Energy Density in Coulomb Gauge I want to show that the energy density
$$\mathcal{H} = \frac{1}{8 \pi}(\vec{E}^2 + \vec{B}^2)$$
of the EM field can be written as the following in the Coulomb gauge:
$$\mathcal{H} = \frac{1}{8\pi}(\frac{1}{c^2}(\frac{\partial \vec{A}}{\partial t})^2 - \vec{A} \cdot \vec{\nabla}^2 \vec{A} ).$$
Any help would be much appreciated!
 A: The Coulomb gauge is
$$
\tag 1 \nabla \cdot \mathbf A = 0
$$
It seems that You study the free theory with no charge density $\rho$, for which
$$
\Delta A_{0} = -4\pi \rho \equiv 0 \Rightarrow A_{0} = \text{const}
$$ 
so this gauge provides also (we may simply set const to zero)
$$
A_{0} = 0
$$
With this condition, the definitions of electric and magnetic fields through 4-potential are
$$
\mathbf E = -\frac{1}{c}\partial_{t}\mathbf A, \quad \mathbf B = \nabla \times \mathbf A
$$
The square of electric field
$$
\tag 2 \mathbf E^{2} = \frac{1}{c^{2}}\partial_{t}(\mathbf A)^{2}
$$ 
immediately gives You the first term in the re-expressed Hamiltonian.
The square of magnetic field is
$$
\mathbf B^{2} = [\nabla \times \mathbf A]\cdot [\nabla \times \mathbf A]
$$
Let's rewrite it by using the property
$$
\epsilon_{ijk}\epsilon^{ilm} = (\delta^{l}_{j}\delta^{m}_{k} - \delta^{l}_{k}\delta^{m}_{j})
$$
We have
$$
\mathbf B^{2} \equiv B_{i}B^{i} = \epsilon_{ijk}\epsilon^{ilm}\partial^{j}A^{k}\partial_{l}A_{m} = (\partial^{l}A^{m}\partial_{l}A_{m} - \partial^{l}A^{m}\partial_{m}A_{l})
$$
By using chain derivative rule, we have
$$
\mathbf B^{2} = \partial^{l}(A^{m}\partial_{l}A_{m}) - \mathbf A \cdot \Delta \mathbf A - \partial^{l}(A^{m}\partial_{m}A_{l}) + (\mathbf A \cdot \nabla)(\nabla \cdot \mathbf A)
$$
This expression may be simplified. First, the first and the third terms are total derivatives, so that vanish after integration over 3-space. Second, the fourth term vanishes due to Coulomb gauge fixing condition. Therefore,
$$
\tag 3 \mathbf B^{2} \simeq -\mathbf A \cdot \Delta \mathbf A
$$
By collecting the results $(2)-(3)$ You obtain
$$
H(t) = \frac{1}{8\pi}\int d^{3}\mathbf r( \mathbf E^{2} +\mathbf B^{2}) = \frac{1}{8\pi} \int d^{3}\mathbf r\left(\frac{1}{c^{2}}(\partial_{t}\mathbf A)^{2} - \mathbf A \cdot \Delta \mathbf A\right)
$$
