I got a task, I don't quite know how to solve.

I've got the following Hamiltonian: $$ \hat H = \frac{B}{\hbar^2}\hat{\mathbf S}_1\cdot \hat{\mathbf S}_2+\frac{C}{\hbar}\left(\hat S_{1z}+\hat S_{2z}\right), \qquad \hat{\mathbf S}_{j=1,2}=(\hat S_{jx},\hat S_{jy},\hat S_{jz}). $$ ($B,c$ constants)

My task is to calculate the Eigenvalues and Eigenstates of $\hat{H}$ for:

  1. Two spin 1/2 particles

  2. One spin 1/2 and one spin 1 particle

I got a Tip. I have to write Hamiltonian with the following operators $\hat{S}^2,\hat{S}_z,\hat{S}_1^2,\hat{S}_2^2$, where $\boldsymbol{\hat{S}}=\boldsymbol{\hat{S}_1}+\boldsymbol{\hat{S}_2}$

This was no Problem: $\hat{H}=\frac{B}{\hbar^2}(\hat{S}^2-\hat{S}_1^2-\hat{S}_2^2)+\frac{c}{\hbar}\hat{S}_z$

Now I'm pretty much stuck. My idea was to give out these operators as matrix and the rest is simple linear algebra. But I quit don't understand that. Furthermore I heard that I need the Clebsch-Gordan coefficients, but don't exactly know where.


closed as off-topic by John Rennie, Emilio Pisanty, Jon Custer, Yashas, Cosmas Zachos Jun 22 '17 at 0:44

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Notice that $S_j = S^1_j\otimes 1_2 + 1_1\otimes S^2_j$, with $S_j$ living in $\mathcal{H_1}\otimes\mathcal{H_2}$.

The Hamiltonian commutes with $(S^2, S_z)$, therefore it can be diagonalised onto their eigenstates; as such, the collection $|S\, M_S\rangle$ (with corresponding angular momentum eigenvalues) spans the set of possible eigenstates of $H$. Using the initial definition of $S$ in terms of $S_1,S_2$ one can turn each $|S\, M_S\rangle$ into an expansion in terms of $|s_1\, m_{s_1}\rangle$ and $|s_2\, m_{s_2}\rangle$ with some more or less complicated coefficients (Clebsch-Gordan).

  • $\begingroup$ Thanks, I think I got some more understanding now. So I basically apply |S $m_s$> and get the Eigenvalues. After that I use the clebsch-gordan coefficients to get |$s_i$ $m_i$> of the particles? How do I do that? We just did the other way round. $\endgroup$ – S. Den Jun 21 '17 at 14:45
  • $\begingroup$ From the sum of angular momenta one know what $|S\,M_S\rangle$ are: you can expand it in terms of the spin $1/2$ particles and then invert such expansion. $\endgroup$ – gented Jun 21 '17 at 15:30
  1. The space of two spin $\frac{1}{2}$ is spanned by the basis $|0,0>, |1,1>, |1,0>, |1,-1>$ (written in $|S,S_z>$ form). Now write down the Hamiltonian in this basis in matrix form and find eigenfunction and eigenvalues of the corresponding matrix. In this case the matrix turns out to be diagonal. $$ H= \left[ {\begin{array}{cccc} -\frac{3B}{2} & 0 & 0 & 0 \\ 0 & \frac{B}{2}+C & 0 & 0 \\ 0 & 0 & \frac{B}{2} & 0 \\ 0 & 0 & 0 & \frac{B}{2}-C\\ \end{array} } \right] $$ So, $|0,0>, |1,1>, |1,0>, |1,-1>$ are the eigenstates with eigenvalue $-\frac{3B}{2}$, $\frac{B}{2}+C$, $\frac{B}{2}$, $\frac{B}{2}-C$ respectively. You will need Clebsh Gordon coefficient if you want to write the states in $|S_{1z},S_{2z}>$ basis.

  2. Follow the same procedure. In this case the basis will be $|\frac{3}{2},\pm\frac{3}{2}>$, $|\frac{3}{2},\pm\frac{1}{2}>$, $|\frac{1}{2},\pm\frac{1}{2}>$

  • $\begingroup$ I don't get the downvote... perfectly legit answer... and correct too. The OP is not asking for the change of basis, nor it is required... $\endgroup$ – ZeroTheHero Jun 21 '17 at 15:40

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