# [Physics] Derivation of the “Bethe sum rule”

quantum mechanicssolid-state-physics

I am trying to work out the steps of the proof of the expression: $$\sum_n (\mathcal{E_n}-\mathcal{E_s})|\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle|^2 = \frac{\hbar^2q^2}{2m}$$ from Eq. (5.48) in the book Principles of the Theory of Solids by Ziman. In the book it is mentioned that this can be shown by expanding out: $$[[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}],e^{-i\mathbf{q}\cdot\mathbf{r}}]$$ where $$\mathcal{H} = -\frac{\hbar^2}{2m}\nabla^2+\mathcal{V}(\mathbf{r})$$ with $\mathcal{V}(\mathbf{r})$ being periodic. What I did is a simple expansion: $$[[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}],e^{-i\mathbf{q}\cdot\mathbf{r}}] = 2\mathcal{H}-e^{i\mathbf{q}\cdot\mathbf{r}}\mathcal{H}e^{-i\mathbf{q}\cdot\mathbf{r}}-e^{-i\mathbf{q}\cdot\mathbf{r}}\mathcal{H}e^{i\mathbf{q}\cdot\mathbf{r}}$$ Then I took the inner product with the eigenstates of the Hamiltonian $\mathcal{H}|s\rangle = E_s |s\rangle$ to get $$\langle s|[[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}],e^{-i\mathbf{q}\cdot\mathbf{r}}]|s\rangle = 2E_s – \langle s|e^{i\mathbf{q}\cdot\mathbf{r}}\mathcal{H}e^{-i\mathbf{q}\cdot\mathbf{r}}|s\rangle – \langle s|e^{-i\mathbf{q}\cdot\mathbf{r}}\mathcal{H}e^{i\mathbf{q}\cdot\mathbf{r}}|s\rangle$$ Now, what's obstructing my calculation is the fact that I cannot justify the last two terms in the above expression being equal. I really need them to be equal to show the top identity (also known as the Bethe sum rule). The main obstacle is the fact that $e^{i\mathbf{q}\cdot\mathbf{r}}$ is non-Hermitian.

I have found this identity in many books and journal articles. But I cannot find a satisfactory proof anywhere. One such example is this article:

Sanwu Wang. Generalization of the Thomas-Reiche-Kuhn and the Bethe sum rules. Phys. Rev. A 60 no. 1, pp. 262–266 (1999).

Th result is stated in Eq. (3) and proof is given in section III. A. At the end of the proof they set $F = e^{i.\mathbf{q}.\mathbf{r}}$ to recover Eq. (3). The fact that they did not assume any form for $F$ means it must hold for any function. There is, however, one step that I cannot justify. In Eq. (9) how can they write: $$\langle 0|F(x)|l \rangle = \langle l|F(x)|0 \rangle$$ If I can even justify the above equality (for my case) then I'm set.

Start with the expression $$\sum_n (\mathcal{E_n}-\mathcal{E_s})|\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle|^2 = \sum_n (\mathcal{E_n}-\mathcal{E_s})\langle s|e^{-i\mathbf{q}\cdot\mathbf{r}}|n \rangle\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle.$$ The first trick is to realize that $(\mathcal{E_n}-\mathcal{E_s})e^{i\mathbf{q}\cdot\mathbf{r}}$ equals $\pm[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}]$ when it is inside either of the two brackets: $$\sum_n (\mathcal{E_n}-\mathcal{E_s})|\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle|^2 = \sum_n \langle s|e^{-i\mathbf{q}\cdot\mathbf{r}}|n \rangle\langle n|[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}]|s \rangle = \langle s|e^{-i\mathbf{q}\cdot\mathbf{r}}[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}]|s \rangle,$$ by summing $|n \rangle\langle n|$ to $1$. Analogously, you can do this on the first factor to get $$\sum_n (\mathcal{E_n}-\mathcal{E_s})|\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle|^2 =- \langle s|[\mathcal{H},e^{-i\mathbf{q}\cdot\mathbf{r}}]e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle.$$
You now need to calculate the commutator. Since $\mathcal{V}(\mathbf{r})$ commutes with $e^{\pm i\mathbf{q}\cdot\mathbf{r}}$, you only need to worry about the kinetic term. Thus $$[\mathcal{H},e^{\pm i\mathbf{q}\cdot\mathbf{r}}]=\frac{1}{2m}[\mathbf{p}^2,e^{\pm i\mathbf{q}\cdot\mathbf{r}}] =\frac{1}{2m}\mathbf{p}\cdot[\mathbf{p},e^{\pm i\mathbf{q}\cdot\mathbf{r}}] +\frac{1}{2m}[\mathbf{p},e^{\pm i\mathbf{q}\cdot\mathbf{r}}]\cdot\mathbf{p},$$ so $$[\mathcal{H},e^{\pm i\mathbf{q}\cdot\mathbf{r}}]= \pm\frac{1}{2m}\mathbf{p}\cdot(\hbar \mathbf{q}e^{\pm i\mathbf{q}\cdot\mathbf{r}}) \pm\frac{1}{2m}(\hbar \mathbf{q}e^{\pm i\mathbf{q}\cdot\mathbf{r}})\cdot\mathbf{p} =\pm\frac{\hbar}{2m}\mathbf{q}\cdot\left(e^{\pm i\mathbf{q}\cdot\mathbf{r}}\mathbf{p}+\mathbf{p}e^{\pm i\mathbf{q}\cdot\mathbf{r}}\right).$$ To phrase this just right, you need to have $e^{+i\mathbf{q}\cdot\mathbf{r}}$ to the left or $e^{- i\mathbf{q}\cdot\mathbf{r}}$ to the right. Using the commutator $[\mathbf{p},e^{\pm i\mathbf{q}\cdot\mathbf{r}}]=\pm \hbar \mathbf{q}e^{\pm i\mathbf{q}\cdot\mathbf{r}}$, as above, you get $$[\mathcal{H},e^{+ i\mathbf{q}\cdot\mathbf{r}}]= +\frac{\hbar}{2m}\mathbf{q}\cdot\left(e^{+ i\mathbf{q}\cdot\mathbf{r}}\mathbf{p}+\mathbf{p}e^{+ i\mathbf{q}\cdot\mathbf{r}}\right) = \frac{\hbar}{2m}e^{+ i\mathbf{q}\cdot\mathbf{r}}\mathbf{q}\cdot\left(2\mathbf{p}+\hbar\mathbf{q}\right)$$ and $$[\mathcal{H},e^{- i\mathbf{q}\cdot\mathbf{r}}]= -\frac{\hbar}{2m}\mathbf{q}\cdot\left(e^{- i\mathbf{q}\cdot\mathbf{r}}\mathbf{p}+\mathbf{p}e^{- i\mathbf{q}\cdot\mathbf{r}}\right) =- \frac{\hbar}{2m}\mathbf{q}\cdot\left(2\mathbf{p}+\hbar\mathbf{q}\right)e^{- i\mathbf{q}\cdot\mathbf{r}}.$$
Putting either of these into the corresponding formula above, you get $$\sum_n (\mathcal{E_n}-\mathcal{E_s})|\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle|^2 = \frac{\hbar}{2m}\langle s|\mathbf{q}\cdot\left(2\mathbf{p}+\hbar\mathbf{q}\right)|s \rangle = \frac{\hbar^2}{2m}\mathbf{q}^2+\frac{\hbar}{m}\langle s|\mathbf{q}\cdot\mathbf{p}|s \rangle.$$ You can then always force the mean value of $\mathbf{p}$ to vanish.