1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Perturbation Theory: Calculating Energy Corrections

  1. Feb 10, 2014 #1
    Hello Everyone. I am very confused on the following questions and have a few confusions about the problem that I hope someone can clear up for me (explained later). Here is the question.

    1. The problem statement, all variables and given/known data
    The paramagnetic resonance of a paramagnetic ion in a crystal lattice is described by
    the spin Hamiltonian

    [tex]\hat{H}=aH\hat{S}_z+bH\hat{I}_z+2D(3\cos^2\theta-1)\left[ \hat{S}_z^2-\frac{1}{3}S(S+1) \right]+\frac{1}{2}D \sin 2\theta \left[ \hat{S}_+\left( \hat{S}_Z + \frac{1}{2} \right) +\hat{S}_-\left( \hat{S}_z-\frac{1}{2} \right) \right]+ \\ \frac{1}{4}D\sin^2 \theta (\hat{S}_+^2+\hat{S}_-^2)+A\hat{S}_z\hat{I}_z+\frac{1}{2}A(\hat{S}_+\hat{I}_-+\hat{S}_-\hat{I}_+)[/tex]

    ##\hat{\vec{S}}## and ##\hat{\vec{I}}## are the electronic and nuclear spin operators, respectively. a, b, A, and D areconstants (a ≪ b), ##\theta## is the angle between the crystal symmetry axis and the direction
    of the magnetic field ##\vec{H}##. For simplicity we take ##\hbar=1##

    (i) Assuming that A,D ≪ a, compute the corrections to the energy levels of the unper-
    turbed Hamiltonian ##\hat{H}_0=aH\hat{S}_z+bH\hat{I}_z## up to second order in perturbation theory.
    Solve this problem for arbitrary values of S and I.
    (ii) Assuming that A,D ∼ a, how do you calculate the first order corrections to the
    energy levels of the unperturbed Hamiltonian ##\hat{H}_0=aH\hat{S}_z##? While you should not
    attempt to solve this problem for arbitrary values of S and I, make sure you report all
    matrix elements involved in the calculation.

    2. Relevant equations

    1. The first order energy correction is:


    Where ##|n,l_n>## are the degenerate kets of the unperturbed Hamiltonian and ##\hat{V}## is the perturbation and must be diagonal.

    2. The second order energy correction is:

    [tex]\Delta^2=\sum_{k \notin S_n}\frac{|<k|\hat{V}|n,l_n>|^2}{E_n^0-E_k^0}[/tex]

    Where ##|k>## are the non-degenerate kets, ##E_n^0## is the nth energy level of the unperturbed Hamiltonian (same deal for ##E_k^0##). And ##S_n## is the degenerate subspace.

    3. The attempt at a solution

    As much as I hate to say this, I seem to be a bit too confused to confidently know where to start. I realize I have a few problems / confusions about some things and I am wondering if any one can clarify. Maybe after I get some clarification on what is confusing me, I can make some real progress on this problem. (I will only deal with part 1 for now.)

    1.)So calculating the first order correction to the energy amounts to using equation 1 from above on the perturbation. I am assuming the perturbation is

    [tex]\hat{V} = 2D(3\cos^2\theta-1)\left[ \hat{S}_z^2-\frac{1}{3}S(S+1) \right]+\frac{1}{2}D \sin 2\theta \left[ \hat{S}_+\left( \hat{S}_Z + \frac{1}{2} \right) +\hat{S}_-\left( \hat{S}_z-\frac{1}{2} \right) \right]+ \\ \frac{1}{4}D\sin^2 \theta (\hat{S}_+^2+\hat{S}_-^2)+A\hat{S}_z\hat{I}_z+\frac{1}{2}A(\hat{S}_+\hat{I}_-+\hat{S}_-\hat{I}_+)[/tex]

    Now this is obviously a mess. My first thought is to find a basis in which this perturbation is diagonal, but I cannot do that because I have ladder operators in there! I feel like my only choice of basis are either the ##|j,m>## basis, or the ##|m_1,m_2>## basis.

    2.) The problem says to solve for any value of S and I. But this is confusing because don't the value of S and I dictate how large the matrix is? How can I find an arbitrarily sized matrix, diagonalize it, and then find the eigenvalues if the size can be anything?

    If anyone can help, I would really appreciate it!
  2. jcsd
  3. Feb 10, 2014 #2
    Given some S and I values, you can determine a total spin and compose appropriate eigenstates using Clebsch-Gordan coefficients: since operators act on electron and nucleus respectively and with the usual eigenvalues of Sz, S± and S^2 (and the I-spin equivalents), the problem seems a little tedious but straightforward...
    Your answer will be a generic equation for the (nn) matrix element, so your second question is not an issue.
  4. Feb 10, 2014 #3
    Thank you for your response.

    But how would I find the first order correction? The matrix needs to be diagonal (in other words, the first order corrections are the eigenvalues of the matrix). There is no way this matrix is diagonal, and I am confused on how to show this for a general nxn matrix that isn't diagonal.
  5. Feb 10, 2014 #4
    Well at first sight i would guess the first order correction will give you back the S^2 and Sz eigenvalues terms and the second-order will take care of the ladder operators, but you have to follow your equations and go through it to see what happens, the 2nd order matrix (which you don't need to focus on) doesn't need to be diagonal...
    ex. with electron eigenstates:
    <S1|S+|S0> = [s(s+1)–m^2–m]^1/2 ≠0
  6. Feb 11, 2014 #5

    I don't think I can do the first order correction unless the matrix is diagonal. I am going through this and I am still confused. So let's take the first term in the perturbation (Excluding the constants).


    Finding the matrix elements of this gives. (with states of j,m and j',m')

    [tex]\left[ (m')^2 -\frac{1}{3}S(S+1) \right]\delta_{jj'}\delta_{mm'}[/tex]

    So this isn't too bad, I can immediately see that this term is diagonal. But the next term is not. Since there are ladder operators I get delta functions involving (m+1) and (m-1) terms.

    Could I just use only the terms that are diagonal and get the first order correction of these? That would give me the correction for the ##\hat{S}_z## and ##\hat{S}^2## eigenenergy corrections. Is this what you meant? Then just use the second order to obtain corrections for the ladder operator terms?
  7. Feb 11, 2014 #6
    You use all the terms all the time. Now some of them return nothing (the term vanishes in that order) and some of them give you a perturbation, but there's nothing wrong with 0. ex: <Sn|S+|Sn> = 0, and there's no perturbation associated in that order.
  8. Feb 11, 2014 #7
    I must be confused about something. It was my understanding that you cannot obtain the first order perturbation using the equation I provided unless the perturbation matrix was diagonal. If it isn't diagonal, then you need to diagonalize it, and the eigenvalues are the perturbation.

    This matrix isn't diagonal, so how can I obtain the first order perturbation?
  9. Feb 11, 2014 #8
    Ok. The 1st order perturbation matrix is diagonal by essence: you construct it this way, just look at the form of your equation. Of course it could turn out to be a zero matrix, which is still diagonal. The 2nd order one is antisymmetric and therefore cannot be diagonal, but again you construct it this way, it's not a pre-existing matrix (again, look at the form of your equation).
    You might be confused because of degenerate perturbation theory, which is a little more subtle since non-degenarate 2nd order would give you infinities if you tried to apply it there...
  10. Feb 11, 2014 #9
    I think I understand what you are saying, but in my notes from class I wrote this.

    Summary for Degenerate Perturbation Theory:

    1. Diagonalize the perturbation matrix in the degenerate subspace.
    2. The roots, or eigenvalues give the first order correction to the energy.

    There is more but it is not relevant. The form of my equation I wrote is of the form of diagonal components.
    So if the form of the perturbation is diagonal by essence, why does the first step say to diagonalize in the degenerate subspace? Or is the act of finding the diagonal component doing what you describe?

    Sorry for my confusion, I very much appreciate your replies thus far, I REALLY want to understand this fully!
  11. Feb 11, 2014 #10
    This is what i just said: the relevant equations (the ones you mentioned in your original post) are the ones pertaining to non-degenerate perturbation. And now you're talking about degenerate theory, which would indeed require you to construct the matrix, get eigenvalues, eigenfunctions etc... but it's not what you need here, do you?
  12. Feb 11, 2014 #11
    The equations I posted are from degenerate perturbation theory (they are very similar to non-deg theory). The system is a system of two spin particles, which would mean it does indeed have degeneracies correct? Since for a single ##j## and ##m## we can have a number of different ##m_s## values that add up to ##m##. So I believe it would be degenerate correct?
  13. Feb 12, 2014 #12
    Hey, sorry i was out a long time.
    Look at your unperturbed Hamiltonian, this is what gives you the degeneracy in energies: for a given S and I and knowing that a<<b, how many degenerate energies do you have?
    Now what states do you use, |jmj> or |sms>|imi>?
  14. Feb 12, 2014 #13
    So if a << b, would that mean that I essentially only have the energies from the term involving b? Meaning that I wouldn't have any degeneracies and can use non-deg theory?

    So I would use |i,mi>???

    I also asked my professor about a problem very similar to this one and he said..

    "You do not need to use degenerate perturbation theory if the perturbation does not mix states."

    I don't know what exactly this means. :\
  15. Feb 12, 2014 #14
    Slow down and look carefully at H0: take a concrete example and work out exactly how many different energies you get, with how many degeneracies...
  16. Feb 12, 2014 #15
    Okay so starting with the unperturbed hamiltonian, I use the |j,mj> basis.

    [tex]aHm_s\delta_{j,j'}\delta_{m,m'}+bHm_i \delta_{j,j'}\delta_{m,m'}[/tex]

    This tells me the matrix is diagonal in this basis, and the energies are..

    [tex]E_n=aHm_s+bHm_i [/tex]

    So I DO get degeneracies correct? Because for a single ##n##, I can have multiple different values of ##m_s## and ##m_i##

    So would the number of degeneracies be 2(2n+1)? Since I get 2n+1 for each different kind of m?
  17. Feb 12, 2014 #16
    Take a state |Sms>|ImI>, apply H0 to a concrete example (S = 1/2, I = 1 for example) and list all the energies you get. how many are degenerate, with a<<b (which doesn't mean you discard anything)?
  18. Feb 12, 2014 #17
    Okay, sorry I misunderstood what you meant by a concrete example. So I took S=1/2 and I=1, and used the |S,I,ms,mi> basis since it is diagonal and got the energies to be the equation I wrote in my last post. I got the energies to be..

    I don't know where the a/2 comes into play. I hope I am doing what you ask correctly :\

    EDIT: I got these energies because ms=1/2,-1/2 and mi=-1,0,1 So I don't see any that are the same..
  19. Feb 12, 2014 #18
    a << b come in to tell you that there's no chance that two combinations above would give you equal energies.
    ex. if a = b = 1 you see that H(a/2) = H(-a/2+b)
    So do you have to deal with degeneracies now?
  20. Feb 12, 2014 #19
    Oh! Now I understand! So I really don't have any degeneracies and can use non-degenerate perturbation theory. Thank you so much, I am going to go through it now and hopefully not get stuck anywhere else.

    Looking at part (ii) though, how do things change when A and D are of the same order as a? It says the unperturbed Hamiltonian is ##\hat{H}_0=aH\hat{S}_z##. Wouldn't that now mean that I need to include ##bH\hat{I}_z## in the perturbation?
  21. Feb 12, 2014 #20
    Well, that's a little more obscure in the way it's formulated.
    I would say that since you still have a<<b (nothing is said about a change there), then if the unperturbed Hamiltonian is taken to be the aH term and the perturbed one is going to include back the bH term along with the rest, it makes sense to consider bH as a new unperturbed Hamiltonian and all the rest as a perturbation. Carry this out and see what happens, but that's the only way to make sense of the question in my opinion... Good luck, i have to go!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted