A. Neumaier said:
Again you are calling the free Maxwell equations the equations in the vector potential A, while Maxwell wrote down equations in E and B, later combined into the electromagnetic field tensor F.
That was long time ago. We now regard Maxwell theory as an abelian gauge theory and study it in terms of A_{\mu}.
... that your equations are completely equivalent to those of Maxwell,
They are not “my equations”. They are the Maxwell equations written in terms of the gauge field. They are
also called Maxwell’s equations in the
literatures.
From the A's you get the F's, yes, but this does not yet make an equivalence. From the F's you get an infinite collection of possible A's,
This is just the poor man version of what I have already said. Let me repeat: Minkowski spacetime M = \mathbb{R}^{(1,3)} is topologically trivial. This means that all de Rham cohomology groups are
trivial:
H^{p}(M) \equiv \frac{\{ \mbox{closed p-forms} \}}{\{ \mbox{exact p-forms} \}} = 0.
Thus,
on M,
a form is exact if and only if it is closed (Poincare Lemma). So,
up to gauge transformation, \{ dF = \delta F = 0 \}
if and only if \{ d^{2}A = \delta dA = 0 \}.
so the equations are already not equivalent classically,
They are, because A and A + d\lambda describe the
same physics. Mathematically, we speak of
equivalence classes with A and A + d\lambda are
identified.
Thus your arguments about the A-equation in QFT do not imply anything about the F-equations. Indeed, these hold in QFT on the operator level, completely unchanged!
Almost all textbooks on QFT quantize the vector potential A_{\mu}
not the field tensor F_{\mu\nu}. Open one of those textbooks and find the expansion A_{\mu}(x) = \int \frac{d^{3}k}{2k_{0}(2 \pi)^{3}} \sum_{\beta = 0}^{3} a^{(\beta)}(k) \epsilon_{\mu}^{(\beta)}(k) e^{-ikx} + \mbox{H.C.} \ . Now, if you calculate F_{\mu\nu} from the above A_{\mu}, you find that \epsilon^{\mu\nu\rho\sigma}\partial_{\nu}F_{\rho\sigma} = 0 holds identically. However, you also find that the remaining Maxwell equations
fail to hold. Indeed, you obtain \partial^{\mu}F_{\mu\nu} = - \partial_{\nu}(\partial \cdot A) \neq 0 \ .
So, I can summarise my “argument” in #8 by the following: The free Maxwell equation \partial^{\mu}F_{\mu\nu}=0 does not hold as operator equation in the usual covariant quantization of the em-field that one can find in almost all usual textbooks. And that is a complete answer to the remarks raised in #1.
The same thing happens in QED. To see that, consider a classical theory described by \mathcal{L}(\varphi_{a} , A_{\mu}) where \varphi_{a} = ( \varphi , \varphi^{\ast}) is a complex scalar field and A_{\mu} is a massless vector field. Assume that our theory is invariant under the local (infinitesimal) transformations \delta \varphi_{a}(x) = i \epsilon (x) \varphi_{a}(x), \ \ \ \delta A_{\mu}(x) = \partial_{\mu}\epsilon (x), with arbitrary spacetime-dependent function \epsilon (x). Now we define the objects J^{\mu} \equiv \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi_{a})} (i\varphi)_{a} \equiv \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi)} (i\varphi) + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi^{\ast})}(-i\varphi^{\ast}) , and F^{\mu\nu} \equiv - \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}A_{\nu})} . With simple algebra, we find
\begin{align*}\delta \mathcal{L} & = \left( \mathcal{E}^{a}(\varphi) i\varphi_{a} + \partial_{\mu}J^{\mu} \right) \epsilon \\ & + \left( \mathcal{E}^{\mu}(A) + J^{\mu} - \partial_{\nu}F^{\nu\mu}\right) \partial_{\mu}\epsilon \\ & - \frac{1}{2} \left( F^{\mu\nu} + F^{\nu\mu}\right) \partial_{\mu}\partial_{\nu}\epsilon , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \end{align*} where \mathcal{E}^{a}(\varphi) and \mathcal{E}^{\mu}(A) are the Euler derivatives of the fields:
\mathcal{E}^{a}(\varphi) = \frac{\partial \mathcal{L}}{\partial \varphi_{a}} - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi_{a})}\right) , \mathcal{E}^{\mu}(A) = \frac{\partial \mathcal{L}}{\partial A_{\mu}} - \partial_{\nu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\nu}A_{\mu})}\right) \equiv \frac{\partial \mathcal{L}}{\partial A_{\mu}} + \partial_{\nu}F^{\nu\mu} . Thus, \delta \mathcal{L} = 0 if and only if each coefficient of \epsilon, \partial_{\mu}\epsilon and \partial_{\mu}\partial_{\nu}\epsilon vanishes identically.
So, from the first line in (1), we obtain the familiar Noether identity \mathcal{E}^{a}(\varphi) \ (i\varphi)_{a} + \partial_{\mu}J^{\mu} \equiv 0, associated with invariance under the global U(1) transformations \delta \varphi_{a}(x) = i\epsilon \varphi_{a}(x) \ , \ \ \ \ \delta A_{\mu}(x) = 0 . Thus, on actual “trajectories”, i.e., when \mathcal{E}^{a}(\varphi) = 0, we have a locally-conserved current \partial_{\mu}J^{\mu} = 0 and time-independent global-charge Q = q \int d^{3}x \ J^{0}(x) \ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
The third line in (1) gives us F_{\mu\nu} = - F_{\nu\mu}, and the second line gives us the identity \mathcal{E}^{\mu}(A) + J^{\mu} - \partial_{\nu}F^{\nu\mu} \equiv \frac{\partial \mathcal{L}}{\partial A_{\mu}} + J^{\mu} \equiv 0 . This shows that our Lagrangian \mathcal{L}(\varphi , A) must contain a term proportional to (-A_{\mu}J^{\mu}) which shows that \mathcal{L}(\varphi_{a} , A_{\mu}) describes an interacting theory of massless vector field A_{\mu} and an electrically charged scalar field \varphi_{a}. Of course, this is just a natural consequence of local gauge invariance. The identity also shows that the equation of motion followed by the field A_{\mu}, \mathcal{E}^{\mu}(A) = 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) is
equivalent to the Maxwell equation \partial_{\nu}F^{\nu\mu} = J^{\mu} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)
With this, we end our discussions of the classical (Lagrangian) scalar electrodynamics. Also, the perturbative solution of (Lagrangian) scalar QED is discussed in many textbooks, so we don’t need to do that here. We will not bother ourselves with the vector potential A_{\mu} or the underline Lagrangian. Instead, we consider a
local QFT consisting of the
electrically charged field \varphi, the
Maxwell’s field tensor F_{\mu\nu} and the conserved (electric)
vector current J^{\mu}. Of course, all these fields are operator-valued
distributions. The question we need to answer is the following: Are these fields together with the “operator” Maxwell’s equations (4)
sufficient for non-trivial QED? The answer is
negative and it is known for long time.
Theorem (Ferrari, Picasso & Strocchi):
In any local QFT in which the Noether charge Q_{R} \equiv q J^{0}(f_{T}, f_{R}) = q \int d^{4}x \ f_{T}(x^{0})f_{R}(\vec{x}) J^{0}(x^{0}, \vec{x}) ,
generates non-trivial automorphism on the local field algebra, \lim_{R \to \infty} [ Q_{R} , \varphi (f)] = q \varphi (f),
the Maxwell equations \partial_{\nu}F^{\nu\mu} = J^{\mu},
cannot be valid.
The proof is very easy and can be done formally with no need for all those smearing test functions. Just substitute J^{0} = \partial^{j}F_{j0} and use Stokes theorem, then the commutator vanishes by
locality. This contradicts the assumption that \varphi is a
charged local field (i.e., transforms non-trivially under the group generated by Q). So, in order to
keep Maxwell’s equations as
valid operator equations, one has to
abandon locality. This is the so called Coulomb gauge quantization. Thus, an
unphysical local field operator, B^{\mu} = \partial_{\nu}F^{\nu\mu} - J^{\mu}, must
necessarily be introduced in the local formulation of QED. When B^{\mu} = - \partial^{\mu}(\partial \cdot A), this gives the usual Gupta-Bleuler formulation.
But I'll try to give one in my next post.
