Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spin 3/2

  1. May 19, 2005 #1
    I've been trying to work on this for a while:

    Let us say I have an [tex]S=1/2[/tex] dimer with [tex]H=JS_{1}\cdot S_{2}[/tex]. With a [tex]\hat{z}[/tex]-diagonal basis, [tex]|\uparrow\uparrow\rangle[/tex], [tex]|\uparrow\downarrow\rangle[/tex], [tex]|\downarrow\uparrow\rangle[/tex], [tex]|\downarrow\downarrow\rangle[/tex], I can easily construct the H-matrix by either using the Pauli matrices or the S-operators. Diagonalizing the matrix gives me the energy eigenvalues and the eigenvectors. Although I can get the energies in an easier way.

    My problem/dilemma/question is this: What if I have an [tex]S=3/2[/tex] dimer (same form on H)? What [tex]\hat{z}[/tex]-diagonal basis (if any) can I use? And am I right in assuming that the matrices to use are the [tex]4\times4[/tex]-matrices listed in e.g. Schiff: Quantum Mechanics (1968), page 203? (Don't feel like typing them right now)

    And a bonus-question: Assuming now [tex]S=1[/tex]. What happens?


    *I feel a bit silly for not knowing this*
     
  2. jcsd
  3. May 20, 2005 #2
    About the basis, you can use the [tex]|00\rangle, |01\rangle, |02\rangle,|03\rangle, |10\rangle,|11\rangle[/tex]... basis and matrices are 4*4 dont know what though, also you shouldnt feel silly for having problems with this problem. I think most people settle for understanding the spin 1/2 problem and then refering a book when the spin gets higher.
     
    Last edited: May 20, 2005
  4. May 20, 2005 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    U can use the 4*4 matrices without any problem for 3/2 spin.I don't have Schiff's 1968 book (i got the incomplete 1949 one),but angular momentum is described in zilllion of books,even special books on angular momentum in QM.

    And for spin 1,there are 3 generators which are 3*3 matrices.

    Daniel.

    The basis is the standard basis:[itex] |j,m\rangle [/itex] which spans the irreducible space [itex] \mathcal{E}_{j} [/itex].
     
  5. May 20, 2005 #4
    Let me get this straight...you are suggesting that I use the the total spin S (and m) as basis, i.e.

    [itex]|3\pm3\rangle, |3\pm2\rangle[/itex] and so on all the way to [itex]|00\rangle[/itex]

    Right?

    But what if would want to start from scratch so to say, i.e. just apply the Hamiltonian to the [itex]|m_{1}m_{2}\rangle[/itex] basis. For the [itex]S=1/2[/itex], I could just apply the [itex]S[/itex]-operators to the kets or the [itex]S[/itex]-matrices to the columns. How would I do something similar in the [itex]S=3/2[/itex] case? I guess my question is how to write the [itex]|m_{1}m_{2}\rangle[/itex] in this case as columns (i.e. in matrix form).

    I guess it is not really necessary to do it this way. I mean, I can always look in a table of Clebsch-Gordan coefficients, but I would like to know.

    Btw, thanks for the help earlier.
     
  6. May 20, 2005 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    So you're talking about composing 2 3/2 spins.Use the C-G theorem to get the irreducible spaces and then the C-G formula to find the vectors in the basis.

    What [itex] |m_{1},m_{2}\rangle [/itex] basis...?There's no such thing. :uhh:


    Daniel.
     
  7. May 20, 2005 #6
    For the [tex]S=1/2[/tex] case I can let the [tex]H=J(S_1^xS_2^x+S_1^yS_2^y+S_1^zS_2^z)[/tex] operate on [tex]|\uparrow\uparrow\rangle=|\frac{1}{2}\frac{1}{2}\rangle[/tex] etc. by:

    1) either using the operations [tex]S^x|\uparrow\rangle=\frac{1}{2}|\downarrow\rangle[/tex] etc.

    2) or by writing [tex]S^x, S^y, S^z[/tex] as Pauli matrices and multiply with [tex]|\uparrow\rangle=\left(\begin{array}{cc}1\\0\end{array}\right)[/tex] and [tex]|\downarrow\rangle=\left(\begin{array}{cc}0\\1\end{array}\right)[/tex].

    So I am wondering: for the [tex]S=3/2[/tex] case, is there an analogous way of doing this?

    *Thanks for your help*
     
  8. May 20, 2005 #7

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Hold on.

    [tex]\hat{S}_{x}|\uparrow\rangle \neq \frac{1}{2}|\downarrow\rangle [/tex]

    Do you see why?

    Daniel.
     
  9. May 20, 2005 #8
    No, I don't. Apart from the fact that I didn't include hbar.
     
  10. May 20, 2005 #9

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Oh,u used [itex] \hbar=1 [/itex].Sorry.

    Daniel.
     
  11. May 20, 2005 #10

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Yes of course.Find the spin matrices for S=3/2 (i guess you have them in Schiff) and the basis vectors (which will be columns of 4 entries).

    Daniel.
     
  12. May 20, 2005 #11
    Ok, thanks. I'll try to find the basis vectors.

    Another question: Do you know of a table of 6j-symbols (I will later go over to work on a trimer)? I haven't been able to find one.
     
  13. May 20, 2005 #12

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Hmm,Rose or Edmonds books on Angular Momentum in QM should have them.

    Daniel.
     
  14. May 20, 2005 #13
    Aargh! I feel stupid.... :frown: the weekend is closing in, I guess......

    I have the [itex]S=3/2[/itex] matrices, but I don't know the basis vectors (the column matrices), and I can't find them. I am assuming they are not as simple as


    [itex]|\frac{3}{2}\rangle=\left(\begin{array}{cc}1\\0\\0\\0\end{array}\right), |\frac{1}{2}\rangle=\left(\begin{array}{cc}0\\1\\0\\0\end{array}\right), |-\frac{1}{2}\rangle=\left(\begin{array}{cc}0\\0\\1\\0\end{array}\right), |-\frac{3}{2}\rangle=\left(\begin{array}{cc}0\\0\\0\\1\end{array}\right)[/itex]

    ...right? :confused:
     
  15. May 20, 2005 #14
    Ok, ignore my last post...I used the basis vectors from above, and after some tedious work, I got the correct energy eigenvalues and eigenvectors. Now off to the trimer. :smile:
     
  16. May 20, 2005 #15

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    I wish i could ignore it,but in fact the standard basis is

    [itex]\left|\frac{3}{2},\frac{3}{2}\right\rangle=\left(\begin{array}{cc}1\\0\\0\\0\end{array}\right), \left|\frac{3}{2},\frac{1}{2}\right\rangle=\left(\begin{array}{cc}0\\1\\0\\0\end{array}\right), \left|\frac{3}{2},-\frac{1}{2}\right\rangle=\left(\begin{array}{cc}0\\0\\1\\0\end{array}\right), \left|\frac{3}{2},-\frac{3}{2}\right\rangle=\left(\begin{array}{cc}0\\0\\0\\1\end{array}\right)[/itex]

    Actually u don't need the exact form of the basis vectors and neither the operators,but that's another story.

    Daniel.
     
  17. May 20, 2005 #16
    I know, that is the basis I used. I meant ignore the fact that I thought this wasn't the basis. I was wrong. :smile:

    There is of course a simpler way of getting the energies. Simply by using

    [itex]H=\frac{1}{2}J(S_{tot}(S_{tot}+1)-S_1(S_1+1)-S_2(S_2+1))[/itex]

    where [itex]S_1=S_2=\frac{3}{2}[/itex] and [itex]S_{tot}[/itex] takes the values 3,2,1, and 0.

    However, doing it this way, how can I find out the degeneracies?
     
  18. May 20, 2005 #17

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    How did u get that Hamiltonian...?

    Daniel.
     
  19. May 21, 2005 #18
    Ok, we have [itex]H=J\bar{S}_1\cdot\bar{S}_2[/itex].

    However, we know that [itex]\bar{S}_{tot}=\bar{S}_1+\bar{S}_2[/itex].

    Squaring both sides: [itex]\bar{S}_{tot}^2=(\bar{S}_1+\bar{S}_2)^2=\bar{S}_1^2+\bar{S}_2^2+2\bar{S}_1\cdot\bar{S}_2[/itex].

    Hence, the Hamiltonian can be written [itex]H=\frac{1}{2}J(\bar{S}_{tot}^2-\bar{S}_1^2-\bar{S}_2^2)[/itex].

    Therefore, the energy eigenvalues (and not the Hamiltonian, sorry) are

    [tex]E=\frac{1}{2}J(S_{tot}(S_{tot}+1)-S_1(S_1+1)-S_2(S_2+1))[/tex]


    where I used [itex]S^2|\Psi\rangle=S(S+1)|\Psi\rangle[/itex] (with [itex]\hbar=1[/itex]).

    Still, doing it like this, I don't know how I can find out the degeneracies.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Spin 3/2
  1. Spin 1/2 vs Spin 2 (Replies: 3)

  2. Spin 3/2 (Replies: 3)

  3. Spin 2 1/2 (Replies: 4)

Loading...