Let \( p\rightarrow a \).
The energies, as we have seen, are independent of \( m_a \) and \( m_i \). We sum now over all \( m_a \) on both sides of the above equation and divide by \( 2j_a+1 \), recalling that \( \sum_{m_a}=2j_a+1 \). This results in $$ \varepsilon_{a}^{\mathrm{HF}}=\varepsilon_a+ \frac{1}{2j_a+1}\sum_{i\le F}\sum_{m_a} \langle ai \vert \hat{V}\vert ai\rangle_{AS}, $$