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Degenerate Perturbation Theory: Two Spin 1/2 Particles

  1. Jan 30, 2014 #1
    So I know this might be a lot to read but I am having a very hard time understanding how to use the formulas in degenerate perturbation theory. Here is the problem I am on.

    1. The problem statement, all variables and given/known data
    A system of two spin-1/2 particles is described by the following Hamiltonian.

    [tex]\hat{H}=\hat{\vec{S}}_1 \cdot \hat{\vec{S}}_2 + B \hat{S}_1^z[/tex]

    For simplicity, let [itex]\hbar =1 [/itex] and the Hamiltonian be dimensionless.

    (i) Compute the exact energy eigenvalues and eigenvectors.
    (ii) Treating the term proportional to B as a perturbation, compute the corrections to the unperturbed energies through second order in B and to the eignkets through first order in B.
    (iii) How do your results in (ii) compare to the Taylor expansion of the exact results for the energy eigenvalues and eigenvectors in (i)?


    2. Relevant equations

    In the following equations, superscripts represent orders of approximation (first order, second order, ect.). [itex]|n^0,l_n>[/itex] represents a ket in the degenerate subspace. S_n is the degenerate subspace. [itex]|k^0>[/itex] represents a ket that is NOT in the degenerate subspace.

    (1): First order Energy Correction
    [tex]\Delta^1_{n,l_n}=<n^0,l_n|\hat{V}|n^0,l_n>[/tex]

    (2): Second order Energy Correction
    [tex]\Delta^2_{n,l_n}= \sum_{k \notin S_n}\frac{|<k^0|\hat{V}|n^0,l_n>|^2}{E_n^0-E_k^0}[/tex]

    (3): First order approximation for the eigenkets in the degenerate subspace
    [tex]|n^1,l_n>=\sum_{l_n \ne l_n'}\frac{|n^0,l_n>}{\Delta_{n,l_n}^1-\Delta_{n,l_n'}^1}\sum_{k \notin S_n}<n^0,l_n|\hat{V}|k^0>\frac{1}{E_n^0-E_k^0}<k^0|\hat{V}|n^0,l_n>[/tex]



    3. The attempt at a solution

    (i) So for this one, I first re-wrote the Hamiltonian by computing the dot product and writing the [itex]S_x[/itex] and [itex]S_y[/itex] terms in terms of the ladder operators. I then wrote all the possible kets I can have in a system with two spin 1/2 particles, then using clebch-gordon coefficients, re-wrote them in the [itex]|s_1,s_2,m_1,m_2>[/itex] basis. I then computed the entire matrix and calculated the eigenvalues. There was a LOT of room for error here but I got..

    [tex]E_1=-\frac{1}{4}[/tex]
    [tex]E_2=\frac{1}{4}-\frac{1}{2}B[/tex]
    [tex]E_{3,4}=\frac{1}{3}B \pm \sqrt{\frac{1}{9}B^2+\frac{5}{6}}[/tex]

    First question: How is it possible I got a negative energy? I then plugged these into the matrix and found the eigevectors.

    (ii) This is the part I am really stuck on. First I computed the matrix form of the perturbation V. Which was NOT diagonal, but only had TWO terms survive. I then diagonalized it and found the eigenvalues to be..

    [tex]\text{Eigenvalues}=0,-\frac{B}{2}[/tex]

    Meaning that, these are the first order corrections to the energies via equation (1) above. This makes sense considering that in the first two energies I listen above for the total Hamiltonian, the term [itex]-\frac{1}{4}[/itex] is not being added to anything (eigenvalue of 0 in the perturbation) and the term [itex]\frac{1}{4}-\frac{B}{2}[/itex], has the original energy of 1/4, plus the perturbation of B/2. I then wanted to find the eigenenergies of the unperturbed Hamiltonian to confirm this, so I did and indeed the eigenenergies are [itex]\pm \frac{1}{4}[/itex] and [itex]\pm \frac{1}{2}\sqrt{\frac{5}{2}}[/itex].

    So here is my second question: Once again I am getting negative energies, why? Also, is this system degenerate? I have four distinct eigenenergies in the unperturbed Hamiltonian, doesn't that mean it is non-degenerate and I can use non-degenerate perturbation theory?

    So now my issue comes with calculating the second order energy level shift and the first order eigenkets. For both equations (2) and (3) above, I don't know how to evaluate the following term in the summation.

    [tex]\frac{1}{E_n^0-E_k^0}[/tex]

    Both of the energy terms in the denominator represent unperturbed energies. But I am just having a really hard time figuring out how to evaluate the summations. I don't know where to go here and I keep trying to look up examples in second order but everything I find only goes to FIRST order.

    Can anyone help??
     
  2. jcsd
  3. Jan 31, 2014 #2

    maajdl

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    For question (i), I didn't get the sames eigenvalues as you.
    I got these:

    (-1 - 2*Sqrt(1 + B²))/4
    (+1 - 2*Sqrt(1 + B²))/4
    (-1 + 2*Sqrt(1 + B²))/4
    (+1 + 2*Sqrt(1 + B²))/4

    I don't see why negative values would be impossible.
    The unperturbed Hamiltonian represents the the scalar product of the spins: it can be positive or negative equally well. The same for the perturbation. You can see the perfect symmetry between positive and negative eigenvalues.

    The unperturbed eigenvalues, which you need for the perturbation calculations, are:

    -3/4
    -1/4
    -1/4
    +3/4

    This unperturbed spectrum shows no degeneracy at all.

    Am I wrong?
     
  4. Jan 31, 2014 #3

    vela

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    If the two spins are anti-parallel, it's not unreasonable to expect their dot product to be negative.

    I got different energies than you did. You should try this part again except use the ##\lvert s, m_s \rangle## basis, where ##\vec{S} = \vec{S}_1 + \vec{S}_2##. The trick is to write
    $$\vec{S}_1\cdot\vec{S}_2 = \frac{\vec{S}^2 - \vec{S}_1^2 - \vec{S}_2^2}{2}.$$ In this basis, the unperturbed Hamiltonian will be diagonal, which is what you want before you start in on the perturbation theory for part (b).

     
  5. Jan 31, 2014 #4
    Thank you everyone for your comments so far and taking the time to help me out.

    But how would I apply [itex]|s,m_s>[/itex] ket to, say, [itex]\vec{S}_1[/itex]? Say for example I had the vector [itex]|0,0>[/itex] in this basis. [itex]s_1[/itex] can either be +1/2 or -1/2 since the sum of the two spins must equal zero. Wouldn't I need to convert this basis to the [itex]|m_1,m_2>[/itex] basis?
     
  6. Jan 31, 2014 #5

    vela

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    Yes, you want to express ##\lvert 0, 0 \rangle## in terms of the ##\lvert m_1, m_2 \rangle## basis, apply ##S_1##, and then express the result in terms of the ##\lvert s, m_s \rangle## basis. It's very straightforward to do. Do you know what the ##\lvert s, m_s \rangle## kets are in terms of the ##\lvert m_1, m_2 \rangle## ket?
     
  7. Jan 31, 2014 #6
    Yes, I got the kets to be..

    [tex]|0,0>=\frac{1}{\sqrt(2)}\left[ |-1/2,1/2>-|1/2,-1/2> \right][/tex]
    [tex]|1,-1>=|-1/2,-1/2>[/tex]
    [tex]|1,0>=\frac{1}{\sqrt{2}}\left[ |-1/2,1/2>+|1/2,-1/2> \right][/tex]
    [tex]|1,1>=|1/2,1/2>[/tex]

    So say for example, I want to find <1,1|H|1,1>
    The [itex]S^2[/itex] term would be 1(1+1)=2
    The [itex]S_1^2[/itex] term would be (1/2)(1/2+1)=3/4
    The [itex]S_2^2[/itex] term would be the same, 3/4
    But this would mean that [itex]S_1 \cdot S_2[/itex] would be 2+3/4+3/4=7/2??

    (EDIT: Oops I forgot to subtract, it would be 2-3/4-3/4=1, but then divide by 2 to get 1/2. And then there is a factor of 1/2 out in front from the normalization constants making the total result 1/4.)

    The other two terms are more straight forward for me since [itex]S_1^z[/itex] on |1,1> would simply be 1/2 (setting h-bar =1).

    Am I on the right track?
     
    Last edited: Jan 31, 2014
  8. Jan 31, 2014 #7

    vela

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    You should check your arithmetic, e.g., 2 - 3/4 - 3/4 = 1/2, not 1.

    You mean ##S_{1z} \lvert 1, 1 \rangle = \frac{1}{2} \lvert 1, 1 \rangle##.

    Looks like you have a handle on it. What do you get when you apply ##S_{1z}## to ##\lvert 1, 0 \rangle##?
     
  9. Jan 31, 2014 #8

    maajdl

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    Vela,

    I am missing something.
    Why not using the |m1,m2> basis?
    This is how I did it, and this was extremely simple.
    Is the use of the total spin basis implicitely required somehow by this exercise?

    Thanks to clarify,

    maajdl
     
  10. Jan 31, 2014 #9

    vela

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    You didn't do it correctly because the energies you mentioned above aren't correct. ##\vec{S}_1\cdot \vec{S}_2## doesn't commute with ##S_{1z}## or ##S_{2z}##, so it's not going to be diagonal in the ##\lvert m_1, m_2 \rangle## basis. To use perturbation theory, however, you want the basis that diagonalizes ##\vec{S}_1\cdot \vec{S}_2##.
     
  11. Jan 31, 2014 #10
    This would be 0 correct? Since one ket has +1/2 and the other has -1/2.
     
  12. Jan 31, 2014 #11

    vela

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    No. That's why it's important to keep the kets around. Applying an operator to a ket results in another ket, not a number. For example, ##S_{1z} \lvert + + \rangle = 1/2 \lvert + + \rangle##. It's not ##S_{1z} \lvert + + \rangle = 1/2##.
     
  13. Jan 31, 2014 #12
    Sorry I was having expectation values stuck in my head and I guess I got confused.

    Just for clarity couldn't I also just re-write [itex]S_1\cdot S_2[/itex] in terms of ladder operators and compute it that way, like I originally did? This just seems like another way
     
  14. Jan 31, 2014 #13

    vela

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    If it's mathematically valid, you can do anything you want. Some ways will be more efficient than others. If you use the ##\lvert s, m_s \rangle## basis, the first term of the Hamiltonian is already diagonalized. Finding the matrix for the second term is like four lines of math. Diagonalizing the full Hamiltonian reduces to solving a 2x2 system. Or you can do it the way you tried, which is more complicated and presents the opportunity for more errors.
     
  15. Jan 31, 2014 #14
    Thank you so much Vela for your help so far.

    I just calculated the diagonal components of the unperturbed Hamiltonian using the basis you suggest and I got.. (starting from the top left corner)

    0,5/4,3/4,1,4

    Is this correct? So the next step would be to find V in this basis, and then add them both together to get the total Hamiltonian matrix? Then find the eigenvalues and eigenvectors.

    EDIT: One thing I am confused about. You say it is easy to find the perturbation matrix using this basis, but is V diagonal in this basis? If it is not diagonal, don't I need to find each term?
     
  16. Jan 31, 2014 #15
    Oh wait. Is the second term also diagonal? Because I notice that it commutes with both S and S_z if I did my math right.
     
  17. Jan 31, 2014 #16

    vela

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    I'm not sure what you're doing. Say your first basis vector is ##\lvert 1,1 \rangle##. Apply the unperturbed Hamiltonian to it. You get, using some of your results from above,
    $$\vec{S}_1\cdot \vec{S}_2 \lvert 1,1 \rangle = \frac{\vec{S}^2 - \vec{S}_1^2 - \vec{S}_2^2}{2} \lvert 1,1 \rangle = \frac{2-\frac{3}{4}-\frac{3}{4}}{2} \lvert 1,1 \rangle = \frac{1}{4}\lvert 1,1 \rangle.$$ So the first column of the matrix would be
    $$\begin{pmatrix}1/4\\0\\0\\0\end{pmatrix}.$$ What do you get for the other columns?
     
  18. Jan 31, 2014 #17
    Yes I did get that. Except I am writing my hamiltonian with the m=0,l=0 state in the top left and the m=1,l=1 in the bottom right, so I got the same thing, except the 1/4 is on the bottom. All the columns I got were..

    [tex]
    \begin{pmatrix}
    0 \\
    0 \\
    0 \\
    1/4
    \end{pmatrix}
    [/tex]
    [tex]
    \begin{pmatrix}
    0 \\
    5/4 \\
    0 \\
    0
    \end{pmatrix}
    [/tex]

    [tex]
    \begin{pmatrix}
    0 \\
    0 \\
    3/4 \\
    0
    \end{pmatrix}
    [/tex]
    [tex]
    \begin{pmatrix}
    0 \\
    0 \\
    0 \\
    0
    \end{pmatrix}
    [/tex]

    I then found all the entries to the second expression in the hamiltonian and found its matrix to be.

    [tex]
    \begin{pmatrix}
    0 & 0 & -\frac{B}{2} & 0 \\
    0 & -\frac{B}{2} & 0 & 0 \\
    -\frac{B}{2} & 0 & 0 & 0 \\
    0 & 0 & 0 & \frac{B}{2}
    \end{pmatrix}
    [/tex]

    I then combined both the matrices of the first and second term in the Hamiltonian, and found the eigenvalues to be..

    [tex]
    E_1=\frac{1}{4}\left( 5-2B \right)
    [/tex]
    [tex]
    E_2=\frac{1}{4}\left( 1+2B \right)
    [/tex]
    [tex]
    E_{3,4}=\frac{3}{8}\left( 1 \pm \sqrt{1-\left( \frac{4}{3}B \right)^2} \right)
    [/tex]
     
  19. Jan 31, 2014 #18

    vela

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    I don't see where you're getting the values like 5/4 in the unperturbed Hamiltonian.
     
  20. Jan 31, 2014 #19
    Hmm. Perhaps I just made another arithmetic mistake.

    5/4 is from the term..

    [tex]<1,-1|H_1|1,-1>[/tex]

    Where [itex]H_1[/itex] represents the first term in the Hamiltonian.
    This becomes..

    [tex]\frac{1}{2}<-\frac{1}{2},-\frac{1}{2}|S^2-S_1^z-S_2^z|-\frac{1}{2},-\frac{1}{2}>[/tex]

    The term involving [itex]S^2[/itex] is 1(1+1)=2

    The term involving [itex]S_1^z[/itex] is, (-1/2)(-1/2+1)=-1/4
    The term involving [itex]S_2^z[/itex] is, (-1/2)(-1/2+1)=-1/4

    So all together I have. (1/2)[2+1/4+1/4]=5/4

    Where did I go wrong?
     
  21. Jan 31, 2014 #20

    vela

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    Take ##\vec{S} = \vec{S}_1 + \vec{S}_2##, square both sides, and then solve for ##\vec{S}_1 \cdot \vec{S}_2##.
     
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