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I know we can express the electric field $\mathbf{E}$ and the magnetic field $\mathbf{B}$ in terms of the electric potential $\phi$ and vector potential $\mathbf{A}$:

$$ \mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t}, $$

$$ \mathbf{B} = \nabla \times \mathbf{A}.$$

I have been reading Gerry and Knight's book on quantum optics and in chapter two they quantise the EM field by starting with Maxwell's theory. They start in the vacuum free of sources of currents and state that

$$ \mathbf{E} = - \frac{\partial \mathbf{A}}{\partial t}, $$

$$ \mathbf{B} = \nabla \times \mathbf{A}.$$

What happened to the $-\nabla \phi$ contribution for $\mathbf{E}$?

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It is called the Gibbs gauge condition (or sometimes Hamiltonian gauge, or temporal gauge in textbooks).

  • J.W. Gibbs, Velocity of Propagation of Electrostatic Forces, Nature 53, 509 (1896)

In electromagnetism, there is a gauge freedom such that transforming the potential fields as $$ \phi \rightarrow \phi - \frac{1}{c} \frac{\partial \lambda}{\partial t} \\ \mathbf{A} \rightarrow \mathbf{A} - \nabla \lambda $$ will not change the Maxwell's equations for arbitrary scalar function $\lambda$.

Gibbs gauge is simply $$ \tag{Gibbs} \phi = 0 $$ exactly as you stated in the question. Nothing more.

For some other cases, if you set a $\lambda$ such that it gives $$ \tag{Coulomb} \nabla \cdot \mathbf{A} =0 $$ then it is called Coulomb gauge, if $$ \tag{Lorenz} \nabla \cdot \mathbf{A} - \frac{1}{c} \frac{\partial \phi}{\partial t} = 0 $$ then it is called Lorenz gauge. There are several gauges that are used in literature. It actually depends on your problem. Just like choosing a coordinate system in order to solve it easier. Since you have a source-free problem, it is better to get rid of the scalar potential and only deal with the vectorial potential.

EDIT: As I stated in the comments below, the Gibbs gauge provides $\nabla \cdot \mathbf{A} =0$ far from charges. However, in the problem above, there are no charges so you have the gauge condition $\nabla \cdot \mathbf{A} =0$ everywhere just like Coulomb gauge. So, Gibbs gauge implies Coulomb gauge in the source-free case.

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    $\begingroup$ That makes sense, however my textbook explicitly stated it was using the Coulomb gauge... maybe this is an error? If you want to look for yourself it is on page 18 of Introductory Quantum Optics by Gerry & Knight. $\endgroup$ – Matt0410 Apr 15 '18 at 11:51
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    $\begingroup$ In page 75 after Eq.(4.9), the authors state the following: "At this point we make a definite choice of gauge, namely the Coulomb (or radiation) gauge, for which $\Phi=0$ and A satisfies the transversality condition $\nabla \cdot A = 0$." So, it seems they define the Coulomb gauge by combining Gibbs and Coulomb together. $\endgroup$ – Oktay Doğangün Apr 15 '18 at 12:16
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    $\begingroup$ Actually, I think I understand what they did. The Gibbs gauge has a property that far from the charges you obtain $\nabla \cdot A = 0$ like the Coulomb gauge. Since in the problem there are no charges and currents, the approximation of being far becomes exact and thus automatically you get the Coloumb gauge for gratis. $\endgroup$ – Oktay Doğangün Apr 15 '18 at 12:28
  • $\begingroup$ I have tried to show that Maxwell's equations in the Gibbs gauge reduce to the wave equation for $\mathbf{A}$ but I can't. If I substitute $\mathbf{E}$ and $\mathbf{B}$ into Maxwell's equations I get $\nabla^2\phi + \partial_t \nabla \cdot \mathbf{A} = -\frac{1}{\epsilon_0}\rho$ and $\Box^2 \mathbf{A} - \nabla(\nabla \cdot A + 1/c^2 \partial_t \phi) = -\mu_0J $. Applying Gibbs gauge and vacuum, these reduce to $\partial_t \nabla \cdot \mathbf{A} =0 $ and $-1/c^2 \partial^2_t \mathbf{A} =0$. Now I am not sure how I get the wave equation for $A$ from this... ($\Box^2$ is the d'Alembertian) $\endgroup$ – Matt0410 Apr 15 '18 at 13:13

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