# Eigenvalues of Hamiltonian of two interacting spins

I want to compute the eigenvalues of the following Hamiltonian for a system of two interacting 1/2 spin particles :

\begin{aligned} \hat{H} & =A \overrightarrow{\hat{S}}_{(1)} \cdot \overrightarrow{\hat{S}}_{(2)} \\ & =A\left(\hat{S}_{(1) x} \hat{S}_{(2) x}+\hat{S}_{(1) y} \hat{S}_{(2) y}+\hat{S}_{(1) z} \hat{S}_{(2) z}\right) \end{aligned}

Since we can show that $$\left[\hat{H}, \hat{S}^2\right]=0$$, and \begin{aligned} \hat{S}^2 & =\left(\hat{S}_{(1) x}+\hat{S}_{(2) x}\right)^2+\left(\hat{S}_{(1) y}+\hat{S}_{(2) y}\right)^2+\left(\hat{S}_{(1) z}+\hat{S}_{(2) z}\right)^2 \\ & =\hat{\mathbf{S}}_{(1)}^2+\hat{\mathbf{S}}_{(2)}^2+2 \hat{S}_{(1) z} \hat{S}_{(2) z}+2 \hat{S}_{(1) x} \hat{S}_{(2) x}+2 \hat{S}_{(1) y} \hat{S}_{(2) y} \\ & =\frac{3}{4} \hbar^2\left(\hat{I}_{(1)}+\hat{I}_{(2)}\right)+\frac{2}{A} \hat{H}, \end{aligned}

we can find the eigenvalues of $$\hat H$$ by using the fact that we know the eigenvalues of $$\hat{S}^2$$, which are $$\hbar^2 s(s+1)$$ for $$s=0,1$$. So we write $$\hat{H}=\frac{A}{2} \hat{S}^2-\frac{3}{8} \hbar^2 A \hat{I}$$ and thus the eigenvalues of the Hamiltonian are \begin{aligned} \lambda_0 & =-\frac{3}{8} A \hbar^2 \\ \lambda_1 & =\frac{5}{8} A \hbar^2 \end{aligned}

My questions is if my approach is correct on computing the eigenvalues?

Because I could also calculate the eigenvalues in the following way:

$$\hat H = \frac{A}{2} (\hat S^2 - \hat S{_1}^2 -\hat S{_2}^2)$$

$$\frac{A}{2}\left(\hat S^2- \hat S_1^2-\hat S_2^2 \right)|\psi\rangle=\frac{A \hbar^2}{2}\left( s(s+1) - s_1(s_1+1)-s_2(s_2+1)\right)|\psi\rangle=\frac{A \hbar^2}{2}( s(s+1) -\frac{3}{2}) = -A\hbar^2 \frac{3}{4}|\psi\rangle \quad \text{for}\quad s=0 \quad \text{and } = A\hbar^2 \frac{1}{4}|\psi\rangle \quad\text{for}\quad s=1$$

with $$s_i = \frac{1}{2}$$

But in this case I don't get the same values as in the first method. Why is that so? What am I missing?

In your first calculation it looks like you used $$\hat{I}_{(1)}+\hat{I}_{(2)}=\hat{I}$$, whereas in the equivalent step in your second calculation you simply replaced $$\hat{S}_i^2$$ with $$\hbar^2s_i(s_i+1)$$, omitting the $$\hat{I}$$. This silently assumes $$\hat{I}_i=\hat{I}$$ (the latter identitiy operator referring to the overall Hilbert space of the 2-particle-system), therefore the difference. The correct way would be the second one.
In this kind of calculations you should be aware that writing $$\hat{\mathcal{O}}_1$$ for some observable really means $$\hat{\mathcal{O}}_1\otimes\hat{I}_2$$. That means, assuming a product state $$\lvert\psi\rangle=\lvert\psi_1\rangle\otimes\lvert\psi_2\rangle$$, measuring $$\mathcal{O}$$ in the first subsystem does not influence (collapse) your state in the second system. Funny things happen when your state is not a product state ("entanglement"), but that is another story.