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

In summary, the conversation discusses a Hamiltonian for 3 spin half particles in a cluster and the possible spin combinations from Clebsch-Gordan coefficients. The question is raised about the degeneracy of the energy eigenvalue and it is mentioned that this can be solved using Racah W-coefficients or Wigner 9-j symbols. The short answer is that the two doublets are not degenerate in energy, but later it is mentioned that the Hamiltonian will give two energy states with a degeneracy of 4. The conversation also briefly mentions the possibility of solving the states by first finding the spin for two particles and then adding the third, and that a matrix approach can also be used.
  • #1
Species8472
4
0
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]

??
 
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  • #2
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."
 
  • #3
kuruman said:
The short answer to your question is "no, I don't believe that the two doublets are degenerate in energy."

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 ...
 

1. What is energy degeneracy in quantum mechanics?

Energy degeneracy refers to the phenomenon where multiple quantum states have the same energy value. This means that the particles in these states will have the same energy level, but may have different momentum or spin values.

2. How does the energy degeneracy in a system with 3 spin half particles at s=1/2 work?

In a system with 3 spin half particles at s=1/2, there are a total of 8 possible quantum states. However, due to the principle of energy degeneracy, only 4 of these states will have unique energy values. This is because the other 4 states will have the same energy values as the first 4, resulting in a degeneracy of 4.

3. What causes energy degeneracy in quantum systems?

Energy degeneracy can occur when there are symmetries in a quantum system that result in different states having the same energy value. This can be due to factors such as spin, angular momentum, or other quantum numbers.

4. How does energy degeneracy impact the behavior of particles in a system?

Energy degeneracy can affect the behavior of particles in a system by limiting the number of distinct energy levels that are available. This can lead to certain particles being more likely to occupy certain states, and can also impact the overall stability and dynamics of the system.

5. Can energy degeneracy be broken in a quantum system?

Yes, energy degeneracy can be broken by introducing external factors such as magnetic or electric fields, which can split the degenerate energy levels. This is known as lifting the degeneracy and can result in a more complex energy spectrum for the system.

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