1
$\begingroup$

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?

|cite|improve this question|||||
$\endgroup$
  • $\begingroup$ the eigenbasis is the tensor product of the two basis $|\pm\rangle$ of each $\sigma_z$ $\endgroup$ – Francesco Bernardini Nov 19 '19 at 21:08
0
$\begingroup$

If this is a homework question you should tag it as such.

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).

I'll leave it at that, since you say you might be able to work out the later parts given a solution to the first part. Good luck!

|cite|improve this answer|||||
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – fielder May 17 '19 at 23:25
  • 1
    $\begingroup$ $\hat{s}_i=\frac{\hbar}{2}\sigma_{z,i}$, 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. $\endgroup$ – Will May 18 '19 at 17:45
  • 1
    $\begingroup$ 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? $\endgroup$ – fielder May 18 '19 at 18:57
  • $\begingroup$ Yes! Exactly :) $\endgroup$ – Will May 18 '19 at 19:32
  • $\begingroup$ Okay, I think I might be starting to understand this. So my problem is what tensor product would produce that hamiltonian. Apparently the exercise wants us to specifically work with the phases after all. $\endgroup$ – fielder May 18 '19 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.