How do you evaluate the canonical ensemble average of a product of spins, e.g.:



$$S_x = \frac{\hbar}{2} \begin{pmatrix} 0 & 1\\ 1 & 0\\ \end{pmatrix}$$ $$S_y = \frac{\hbar}{2} \begin{pmatrix} 0 & -i\\ i & 0\\ \end{pmatrix}$$ $$S_z = \frac{\hbar}{2} \begin{pmatrix} 1 & 0\\ 0 & -1\\ \end{pmatrix}$$

The matrices are too simple, but then how is the ensemble average of the resultant matrix defined? EDIT: I knew it is about the density matrix, but I guess I was confuseing this with something else. Thanks anyway.

  • $\begingroup$ Do you have the hamiltonian? $\endgroup$ Dec 12, 2013 at 4:44
  • $\begingroup$ Hamiltonian is just Ising model: $$H = A\sum_{i=1}^N S_i^zS_{i+1}^z$$, A is constant. $\endgroup$
    – student1
    Dec 12, 2013 at 14:44

1 Answer 1


The term canonical gives it away.

The canonical ensemble density matrix $\rho$ is defined as follows in terms of the Hamiltonian $H$ and inverse temperature $\beta = 1/kT$: \begin{align} \rho(\beta) = \frac{1}{Z(\beta)}e^{-\beta H}, \qquad Z(\beta) = \mathrm{tr}(e^{-\beta H}) \end{align} Then the canonical ensemble average of any observable $O$ is given by \begin{align} \langle O\rangle = \mathrm{tr}(\rho\, O). \end{align} In your case, simply use $H = H_{\mathrm{Ising}}$ and $O = S_zS_x$.

  • $\begingroup$ Also, do I need to diagonalize the Hamiltonian first before plugging in these equations? $\endgroup$
    – student1
    Dec 13, 2013 at 1:52
  • $\begingroup$ @student1 I haven't done the computation myself. You do not need to diagonalize the Hamiltonian although that will probably make it significantly easier to compute the density matrix unless perhaps you're using a computer algebra system. $\endgroup$ Dec 13, 2013 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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