Consider the ground state \( |\Phi\rangle \) of a bound many-particle system of fermions. Assume that we remove one particle from the single-particle state \( \lambda \) and that our system ends in a new state \( |\Phi_{n}\rangle \). Define the energy needed to remove this particle as $$ E_{\lambda}=\sum_{n}\vert\langle\Phi_{n}|a_{\lambda}|\Phi\rangle\vert^{2}(E_{0}-E_{n}), $$ where \( E_{0} \) and \( E_{n} \) are the ground state energies of the states \( |\Phi\rangle \) and \( |\Phi_{n}\rangle \), respectively.

• Show that

$$ E_{\lambda}=\langle\Phi|a_{\lambda}^{\dagger}\left[ a_{\lambda},H \right]|\Phi\rangle, $$ where \( H \) is the Hamiltonian of this system.

• If we assume that \( \Phi \) is the Hartree-Fock result, find the

relation between \( E_{\lambda} \) and the single-particle energy \( \varepsilon_{\lambda} \) for states \( \lambda \leq F \) and \( \lambda >F \), with $$ \varepsilon_{\lambda}=\langle\lambda|\hat{t}+\hat{u}|\lambda\rangle, $$ and $$ \langle\lambda|\hat{u}|\lambda\rangle=\sum_{\beta \leq F} \langle\lambda\beta|\hat{v}|\lambda\beta\rangle. $$ We have assumed an antisymmetrized matrix element here. Discuss the result.

The Hamiltonian operator is defined as $$ H=\sum_{\alpha\beta}\langle\alpha|\hat{t}|\beta\rangle a_{\alpha}^{\dagger}a_{\beta}+ \frac{1}{2}\sum_{\alpha\beta\gamma\delta}\langle\alpha\beta|\hat{v}|\gamma\delta\rangle a_{\alpha}^{\dagger}a_{\beta}^{\dagger}a_{\delta}a_{\gamma}. $$