= Orthogonality does not diagonalize a varying potential
{title2=$\langle\phi_1,\phi_2\rangle=0\ \nRightarrow\ \langle\phi_1,V\phi_2\rangle=0$}
In the <Hilbert space> $L^2([0,2\pi])$, choose <orthonormal> functions $\phi_1(x)=\cos x/\sqrt\pi$ and $\phi_2(x)=\sin x/\sqrt\pi$. Their <inner product> is zero, but for the real <potential energy> $V(x)=V_0\sin2x$,
$$
\langle\phi_1,V\phi_2\rangle
=\frac{V_0}{\pi}\int_0^{2\pi}\cos x\sin x\sin2x\,dx
=\frac{V_0}{2\pi}\int_0^{2\pi}\sin^2(2x)\,dx
=\boxed{\frac{V_0}{2}}.
$$
Thus ordinary <orthogonality> alone does not remove an off-diagonal <matrix element> of a spatially varying <potential energy>. A constant scalar potential does give zero between distinct <orthonormal> states. The example separates these two claims; it is not a model of localized atomic orbitals in a particular crystal.
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