1d Ising model: Transfer matrices

Question

we came across a peculiarity when calculating the partition function of $N$ spins $s_i=\pm1$ with Hamiltonian $$H=-J\sum_{i=1}^Ns_is_{i+1}$$ where we impose periodic boundary conditions such that $s_{N+1}=s_1$.
Now we can calculate the partition function as $$Z=\sum_{spins}e^{-\beta H[s]}=\sum_{s_1=\pm1}...\sum_{s_N=\pm1}\prod_{i=1}^Ne^{\beta Js_is_{i+1}}=\prod_{i=1}^N(\sum_{s_i=\pm1}e^{\beta Js_is_{i+1}})$$
The sum over $s_i=\pm1$ on the RHS can be performed without specifying the value of $s_{i+1}$ since $$\sum_{s_i=\pm1}e^{\beta Js_is_{i+1}}=e^{\beta Js_{i+1}}+e^{-\beta Js_{i+1}}=2\cosh(\beta J)$$ for $s_{i+1}=\pm1$. This leaves us with $$Z=(2\cosh(\beta J))^N$$ But we could have used a different technique to calculate the partition function:
Define $$h_1(s)=\cfrac{1+s}{2} \quad h_2(s)=\cfrac{1-s}{2}$$ such that $$\sum_{s=\pm1}h_\tau(s)=1$$ $$h_\tau(s)h_\rho(s)=\delta_{\tau\rho} h_\rho(s)$$ With that we can expand any function $f(s_i,s_{i+1})$ in terms of $h_\tau(s_i)$ and $h_\rho(s_{i+1})$. In particular:$$k_{i}=:e^{\beta Js_is_{i+1}}=T_{\tau\rho}h_\tau(s_i)h_\rho(s_{i+1})$$ Using the properties from above, we can see that $$\sum_{s_{i+1}=\pm1}k_{i}k_{i+1}=\sum_{s_{i+1}=\pm1}T_{\tau\rho}T_{\mu\nu}h_\tau(s_i)h_\rho(s_{i+1})h_\mu(s_{i+1})h_\nu(s_{i+2})=h_\tau(s_i)T_{\tau\rho}T_{\rho\nu}h_\nu(s_{i+2})$$ Since $$Z=\sum_{s_1=\pm1}...\sum_{s_N=\pm1}\prod_{i=1}^Nk_i$$ we get $$Z=\sum_{s_1=\pm1}h_\tau(s_1)[T...T]_{\tau\rho}h_\rho(s_{N+1})=Tr(T^N)$$ The transfer Matrix $T$ has the form $$T=\begin{bmatrix} e^{\beta J} & e^{-\beta J} \\ e^{-\beta J} & e^{\beta J} \end{bmatrix}$$ it is symmetric and can therefore be diagonalized with orthogonal matrices which gives us for the partition function $$Z=Tr(\mathcal O^\top \hat{T}^N \mathcal O)=\sum_{i}\lambda_i^N$$ where we used the cyclic propertry of the trace and the $\lambda_i$ are the Eigenvalues of $T$ The result of this calculation gives $$Z=(2\cosh(\beta J))^N+(2\sinh(\beta J))^N=(2\cosh(\beta J))^N[1+(\tanh(\beta J))^N]$$ And this is where the problem arises: The two results are different. For large $N$ we obtain our old result since $|\tanh(x)|<1$.
But for any finite $N$ there is an additional contribution of $(2\sinh(\beta J))^N$
Did we somehow overcount in our second method or did the first method not capture the system correctly?
Furthermore, what is the physical significance of "small" eigenvalues? I heard in a lecture that if one unique, largest eigenvalue of a transfer matrix exists, then the corresponding eigenvector is the state of thermal equilibrium. This makes sense to me since one could consider a normalized transfer matrix (in a sense of $\lambda_{max}=1$) and repeatedly apply it to some initial state. But what do the other eigenvectors represent? Which information is lost when we neglect the small eigenvalues in our partition sum?
I would be very happy if someone could demystify this issue.

Answer 1

I think the first way in which you do it has a flaw. Suppose that the single term did depend on both $s_i$ and $s_{i+1}$, then $$ \sum_{s_1=\pm 1}\cdots\sum_{s_N=\pm 1}\prod_{i=1}^N f(s_i,s_{i+1}) = \prod_{i=1}^N\left(\sum_{s_i=\pm 1}f(s_i,s_{i+1})\right)\,, $$ wouldn't make sense because all $s_{i+1}$ would survive even after the sum has been performed. If you had a function of $s_i$ only, then such an identity would hold $$ \sum_{s_1=\pm 1}\cdots\sum_{s_N=\pm 1}\prod_{i=1}^N f(s_i) = \prod_{i=1}^N\left(\sum_{s_i=\pm 1}f(s_i)\right)\,. $$ However I can't figure out why you still get a correct result in the $N\to\infty$ limit.