1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum time dependant perturbation HW

  1. Mar 23, 2007 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?
     
  2. jcsd
  3. Mar 23, 2007 #2

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. Mar 23, 2007 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Okay!

    -----------
     
  5. Mar 23, 2007 #4

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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???
     
    Last edited: Mar 23, 2007
  6. Mar 25, 2007 #5

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?
     
  7. Mar 28, 2007 #6

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The problem was that the perturbation is not E itself; it's the electric potential associated with E(t)! Duh!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Quantum time dependant perturbation HW
Loading...