Ground state of BCS mean field Hamiltonian

Question

I have question following the logics of BCS Theory regarding the ground state. First let me recap the logics of textbooks, for example, by Carsten Timm . After obtaining the interacting BCS Hamiltonian $$H=\sum\limits_{\bf{k}\sigma}\xi_{\bf{k}}c^\dagger_{\bf{k}\sigma}c_{\bf{k}\sigma}+\frac{1}{N}\sum\limits_{\bf{kk'}}V_{\bf{kk'}}c^\dagger_{\bf{k\uparrow}}c^\dagger_{-\bf{k}\downarrow}c_{-\bf{k}'\downarrow}c_{\bf{k}'\uparrow},$$ the next step is to use the BCS ansatz, stating that the superconducting ground state has the form $$|GS\rangle=\prod\limits_{k}(u_{\bf{k}}+v_{\bf{k}}c^\dagger_{\bf{k}\uparrow}c^\dagger_{-\bf{k}\downarrow} )|0\rangle,$$ to find the correct expression for $u_{\bf{k}}$ and $v_{\bf{k}}$ by lowering the ground state energy variation $\langle GS|H|GS \rangle$. But the ansatz is not justified, right? So is $|GS\rangle$ the genuine ground state of $H$(My calculation seems to deny this)? If not, then is it possible to get the ground state of $H$?

After this part, Carsten Timm studied the mean field BCS Hamiltonian $$H_{MF}=\sum\limits_{\bf{k}\sigma}\xi_{\bf{k}}c^\dagger_{\bf{k}\sigma}c_{\bf{k}\sigma}-\sum\limits_{\bf{k}}\Delta^*_{\bf{k}}c_{-\bf{k}\downarrow}c_{\bf{k}\uparrow}-\sum\limits_{\bf{k}}\Delta_{\bf{k}}c^\dagger_{\bf{k\uparrow}}c^\dagger_{-\bf{k}\downarrow}+const,$$ obtaining its excitation spectrum through the usual Bogoliubov transformation, and the diagonalized Hamiltonian reads $$H_{MF}=\sum\limits_{\bf{k}}\sqrt{\xi^2_{\bf{k}}+|\Delta_{\bf{k}}|^2}(\gamma^\dagger_{\bf{k}\uparrow}\gamma_{\bf{k}\uparrow}+\gamma^\dagger_{-\bf{k}\downarrow}\gamma_{-\bf{k}\downarrow})$$ where $\gamma$'s are Bogoliubov operators.

However, the ground state (let it be $|GSMF\rangle$) for $H_{MF}$ is not mentioned. What I know is that the ground state for $H_{MF}$ must satisfy $\gamma|GSMF\rangle=0$ for all $\gamma$'s, according to the free quasi-particle picture in this mean-field Hamiltonian. but is $|GSMF\rangle$ exactly the $|GS\rangle$ mentioned before? If not, then what is the explicit form of $|GSMF\rangle$? Why $|GSMF\rangle$ is not solved (or even mentioned} from $H_{MF}$?

Answer 1

$|GS\rangle$ is the ground-state of $H_{MF}$. As can be verified: $$\sum\limits_{\bf{k}\sigma}\xi_{\bf{k}}c^\dagger_{\bf{k}\sigma}c_{\bf{k}\sigma}\prod\limits_{\bf{k'}}(u_{\bf{k'}}+v_{\bf{k'}}c^\dagger_{\bf{k'}\uparrow}c^\dagger_{-\bf{k'}\downarrow})|0\rangle=\sum\limits_{\bf{k}}(\xi_{\bf{k}}+\xi_{-\bf{k}})v_{\bf{k}}c^\dagger_{\bf{k}\uparrow}c^\dagger_{-\bf{k}\downarrow}\prod\limits_{\bf{k'}}'(u_{\bf{k'}}+v_{\bf{k'}}c^\dagger_{\bf{k'}\uparrow}c^\dagger_{-\bf{k'}\downarrow})|0\rangle$$ where the superscript $'$ on $\prod$ means that the product excludes the $\bf{k=k'}$ term. And $$-\sum\limits_{\bf{k}}\Delta^*_{\bf{k}}c_{-\bf{k}\downarrow}c_{\bf{k}\uparrow}\prod\limits_{\bf{k'}}(u_{\bf{k'}}+v_{\bf{k'}}c^\dagger_{\bf{k'}\uparrow}c^\dagger_{-\bf{k'}\downarrow})|0\rangle=-\sum\limits_{\bf{k}}\Delta^*_{\bf{k}}v_{\bf{k}}\prod\limits_{\bf{k'}}'(u_{\bf{k'}}+v_{\bf{k'}}c^\dagger_{\bf{k'}\uparrow}c^\dagger_{-\bf{k'}\downarrow})|0\rangle$$ $$-\sum\limits_{\bf{k}}\Delta_{\bf{k}}c^\dagger_{\bf{k}\uparrow}c^\dagger_{\bf{-k}\downarrow}\prod\limits_{\bf{k'}}(u_{\bf{k'}}+v_{\bf{k'}}c^\dagger_{\bf{k'}\uparrow}c^\dagger_{-\bf{k'}\downarrow})|0\rangle=-\sum\limits_{\bf{k}}\Delta_{\bf{k}}u_{\bf{k}}c^\dagger_{\bf{k}\uparrow}c^\dagger_{-\bf{k}\downarrow}\prod\limits_{\bf{k'}}'(u_{\bf{k'}}+v_{\bf{k'}}c^\dagger_{\bf{k'}\uparrow}c^\dagger_{-\bf{k'}\downarrow})|0\rangle$$ Assume $\xi_{\bf{k}}=\xi_{-\bf{k}}$. Combining these terms and using $$u_{\bf{k}}=\sqrt{\frac{E_{\bf{k}}+\xi_{\bf{k}}}{2E_{\bf{k}}}}e^{-i\phi_{\bf{k}}/2}$$ $$v_{\bf{k}}=\sqrt{\frac{E_{\bf{k}}-\xi_{\bf{k}}}{2E_{\bf{k}}}}e^{i\phi_{\bf{k}}/2}$$ $E_{\bf{k}}=\sqrt{\xi^2_{\bf{k}}+|\Delta_{\bf{k}}|^2}$,$\phi_{\bf{k}}=\arg\Delta_{\bf{k}}$, we get $$H_{MF}|GS\rangle=-\sum\limits_{\bf{k}}(\sqrt{\xi^2_{\bf{k}}+|\Delta_{\bf{k}}|^2}-\xi_{\bf{k}})|GS\rangle$$ The ground-state energy of mean-field BCS Hamiltonian is $-\sum\limits_{\bf{k}}(\sqrt{\xi^2_{\bf{k}}+|\Delta_{\bf{k}}|^2}-\xi_{\bf{k}})$.