<\psi_b | z \dot{x} | \psi_a> (magnetic dipole & quadrupole transitions)

[andresordonez]

This is part 2 of problem 4.8 from "Physics of Atoms and Molecules - Bransden, Joachain"

Homework Statement

Deduce [4.106] from [4.105]

Homework Equations

$$\tilde{M}_{ba} = - \frac{m\omega_{ba}}{\hbar c} \langle \psi _b \vert z \dot{x} \vert \psi _a \rangle$$ [4.105]

$$\tilde{M}_{ba} = -\frac{\omega_{ba}}{2\hbar c} \langle \psi_b \vert L_y \vert \psi_a \rangle - \frac{i m \omega_{ba}^2}{2\hbar c} \langle \psi_b \vert zx \vert \psi_a \rangle$$ [4.106]

The Attempt at a Solution

$$-\frac{m\omega_{ba}}{\hbar c}\left\langle \psi_{b}\vert z\dot{x}\vert\psi_{a}\right\rangle =-\frac{m\omega_{ba}}{i\hbar^{2}c}\left\langle \psi_{b}\vert z\left[x,H_{0}\right]\vert\psi_{a}\right\rangle =i\frac{m\omega_{ba}}{\hbar^{2}c}\left\langle \psi_{b}\vert\left(\left[zx,H_{0}\right]-\left[z,H_{0}\right]x\right)\vert\psi_{a}\right\rangle =i\frac{m\omega_{ba}}{\hbar^{2}c}\left\langle \psi_{b}\vert\left(\left[zx,H_{0}\right]-i\hbar\dot{z}x\right)\vert\psi_{a}\right\rangle$$

$$=-i\frac{m\omega_{ba}^{2}}{\hbar c}\left\langle \psi_{b}\vert zx\vert\psi_{a}\right\rangle +\frac{\omega_{ba}}{\hbar c}\left\langle \psi_{b}\vert p_{z}x\vert\psi_{a}\right\rangle =-\frac{\omega_{ba}}{\hbar c}\left\langle \psi_{b}\vert\left(L_{y}-zp_{x}\right)\vert\psi_{a}\right\rangle -i\frac{m\omega_{ba}^{2}}{\hbar c}\left\langle \psi_{b}\vert zx\vert\psi_{a}\right\rangle$$

$$=-\frac{\omega_{ba}}{2\hbar c}\left\langle \psi_{b}\vert L_{y}\vert\psi_{a}\right\rangle -i\frac{m\omega_{ba}^{2}}{2\hbar c}\left\langle \psi_{b}\vert zx\vert\psi_{a}\right\rangle +\left[-\frac{\omega_{ba}}{2\hbar c}\left\langle \psi_{b}\vert L_{y}\vert\psi_{a}\right\rangle -i\frac{m\omega_{ba}^{2}}{2\hbar c}\left\langle \psi_{b}\vert zx\vert\psi_{a}\right\rangle +\frac{\omega_{ba}}{\hbar c}\left\langle \psi_{b}\vert zp_{x}\vert\psi_{a}\right\rangle \right]$$

But then:

$$-\frac{\omega_{ba}}{2\hbar c}\left\langle \psi_{b}\vert L_{y}\vert\psi_{a}\right\rangle -i\frac{m\omega_{ba}^{2}}{2\hbar c}\left\langle \psi_{b}\vert zx\vert\psi_{a}\right\rangle +\frac{\omega_{ba}}{\hbar c}\left\langle \psi_{b}\vert zp_{x}\vert\psi_{a}\right\rangle =0?$$

I tried to prove the last equation without success. Any help would be much appreciated.

 

[dextercioby]

Care to explain where the two 2's in the denominator came from ? I think you need to use that [z,p_z]=0 on commutator's domain.

 

[andresordonez]

I guess you mean the 2's in this expression

$$-\frac{\omega_{ba}}{2\hbar c}\left\langle \psi_{b}\vert L_{y}\vert\psi_{a}\right\rangle -i\frac{m\omega_{ba}^{2}}{2\hbar c}\left\langle \psi_{b}\vert zx\vert\psi_{a}\right\rangle +\left[-\frac{\omega_{ba}}{2\hbar c}\left\langle \psi_{b}\vert L_{y}\vert\psi_{a}\right\rangle -i\frac{m\omega_{ba}^{2}}{2\hbar c}\left\langle \psi_{b}\vert zx\vert\psi_{a}\right\rangle +\frac{\omega_{ba}}{\hbar c}\left\langle \psi_{b}\vert zp_{x}\vert\psi_{a}\right\rangle \right]$$

I was just trying to get [4.106]

$$a = \frac{a}{2} +\frac{a}{2}$$

I think you mean $$[z,p_z]=i\hbar$$ or $$[z,p_x]=0$$. Anyway I don't see how that would help