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Hydrogen - perturbation theory question

  1. Apr 18, 2014 #1
    1. The problem statement, all variables and given/known data

    2m6ljk2.jpg

    Part (a): Explain origin of each term in Hamiltonian. What does n, l, m mean?

    Part (b): Identify which matrix elements are non-zero

    Part (c): Applying small perturbation, find non-zero matrix elements

    Part (d): Find combinations of n=2 states and calculate change in energies. Sketch energies before and after perturbation.



    2. Relevant equations



    3. The attempt at a solution

    Part (a)

    First term is KE, second term is PE.

    n: energy levels, l: eigenvalues of L2, m: eigenvalues of Lz.

    For n =2, 0 ≤ l ≤ 1 and m ≤|l|.

    Part (b)

    Electric dipole selection rules:

    ##\Delta l = \pm 1## and ##\Delta m = 0, \pm 1## for l' and m'.

    Thus non-zero elements are: ##<2,0,0|z|2,1,0>## and ##<2,1,0|z|2,0,0>##, ##<2,0,0|z|2,1,1>## and ##<2,1,1|z|2,0,0>## and finally ##<2,1,-1|z|2,0,0>##.

    You can see that l' on the bra vectors differ by l in the ket by ##\pm1##.

    Part(c)

    The perturbation is ##eEz##.

    I have found that ##<2,0,0|z|2,1,0> = <2,1,0|z|2,0,0> = -3a_z##.

    But, the rest give zero values, simply by observing the factor in ##e^{i\phi} d\phi##.

    [tex]<2,0,0|z|2,1,1> = \frac{1}{8a_z^4} \frac{-1}{\pi \sqrt{8}} \int_0^{\infty}r^4\left(1 - \frac{r}{2a_z}\right)e^{-\frac{r}{a_z}} dr \int_0^{\pi} sin^2 \theta cos \theta d\theta \int_0^{2\pi} e^{i\phi} d\phi[/tex]

    Which is zero since ##\int_0^{2\pi} e^{i\phi} d\phi = 0##.

    Same with finding ##<2,1,-1|z|2,1,1>##.

    Part(d)

    What do they mean linear combination of n=2 states? By perturbation theory, the first order correction to perturbed state ##|E_2'> = |E_2> + \beta |b>##.

    Then by comparing the powers of ##\beta## we get:

    [tex]|b> = eE \sum_{m\neq 2} \frac{<E_m|z|E_2>}{E_2 - E_m} |E_m>[/tex]
     
  2. jcsd
  3. Apr 18, 2014 #2
    You need to diagonalize the degenerate subspace. Note that the unperturbed energies among all n=2 states are the same, so there would be terms where the denominator of your last equation is zero!

    This problem is solved by diagonalizing the perturbation within the degenerate subspace. You should work through it and see why this is.

    Also for part (b), consider that $$[z, Lz] = 0 $$
    thus
    $$\langle n'\,\ell' m' \mid [z, L_z] \mid n\,\ell\, m \rangle= 0 $$
    $$\langle n'\,\ell' m' \mid z L_z - L_z z\mid n\,\ell\, m \rangle= 0 $$
    $$(m - m')\langle n'\,\ell' m' \mid z\mid n\,\ell\, m \rangle= 0 $$

    What does this say about [itex](m - m')[/itex] and [itex]\langle n'\,\ell' m' \mid z\mid n\,\ell\, m \rangle[/itex] ?
     
  4. Apr 18, 2014 #3
    What do they mean by 'n=2' states? Do they mean states like |2,1,0>, |2,1,1>, |2,1,-1> and |2,0,0>? How do I know which diagonalizes perturbation hamiltonian? If any of the kets diagonalize the hamiltonian it means that it is an eigenket of the hamiltonian ##eEz##?

    It says that ##m = m'## or ##\langle n'\,\ell' m' \mid z\mid n\,\ell\, m \rangle = 0##.
     
  5. Apr 18, 2014 #4
    Yes, you are looking for four combinations of those four kets which are eigenkets of ##e\mathcal{E}z##.
    Yes, this is correct.
     
  6. Apr 18, 2014 #5
    So ##\Delta l = \pm 1## but they must have the same values of m. So the only non-zero value is ##<2,1,0|z|2,0,0>##.

    For part (d), how do I know which of the n=2 states are eigenkets of ##eEz##? I know that if any one of the states are eigenkets, then a linear combination of them are also eigenkets. I guess I have to try out each one.

    Suppose ##z|2,0,0> = \lambda|2,0,0>##: How do we show that this is right?
     
  7. Apr 18, 2014 #6
    I suggest you utilize your answers for part (b)/part (c) to answer that question, and to solve part (d).
     
  8. Apr 19, 2014 #7
    Is the only non-zero element ##<2,1,0|2,0,0>## for part (b)?
     
  9. Apr 19, 2014 #8
  10. Apr 21, 2014 #9
  11. Apr 22, 2014 #10
    I think I got it. To diagonalize the perturbation Hamiltonian, we must find its eigenvalue equation. The elements of the eigenvector gives the amplitude of each of the original eigenbasis ##(\phi_{200},\phi_{211}, \phi_{210}, \phi_{21-1})## (defined in that order in the matrix)

    It turns out that the only non-zero matrix elements in the Hamiltonian matrix are ##\langle\phi_{200}|z|\phi_{210}\rangle = \langle\phi_{210}|z|\phi_{200}\rangle = -3a_0##

    Solving for the eigenvalues and eigenvectors, the states that were once degenerate but not any longer are:

    [tex]|n=2,\pm\rangle = \frac{1}{2}\left(|\psi_{200}\rangle \mp |\psi_{210}\rangle\right)[/tex]
     
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