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QM: Energy degeneracy at s=1/2 with 3 spin half particles

  1. Oct 12, 2009 #1
    I have a hamiltonian:

    [itex]H = J\left(\frac{1}{2\hbar^2}S_{tot}^2 - \frac{9}{8}\right)[/itex]

    where [itex]S_{tot} = S_1 + S_2 + S_3[/itex]

    for 3 spin half particles in a cluster. The magnetic properties are governed entirely by their spin.

    I have set up the possible spin combinations from Clebsch-Gordan to be:
    From extending the triplet [itex]|1\;1\rangle, |1\;0\rangle, |1\;-1\rangle[/itex] in two spin half particles with an additional particle I get:
    [itex]
    \newcommand{\ket}[1]{\left|#1 \right\rangle}
    \begin{array}{lll}
    \left.
    \begin{array}{l}
    \ket{\frac{3}{2}\;\frac{3}{2}} \\
    \ket{\frac{3}{2}\;\frac{1}{2}} \\
    \ket{\frac{3}{2}\;-\frac{1}{2}} \\
    \ket{\frac{3}{2}\;-\frac{3}{2}}
    \end{array}
    \right\}
    &\mbox{quadruplet} & s=3/2 \\
    \\
    \left.
    \begin{array}{l}
    \ket{\frac{1}{2}\;\frac{1}{2}}_1 \\
    \ket{\frac{1}{2}\;-\frac{1}{2}}_1
    \end{array}
    \right\}
    &\mbox{dublet} & s=1/2
    \end{array}
    [/itex]

    From extending the singlet [itex]|0\;0\rangle[/itex] in two particles with another particle I get:
    [itex]
    \newcommand{\ket}[1]{\left|#1 \right\rangle}
    \begin{array}{lll}
    \left.
    \begin{array}{l}
    \ket{\frac{1}{2}\;\frac{1}{2}}_2 \\
    \ket{\frac{1}{2}\;-\frac{1}{2}}_2
    \end{array}
    \right\}
    &\mbox{dublet} & s=1/2
    \end{array}
    [/itex]

    Now as you can see I have 4 states for S=3/2 and 4 states for S=1/2. Now my question is, does the 4 states for S=1/2 translate into a degeneracy of 4 for the energy eigenvalue in the equation:

    [itex]H|s\;m\rangle = E|s\;m\rangle[/itex]

    ??
     
  2. jcsd
  3. Oct 20, 2009 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    When you are adding three angular momenta and you wish to go the "high road" you can't use Clebsch-Gordan coefficients; they are used to couple two angular momenta. For three you need to use Racah W-coefficients or Wigner 9-j symbols (look them up).

    For the "low road", you need to expand the total spin operator in terms of one-electron operators

    [tex]S^{2}=(s_1+s_2+s_3)^2=s_1^2+s_2^2+s_3^3+2s_1\cdot s_2+2s_2 \cdot s_3+2s_3\cdot s_1[/tex]

    replace the operator dot products with

    [tex]s_i\cdot s_j=\frac{1}{2}(s_{i+}s_{j-}+s_{i-}s_{j+})+s_{iz}s_{jz} [/tex]

    calculate the matrix elements of the 8x8 Hamiltonian using the basis states |s1z>|s2z>|s3z>

    diagonalize the Hamiltonian (you should be able to reduce it to block diagonal form of a 4x4 and two 2x2 submatrices) to get the eigenvalues that should give you the splittings and the eigenvectors.

    The short answer to your question is "no, I don't believe that the two doublets are degenerate in energy."
     
  4. Oct 20, 2009 #3
    It turned out they are actually, I solved the problem later on. The hamiltoinian will give you two energystates both with degeneracy 4.

    It is also fully possible to solve the states by first finding the spin for two particles and then add the third. I was not looking for the coefficients, so all I had to do was read the 6 J,M states from the 1x1/2 table. The singlet state was easy to solve.

    The matrix approach is beyond the scope of this problem really, but thanks for replying :)

    I handed the solutions in the day after i posted anyway. Was a home exam actually ...
     
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