the other hand, any chosen EM field is unique; different fields lead to different results (forces).

Ján Lalinský
  • 37,229
  • 2
    Or a 4-vecotr field of three spatial and one time coordinate, but that doesn't effect the argument at all. – dmckee --- ex-moderator kitten Mar 18 '15 at 20:11
  • 1
    I'm not sure how being real or complex prevents the Maxwell field from being the wavefunction of a photon. Also it seems one can represent any EM field as a sum of complex exponentials. Perhaps you can explain what you meant in further detail? – sbphysics Mar 18 '15 at 23:26
  • @shphysics: The time evolution of a wave function is given by $\mathrm{e}^{\mathrm{i}Ht}$, which is a unitary operator which will turn even an initially real wavefunction into a complex one. Wavefunctions are complex, physical fields like the electric or magnetic field are not, there isn't really more to say. – ACuriousMind Mar 18 '15 at 23:53
  • @ACuriousMind You can quite easily define a complex version of the EM field with the correct time evolution properties. It's in fact the best way to discuss most nontrivial optics problems. Furthermore, it's a little bit meaningless to say a quantity "is" real or "is" complex. We choose to define it as complex because it's computationally useful to do so (we don't even have to talk about amplitudes, as I explain in http://physics.stackexchange.com/questions/166333/why-do-we-need-a-wave-function/166361#166361 ). Only relative phases are observable and classical optics gives those in spades. – Leandro M. Mar 19 '15 at 00:05
5

I believe Weinberg is trying to make a distinction between two viewpoints of quantum particles, one historical (although we still use it when thinking about and teaching non-relativistic quantum mechanics) and one modern.

In non-relativistic quantum mechanics, we typically start by assuming the existence of "particles" familiar to us from classical mechanics -- electrons, for example. We then associate a wavefunction with each such particle in order to describes its quantum behavior. In some vague sense, in regular quantum mechanics the particle "is" its wavefunction.

However in quantum field theory, we no longer start from the particle picture. The fundamental objects are quantum fields. These fields are not wavefunctions, but rather operators that act on states. When we quantize these fields, we can describe excitations of the fields in terms of particles with which we are familiar. Particles are not distinct objects in their own right, but rather localized excitations of the fields.

mpv
  • 5,220
d_b
  • 8,113
  • 1
  • 21
  • 46
  • 1
    So based on what you wrote, you would say, in the context of QFT, that the 'Maxwell field' is an operator. That the Electromagnetic field in NRQM is now called a 'wavefunction' in QFT. And that the 'photon' is an excitation of the Maxwell field operator coming from its quantization. I admit that quantizing an operator confuses me. Also Where does the EM field/wavefunction enter into this? – sbphysics Mar 18 '15 at 18:57
  • Did you want to say "These fields are not wavefunctions, but rather operators that act on states"? – PhotonBoom Mar 18 '15 at 22:25