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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): $$ \sum_n f_{ni} = 1. $$

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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 $$ \sum_n\frac{2m(E_n-E_i)}{\hbar^2}\big|\langle n | x |i\rangle\big|^2=1. $$

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    $\begingroup$ thanks for this! helped with a derivation in Sakurai he skips over $\endgroup$ – nickodel Jan 23 '15 at 11:52

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