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Time independent perturbation

  1. Apr 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Given:
    [tex]\mathbf{H}=V_0\begin{bmatrix}
    1-\epsilon & 0 & 0\\
    0 & 1& \epsilon\\
    0 & \epsilon & 2
    \end{bmatrix}[/tex]

    [tex]\epsilon<<1[/tex]

    a) Find eigenvalues and eigenvectors of the unperturbed Hamiltonian [tex](\epsilon=0)[/tex]

    b) Solve for eigenvalues of the Perturbed Hamiltonian then expand them up to the second order in epsilon.

    c) Use 1st and 2nd order non-degenerate perturbation theory to approximate the eigenvalue that grows out from the non-degenerate state of the unperturbed Hamiltonian. Compare to part b.

    d) Use degenerate perturbation theory to find first order correction of the eigenvalue of the initially degenerate states. Compare to part b.

    2. Relevant equations

    Non-degenerate perturbation:
    [tex]E_n \approx E_n^0+<\psi_n^0|\hat{H_1}|\psi_n^0>+\sum_{m \neq n} \frac{|<\psi_m^0|\hat{H_1}|\psi_n^0>|^2}{E_n^0-E_m^0}[/tex]

    [tex]E_n^0[/tex] are the unperturbed eigenvalues
    [tex]|\psi_n^0>[/tex] are the unperturbed eigenstates


    3. The attempt at a solution

    a)
    [tex]\mathbf{H_0}=V_0\begin{bmatrix}
    1 & 0 &0 \\
    0& 1&0 \\
    0 &0 & 2
    \end{bmatrix}[/tex]

    Therefore:

    [tex]E_1^0=V_0[/tex]
    [tex]E_2^0=V_0 [/tex]
    [tex]E_3^0=2V_0 [/tex]
    [tex]|\psi_1^0>=\begin{bmatrix}
    1\\ 0
    \\ 0

    \end{bmatrix}[/tex]
    [tex]|\psi_2^0>=\begin{bmatrix}
    0\\ 1
    \\ 0

    \end{bmatrix}[/tex]
    [tex]|\psi_3^0>=\begin{bmatrix}
    0\\ 0
    \\ 1

    \end{bmatrix}[/tex]


    b) I subtracted lamda times the identity motrix from the hamiltonian matrix and set its determinant equal to 0 to find the eigenvalues the usual way. After the series approximation, I got:

    [tex]\lambda_1=V_0-V_0 \epsilon [/tex]
    [tex]\lambda_2=V_0-V_0 \epsilon^2 [/tex]
    [tex]\lambda_3=2V_0+V_0 \epsilon^2 [/tex]

    c) I used the non-degenerate perturbation above and got recover what I got in part b for the 3rd lambda.

    d) Here's the problem, I had no clue what to do. I know that I am suppose to set up a matrix and its eigenvalues are the first order correction. And I know how to pick out the matrix elements. But what is the size of my matrix? 2 by 2 because of 2-fold degeneration? Or 3 by 3 since the hamiltonian matrix is 3 by 3? And after I got the eigenvalues, how do I compare them with part b? Which eigenvalue here corresponds to which eigenvalue in part b?
     
    Last edited: Apr 3, 2009
  2. jcsd
  3. Apr 3, 2009 #2

    turin

    User Avatar
    Homework Helper

    You have correctly identified the degenerate subspace: {|ψ1>(0),|ψ2>(0)}. You must diagonalize the perturbing Hamiltonian, H-H(0), in this subspace, and then use the resulting eigenbasis.
     
  4. Apr 3, 2009 #3
    Well since the basis vectors just forms the identity matrix after diagonalizing I will still get:
    [tex]\mathbf{H-H_0}=\mathbf{H_1}=V_0 \begin{bmatrix}
    -\epsilon & 0 & 0 \\
    0 & 0 & \epsilon\\
    0 & \epsilon & 0
    \end{bmatrix}[/tex]

    This will lead to the 3 eigenvalues with the first 2 repeating:
    [tex] \lambda_1 =V_0 \epsilon[/tex]
    and
    [tex] \lambda_2 =V_0 \epsilon[/tex]
    and
    [tex] \lambda_3 =-V_0 \epsilon[/tex]

    But which of these corresponsds to which unperturbed eigenvalue? In other words, how can I construct the followings?

    [tex]E_1=E_1^0+\lambda_?[/tex]
    [tex]E_2=E_1^0+\lambda_?[/tex]
    [tex]E_3=E_1^0+\lambda_?[/tex]

    Which lambda goes where?
    And even more troubling, none gives the lambda squared that is present part b. What am I doing wrong?
     
    Last edited: Apr 3, 2009
  5. Apr 3, 2009 #4

    turin

    User Avatar
    Homework Helper

    Form a matrix from the matrix elements:

    f|(0) H(1)i>(0)

    where {i,f} = {1,2} (the degenerate subspace). Then, diagonalize this matrix. Use the basis that diagonalizes this matrix as the basis for the degenerate subspace. This will guaruntee that perturbed transitions to different states in the degenerate subspace vanish, so that the divergent energy denominators will no longer be a problem.
     
  6. Apr 4, 2009 #5
    Are you saying that my matrix elements should be:

    [tex]\begin{bmatrix}
    1 & 0 & 0
    \end{bmatrix}V_0\begin{bmatrix}
    -\epsilon & 0 & 0\\
    0 &0 & \epsilon\\
    0& \epsilon & 0
    \end{bmatrix}\begin{bmatrix}
    1\\ 0
    \\
    0
    \end{bmatrix} =-V_0 \epsilon[/tex]
    [tex]\begin{bmatrix}
    1 & 0 & 0
    \end{bmatrix}V_0\begin{bmatrix}
    -\epsilon & 0 & 0\\
    0 &0 & \epsilon\\
    0& \epsilon & 0
    \end{bmatrix}\begin{bmatrix}
    0\\ 1
    \\
    0
    \end{bmatrix} =0[/tex]
    [tex]\begin{bmatrix}
    0 & 1 & 0
    \end{bmatrix}V_0\begin{bmatrix}
    -\epsilon & 0 & 0\\
    0 &0 & \epsilon\\
    0& \epsilon & 0
    \end{bmatrix}\begin{bmatrix}
    1\\ 0
    \\
    0
    \end{bmatrix} =0[/tex]
    [tex]\begin{bmatrix}
    0 & 1 & 0
    \end{bmatrix}V_0\begin{bmatrix}
    -\epsilon & 0 & 0\\
    0 &0 & \epsilon\\
    0& \epsilon & 0
    \end{bmatrix}\begin{bmatrix}
    0\\ 1
    \\
    0
    \end{bmatrix} =0[/tex]

    Therefore forming the matrix:
    [tex]\mathbf{A}=\begin{bmatrix}
    -V_0\epsilon & 0\\
    0& 0
    \end{bmatrix}[/tex]
    Then using:
    [tex]|\mathbf{A}-\lambda\mathbf{I}|=\begin{vmatrix}
    -V_0\epsilon-\lambda & 0\\ 0
    & -\lambda
    \end{vmatrix}=(V_0\epsilon+\lambda)\lambda=0[/tex]
    to get the eigenvalues:
    [tex]\lambda_1=-V_0\epsilon[/tex]
    [tex]\lambda_2=0[/tex]
    With only 2, I know that they correspond to the initially degenerate eigenvalues right? But neither of them recover my answers in part b, why is this so?
     
    Last edited: Apr 4, 2009
  7. Apr 4, 2009 #6

    turin

    User Avatar
    Homework Helper

    Think about the meaning of what you are doing. What is the physical meaning of a QM eigenvector? What is the meaning of a QM operator? What happens to the state if it is an eigenstate of an operator, and that operator is solely responsible to perturb the state?
     
  8. Apr 4, 2009 #7
    This I really don't know. We ran out of time due to some bad weather days on which classes were canceled. Perturbation theory wasn't really covered completely in class. I've got the equation (from the text) of dealing with non-degenerate perturbation. I know that the equation won't work for degenerate because we'll get a zero in the denominator, so we need to set up a matrix (n by n for n-fold degeneracy) for degenerate perturbation. I think we pick out the entries of the matrix as I did above. Then the eigenvalues of that matrix will be the first order correction. Am I correct?
     
  9. Apr 5, 2009 #8

    turin

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    Homework Helper

    I believe so.
     
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