|1s> -> |2p> transition probabilities

[cscott]

Homework Statement

A hydrogen atom is placed in a time-dependent homogeneous electric field given by $\epsilon = \epsilon_0 (t^2 + \tau^2)^{-1}$ where $\epsilon_0,\tau$ are constants. If the atom is in the ground state at $t=-\inf$, obtain the probability that it ill be found in a 2p state at $t=\inf$. Assume that the electric field is along the $\hat{z}$ direction.

Homework Equations

$$\dot{|2p>} = \int_{\inf}^{\inf} H'_{2p,1s} e^{-i \omega_{2p,1s}} dt$$

$$H'_{2p,1s} = <2p|H'|1s>$$

The attempt at a solution

Firstly I'm confused with the assumption of field direction and how I get it into the solution. Maybe this is the source of my troubles...

By taking $H'(t)= q\epsilon_0 (t^2 + \tau^2)^{-1}$ I get zero for a transition between $|1s>$ and $|2p0>, |2p\pm1>$ because I get Hamiltonian elements of zero.

$|2p0>$ has a $\cos \theta$ factor so upon integration in spherical,

$$\int_0^\pi \sin \theta d\theta \cos \theta = 0$$

and $|2p\pm 1>$ has the $e^{\pm i\phi}$ factor,

$$\int_0^{2\pi} e^{\pm i\phi} d\phi = 0$$

Am I simply right or really missing something? The question gives the time integral result so it seems like Hamiltonian elements shouldn't be zero...

I'm using the wavefunctions from here: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html#c3

 

[fzero]

Firstly I'm confused with the assumption of field direction and how I get it into the solution. Maybe this is the source of my troubles...

By taking $H'(t)= q\epsilon_0 (t^2 + \tau^2)^{-1}$ I get zero for a transition between $|1s>$ and $|2p0>, |2p\pm1>$ because I get Hamiltonian elements of zero.

The interaction term in the Hamiltonian should be such that the electron couples to the electric potential. The resulting matrix elements will no longer vanish.