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I am going through Faist's Quantum Cascade Lasers book.

In the first section of the fourth chapter, he asserts that (in the context of a perturbative analysis of electronic transitions) the following can be shown: $$\langle\chi_n|p_z|\chi_m\rangle = im_0\omega_{nm}\langle\chi_n|z|\chi_m\rangle$$ Using: $$H_0=\frac{p^2}{2m}+V(r)$$ $$[z,p_z]=i\hbar.$$

Here is the progress I've made, but I seem to be proving myself into a circle.

$$\langle\chi_n|[H_0,z]|\chi_m\rangle = 0$$ $$0=\langle\chi_n|H_0\,\,z|\chi_m\rangle-\langle\chi_n|z\,\,H_0|\chi_m\rangle$$ $$0=\langle\chi_n|\frac{p^2}{2m}\,z|\chi_m\rangle+\langle\chi_n|\,V(r)\,\,z|\chi_m\rangle-\langle\chi_n|z\frac{p^2}{2m}|\chi_m\rangle-\langle\chi_n|z\,\,V(r)|\chi_m\rangle$$ Let V(r) be slowly varying (dipole approximation.) $$0=\langle\chi_n|\frac{p^2}{2m}\,z|\chi_m\rangle+V(r)z_{nm}-\langle\chi_n|\,z\,\frac{p^2}{2m}|\chi_m\rangle-V(r)z_{nm}$$ $$0=\langle\chi_n|\frac{p\,p\,z}{2m}|\chi_m\rangle-\langle\chi_n|\frac{z\,p\,p}{2m}|\chi_m\rangle$$ $$0=\langle\chi_n|\frac{p\,z\,p-pi\hbar}{2m}|\chi_m\rangle-\langle\chi_n|\frac{z\,p\,p}{2m}|\chi_m\rangle$$ $$0=\langle\chi_n|\frac{z\,p\,p-i\hbar p-pi\hbar}{2m}|\chi_m\rangle-\langle\chi_n|\frac{z\,p\,p}{2m}|\chi_m\rangle$$ $$0=\langle\chi_n|\frac{-i\hbar p-pi\hbar}{2m}|\chi_m\rangle$$ $$0=\frac{-i\hbar}{2m}\langle\chi_n|p-p|\chi_m\rangle$$ $$0=0$$

EDIT 1:

I believe the term $\omega_{nm}$ should be coming from the transition energy: $$\langle\chi_n|H_0|\chi_m\rangle = \hbar \omega_{nm}$$

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  • $\begingroup$ Why do you think that $$\langle\chi_n|[H_0,z]|\chi_m\rangle = 0$$ ? $\endgroup$ – Ján Lalinský Jul 25 '16 at 18:11
  • $\begingroup$ Shouldn't energy and position commute to 0? $\endgroup$ – Fariman Jul 25 '16 at 18:59
  • $\begingroup$ Why should they? The Hamiltonian contains momentum $p_z$ corresponding to coordinate $z$. $\endgroup$ – Ján Lalinský Jul 25 '16 at 20:16
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Thanks to Ján Lalinský for getting me to double check my faulty memory. My assumption of $\langle\chi_n|[H_0,z]|\chi_m\rangle = 0$ was wrong. Here is the solution after double checking my work.

$$\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\chi_n|H_0\,\,z|\chi_m\rangle-\langle\chi_n|z\,\,H_0|\chi_m\rangle$$ $$\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\chi_nH_0^\dagger|\,\,z|\chi_m\rangle-\langle\chi_n|z\,\,|H_0\chi_m\rangle$$ $$\langle\chi_n|[H_0,z]|\chi_m\rangle=E_n z_{nm}-E_m z_{nm} = E_{nm} z_{nm}$$ $$\langle\chi_n|[H_0,z]|\chi_m\rangle=\hbar \omega_{nm} z_{nm}$$

$$\hbar \omega_{nm} z_{nm}=\langle\chi_n|\frac{p^2}{2m}\,z|\chi_m\rangle+\langle\chi_n|\,V(r)\,\,z|\chi_m\rangle-\langle\chi_n|z\frac{p^2}{2m}|\chi_m\rangle-\langle\chi_n|z\,\,V(r)|\chi_m\rangle$$

Let V(r) be slowly varying (dipole approximation.)

$$\hbar \omega_{nm} z_{nm}=\langle\chi_n|\frac{p^2}{2m}\,z|\chi_m\rangle+V(r)z_{nm}-\langle\chi_n|\,z\,\frac{p^2}{2m}|\chi_m\rangle-V(r)z_{nm}$$ $$\hbar \omega_{nm} z_{nm}=\langle\chi_n|\frac{p\,p\,z}{2m}|\chi_m\rangle-\langle\chi_n|\frac{z\,p\,p}{2m}|\chi_m\rangle$$ $$\hbar \omega_{nm} z_{nm}=\langle\chi_n|\frac{p\,z\,p-pi\hbar}{2m}|\chi_m\rangle-\langle\chi_n|\frac{z\,p\,p}{2m}|\chi_m\rangle$$ $$\hbar \omega_{nm} z_{nm}=\langle\chi_n|\frac{z\,p\,p-i\hbar p-pi\hbar}{2m}|\chi_m\rangle-\langle\chi_n|\frac{z\,p\,p}{2m}|\chi_m\rangle$$ $$\hbar \omega_{nm} z_{nm}=\langle\chi_n|\frac{-i\hbar p-pi\hbar}{2m}|\chi_m\rangle$$ $$\hbar \omega_{nm} z_{nm}=\frac{-i\hbar}{2m}\langle\chi_n|p+p|\chi_m\rangle$$ $$i\omega_{nm} z_{nm}=\frac{1}{2m}2p_{nm}$$ $$i m\omega_{nm} z_{nm}=p_{nm}$$

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