= First-order cancellation for an exchange-odd perturbation
{title2=$PVP=0$}
Let $X$ be a <unitary operator> with $X^2=I$, and let $P$ be the orthogonal <projection operator> onto an unperturbed <energy eigenspace>. Suppose $X$ acts as the same sign $\varepsilon=\pm1$ throughout that <eigenspace> and $XVX=-V$. For any $u,v$ in the <eigenspace>, preservation of the <inner product> gives $\langle u,Vv\rangle=\langle Xu,XVv\rangle=-\langle\varepsilon u,V\varepsilon v\rangle=-\langle u,Vv\rangle$. Hence every projected <matrix element> vanishes, so $PVP=0$ and <degenerate perturbation theory> gives zero first-order <energy> corrections. A zero diagonal alone would be insufficient. For equal <spins>, the <exchange parity of two coupled identical spins> is constant on each total-<spin> sector, while their spin-projection difference changes sign under exchange.
Back to article page
