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

$$[S_zS_x]$$

Where:

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

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.

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  • $\begingroup$ 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

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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 $O = S_zS_x$.

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  • $\begingroup$ 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

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