# Eigenstates of coupled Angular Momentum

SO I have a hamiltonian: $$H=\alpha J_1\cdot J_2$$ And I am asked to find the eigenstates and eigenvalues of this Hamiltonian in terms of products of the eigenstates of the z components of the individual spins.

The wording of this question is kind of confusing me. I could solve it by rewriting it in terms of $(J_1+J_2)^2,J_1^2,J_2^2$ and then find the eigenvalues, but that is not in terms of the z components. How should I go about doing this? Thanks

The question is essentially just asking you to perform Clebsch-Gordan (CG) decomposition, namely a change of basis on the Hilbert space $\mathcal H_{j_1}\otimes\mathcal H_{j_2}$ of the composite system of spins.

Recall that for each of the Hilbert spaces $\mathcal H_{j_1}$ and $\mathcal H_{j_1}$ there exist orthonormal bases of eigenvectors of $\mathbf J_1^2, J_1^z$ and $\mathbf J_2^2, J_2^z$ respectively, and these bases are \begin{align} \text{for $\mathcal H_{j_1}$}&:\qquad \{|j_1, m_1\rangle\,|\,m_1=-j_1,-j_1+1,\dots,j_1-1, j_1\} \\ \text{for $\mathcal H_{j_2}$}&:\qquad \{|j_2, m_2\rangle\,|\,m_2=-j_2,-j_2+1,\dots,j_2-1, j_2\} \end{align} It follows that an orthonormal basis of eigenvectors for the composite Hilbert space $\mathcal H_{j_1}\otimes\mathcal H_{j_2}$ consists of the set of all tensor products of these basis elements; \begin{align} \text{for $\mathcal H_{j_1}\otimes\mathcal H_{j_2}$}:\qquad \{|j_1, m_1\rangle|j_2,m_2\rangle\,|\,-j_1\leq m_1\leq j_1, -j_2\leq m_2\leq j_2\} \tag{$\star$} \end{align} This is what being referred to as

"...products of eigenstates of the z components of individual spins."

So once you've found the eigenstates of the Hamiltonian, which you will presumably write in terms of the basis \begin{align} \{|j,m,j_1,j_2\rangle\}, \end{align} you just need to decompose them into the tensor product basis in $(\star)$ above.

• So how could I do this generally, given J. I do not know what spin they have. The coupled would go from $j_1+j_2,...,|j_1-j_2|$. – yankeefan11 Mar 30 '14 at 1:15
• @yankeefan11 To do this for a given $j_1$ and $j_2$, you either look in a CG table, or you compute the CG coefficients yourself. – joshphysics Mar 30 '14 at 1:16
• Yeah, but I am not given $j_1,j_2$ – yankeefan11 Mar 30 '14 at 1:17
• @yankeefan11 As far as I am aware, no closed form expression for CG coefficients for general $j_1$ and $j_2$ exists, hence the need for a table. – joshphysics Mar 30 '14 at 1:22
• Thats why I am confused on how to answer this question. I could solve it given $J_1,J_2$, but I'm not given that – yankeefan11 Mar 30 '14 at 1:25