How to introduce the electromagnetic field in Quantum Field Theory?

There are many ways to introduce the electromagnrtic field in Quantum Field Theory(QFT), such as canonical quantization method which introduces the creation and annihilation oprators by treating the amplitudes of electromagnetic waves as the operators.

One way I have read in book is different, but I don't understand.

At first, the author introduces a transformation to change the quantum field $\psi$ $$\psi\rightarrow\psi'=\psi \ e^{-i\alpha(x)}$$ And then, the former covariant derivative $\partial_{\mu}$ is no longer covariant $$\partial_\mu(\psi \ e^{-i\alpha(x)}) = e^{-i\alpha(x)}(\partial_\mu \psi-i\psi \ \partial_\mu\alpha(x) )$$ There is an inhomogeneous term that make $\partial_\mu$ is not conariant.

So, the author has to introduce a new definition of covariant derivative, using another symbol $D_\mu$, by introducing a vector field $A_\mu$, to make $D_\mu$ covariant $$\partial_\mu\psi\rightarrow D_\mu\psi=\partial_\mu\psi+ \frac{ie}{\hbar c}A_\mu\psi$$ $$\partial_\mu\psi^\dagger\rightarrow D_\mu\psi^\dagger=\partial_\mu\psi^\dagger- \frac{ie}{\hslash c}A_\mu\psi^\dagger$$ and check that $D_\mu$ is covariant $$D'_\mu\psi'=e^{-i\alpha(x)}D_\mu\psi$$ at the same time $$A'_\mu=A_\mu+ \frac{\hslash c}{e}\partial_\mu\alpha(x)$$ And then, the author says the vector field $A_\mu$ is the vector potential of electromagnetic field, and writes down the Langrangian $$L_{matter+em}=L_{matter}(\partial_\mu\psi\rightarrow D_\mu\psi)+L_{em}$$

I am confused what the author does. In this part, I can follow the derivation, but I can not understand.

Why need we introduce the transformation at first? What does the idea base on?

I don't know why $A_\mu$ is the vector potential of eletromagnetic field. Is it because the propeties that $A_\mu$ has is the same as the vector potential of eletromagnetic field by after calculating?

I mean, we should have not known what the $A_\mu$ is when we introduced it to make $D_\mu$ covariant. So, how do we know the $A_\mu$ is? What is the reason that we treat the $A_\mu$ as the vector potential of electromagnetic field?

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Try this explanation on for size. I emphasise that it is my own way of understanding the $U(1)$ Gauge Invariance of electrodynamics and I haven't seen it elsewhere in exactly the same words.

The derivation you cite is wontedly given in the context of the semi-classical (i.e. first quantized, or before the quantum field is introduced) Dirac or Schrödinger equation for the electron: the same reasoning applies to both. These equations describe a fermion, so one cannot re-interpret their particle fields $\psi$ as a macroscopic, classically-measureable field: the Pauli exclusion principle means that you cannot copy your fermion and have $N$ (where $N$ is very big) particles in the same quantum state. Otherwise, you could in principle measure the full complex value of $\psi(x,y,z,t)$ to arbitrary accuracy by copying $\psi$ in this way and then doing a classical measurement - something that you can do with bosons (see below).

What does this mean at the one particle level? It means that only $|\psi|^2$ is experimentally meaningful for one particle: you can measure the probability of finding the particle at a position in space, but the phase of $\psi$ has no such meaning. So we should be able to multiply $\psi$ by an arbitrary phasing function $e^{i\,\alpha(x,y,z,t)}$ and get something that means physically the same thing. But, in the light of the structure of the Dirac or Schrödinger equations this seems absurd: most assuredly:

$\partial_j \left[e^{i\,\alpha(x,y,z,t)}\,\psi(x,y,z,t)\right]\neq e^{i\,\alpha(x,y,z,t)}\, \partial_j\,\psi(x,y,z,t)$

but rather

$\partial_j\left[ e^{i\,\alpha(x,y,z,t)}\,\psi(x,y,z,t)\right]= e^{i\,\alpha(x,y,z,t)}\, \left[\partial_j\,\psi(x,y,z,t) + i\, \psi(x,y,z,t)\,\partial_j\,\alpha(x,y,z,t)\right]$

and so such an assumption of invariance with respect to arbitrary phasing is invalidated because phasing patently "ruins" the structure of the Dirac or Schrödinger equations unless the phase factor $\alpha$ is globally constant. This is reasonable: the phase of $\psi$ is most definitely part of the solutions of the equations and plays a definite role in diffraction and other wave effects that have a bearing on the intensity field $|\psi|^2$.

But one can "retrieve" the situation by postulating that the single particle is coupled to some outside field, so that the Dirac or Schrödinger equations now have the terms involving this coupled-in field: if so, we can add an arbitrary phase $\alpha(x,y,z,t)$ and keep the global equation the same by saying that, whenever we do this, we must take away a balancing $i \partial_j\,\alpha(x,y,z,t)$ from the coupled in field. Thus, we conclude, if the coupled-in field $A_j$ is at all physical, it has to give the same measurements as the field $A_j + \partial_j\,\alpha(x,y,z,t)$: the original plus any arbitrary field of the form $\partial_j\,\alpha(x,y,z,t)$, where $\alpha(x,y,z,t)$ is some suitably well-defined scalar field. But there is a field that behaves exactly like this: the electromagnetic four-potential. We can add a spatial gradient $\nabla \alpha$ to the vector part and at the same time add the scalar $\partial_t\alpha$ to the scalar electric potential without affecting the electric $\mathbf{E}$ and magnetic $\mathbf{B}$ fields.

So there we have it: the assumed "gauge invariance" suggests the electromagnetic field, because the balancing field behaves the same way as gauge-transformed vector magnetic and electric potentials. So the essential reasoning here is really a hunch: if it looks like a duck and quacks, maybe it is a duck. So too with the outside field coupling into the Dirac equation. It behaves like the potential fields of Maxwell's electromagnetism, so "we" (or rather the physicist who first thought of this - my ignorance sadly hinders my telling you who) go with our hunch and see what happens when we assume the field is the electromagnetic field.

Note that the same ideas do not apply to bosons. We can think of Maxwell's equations as the first-quantized equations for the photon. No one speaks of making Maxwell's equations invariant with respect to multiplication by an arbitrary phase function $e^{i\,\alpha(x,y,z,t)}$ in the same way as is done for the Dirac equation. This is even though the Maxwell equations can be cast in a quaternionic form that is identical to a zero-mass Dirac equation when the latter is thought of as two quaternionic equations coupled by a mass term - so the same mathematical trick would be just as valid with the Maxwell equations as it would with the Dirac equation. My interpretation is this: the photon's phase is classically meaningful: photons are bosons, so in principle we can get as many as we like in the same state: so many indeed that we can measure the phase of their common "wave function" (which is now the electromagnetic field - see caveats in the afternotes below) to arbitrary accuracy with classical measurement devices such as interferometers. Every one-particle photon state corresponds EXACTLY to a macroscopic, classical electromagnetic field that we can set up and measure in detail to arbitrary accuracy in the laboratory. So, even though the phase of one photon is "hidden" just as is the phase of one electron above - there are no "phase" eigenstates and no "phase" observable - it must still be "absolute" in the "prolifically-copy-and-classically-measure" sense just described.

This ends my answer, but I add some interesting related things below.

Maxwell's Equations from $U(1)$ Gauge Invariance

Incidentially, one can use this gauge invariance thinking to motivate or "derive" the Maxwell equations. Given suitable differentiability assumptions on the field, the simplest way to derive fields that are exactly unaffected by the gauge transformations is to form the tensor curl:

$F_{\mu\,\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$

(try this if you've not before: it leaves nonzero $F_{\mu\,\nu}$ that are unaffected by the gauge transformation). Now we postulate a Lorentz-invariant wave equation for a massless field: $\Box \mathbf{A} = \mathbf{0}$ is the obvious one (let's just go all the way back to D'Alembert!). You straight away have get the freespace Maxwell equations as equations fulfilled by the "physical", unaffected by gauge $F_{\mu,\nu}$.

How ironic that the Pauli Exclusion Principle, which was first postulated to "stop" electrons in the Bohr atom from radiating and thus told against behavior seemingly foretold by the Maxwell equations can be used to motivate a gauge transformation that pretty much leads from the Dirac equation back to the Maxwell equations!

An intriguing and uncluttered "first quantized" or "baby" formulation of QED (appealing especially to non quantum field theorists like me) can be gotten by postulating the Dirac-Maxwell equation:

$\gamma^\mu\left(i \partial_\mu - q A_\mu\right) \psi + V \psi - \psi = 0$

$\partial_\nu F^{\nu\,\mu} = q\,\bar{\psi} \gamma^\mu \psi$

with the Lorenz Gauge $\partial_\mu A^\mu = 0$

i.e. intuitively, the source for the Maxwell equations is $q$ times the probability current density (here $\bar{\psi}$ stands for charge conjugated $\psi$). A first quantized electron field is now nonlinearly coupled to a first quantized photon field. This nonlinear system can be solved exactly for the hydrogen atom potential $V$ and in other situations

A. O. Barut and J. Kraus, "Nonperturbative Quantum Electrodynamics: The Lamb Shift", Foundations of Physics, Vol. 13, No. 2, 1983

and this solution can indeed model the Lamb shift and spontaneous emission. The series solution comes out to something very like the standard QED perturbation terms and indeed one must step in and "renormalize" this solution too.

Maxwell's Equations as the Propagation Equations Photon Wave Function

For various reasons, a first quantized "photon wave function" has some different behaviours and interpretations from the fermion field $\psi$ in the Dirac equation. It is perfectly valid to treat the solutions of Maxwell's equations as one-photon quantum states, but one must understand that there are difficulties in handling localized position eigenstates for the photon precisely analogous to those for fermions. See the works of Iwo Bialynicki-Birula, Margaret Hawton in the late 1990s and early 2000s, for example:

Iwo Bialynicki-Birula, "On the Wave Function of the Photon" Acta Physica Polonica 86, 97-116 (1994)

Margaret Hawton and William E. Baylis, "Angular momentum and the geometrical gauge of localized photon states", Phys. Rev. A 71, 033816 (2005)

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Maybe I get your point. You say that the electromagnetic field is the reason why the phase of matter field is not constant, right? All the effects or information that matter interact with electromagnetic field is the changing of phase of matter field, right? – qfzklm Aug 6 '13 at 13:17
But I still have a question. All the field interacting with matter maybe change the phase and amplitude of matter field. So, does only electromagntic field change the phase only? If not, how we identify the $A_\mu$ is electromagnetic field? Maybe it is the superposition of electromagnetic field and some other field... – qfzklm Aug 6 '13 at 13:26
The electromagnetic field can absorb any phase you put onto the fermion's wave function in such a way that the physical fields $\mathbf{E}$, $\mathbf{B}$ ... are not affected. Or, the other way around, when you make a gauge transformation on the electromagnetic field (which doesn't affect your physical fields), the matter's phase is "where" the added gauge $\mathbf{A},\phi$ "go", and, because this matter is fermionic, there is no chance of an experimenter being able to "amplify" the wavefunction to see the phase as they in principle can with bosons. – WetSavannaAnimal aka Rod Vance Aug 6 '13 at 13:30
As to your second question: as I said - the whole thing is simply a "hunch" - just as with many things in physics - like Einstein modeling his field equations on the poisson equation and so on. Also, we are talking "electrons" here, so it's not unreasonable to assume it's the electromagnetic field!! You can then go and test this hunch with, say, the Aharonov-Bohm effect, which suggests strongly that the $\mathbf{A}$ field is the appropriate one to put into the electromagnetic momentum $p + q \mathbf{A}$. Nothing proves that it's not some mixture of fields like you say- you just "suck and see"! – WetSavannaAnimal aka Rod Vance Aug 6 '13 at 13:34
Nice answer. Just one minor thing: when you said "the photon's phase is classically meaningful" I'm sure you meant the phase of the corresponding coherent state. It may be worth a caveat to the effect that the phase of a single photon isn't meaningful, just to avoid possible misunderstanding? – twistor59 Aug 6 '13 at 14:36