# Grand potential $\leftrightarrow$ ground state energy of interacting electrons in a solid

I want to calculate the ground state energy $E_0$ of interacting electrons in a solid at $T=0$ via pertubation theory and Feynman diagrams, i.e. I want to understand the connection between the pertubation series and the diagrams in the case of the ground state energy.

I found a nice introduction where the grand potential at $T>0$ is derived. First algebraical: $$\Omega-\Omega_0 = - \frac 1 \beta \ln\frac{\mathcal Z}{\mathcal Z_0} = -\frac 1 \beta \ln\left[ \sum_{n=0}^\infty \frac{(-1)^n}{n!} \int_0^\beta \text d \tau_1 \dots \int_0^\beta \text d \tau_n \; \langle \text T\{V(\tau_1)\dots V(\tau_n)\}\rangle_0 \right]\,,$$ secondly in terms of Feynman diagrams which just represent terms of this series. Here $V(\tau)$ is the interaction potential in the interaction picture, $\langle ... \rangle_0$ is the expectation value of the non-interacting system, $\text T$ is the Dyson time-ordering operator, and $\beta=\frac 1 {k_B T}$.

But what I want is the ground state energy $E_0$ at $T=0$ and not the grand potential at $T>0$. So I thought I can obtain the ground state energy $E_0$ by setting the chemical potential $\mu=0$ and performing the limit $T\rightarrow 0$ in the grand potential: $$E_0 = \lim_{T\rightarrow 0} (\Omega-\Omega_0)|_{\mu=0}$$

Is that correct? Or are there better ways?

• It doesn't seem wise to use the formula you propose for your problem. Gell Mann-Low theorem seems more suited doesn't it? physics.stackexchange.com/questions/58027/… May 12 '15 at 10:42
• Ok, I understand. But is my way wrong? May 12 '15 at 11:54
• Setting $\mu=0$ should lead to an empty system at $T=0$. Rather, I think you should calculate the grand potential for finite $\mu$ and then just subtract the contribution from the finite density, i.e. you want $E_0 = \Omega + \mu \langle N \rangle$. May 12 '15 at 12:07