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Time-dependent perturbation theory: hydrogen atom in electric field

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data

    A Hydrogen atom in its ground state (n,l,m) = (1,0,0) is placed in a weak electric field

    E(t) = 0 if t < 0
    [tex]Eo *e^{\frac{-t}{\tau}} [/tex]if t > 0

    E is in the positive z direction

    What is the probability that it will be found in any of the n=2 states at time t > 0 ? use first order perturbation theory

    2. Relevant equations

    [tex]C^{(1)}_{ba} = \frac{1}{(i\hbar)} \int^t_0 H'_{ba} e^{iw_{ba}t'}dt' [/tex]

    where [tex] H'_{ba}[/tex] is the (b,a) matrix element of the perturbation and [tex]w_{ba} = \frac{(E^{(0)}_b - E^{(0)}_a)}{\hbar} [/tex]

    3. The attempt at a solution

    so basically i set up the schrodinger equation in spherical coordinates and added the perturbation [tex]H' = e*Eo *e^{\frac{-t}{\tau}}*rcos{\theta} [/tex]

    where e is the magnitude of the charge charge of an electron. i then evaluated [tex]C_{ba} [/tex] for a = 1s and b = 2s, 2p0, 2p1, 2p-1 all separately.

    i got [tex] H'_{ba} = 0 [/tex] for everything except 2p0, so the coefficients will be 0, for 2p0 i got this for the coefficient:

    [tex] (\frac{e*Eo}{i\hbar3\sqrt{2}a^4_{\mu}})(\int^{\infty}_0 r^4 e^{\frac{-3r}{2a_{\mu}}}dr)(\frac{1}{\frac{i3\mu e^4}{128\pi^2 \epsilon^2_o\hbar^3} - \frac{1}{\tau}})(e^{t(\frac{i3\mu e^4}{128\pi^2\epsilon^2_0\hbar^3}-\frac{1}{\tau}]}-1) [/tex]
    a_{\mu} =\frac{4\pi\epsilon_o\hbar^2}{\mu e^2}
    \mu = [/tex] reduced mass

    and then just take the absolute value squared to get the probability of being found in that state at time t. i was wondering if i did this right? also sorry about using e for the exponential and the charge, if it has variables in its power then its the exponential, otherwise its charge.
    Last edited: Oct 30, 2008
  2. jcsd
  3. Oct 30, 2008 #2


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    Homework Helper

    I have not checked all your algebra, but the fact that the only n=2 transition is to 2p is correct (given the dipole form of the perturbation).

    This is more generally called the "dipole selection rule" and it is comes about because of the cos(theta) in the perturbation which is proportional to Y_{10}.

    Thus transitions from the ground-state must go to states with [itex]\ell=1[/itex].
  4. Oct 30, 2008 #3
    oh it was actually asking for the probabilities for the individual n=2 orbitals
    i received probabilities of 0 for 2s and the 2p orbitals corresponding to the quantum numbers m = +/- 1. the only non zero probability transition i recieved was for 2p corresponding to m = 0.

    the field is just a uniform electric field in the z direction that decays with time so i'm not sure if the selection rules are the same for the dipole transition.
  5. Oct 30, 2008 #4


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    Homework Helper

    ...And I am telling you that what you have found is correct and the this *is* the dipole selection rule.

    Your perturbation is basically
    which *is* the dipole operator [itex]-e\bold{x}[/itex] dotted with the field vector.

    You have chosen the field to be in the z-direction and thus, like I said, the spherical harmonic (1,0) is the perturbation. thus the final state must have [itex]\ell=1[/itex] and [itex]m=0[/itex] for transtions from the ground state. More generally (e.g., for transitions from states other than ground) dipole selection rules hold.
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