0
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Hint, what is $A^0$? $\endgroup$ – Lewis Miller Jul 25 at 1:55
1
$\begingroup$

There are a couple of issues with the phrasing of the question:

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 $H = H_{Coulomb} + H_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.