- #1
Species8472
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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]
??
[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]
??