Assignment 18 - Interacting Electron Gas - Classes of 27/10 and 29/10 (Due: 09/11 23h55)
Conditions d’achèvement
\(
\hat{U}=\frac{1}{2}\frac{e^2}{4\pi\epsilon_0}\sum_{\sigma_1,\sigma_2}\int d{\bf r} d{\bf r}^\prime \frac{\Psi^{\dagger}_{\sigma_1}({\bf r})\Psi^{\dagger}_{\sigma_2}({\bf r}^\prime)\Psi_{\sigma_2}({\bf r}^\prime)\Psi_{\sigma_1}({\bf r})}{|{\bf r} - {\bf r}^{\prime}|} \; ,
\)
where \(\Psi^{\dagger}_{\sigma_i}({\bf r})\) and \(\Psi_{\sigma_i}({\bf r})\) are field operators (sum over repeated indices \(\sigma_{i}\)).
\(
\Pi^{(0)}({\bf q}, \omega_{\bf q}) = -2i \int \frac{d^4 p}{(2 \pi)^4} \mathcal{G}^{(0)}(p+q) \mathcal{G}^{(0)}(p)
\)
where \(p=({\bf p},\omega_{\bf p})\) and \(q=({\bf q},\omega_{\bf q})\) are quadri-momenta (\(d^4 p = d^3 {\bf p} d \omega_{\bf p}\)).
Using the expression of the free propagator \(\mathcal{G}^{(0)}(p)\) derived in Assignment 17, show that:
\(
\Pi^{(0)}({\bf q}, \omega_{\bf q}) = -2 \displaystyle \int \frac{d^3 {\bf p}}{(2 \pi)^3} \frac{\theta(k_F-|{\bf p}|)\theta(|{\bf p}+{\bf q}|-k_F)}{\omega_{\bf q}+\epsilon_{|{\bf p}|}-\epsilon_{|{\bf p}+{\bf q}|}+ i \eta} +2 \displaystyle \int \frac{d^3 {\bf p}}{(2 \pi)^3} \frac{\theta(|{\bf p}|-k_F)\theta(k_F-|{\bf p}+{\bf q}|)}{\omega_{\bf q}+\epsilon_{|{\bf p}|}-\epsilon_{|{\bf p}+{\bf q}|}- i \eta}
\)
Ouvert le : lundi 10 août 2020, 00:00
À rendre : lundi 9 novembre 2020, 23:59
Problem 1 - Ground-state energy of the interacting electron gas
Consider the electron-electron interaction term:\(
\hat{U}=\frac{1}{2}\frac{e^2}{4\pi\epsilon_0}\sum_{\sigma_1,\sigma_2}\int d{\bf r} d{\bf r}^\prime \frac{\Psi^{\dagger}_{\sigma_1}({\bf r})\Psi^{\dagger}_{\sigma_2}({\bf r}^\prime)\Psi_{\sigma_2}({\bf r}^\prime)\Psi_{\sigma_1}({\bf r})}{|{\bf r} - {\bf r}^{\prime}|} \; ,
\)
where \(\Psi^{\dagger}_{\sigma_i}({\bf r})\) and \(\Psi_{\sigma_i}({\bf r})\) are field operators (sum over repeated indices \(\sigma_{i}\)).
- Write the related two-body (quartic) operator \(\hat{U}\) in momentum space in terms of the Fourier transform \(U({\bf q})\) (calculated in Assignment 13) and in terms of creation and destruction operators \(c^{\dagger}_{{\bf k}, \sigma}\) (\(c_{{\bf k}, \sigma}\)) of electrons with momentum \({\bf k}\) and spin \(\sigma\).
- Show that the 1$^{\rm st}$ order contribution to the ground-state energy per particle of the jeliium model is given by:
\( \frac{E^{(1)}}{N} = -\frac{0.916}{r_s} \mbox{ Ryd .} \)
Hint: Calculate the first-order correction \(\langle \Phi_0 | \hat{U}| \Phi_0\rangle\) (where \( | \Phi_0\rangle\) is the electron gas ground state) in momentum representation. Use Wick's theorem to express it in terms of pairings of operators and draw Feynman diagrams representing each term (there should be two). Argue that one of them is constant and should cancel the positive back-ground in the Jellium model and calculate the contribution from the other one. - Now consider the second order correction. Show that the contribution to the ground-state energy is given by:
\( \frac{E^{(2)}}{N} = (\epsilon_d + \epsilon_x) \mbox{ Ryd} \; , \)
where
\( \epsilon_d = \frac{-3}{8\pi^5} \int \frac{d \bar{\bf q}}{\bar{q}^4} \int_{|\bar{\bf k}+\bar{\bf q}|>1} d\bar{\bf k} \int_{|\bar{\bf p}+\bar{\bf q}|>1} d\bar{\bf p} \frac{\theta(1-\bar{k})\theta(1-\bar{p})}{\bar{\bf q} \cdot (\bar{\bf q} + \bar{\bf p} + \bar{\bf k})} \; , \)
\( \epsilon_x = \frac{3}{16\pi^5} \int \frac{d \bar{\bf q}}{\bar{q}^2} \int_{|\bar{\bf k}+\bar{\bf q}|>1} d\bar{\bf k} \int_{|\bar{\bf p}+\bar{\bf q}|>1} d\bar{\bf p} \frac{\theta(1-\bar{k})\theta(1-\bar{p})}{[\bar{\bf q} \cdot (\bar{\bf q} + \bar{\bf p} + \bar{\bf k})]\;|\bar{\bf q} + \bar{\bf p} + \bar{\bf k}|^2} \; . \)
(the momenta in the integrals are re-scaled, in units of \(k_F\): \(\bar{\bf k}={\bf k}/k_F\), etc.) - Does any of these integrals diverge? Why?
Problem 2 - Bubble diagram
Consider now the zero-order polarizability for the electron gas:\(
\Pi^{(0)}({\bf q}, \omega_{\bf q}) = -2i \int \frac{d^4 p}{(2 \pi)^4} \mathcal{G}^{(0)}(p+q) \mathcal{G}^{(0)}(p)
\)
where \(p=({\bf p},\omega_{\bf p})\) and \(q=({\bf q},\omega_{\bf q})\) are quadri-momenta (\(d^4 p = d^3 {\bf p} d \omega_{\bf p}\)).
Using the expression of the free propagator \(\mathcal{G}^{(0)}(p)\) derived in Assignment 17, show that:
\(
\Pi^{(0)}({\bf q}, \omega_{\bf q}) = -2 \displaystyle \int \frac{d^3 {\bf p}}{(2 \pi)^3} \frac{\theta(k_F-|{\bf p}|)\theta(|{\bf p}+{\bf q}|-k_F)}{\omega_{\bf q}+\epsilon_{|{\bf p}|}-\epsilon_{|{\bf p}+{\bf q}|}+ i \eta} +2 \displaystyle \int \frac{d^3 {\bf p}}{(2 \pi)^3} \frac{\theta(|{\bf p}|-k_F)\theta(k_F-|{\bf p}+{\bf q}|)}{\omega_{\bf q}+\epsilon_{|{\bf p}|}-\epsilon_{|{\bf p}+{\bf q}|}- i \eta}
\)