# How to show Thomas-Reiche-Kuhn sum rule

Given: $$f_{ni} = \frac{2m\omega_{ni}}{\hbar}\big|\langle n | x |i\rangle\big|^2,$$ and a Hamiltonian in the form $$H_0=\frac{p^2}{2m}+V(x),$$ I would like to show the following sum rule (known as Thomas-Reiche-Kuhn from Sakurai's book Modern Quantum Mechanics): $$\sum_n f_{ni} = 1.$$

We have to consider: $$\sum_n f_{ni}=\sum_{n}\frac{2m(E_n-E_i)}{\hbar^2} \big| \langle n | x |i\rangle\big|^2.$$ It's convenient to start from: $$[x,[x,H_0]]=x[x,H_0]-[x,H_0]x=x^2H_0-2xH_0x+H_0x^2$$ but on the other hand $$[x,[x,H_0]]=\left[x,\left[x,\frac{p^2}{2m}+V(x)\right]\right]=\left[x,\left[x,\frac{p^2}{2m}\right]\right]=\frac{1}{2m}[x,2i\hbar p]=-\frac{\hbar^2}{m}.$$ Now: $$\langle i | [x,[x,H_0]]|i\rangle = -\langle i|\frac{\hbar^2}{m} |i\rangle$$ $$2E_i\langle i|x^2|i\rangle -2\langle i|xH_0x|i\rangle =-\frac{\hbar^2}{m}$$ hence using the completeness relation $\sum_n |n\rangle\langle n|=1$ $$\sum_n(E_n-E_i)\big|\langle n | x |i\rangle\big|^2=\frac{\hbar^2}{2m},$$ which can be recast as $$\sum_n\frac{2m(E_n-E_i)}{\hbar^2}\big|\langle n | x |i\rangle\big|^2=1.$$