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Quantum time dependant perturbation HW

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  • #1
quasar987
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Homework Statement


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]?


The Attempt at a Solution



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

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  • #2
nrqed
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Homework Statement


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]?


The Attempt at a Solution



I thought time dependant perturbation theory failed for large time.... So what is going on here?
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
quasar987
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Okay!

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  • #4
quasar987
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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???
 
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  • #5
quasar987
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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?
 
  • #6
quasar987
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The problem was that the perturbation is not E itself; it's the electric potential associated with E(t)! Duh!
 

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