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.


3 Answers 3


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.


I) Let us work in the Schrödinger picture. Assume that the Hamiltonian operator

$$\tag{1} \hat{H}~=~T(\hat{\bf p})+ V(\hat{\bf r}), \qquad T(\hat{\bf p})~:=~ \frac{\hat{\bf p}^2}{2m}, $$

does not depend explicit on time. In particular, the point spectrum of energy eigenvalues $E_n$ does not depend on time. Let $\psi_n({\bf r})$ be the corresponding energy eigen-functions in position space, which satisfies the TISE

$$\tag{2} \hat{H}\psi_n({\bf r})~=~E_n \psi_n({\bf r}).$$

We may assume that $\psi_n({\bf r})\in{\mathbb{R}}$ is a real function, see also this Phys.SE post.

Let us consider a fixed time $t_0$. Let us pick a basis of time-dependent solutions

$$\tag{3} \langle n, t |{\bf r} \rangle~=~ \Psi_n({\bf r},t)~=~ \psi_n({\bf r})e^{-\frac{i}{\hbar}E_n (t-t_0)} $$

that becomes real at the time $t=t_0$. It follows that for any (possibly complex) function $F=F({\bf r})$ of position ${\bf r}$, that the matrix element

$$\tag{4} \langle n, t_0 |F(\hat{\bf r}) | m, t_0 \rangle ~=~ \int \! d^3r ~ \psi^{*}_n({\bf r}) ~F({\bf r})~\psi_m({\bf r})~=~ n \leftrightarrow m $$

is symmetric in $n\leftrightarrow m$ at $t=t_0$. The time dependence of the matrix element is just a phase factor. Hence the absolute value

$$\tag{5} |\langle n, t |F(\hat{\bf r}) | m, t \rangle|~=~ n \leftrightarrow m $$

of the matrix element is symmetric in $n\leftrightarrow m$ at any time $t$.

II) Next

$$2T(\hbar{\bf k})~\stackrel{(1)}{=}~T(\hat{\bf p}+\hbar{\bf k})+T(\hat{\bf p}-\hbar{\bf k})-2T(\hat{\bf p}) $$ $$~=~e^{-i{\bf k}\cdot\hat{\bf r}}T(\hat{\bf p}) e^{i{\bf k}\cdot\hat{\bf r}} +e^{i{\bf k}\cdot\hat{\bf r}}T(\hat{\bf p}) e^{-i{\bf k}\cdot\hat{\bf r}} -2T(\hat{\bf p}) $$ $$\tag{6} ~=~e^{-i{\bf k}\cdot\hat{\bf r}}\hat{H} e^{i{\bf k}\cdot\hat{\bf r}} +e^{i{\bf k}\cdot\hat{\bf r}}\hat{H} e^{-i{\bf k}\cdot\hat{\bf r}}-2\hat{H} .$$

In the last step we used that $[\hat{\bf r},V(\hat{\bf r})]=0$. Hence we get the Bethe sum rule

$$2T(\hbar{\bf k}) ~=~ 2\langle m, t |T(\hbar{\bf k}) | m, t \rangle $$ $$~\stackrel{(6)}{=}~\langle m, t |e^{-i{\bf k}\cdot\hat{\bf r}}\hat{H} e^{i{\bf k}\cdot\hat{\bf r}}| m, t \rangle +\langle m, t | e^{i{\bf k}\cdot\hat{\bf r}}\hat{H} e^{-i{\bf k}\cdot\hat{\bf r}}| m, t \rangle-2\langle m, t |\hat{H}| m, t \rangle $$ $$~=~\sum_{n} (E_n-E_m) |\langle n, t |e^{i{\bf k}\cdot\hat{\bf r}}| m, t \rangle |^2 + \sum_{n} (E_n-E_m) |\langle m, t |e^{i{\bf k}\cdot\hat{\bf r}}| n, t \rangle |^2$$ $$\tag{7} ~\stackrel{(5)}{=}~2\sum_{n} (E_n-E_m) |\langle n, t |e^{i{\bf k}\cdot\hat{\bf r}}| m, t \rangle |^2 .$$

  • $\begingroup$ Yes, I know. I do in fact use the completeness relation as well as the fact that $\langle s|e^{i\mathbf{q}.\mathbf{r}}|n \rangle = \langle n|e^{-i\mathbf{q}.\mathbf{r}}|s \rangle^*$. That is not what is causing the problem. The problem arises from the fact that $|\langle s|e^{i\mathbf{q}.\mathbf{r}}|n \rangle|^2 \ne |\langle s|e^{-i\mathbf{q}.\mathbf{r}}|n \rangle|^2$ in general. $\endgroup$
    – PhHEP
    Dec 2, 2012 at 12:37
  • $\begingroup$ @PhHEP: I updated the answer. $\endgroup$
    – Qmechanic
    Dec 8, 2012 at 22:02

Using $[p,f(r)] = - i \hbar \nabla f(r)$ we find \begin{align} [[p^2,e^{i q \cdot r}],e^{-i q \cdot r}] &= 2\hbar q \cdot[ p,e^{-i q \cdot r}] \\ &= - 2 \hbar^2 q^2 \; . \end{align} On the other hand,

$$[[H,e^{i q \cdot r}],e^{-i q \cdot r}]= 2H -e^{-i q \cdot r}He^{iq \cdot r} -e^{-i q \cdot r}He^{iq \cdot r} $$

Also note that,

$$1 = \left<s|e^{-iq \cdot r}e^{+iq \cdot r} |s\right>= \sum_n \left<s \right|e^{-iq \cdot r}\left|n\right>\left<n \right|e^{+iq \cdot r} \left|s \right> = \sum_n \left| \left<n \right|e^{iq \cdot r}\left|s\right> \right|^2.$$

Now take expectation values of either side to get. Let $E_s = \left<s|H|s\right>$, then \begin{align} -2\frac{\hbar^2 q^2}{2m} &= 2E_s - \left<s \right| \left( e^{-i q \cdot r}He^{iq \cdot r} + h.c. \right) \left|s \right> \; , \\ &= 2E_s -\left( \sum_n \left<s \right| e^{-i q \cdot r}E_n \left|n \right> \left<n \right|e^{iq \cdot r} \left|s \right> + c.c. \right) \; , \\ &=2E_s -\left( \sum_n E_n \left| \left<n \right|e^{iq \cdot r} \left|s \right> \right|^2 + c.c. \right) \; , \\ &=2E_s -2 \sum_n E_n \left| \left<n \right|e^{iq \cdot r} \left|s \right> \right|^2 \; , \\ \end{align} Using the identity and canceling the factor of $2$ we find $$\frac{\hbar^2 q^2}{2m} = \sum_{n} \left( E_n - E_s\right) \left|\left<n \right|e^{i q \cdot r} \left| s \right> \right|^2 \; .$$

  • $\begingroup$ Your second step (second last line) does not make any sense. You just pulled $e^{-i\mathbf{q}.\mathbf{r}}$ out of $\langle n|e^{-i\mathbf{q}.\mathbf{r}}| s \rangle$ and made it pass the $|n\rangle\langle n|$ $\endgroup$
    – PhHEP
    Dec 6, 2012 at 3:29
  • $\begingroup$ Sorry, that wasn't clear. $E_n$ commutes with $e^{i q \cdot r} $. So I just put it beside the eigenstate $\left|n \right>$ and then re-wrote $E_n\left|n \right>$ as $H\left|n \right>$. I also pulled out the $\left<s \right|$ on the left and $\left|s \right>$ on the right by linearity. $\endgroup$ Dec 6, 2012 at 3:37
  • $\begingroup$ Nope. That's not the problem. What you did was: $E_n \langle s | e^{i\mathbf{q}.\mathbf{r}} | n \rangle \langle n | e^{-i\mathbf{q}.\mathbf{r}} | s \rangle = \langle s | e^{i\mathbf{q}.\mathbf{r}} H | n \rangle \langle n | e^{-i\mathbf{q}.\mathbf{r}} | s \rangle \ne \langle s | e^{i\mathbf{q}.\mathbf{r}} H e^{-i\mathbf{q}.\mathbf{r}} | n \rangle \langle n | s \rangle$ $\endgroup$
    – PhHEP
    Dec 6, 2012 at 5:50
  • $\begingroup$ You're forgetting that there is a sum over n. i.e. $\sum_n \left<s \right| e^{iq \cdot r}H \left|n \right> \left<n \right| e^{i q \cdot r} \left| \right> =\left<s \right| e^{iq \cdot r}H \sum_n\left|n \right> \left<n \right| e^{i q \cdot r} \left|s \right> = \left<s \right| e^{iq \cdot r}H e^{i q \cdot r} \left|s \right>$ now commute H through and then expand in your basis functions again. $\endgroup$ Dec 6, 2012 at 5:54
  • $\begingroup$ Okay, I see the missing steps. However, you are making use of $[H,e^{-i\mathbf{q}.\mathbf{r}}] = \hbar^2q^2/2m$. This is incorrect. The correct one is: $[[H,e^{-i\mathbf{q}.\mathbf{r}}],e^{i\mathbf{q}.\mathbf{r}}] = \hbar^2q^2/2m$. I can't believe I didn't notice this before $\endgroup$
    – PhHEP
    Dec 6, 2012 at 6:07

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