Quantum time dependant perturbation HW

1. The problem statement, all variables and given/known data
We consider an hydrogen atom in its fundamental state. At t=0, we apply an electric field in the z direction,

[tex]\mathcal{E}(t)=\mathcal{E}_0e^{-t/\tau}[/tex]

What is the probability that the atom be in the state 2p at [tex]t>>\tau[/tex]?


3. The attempt at a solution

I thought time dependant perturbation theory failed for large time.... So what is going on here?

 

Okay!

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But wait a second.

I said that the desired probability is equal to the sum of the 3 probabilities corresponding to each of the 2p state, i.e. for m=-1,0,1:

[tex]\mathcal{P}_{1s\rightarrow 2p}(t)=\sum_{m=-1}^1\mathcal{P}_{1s\rightarrow 2p}^{(m)}(t)[/tex]

Now, to first order,

[tex]\mathcal{P}_{1s\rightarrow 2p}^{(m)}(t)=\frac{1}{\hbar^2}|\int_0^t dt' e^{i\omega_{1s\rightarrow 2p}t'}W_{1s\rightarrow 2p,m}(t')|^2[/tex]

where

[tex]W_{1s\rightarrow 2p,m}(t')=<1,0,0|\mathcal{E}(t')|2,1,m>=\mathcal{E}_0e^{-t'/\tau}<1,0,0|2,1,m>[/tex]

But |1,0,0> and |2,1,m> are both eigenstates of the hydrogen atom hamiltonian, so they are orthogonal and each of my probabilities are 0, no???

 

It is also zero also to second order:

[tex]\mathcal{P}^{(m)}_{1s\rightarrow 2p,m}(t)=\frac{\mathcal{E}_0^4}{\hbar^4}|\sum_{k}\left(\int_0^te^{i\omega_{k\rightarrow 2p}t'}\hat{W}_{k\rightarrow 2p,m}(t')\left(\int_0^{t'}e^{i\omega_{1s\rightarrow k}-t''}\hat{W}_{1s\rightarrow k}(t'')dt''\right)dt'\right)|^2[/tex]

But, because of the same orthonormality thing as with fist order,

[tex]\hat{W}_{k\rightarrow 2p,m}(t')=e^{-t'/\tau}\delta_{k;2p,m}[/tex]

and

[tex]\hat{W}_{1s\rightarrow k}(t'')=e^{-t''/\tau}\delta_{k;1s}[/tex]

So

[tex]\mathcal{P}^{(m)}_{1s\rightarrow 2p,m}(t)=0[/tex]

to second order also.

Am I doing something wrong?

 

The problem was that the perturbation is not E itself; it's the electric potential associated with E(t)! Duh!