Properties of transfer matrices and their traces

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.

For your first question, use the fact that $$(V^L)_{\sigma\sigma'} = \sum_{\sigma_1,\ldots,\sigma_{L-1}} V_{\sigma\sigma_1} V_{\sigma_1\sigma_2} \ldots V_{\sigma_{L-1}\sigma'}.$$ (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: $$\operatorname{Tr}(V^L) = \sum_{\sigma_L=\{\pm1\}} (V^L)_{\sigma_L\sigma_L}.$$
Regarding your second question, we start with the fact (proved as above) that $$Z \langle \sigma_i \sigma_j \rangle = \sum_{\sigma_1,\ldots,\sigma_L} \sigma_i \sigma_j V_{\sigma_1\sigma_2} \ldots V_{\sigma_L\sigma_1}.$$ Assume without loss of generality that $i \leq j$. It is more suggestive to write the above as $$Z \langle \sigma_i \sigma_j \rangle = \sum_{\sigma_1,\ldots,\sigma_L} V_{\sigma_1\sigma_2} \ldots V_{\sigma_{i-1}\sigma_i} \sigma_i \ldots V_{\sigma_{j-1}\sigma_j} \sigma_j \ldots V_{\sigma_L\sigma_1}.$$
How does this relate to the trace expression given for the correlations? Well by expanding all products inside the trace, $$\operatorname{Tr}(V^{j-i} P V^{L+j-i} P) = \sum_{\sigma_1\ldots\sigma_L,\sigma,\sigma'} V_{\sigma_1\sigma_2} \ldots V_{\sigma_{j-i}\sigma_{j-i+1}} P_{\sigma_{j-i+1}\sigma} V_{\sigma\sigma_{j-i+1}} \ldots V_{\sigma_L\sigma'} P_{\sigma'\sigma_1}$$ 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 $$\operatorname{Tr}(V^{j-i} P V^{L+j-i} P) = \sum_{\sigma_1\ldots\sigma_L} V_{\sigma_1\sigma_2} \ldots V_{\sigma_{j-i}\sigma_{j-i+1}} \sigma_{j-i+1} V_{\sigma\sigma_{j-i+1}} \ldots V_{\sigma_L\sigma'} \sigma_1.$$ By the expression for the correlation function above, and by translation invariance, $$\operatorname{Tr}(V^{j-i} P V^{L+j-i} P) = \langle \sigma_{i-j+1} \sigma_1 \rangle = \langle \sigma_i \sigma_j \rangle.$$