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.
$$
E_{\lambda}=\langle\Phi|a_{\lambda}^{\dagger}\left[
a_{\lambda},H \right]|\Phi\rangle,
$$
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
$$
H=\sum_{\alpha\beta}\langle\alpha|\hat{t}|\beta\rangle a_{\alpha}^{\dagger}a_{\beta}+
\frac{1}{2}\sum_{\alpha\beta\gamma\delta}\langle\alpha\beta|\hat{v}|\gamma\delta\rangle a_{\alpha}^{\dagger}a_{\beta}^{\dagger}a_{\delta}a_{\gamma}.
$$