## Definitions and notations

The integral vanishes if two or more particles are permuted in only one of the Hartree-functions $$\Phi_H$$ because the individual single-particle wave functions are orthogonal. We obtain then $$\int \Phi^*\hat{H}_0\Phi d\mathbf{\tau}= \sum_{i=1}^A \int \Phi_H^*\hat{h}_0\Phi_H d\mathbf{\tau}.$$ Orthogonality of the single-particle functions allows us to further simplify the integral, and we arrive at the following expression for the expectation values of the sum of one-body Hamiltonians $$$$\int \Phi^*\hat{H}_0\Phi d\mathbf{\tau} = \sum_{\mu=1}^A \int \psi_{\mu}^*(x)\hat{h}_0\psi_{\mu}(x)dx. \tag{7}$$$$