# Maxwell's equations and QFT

• I
• Featured
Hello,

I have been wondering about the validity of Maxwell's equations in quantum physics. I looked in the internet and it seems from what I understood that: Maxwell's equations are valid for any situation, classical or quantum. In fact, maybe it holds more legitimacy than Schroedinger equation since it is a relativistic (invariant) set of equations.

Yet, I am really baffled! The equations said to be valid, yet I don't see any wave function in it. (Ok, this might be hilarious. But, any equation I see in quantum physics have a wave function and probability distribution!). Are the electric and magnetic fields alongside the functions ##\rho \ \& \ J## probabilistic!

My question is, how Maxwell's equations are implemented/related to quantum physics? Why it is not usually in use?

*Disclaimer: I am undergraduate student, and I don't have that much experience with quantum physics.
**I would appreciate it if you supported your replies with references of books and papers.

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## Answers and Replies

hilbert2
Science Advisor
Gold Member
The same Lagrangian density function ##\mathcal{L}## that is used in deriving the classical Maxwell equations is also the basis of quantum electrodynamics. A "wave function" describing a quantum field would actually need to be a set of infinitely many functions, one for every point in space, so that you could form a probability density for the field strengths at any point.

• Phylosopher
"But it wasn't until 1884 that Oliver Heaviside, concurrently with similar work by Josiah Willard Gibbs and Heinrich Hertz, grouped the twenty equations together into a set of only four, via vector notation. This group of four equations was known variously as the Hertz–Heaviside equations and the Maxwell–Hertz equations, but are now universally known as Maxwell's equations. Heaviside's equations, which are taught in textbooks and universities as Maxwell's equations are not exactly the same as the ones due to Maxwell, and, in fact, the latter are more easily forced into the mold of quantum physics. This very subtle and paradoxical sounding situation can perhaps be most easily understood in terms of the similar situation that exists with respect to Newton's second law of motion. In textbooks and in classrooms the law F = ma is attributed to Newton, but his second law was in fact F = p', where p' is the time derivative of the momentum p. This seems a trivial enough fact until you realize that F = p' remains true in the context of Special relativity." (my italics)
en.wikipedia.org/wiki/History_of_Maxwell's_equations (The term Maxwell's equations)

• David Lewis, dextercioby and Phylosopher
A. Neumaier
Science Advisor
The equations said to be valid, yet I don't see any wave function in it.
The Maxwell equations are equations both for classical fields and for quantum fields. The latter represent observables, while wave functions represent states.

Both classically and in quantum mechanics, the free Maxwell equations (for electromagnetic fields without sources) are linear, hence free electromagnetic waves and free photons don't interact.

In the presence of matter, one has to use a bigger coupled system consisting of the Maxwell equations coupled to moving sources, which are themselves coupled to the electromagnetic field, and these coupled equations are nonlinear, hence interacting - again both in classical and in quantum mechanics.

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• bhobba, Spinnor and Phylosopher
DrDu
Science Advisor
The point is that up to now, satisfactory QFT's for which you can write down a Hamiltonian and a Schrödinger equation, have only been found in 0, 1 and 2 dimensions. In 3 dimensions, QFT is little more than a rather ill defined set of rules to do perturbation expansions.

• Phylosopher
A. Neumaier
Science Advisor
• bhobba
DrDu
Science Advisor
It is a mathematically well defined set of rules. Yet it is not clear at all whether this perturbation series is even an asymptotic expansion of some more fundamental underlying theory. Of course it is a very successful construct. However, I think that many, especially non-specialist have the idea that QED or QFT is an albeit very complex but complete theory which allows to derive all other observations in nature from it.

samalkhaiat
Science Advisor
Hello,
I have been wondering about the validity of Maxwell's equations in quantum physics. I looked in the internet and it seems from what I understood that: Maxwell's equations are valid for any situation, classical or quantum.
I am afraid, your understanding is not correct. The classical Maxwell equations (like the equations of all other gauge field theories) must be modified at quantization if we want to maintain Poincare’ covariance. Bellow, I will provide two proofs for my statement. The first one is at the undergraduate level (i.e., for you), and the second proof is for those who know few things about the Wightman formalism of QFT.

1) Recall that in the (covariant) Lorenz gauge, the free Maxwell equations are $$\partial^{2}A_{\mu}(x) = 0 , \ \ \ \ \ \ \ \ \ \ (1.1)$$$$\partial^{\mu}A_{\mu}(x) = 0 . \ \ \ \ \ \ \ \ \ \ (1.2)$$

Now, suppose that eq(1.1) and eq(1.2) are valid operator equations in the quantized theory. Also notice that eq(1.1) is a set of 4 massless Klein-Gordon operator field equations. So, we may use the covariant (4-dimentional) commutation relations of the Klein-Gordon operators to set up the following commutation relations for the Maxwell’s field operator $A_{\mu}(x)$:

$$\left[ A_{\mu}(x) , A_{\nu}(y) \right] = - i \eta_{\mu\nu} \ \Delta (x - y \ ; 0) , \ \ \ \ \ \ \ (1.3)$$ where $\Delta (x ; 0)$ is the massless limit of the Lorentz invariant $\Delta$-function $$\Delta (x ; 0) = \lim_{m \to 0} \left( \frac{-1}{(2 \pi )^{3}} \int d^{3} \vec{k} \ \frac{\sin kx}{\omega_{\vec{k}}}\right) .$$ Now, you can see at once that eq(1.2) can not hold as operator equation in the quantized theory because it is incompatible with the covariant commutation relations (1.3): $$\left[ \partial^{\mu}A_{\mu}(x) , A_{\nu}(0) \right] = - i \ \partial_{\nu} \Delta (x ; 0) . \ \ \ \ \ \ \ (1.4)$$ If you interpret the Lorenz condition (1.2) as operator equation, then the LHS of (1.4) is identically zero while the RHS is certainly not identically zero. This inconsistency can be resolved by replacing the Lorenz condition (1.2) (hence modifying Maxwell’s equations) by the weaker (Gupta-Bleuler) condition $$\partial^{\mu}A^{+}_{\mu}(x) | \Psi \rangle = 0 , \ \ \ \ \ \ \ \ \ \ (1.5)$$ involving annihilation operators only. So, the allowed states of the quantized Maxwell theory must be restricted to satisfy the condition (1.5) instead of (1.2). From the condition (1.5) and its adjoint, it follows that the Lorenz condition (1.2) holds (as it should) for expectation values $$\langle \Psi | \partial^{\mu}A^{+}_{\mu} + \partial^{\mu}A^{-}_{\mu}| \Psi \rangle = \langle \Psi | \partial^{\mu}A_{\mu} | \Psi \rangle = 0 .$$ This ensures that the Lorenz condition and hence the full Maxwell’s equations hold as the classical limit of the quantum theory of electromagnetic field.

****

2) Proposition: The free Maxwell equation $$\left( \delta^{\nu}_{\mu} \partial^{2} - \partial_{\mu}\partial^{\nu}\right) A_{\nu}(x) = 0 , \ \ \ \ (2.1)$$ is incompatible with manifest Poincare’ covariance if $A_{\mu}(x)$ is a nontrivial quantized vector field.

Proof: Consider the 2-point function $\langle 0 | A_{\nu}(x) A_{\rho}(y) | 0 \rangle$. Covariance with respect to the Poincare’ group allows us to express the 2-point function in the form $$\langle 0 | A_{\nu}(x) A_{\rho}(y) | 0 \rangle = \eta_{\nu\rho} F_{1}(x-y) + \partial_{\nu}\partial_{\rho}F_{2}(x-y) , \ \ \ \ \ (2.2)$$ for some Lorentz invariant functions $F_{1}$ and $F_{2}$. Applying the Maxwell differential operator $\left( \delta^{\nu}_{\mu} \partial^{2} - \partial_{\mu}\partial^{\nu}\right)$ to both sides of (2.2) and using (2.1), we get $$\eta_{\mu\rho}\partial^{2}F_{1}(x) - \partial_{\mu}\partial_{\rho}F_{1}(x) = 0 . \ \ \ \ \ \ \ \ (2.3)$$ Contracting $\mu$ and $\rho$, we obtain $$\partial^{2}F_{1}(x) = 0 . \ \ \ \ \ \ \ \ \ \ \ \ \ (2.4)$$ Substituting (2.4) back in (2.3), we find $$\partial_{\mu}\partial_{\rho}F_{1}(x) = 0 . \ \ \ \ \ \ \ \ \ \ \ \ \ (2.5)$$ Since $F_{1}(x)$ is Lorentz invariant, a constant $C$ is the only solution of (2.5). Thus, the form of the 2-point function (2.2) reduces to $$\langle 0 | A_{\nu}(x) A_{\rho}(y) | 0 \rangle = \eta_{\nu\rho} C + \partial_{\nu}\partial_{\rho}F_{2}(x-y) . \ \ \ \ \ \ (2.6)$$ Now, from (2.6) and the definition of the physical field $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$, we obtain $$\langle 0 | F_{\mu\nu}(y) F_{\rho\sigma}(x) | 0 \rangle = 0 . \ \ \ \ \ \ \ \ \ \ \ \ (2.7)$$ Smearing the local physical field $F_{\mu\nu}$ with an arbitrary real test function $f^{\mu\nu}(x)$, we conclude that the state $$\int d^{4}x \ f^{\rho\sigma}(x) F_{\rho\sigma}(x) \ | 0 \rangle , \ \ \ \ \ \ \ \ \ \ \ (2.8)$$ must belong to a positive-metric Hilbert space. Thus, eq(2.7) implies that (2.8) vanishes identically. Then, due to the arbitrariness of the test function $f^{\mu\nu}(x)$, we get $$F_{\rho\sigma}(x) | 0 \rangle = 0 .$$ Now, using the separating property of the vacuum with respect to local fields, we conclude that $F_{\mu\nu}(x) = 0$, meaning that $A_{\mu}(x)$ is trivial. As we have seen in part(1), this difficulty cannot be resolved by adding the Lorenz condition (1.2). Thus, the classical gauge field equation cannot survive in quantum theory as a local operator equation.

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• bhobba, nrqed, Delta2 and 4 others
A. Neumaier
Science Advisor
1) Recall that in the (covariant) Lorenz gauge, the free Maxwell equations are $$\partial^{2}A_{\mu}(x) = 0 , \ \ \ \ \ \ \ \ \ \ (1.1)$$$$\partial^{\mu}A_{\mu}(x) = 0 . \ \ \ \ \ \ \ \ \ \ (1.2)$$
No. The free Maxwell equations are not about the vector potential but about the electromagnetic field, and these hold also on the operator level!

• Peter Morgan
samalkhaiat
Science Advisor
No. The free Maxwell equations are not about the vector potential but about the electromagnetic field, and these hold also on the operator level!
One solves the Maxwell equation in terms of $A_{\mu}$; quantizes $A_{\mu}$; classical and quantum interaction is given by $A_{\mu}J^{\mu}$; the successes of QED is achieved by imposing subsidiary condition on $A_{\mu}$, and you now come and tell me it is “ not about the vector potential”.

• protonsarecool and Demystifier
Demystifier
Science Advisor
Gold Member
The free Maxwell equations are not about the vector potential but about the electromagnetic field
Then what is classical general relativity about? Perhaps only about Ricci scalar ##R##, because tensors like ##R_{\mu\nu}## and ##g_{\mu\nu}## are not invariant under diffeomorphisms?

DrDu
Science Advisor
Then what is classical general relativity about? Perhaps only about Ricci scalar ##R##, because tensors like ##R_{\mu\nu}## and ##g_{\mu\nu}## are not invariant under diffeomorphisms?
But they are observables, in contrast to the vectorpotential.

Demystifier
Science Advisor
Gold Member
But they are observables, in contrast to the vector potential.
They are observables only when one fixes coordinates. Similarly, the vector potential is also an observable when one fixes a gauge.

• bhobba
DrDu
Science Advisor
First there is a difference between coordinates and gauge. Second, a tensor is not coordinate dependent, only its components with respect to a basis are.
What would be more appropriate to compare is the vector potential A in QED and the Christoffel connection in GR.

A. Neumaier
Science Advisor
One solves the Maxwell equation in terms of $A_{\mu}$; quantizes $A_{\mu}$; classical and quantum interaction is given by $A_{\mu}J^{\mu}$; the successes of QED is achieved by imposing subsidiary condition on $A_{\mu}$, and you now come and tell me it is “ not about the vector potential”.
The fields that occur in the Maxwell equations do not determine the vector potential. Thus the equations for the vector potential are not the same as those for the electromagnetic field. The latter are called the Maxwell equations.The free massless spin 1 Wightman theory is about the free electromagnetic field, and it satisfies precisely the Maxwell equations in operator form.

QED is not about the Maxwell equations.

samalkhaiat
Science Advisor
The fields that occur in the Maxwell equations do not determine the vector potential. Thus the equations for the vector potential are not the same as those for the electromagnetic field.The latter are called the Maxwell equations.
Maxwell’s electromagnetism is an Abelian gauge theory. So, up to gauge transformation $\mbox{A} \to \mbox{A} + \mbox{d}\lambda$, the free Maxwell equations $\mbox{d}\mbox{F} = 0, \ \ \delta \mbox{F} \equiv {}^{*} \mbox{d}^{*}\mbox{F} = 0$ (on the topologically trivial spacetime $\mathbb{R}^{4}$) are completely equivalent to the free Maxwell equations $\mbox{d}^{2}\mbox{A} = 0, \ \ \delta \mbox{d}\mbox{A} = 0$. In other words, All electromagnetic phenomena are describable by some $A_{\mu}(x)$. Is this news to you?
The free massless spin 1 Wightman theory is about the free electromagnetic field, and it satisfies precisely the Maxwell equations in operator form.
Therefore, by the above equivalence, what is true for the Maxwell equation $\partial^{\nu}F_{\mu\nu} = 0$ is also true for the Maxwell equation $( \eta^{\mu\nu}\partial^{2} - \partial^{\mu}\partial^{\nu})A_{\nu} = 0$. Thus, by proposition (2) of my previous post, neither the latter nor the former form of Maxwell equation can survive quantization. A modification of the Maxwell equation, such as for example, $$\partial_{\nu}F^{\mu\nu} + \partial^{\mu}B = J^{\mu}, \ \ \ \ \ (1)$$$$\partial^{\mu}A_{\mu} + \alpha B = 0, \ \ \ \ \ (2)$$ is indispensable in quantized theory. In this B-field formalism, (1) is called the quantum Maxwell equation. And, eq(2) tells you that the Lorenz condition holds as operator equation only in the Landau-gauge $\alpha = 0$.
So, In any (non-abelian) gauge theory, the classical (non-abelian) Gauss’ law (constraint) $$G^{a}(x) \equiv - (D_{i}E_{i})^{a}(x) = 0,$$ which is one of the (non-abelian) Maxwell equations, cannot hold as an operator equation since it cannot be made compatible with any acceptable commutation relations, rather it is implemented as a restriction on the allowed physical states: $G^{a}| \Psi \rangle = 0$.
Rigorous treatment of this subject can be found in the good book of F. Strocchi, “An introduction to non-perturbative foundations of quantum field theory”, Oxford, 2013.

QED is not about the Maxwell equations.
Don’t throw ambiguous statements at me. Write down for us a classical, Poincare invariant action integral describing the interaction between Maxwell’s field and some charged matter field, then answer the following: (i) Can you do it without using the vector potential? (ii) What is the name of the equation which follows from the action principle? Do you call it The Cucumber equation, or the Maxwell equation?

• dextercioby
A. Neumaier
Science Advisor
are completely equivalent to the free Maxwell equations [...] Is this news to you?
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.

It is indeed news to me that your equations are completely equivalent to those of Maxwell, since they aren't. 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, so the equations are already not equivalent classically, but the A's contain redundant information that must be gauged away. A is not a field, as - in your description - it is determined by the observable fields only up to a residual gauge freedom.

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!

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A. Neumaier
Science Advisor
Write down for us a classical, Poincare invariant action integral describing the interaction between Maxwell’s field and some charged matter field
I didn't claim anything about your setup. QED is not about the Maxwell equations but about a bigger system of equations involving a fermionic field not known before 1925. Such a field does not figure in Maxwell's equations. Neither do Maxwell's equations demand a derivation from an action principle; they stand for themselves.

• Demystifier
A. Neumaier
Science Advisor
Not an expert in QM/QFT but the Aharonov–Bohm effect certainly illustrates that the EM vector A exists and is physically relevant.
https://en.wikipedia.org/wiki/Aharonov–Bohm_effect
We were discussing primarily the free electromagnetic field in Minkowski space, and in that case there is no Aharonov–Bohm effect.

In any case, physical information is only in the corresponding Wilson loop, not in A itself.

We were discussing primarily the free electromagnetic field in Minkowski space, and in that case there is no Aharonov–Bohm effect.

In any case, physical information is only in the corresponding Wilson loop, not in A itself.
I'll have to think about that, and what "physical information" means. It seems to me that all "physical" effects are indirectly measured.

A. Neumaier
Science Advisor
I'll have to think about that, and what "physical information" means. It seems to me that all "physical" effects are indirectly measured.
A is defined only up to a gauge transformation, and all physical information must be gauge invariant.

The same Lagrangian density function ##\mathcal{L}## that is used in deriving the classical Maxwell equations is also the basis of quantum electrodynamics. A "wave function" describing a quantum field would actually need to be a set of infinitely many functions, one for every point in space, so that you could form a probability density for the field strengths at any point.
I don't think he was looking for an exact answer. :-)

samalkhaiat
Science Advisor
Not an expert in QM/QFT but the Aharonov–Bohm effect certainly illustrates that the EM vector A exists and is physically relevant.
https://en.wikipedia.org/wiki/Aharonov–Bohm_effect
The dilemma is the following:
A) The tensor $F_{\mu\nu}$ is observable physical field. However, as dynamical variables $F_{\mu\nu}$ gives incomplete description in the quantum theory.
B) The vector potential $A_{\mu}$ is not an observable. But, as dynamical variables, it was found to give a full (classical and quantum) description of the physical phenomena.
Indeed, this state of affair was demonstrated nicely by the Aharonov-Bohm effect:
Classical electrodynamics can be described entirely in terms of $F_{\mu\nu}$: Once the value of $F_{\mu\nu}(x)$ at a point $x$ is given, we know exactly how a charged particle placed at $x$ will behave. We simply solve the Lorentz force equation. This is no longer the case in the quantum theory. Indeed, in the A-B effect, the knowledge of $F_{\mu\nu}$ throughout the region traversed by the electron is not sufficient for determining the phase of the electron wave function, without which our description will be incomplete. In other words, $F_{\mu\nu}$ under-describes the quantum theory of a charged particle moving in an electromagnetic field. This is why we use the vector potential $A_{\mu}$ as dynamical variable in the A-B effect as well as in QFT. However, the vector potential has the disadvantage of over-describing the system in the sense that different values of $A_{\mu}$ can describe the same physical conditions. Indeed, if you replace $A_{\mu}$ by $A_{\mu} + \partial_{\mu}f$ for any function $f$, you will still see the same diffraction pattern on the screen in the A-B experiment. This shows that the potentials $A_{\mu}(x)$, which we use as dynamical variables, are not physically observable quantities. In fact, even the phase difference at a point is not an observable, a change by an integral multiple of $2\pi$ leaves the diffraction pattern unchanged. The real observable in the A-B effect is the Dirac phase factor $$\Phi (C) = \exp \left( i e \oint_{C} dx^{\mu} A_{\mu}(x) \right) .$$ Just like $F_{\mu\nu}$, $\Phi$ is gauge invariant, but unlike $F_{\mu\nu}$, it gives correctly the phase effect of the electron wave function.

• bhobba, dextercioby and Boing3000
samalkhaiat
Science Advisor
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.

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• Spinnor, bhobba, protonsarecool and 1 other person