# Modifying the Hamiltonian when there is a presence of the Coulomb interaction

Referring to the Hamiltonian of a system of free electrons,

$$H_0= \sum_{\sigma} \int d^3rd^3r' \psi_{\sigma}^{\dagger}(\mathbf{r})\left(- \frac{\hbar^2}{2m}\nabla^2\right)\delta(\mathbf{r}-\mathbf{r'})\psi_\sigma (\mathbf{r'})$$

When the Coulomb interaction is turn on, we can modify this Hamiltonian by imposing

$$\partial^\mu\rightarrow\partial^\mu + i \frac{q}{\hbar c} A^\mu$$

I expected the new Hamiltonian to contain the term like

$$H_{\mathrm{Coulomb}} = \frac{1}{2}\sum_{\sigma,\sigma'}\int d^3rd^3r' \psi_\sigma^\dagger(\mathbf{r})\psi_\sigma(\mathbf{r})\left( \frac{q^2}{4\pi\epsilon_0|\mathbf{r}-\mathbf{r'}|}\right)\psi_{\sigma'}^\dagger(\mathbf{r'})\psi_{\sigma'}(\mathbf{r'})$$

However, when I substituted this to the free Hamiltonian, I could not see anyway to obtain this result at all.

• Hint, what is $A^0$? – Lewis Miller Jul 25 '19 at 1:55

First: Coulomb interaction only affects the temporal component of the derivative operator (i.e. you need to focus on only $$\mu = 0$$). The general form you wrote is useful if you consider the full electromagnetic field (i.e. in the presence of both $$A_0$$ and $$\mathbf A$$).
Second: The Peierls substitution for $$\partial^0$$ leads to a minimal coupling between the electric potential ($$A_0$$) and electron density, which is unavailable in the hamiltonian formalism. To see this the action is required.
Within the action formalism the $$A_0$$ field can be integrated out, which will lead to the effective long-range interaction you have quoted. To demonstrate, I use the imaginary time ($$\tau$$) formalism and slightly modify the parameters in your equations to write the action, $$S = \sum_\sigma \int d\tau d\mathbf{r} ~\Psi_\sigma^\dagger(\tau, \mathbf r) \left[\partial_\tau - i g A_0(\tau, \mathbf r) \right] \Psi_\sigma(\tau, \mathbf r) - \int d\tau H_0 + \frac{1}{2}\int d\tau d\mathbf{r} |\mathbf \nabla A_0(\tau, \mathbf r)|^2.$$
Now from within the partition function $$Z = \int DA_0 ~ D\Psi ~ D\Psi^{\dagger} ~ e^{-S},$$ one may integrate out $$A_0$$ (easier to do that if you Fourier transform and write $$S$$ in momentum space) using properties of Gaussian integration. This will result in $$S = \sum_\sigma \int d\tau d\mathbf{r} ~\Psi_\sigma^\dagger(\tau, \mathbf r) \partial_\tau \Psi_\sigma(\tau, \mathbf r) - \int d\tau (H_{Coulomb} + H_0),$$ where $$H_{Coulomb}$$ is the equation noted in the question. Therefore, the total hamiltonian is $$H = H_{Coulomb} + H_0$$.