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I am calculating different things in Ising 1D ring model using Transfer matrix method: $$Z_{N}=\sum_{\left\{\sigma_{i}=\pm 1\right\}} \exp \left(K \sum_{i=1}^{N} \sigma_{i} \sigma_{i+1}+h \sum_{i=1}^{N} \sigma_{i}\right);\quad \mathbf{T}=\left(\begin{array}{cc} e^{K+h} & e^{-K} \\ e^{-K} & e^{K-h} \end{array}\right);\quad\lambda_{12}=e^{K} ch (h) \pm\left(e^{2 K} sh(h)^{2} +e^{-2 K}\right)^{\frac{1}{2}}$$ I easily found the partition function $Z_N$ and then $$M=\langle\sigma_i\rangle=\dfrac{1}{N}\left\langle\sum_i\sigma_i\right\rangle=tr\left(\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)T^N\right)$$ It was very hard to find eigenvalues of this matrix, but I solved the equation, knowing eigenvalues of matrix $T^N$: $$ \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)T^N\left(\begin{array}{c} k+l \\ c_1k+c_2l \end{array}\right)=\left(\begin{array}{c} \lambda_1^Nk+\lambda_2^Nl \\ -\lambda_1^Nc_1k-\lambda_2^Nc_2l \end{array}\right) $$ where $(k,c_1k),(l,c_2l)$ are eigenvectors of $T^N$. Now I am calculating $\langle\sigma_i\sigma_j\rangle$ and got this result: $$\langle\sigma_i\sigma_j\rangle=\frac{1}{Z}tr(\mathbf{T}^{N-(i-j)}\mathbf{S}\mathbf{T}^{i-j}\mathbf{S});\quad S=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) $$ How to calculate trace of this monster matrix? And I am also not very sure in formula $tr(\mathbf{T}^{N-(i-j)}\mathbf{S}\mathbf{T}^{i-j}\mathbf{S})$ that I got.

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  • $\begingroup$ on some sites people simply multiply S by a diagonalized T, but this is a completely different matrix in my opinion $\endgroup$ Commented Apr 17, 2022 at 19:50

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I decided to calculate this trace in basis where T is diagonalised, found new matrix S: $$\mathbf{P}^{-1}\mathbf{SP}=\dfrac{1}{\lambda_1-\lambda_2}\left(\begin{array}{cc} 2e^{K+h}-(\lambda_2+\lambda_1) & 2e^{K+h}-2\lambda_2\\ 2\lambda_1-2e^{K+h} & (\lambda_1+\lambda_2)-2e^{K+h} \end{array}\right)$$ and then wolfram mathematica finally showed me something normal: $$tr(\mathbf{T}^{N-(i-j)}\mathbf{S}\mathbf{T}^{i-j}\mathbf{S})=\dfrac{(\lambda_1^N+\lambda_2^N)(\lambda_1+\lambda_2-2e^{h+k})^2+(\lambda_1^{N-i+j}\lambda_2^{i-j}+\lambda_2^{N-i+j}\lambda_1^{i-j})(2\lambda_1-2e^{h+k})(2e^{h+k}-2\lambda_2)}{(\lambda_1-\lambda_2)^2}$$

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  • $\begingroup$ Indeed, calculating trace of multiple powers is usually easiest (and is usually done) in the matrix eigenbasis. Not sure why one needs Wolfram mathematica here - these are 2-by-2 matrices. $\endgroup$
    – Roger V.
    Commented Dec 31, 2022 at 7:23

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