Is Maxwell's field the wave function of the photon? 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!
 A: 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).
A: 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. 
A: 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.
