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Quantizing the free electromagnetic field in the Lorenz gauge, $\partial_\mu A^\mu=0$, is subtle. We must add a gauge-fixing term to the action so that $\pi^0$ does not vanish identically. Also, we cannot impose $\partial_\mu A^\mu=0$ directly as an operator equation because again the commutator relations cannot be satisfied even with $\pi^0\neq 0$. So we have to implement it via Gupta-Bleuler's suggestion.

But in the Coulomb gauge $A^0=\vec{\nabla}\cdot{\vec A}=0$, the quantization proceeds in a rather straightforward manner.

What is the root cause of this problem? Does the Lorenz gauge condition fail to remove all the gauge redundancies? Is a simple way to understand this?

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    $\begingroup$ As a side comment, it is called the Lorenz gauge, after Ludvig Lorenz, not to be confused with Hendrik Lorentz $\endgroup$
    – Níckolas Alves
    Commented Jun 13, 2022 at 15:58
  • $\begingroup$ I never thought about this in the context of quantizing the EM field, but the Lorenz gauge condition does fail to remove all gauge freedom. In fact, in vacuum, the Coulomb gauge is often taken as a particular case of the Lorenz gauge. This is discussed, for example, in Wald's Advanced Classical Electromagnetism, Sec. 5.5, among other books on E&M. $\endgroup$
    – Níckolas Alves
    Commented Jun 13, 2022 at 16:01
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    $\begingroup$ I guess a way of perceiving this is noticing that the four-potential initially has four degrees of freedom, only two of which are physical (corresponding to the photon helicities or the classical wave polarizations). The Coulomb gauge does manage to impose two constraints, hence reducing the number of dof's to two. The Lorenz gauge imposes only one constraint, and hence there is still one unphysical degree of freedom left $\endgroup$
    – Níckolas Alves
    Commented Jun 13, 2022 at 16:05

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The thing about the Coulomb gauge is that, in vacuum, you get both $A_0 = 0$ and $\nabla \cdot \vec A = 0$, so it eliminates the $(A_0,\pi^0)$ d.o.f. right out of the gate - we just have $A_0 = 0,\pi^0 = 0$ consistently and only need to care about the spatial parts, and due to $\nabla\cdot \vec A= 0$, the whole thing is just a bunch of oscillators in Fourier space, where the condition $\vec p \cdot \vec A(\vec p) = 0$ then eliminates one of the three remaining degrees of freedom. Note that because the Coulomb gauge has broken Lorentz covariance anyway, we don't care that this elimination of d.o.f. is not stable under Lorentz transformations.

The Lorenz gauge, in contrast, is a Lorentz invariant gauge condition and hence we're not allowed to just "eliminate $A_0$" or something because what's $A_0$ in one frame is a linear combination of all the $A_\mu$ in another - the point of picking the Lorenz gauge is precisely to preserve Lorentz covariance. So in a way the "difficulty" isn't as much down to specifics of the gauge condition as it is to us trying to do things covariantly vs. non-covariantly, but either way the core difference is that we're not allowed to drop the 0-th d.o.f., which really make the whole thing much simpler in the Coulomb gauge.

Additionally, the Lorenz gauge indeed does not completely remove gauge freedom since it remains invariant under harmonic gauge functions, see this question and its answers.

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  • $\begingroup$ "Lorenz gauge indeed does not completely remove gauge freedom since it remains invariant under harmonic gauge functions" but isn't this residual gauge freedom also present in the Coulomb gauge? Even after arranging $\nabla\cdot\vec A=0$, the vector $\vec A$ does not become unique. We can add a $\nabla\theta(x)$ to $\vec A$ to define a new vector field $\vec A^\prime=\vec A+\nabla\theta$ which still satisfies $\nabla\cdot\vec A^\prime=0$ and leads to same $E,B$ fields. $\endgroup$
    – Solidification
    Commented Jun 13, 2022 at 17:05
  • $\begingroup$ @Solidification Sure, there's also non-uniqueness in the Coulomb gauge, see e.g. physics.stackexchange.com/q/558981/50583. That's why nothing in my answer relies on discussing the uniqueness of the conditions - uniqueness is not the problem here. I just added that last sentence because your question sounded unsure whether or not the Lorenz gauge was unique. $\endgroup$
    – ACuriousMind ♦
    Commented Jun 13, 2022 at 17:11
  • $\begingroup$ Edit to above comment: provided $\theta(x)$ obeys Laplace's equation $\nabla^2\theta=0$. $\endgroup$
    – Solidification
    Commented Jun 13, 2022 at 17:59

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