I'm sorry if this is trivial, I've been stuck on a definition in Yeomans, Statistical mechanics of phase transitions. In chapter five she describes the transfer matrix of the 1D Ising model with nearest neighbor interaction of hamiltonian

$$\mathcal{H}=-J\sum_{j=0}^N s_is_{i+1}-H\sum_{j=0}^{N-1}s_i $$

where $-J$ is an interaction energy, $\mu_BH$ is a magnetic field and $s_i=\pm 1$ are spins. She goes on and writes the partition function with periodic boundary conditions arranging it as

$$ \mathcal{Z}=\sum_{\{s\}}e^{\beta Js_0s_1+\beta H(s_0+s_1)/2}e^{\beta Js_1s_2+\beta H(s_1+s_2)/2}e^{\beta Js_2s_3+\beta H(s_2+s_3)/2}\dots e^{\beta Js_{N-1}s_0+\beta H(s_{N-1}+s_0)/2}=\\=\sum_{\{s\}}T_{0,1}T_{1,2},T_{2,3}\dots T_{N-1,0}$$

defining $$T_{i,i+1}=e^{\beta Js_is_{i+1}+\beta H(s_i+s_{i+1})/2}$$ No problem up to this point, she then defines a matrix $T$ the entry of which are the 4 possibilities for $s_i$ and $s_{i+1}$

$$ T=\begin{pmatrix} e^{\beta(J+H)}&e^{-\beta J}\\ e^{-\beta J}&e^{-\beta(J+H)} \end{pmatrix}= \begin{pmatrix} T_{i,i+1}(+,+)&T_{i,i+1}(+,-)\\ T_{i,i+1}(-,+)&T_{i,i+1}(-,-) \end{pmatrix}$$

And goes on saying that the partition function can be written as the trace of $T^N$.

At this point I start to lose track of what the indices mean, because right after she writes that we can write the partition function as

$$\sum_{s_0=\pm 1}(T^N)_{0,0} $$

I'm confused because before the index represented to which spin $T_{i,i+1}$ was referring to, now it is applied to the $N$th power of the matrix, not to a scalar like before. What does $(T^N)_{0,0}$ mean? is it the first element of the matrix $T^N$? It doesn't seem right to me, because I don't see how that would be a trace. Is it referring to the spin $s_0$? Can't be, because the matrix $T$ is not dependent on which spin we are considering. What do those indices mean?

  • 3
    $\begingroup$ The notation is pretty bad, I agree. You should read it as $\sum_{s_0=\pm 1}(T^N)_{s_0,s_0} = (T^N)_{+1,+1} + (T^N)_{-1,-1} = {\rm Tr} (T^N)$. $\endgroup$ – Yvan Velenik Jun 3 '18 at 10:35
  • $\begingroup$ Thank you, that made it clearer, I figured it out. It is useful to start the process by noticing that the sum over the $N-1$th spin yields the element $S_{N-2},S_{1}$ of the matrix product $T^2$, and working backwards to the last sum. $\endgroup$ – user2723984 Jun 3 '18 at 11:17

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