The expectation value of \( \hat{H}_0 \) $$ \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} = A! \int \Phi_H^*\hat{A}\hat{H}_0\hat{A}\Phi_H d\mathbf{\tau} $$ is readily reduced to $$ \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} = A! \int \Phi_H^*\hat{H}_0\hat{A}\Phi_H d\mathbf{\tau}, $$ where we have used Eqs. (5) and (6). The next step is to replace the antisymmetrization operator by its definition and to replace \( \hat{H}_0 \) with the sum of one-body operators $$ \int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} = \sum_{i=1}^A \sum_{p} (-)^p\int \Phi_H^*\hat{h}_0\hat{P}\Phi_H d\mathbf{\tau}. $$