# Homework Help: Quantum time dependant perturbation HW

1. Mar 23, 2007

### quasar987

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,

$$\mathcal{E}(t)=\mathcal{E}_0e^{-t/\tau}$$

What is the probability that the atom be in the state 2p at $$t>>\tau$$?

3. The attempt at a solution

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

2. Mar 23, 2007

### nrqed

There is no problem because the perturbation dies off exponentially with time. So this has the same net result as turning on a perturbation for a finite time and then turning it off.

3. Mar 23, 2007

### quasar987

Okay!

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4. Mar 23, 2007

### quasar987

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:

$$\mathcal{P}_{1s\rightarrow 2p}(t)=\sum_{m=-1}^1\mathcal{P}_{1s\rightarrow 2p}^{(m)}(t)$$

Now, to first order,

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

where

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

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???

Last edited: Mar 23, 2007
5. Mar 25, 2007

### quasar987

It is also zero also to second order:

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

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

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

and

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

So

$$\mathcal{P}^{(m)}_{1s\rightarrow 2p,m}(t)=0$$

to second order also.

Am I doing something wrong?

6. Mar 28, 2007

### quasar987

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