# Quantum Mechanics - Pauli matrix eigenbasis

I'm trying to solve this 3 part question.

1)Suppose we have two 1/2 spin particles. The total spin vector operator for the pair is $$\hat{S} = \hat{s_1} +\hat{s_2}$$. Write down a representation of the product eigenbasis of $$\hat{\sigma_z}\otimes \hat{\sigma_z}$$ for their Hilbert space $$\mathbb{C}^2 \otimes \mathbb{C}^2$$. Show that the $$\hat{\sigma_z}\otimes \hat{\sigma_z}$$ product basis states are eigenstates of $$\hat{S_z}$$ and state their eigenvalues.

2)The Hamiltonian of the two spins is

$$\hat{H} = J\hat{s_1} . \hat{s_2}$$

Where J the coupling strength. Show that $$[\hat{H},\hat{S_z}] = 0$$.

3)Take the tensor product of the Pauli matrices to show that the corresponding matrix representation of $$\hat{H}$$ in the $$\hat{\sigma_z}\otimes \hat{\sigma_z}$$ product basis is $$\hat{H} = \frac{{\hbar}^2J}{4} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$

1)I know that

$$\hat{\sigma_z}\otimes \hat{\sigma_z} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$ But how can we use that as the eigenbasis. And for that matter how would it look like in any pauli direct product matrix representation. Is it just the vectors shown on each of the columns/rows of this matrix?

2)Given $$\hat{S_z}$$ I believe I can compute this part.

3)Again given the previous parts maybe I can do it?

Assuming it is, I don't want to give too much away, but for your first question, in general to find an eigenbasis of an observable it is enough to find its eigenvectors. In your specific case, the observable $$\sigma_z\otimes\sigma_z$$ is diagonal, so finding the eigenvectors is straightforward (the answer to "Is it just the vectors shown on each of the columns/rows of this matrix?" is basically yes, but conventionally you wouldn't include the phases, and more generally if the diagonal entries didn't all have norm 1 it wouldn't be quite right).
• Assuming I use the basis $\hat{\sigma_z}\otimes \hat{\sigma_z}$ without the phases. What would $\hat{s_1}$ and $\hat{s_2}$ be? And isn't $\hat{S_z}$ equal to $\hat{\sigma_z}\otimes \hat{\sigma_z}$? So wouldn't that automatically mean the eigenvalues are all 1 with the eigenvectors being the basis? – fielder May 17 at 23:25
• $\hat{s}_i=\frac{\hbar}{2}\sigma_{z,i}$, but $\hat{S}_z\neq\sigma_z\otimes\sigma_z$. Why do you think that all the eigenvalues should be 1? The only operator whose eigenvalues are all 1 is the identity. – Will May 18 at 17:45
• Oh, so $\hat{S_z}$ is the total spin of the system? That would make it 0 if the spins are opposite and $\pm \hbar$ if they are aligned(0, 0, $\hbar$, $-\hbar$ being the eigenvalues), correct? – fielder May 18 at 18:57