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I am learning quantum field theory, specifically the quantization of the electromagnetic field. We have this Laplacian $$ \mathcal{L} = -\frac{1}{2} \partial_\mu A_\nu \partial^\mu A^\nu -j_\mu A^\mu $$ and I understand why we impose the condition that $$ \langle \chi|\partial_\mu A^\mu |\chi \rangle = 0 \, , $$ but what does it mean? $\partial_\mu A^\mu$ is an operator (right?), but of what? What does it measure?

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Let's take the four-vector apart, decomposing it into its components, the electric potential $\phi$ and the vector potential $\mathbf{A}$:

$$A^\mu=(\phi/c,\mathbf{A})$$

Then, contracting with the derivative (and assuming the $(+,-,-,-)$ signature for the metric), one obtains:

$$\partial_\mu A^\mu = \frac{1}{c}\frac{\partial \phi}{\partial t}-\frac{\partial A_x}{\partial x}-\frac{\partial A_y}{\partial y}-\frac{\partial A_z}{\partial z}=\frac{1}{c}\frac{\partial \phi}{\partial t}-\nabla\cdot\mathbf{A}$$

So the physical interpretation is, in some sense, that this operator's observable gives you a comparison of how fast the electric potential is changing as a function of time versus how much the vector potential is diverging in space.

But this calls into question what we actually mean by "physical meaning" when it comes to gauge fields, because the values of both the electric potential and the vector potential depend on the choice of gauge that we apply.

Note that, for one of the common gauges (the Lorenz gauge), we impose the condition $\frac{1}{c}\frac{\partial \phi}{\partial t}=\nabla\cdot\mathbf{A}$, or equivalently that $\partial_\mu A^\mu=0$. As such, in the Lorenz gauge this expression's physical interpretation is trivial - it's simply the statement of the choice of gauge that we've applied. For this reason, probably the most succinct way to interpret this operator is as follows: the expectation value of the operator at a particular point in spacetime gives you a measure of how different your choice of gauge is from the Lorenz gauge at that point.

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