- #1
MisterX
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Homework Statement
Show that the matrix element of [itex]\mathbf{p} \cdot \mathbf{A} [/itex] between 1S and 2S vanishes to all orders.
I think I need to show the the following
[itex]\langle 2\,0\,0 \mid \boldsymbol{\epsilon}^*(\mathbf{k}, \lambda) \cdot \mathbf{p} e^{-i\mathbf{k}\cdot \mathbf{r}} \mid 1\,0\,0 \rangle = 0[/itex]
Homework Equations
By the orthogonality of [itex]\boldsymbol{\epsilon}^*[/itex] and [itex]\mathbf{k}[/itex] we can change the order
[itex]\left[ \boldsymbol{\epsilon}^*(\mathbf{k}, \lambda) \cdot \mathbf{p},\, e^{-i\mathbf{k}\cdot \mathbf{r}} \right] = 0[/itex]
The Attempt at a Solution
I had thoughts along parity lines, but I haven't been successful so far. Since the states are spherically symmetric, I can choose [itex]\mathbf{k}[/itex] to be along the z axis, which might lead to some clarity. Then we could have [itex]\boldsymbol{\epsilon}^*(k\hat{\mathbf{z}}, \pm) = \frac{1}{\sqrt{2}}\left(\hat{x} \mp i \hat{y} \right)[/itex].
[itex]\langle 2\,0\,0 \mid \left(p_x \mp ip_y\right) e^{-ikz} \mid 1\,0\,0 \rangle[/itex]
[itex]\langle 2\,0\,0 \mid \left(p_x \mp ip_y\right) \left(\cos(kz) - i \sin(kz)\right) \mid 1\,0\,0 \rangle[/itex]
[itex]\langle 2\,0\,0 \mid p_x\cos(kz) \pm p_y\sin(kz) \mid 1\,0\,0 \rangle + i\dots[/itex]
So it seems then I must have independently
[itex]\langle 2\,0\,0 \mid p_x\cos(kz)\mid 1\,0\,0 \rangle =0 [/itex] and [itex]\langle 2\,0\,0 \mid p_y\sin(kz)\mid 1\,0\,0 \rangle =0 [/itex]
I'm really not sure why this would be.
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