I'm having difficulties understanding some arguments in my statistical mechanics lecture and would like to make them more rigorous by proving some properties. For the Ising model on a lattice we defined the transfer matrix $V$ elementwise by (for simplicity's sake consider only the 1D case with spin up and spin down and temperature 1): $$V_{\sigma,\sigma'}=e^{J\sigma\sigma'+\frac{h}{2}(\sigma+\sigma')},$$ for $\sigma,\sigma'\in\{\pm 1\}$.

Now writing the partition function $Z$ as $$Z:=\sum_{\sigma_1,\dots,\sigma_L\in\{\pm 1\}}e^{\sum_{i=1}^LJ\sigma_i\sigma_{i+1}+\frac{h}{2}(\sigma_i+\sigma_{i+1})}=\sum_{\sigma_1,\dots,\sigma_L\in\{\pm 1\}}V_{\sigma_1\sigma_2}V_{\sigma_2\sigma_3}\dots V_{\sigma_L\sigma_1},$$ where we assumed periodic boundary conditions, we concluded that $Z=\text{Tr}(V^L)$.

First question: Could you give me a hint how to show that $Z=\text{Tr}(V^L)$?

My attempts so far:

  • I used that $V=V^*$ giving me that $\text{Tr}(V^2)$ is the sum over the entry-wise product of elements of $V$ and thought about continuing by induction, but this doesn't really provide additional insight as long as I don't know how to interpret the off-diagonal terms of $V^2$ that only contribute for higher powers.
  • I also had this idea of showing it in a more abstract way:
    • Periodic boundary conditions and translation invariance (setting $L+1=1$) correspond to the cyclic permutation invariance of the trace.
    • Now (maybe) one could consider other properties of $Z$ or the Hamiltonian and conclude that the trace is the only function that fulfils those properties.

For two point correlation functions we wrote $$\langle \sigma_i\sigma_j\rangle_L=\frac{1}{Z}\text{Tr}(V^{|i-j|}PV^{L-|i-j|}P),$$ for $P=\text{diag}(1,-1)$.

Second question: How do I make sense of this choice of $P$?

My thoughts:

  • For the one point correlation function (or just expectation at site $i$) this would introduce weights (which seems to work out in some simple, finite scenarios I played around with), but, again I'm lacking a deeper understanding why/how to interpret the those terms and what's the importance of the position of the $P$s.

1 Answer 1


For your first question, use the fact that \begin{equation} (V^L)_{\sigma\sigma'} = \sum_{\sigma_1,\ldots,\sigma_{L-1}} V_{\sigma\sigma_1} V_{\sigma_1\sigma_2} \ldots V_{\sigma_{L-1}\sigma'}. \end{equation} (All my sums will run over $\sigma_i = \{\pm1\}$.) You can prove this by induction if you like.

Then just combine this with the definition of the trace: \begin{equation} \operatorname{Tr}(V^L) = \sum_{\sigma_L=\{\pm1\}} (V^L)_{\sigma_L\sigma_L}. \end{equation}

Regarding your second question, we start with the fact (proved as above) that \begin{equation} Z \langle \sigma_i \sigma_j \rangle = \sum_{\sigma_1,\ldots,\sigma_L} \sigma_i \sigma_j V_{\sigma_1\sigma_2} \ldots V_{\sigma_L\sigma_1}. \end{equation} Assume without loss of generality that $i \leq j$. It is more suggestive to write the above as \begin{equation} Z \langle \sigma_i \sigma_j \rangle = \sum_{\sigma_1,\ldots,\sigma_L} V_{\sigma_1\sigma_2} \ldots V_{\sigma_{i-1}\sigma_i} \sigma_i \ldots V_{\sigma_{j-1}\sigma_j} \sigma_j \ldots V_{\sigma_L\sigma_1}. \end{equation}

How does this relate to the trace expression given for the correlations? Well by expanding all products inside the trace, \begin{equation} \operatorname{Tr}(V^{j-i} P V^{L+j-i} P) = \sum_{\sigma_1\ldots\sigma_L,\sigma,\sigma'} V_{\sigma_1\sigma_2} \ldots V_{\sigma_{j-i}\sigma_{j-i+1}} P_{\sigma_{j-i+1}\sigma} V_{\sigma\sigma_{j-i+1}} \ldots V_{\sigma_L\sigma'} P_{\sigma'\sigma_1} \end{equation} This is very similar to the expression for the correlation function, the main difference being that we want to replace the $P$ matrix entries by $\sigma_i$ and $\sigma_j$.

But this is achieved by setting $P_{\sigma\sigma'} = \sigma \delta_{\sigma\sigma'}$, where $\delta_{\sigma\sigma'}$ is $1$ if $\sigma = \sigma'$ and $0$ otherwise. Indeed, doing this forces $\sigma = \sigma_{j-i+1}$ and $\sigma' = \sigma_1$ in the sum (this is the importance of this choice of $P$). This gives \begin{equation} \operatorname{Tr}(V^{j-i} P V^{L+j-i} P) = \sum_{\sigma_1\ldots\sigma_L} V_{\sigma_1\sigma_2} \ldots V_{\sigma_{j-i}\sigma_{j-i+1}} \sigma_{j-i+1} V_{\sigma\sigma_{j-i+1}} \ldots V_{\sigma_L\sigma'} \sigma_1. \end{equation} By the expression for the correlation function above, and by translation invariance, \begin{equation} \operatorname{Tr}(V^{j-i} P V^{L+j-i} P) = \langle \sigma_{i-j+1} \sigma_1 \rangle = \langle \sigma_i \sigma_j \rangle. \end{equation}


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