I try to calculate the correlation function $<\sigma_i \sigma_j>$ with the method of transfer matrices. I do understand how to use this method with PBC.

But how can I do it without PBC?

My hamiltonian looks like this where $\sigma_i = \pm 1$, $i \in [1,2,...,N]$ and we only look at the nearest neighbour. So $\sum_{ij} \sigma_i \sigma_j = \sigma_1 \sigma_2 + \sigma_2 \sigma_3 + ... + \sigma_{N-1} \sigma_{N}$

$$ H = -J \sum_{ij} \sigma_i \sigma_j - \gamma B \sum_{i=1}^N \sigma_i$$ So $$ \beta H = -\beta J \sum_{ij} \sigma_i \sigma_j - \beta \gamma \sum_i \sigma_i \overset{!}{=} \sum_i u(\sigma_i, \sigma_{i+1})$$

I can't figure out how $u(\sigma, \sigma')$ has to look without PBC. Has anyone a hint?

Edit: I made the mistake that I thought I only can use one transfer matrix, so $Z = Tr(T^N)$. This is not true. So I split the hamiltonian in

$$ H = \sum_{i=1}^{N-1} \left( - J \sigma_i\sigma_{i+1} - \gamma B (\sigma_i + \sigma_{i+1} \right) - \frac{\gamma B}{2} \left(\sigma_1 + \sigma_N\right)$$

My partition function

$$Z = Tr\left(\exp{(-\beta H)}\right) = Tr( T_{\sigma_i \sigma_{i+1}} \cdot \tilde{T}_{\sigma_1, \sigma_N}) $$ With $T_{\sigma \sigma'} = \exp{\left(\beta J \sigma \sigma' + \beta \gamma B (\sigma + \sigma')\right)}$ and $\tilde{T}_{\sigma_1, \sigma_N} = \exp{\left(\frac{\gamma B}{2} (\sigma_1 + \sigma_N)\right)}$

I know that $\sigma_i = \pm 1$, so I can calculate the matrix for both. As far as I know this leads to $$Z = Tr(T^{N-1} \cdot T_{2})$$

  • 2
    $\begingroup$ What boundary conditions do you have? $\endgroup$ Commented May 3, 2022 at 11:03
  • 1
    $\begingroup$ None so far. I only got the information that the system has N spins and $\sigma_1$ is not connected to $\sigma_N$. So it seems like a spin chain to me. $\endgroup$
    – Lie
    Commented May 3, 2022 at 11:06
  • 1
    $\begingroup$ I thought about it and the boundary conditions are Free boundary conditions due to being a chain. I also know that I can use more than one transfer matrices. $\endgroup$
    – Lie
    Commented May 4, 2022 at 12:46

1 Answer 1


To lighten notation, I'll just write $\beta$ instead of $\beta J$ and $h$ instead of $\beta\gamma B$. I'll also write $Z_N$ for the partition function and $\langle\cdot\rangle_N$ for the expectation value with respect to the associated Gibbs measure.

Now, observe that \begin{eqnarray} Z_N \langle\sigma_u\sigma_v\rangle_N &=& \sum_{\sigma_1,\dots,\sigma_N} \sigma_u \sigma_v \exp \left( \sum_{i=1}^{N-1} \beta\sigma_i\sigma_{i+1} + \sum_{i=1}^N h\sigma_i \right) \\ &=& \sum_{\sigma_1,\dots,\sigma_N} e^{\frac{h}{2}\sigma_1 + \frac{h}{2}\sigma_N} \sigma_u \sigma_v \exp \left( \sum_{i=1}^{N-1} \Bigl(\beta\sigma_i\sigma_{i+1} + \frac{h}{2}(\sigma_i + \sigma_{i+1}) \Bigr)\right) \\ &=& \sum_{\sigma_1,\dots,\sigma_N} e^{\frac{h}{2}\sigma_1} \exp \left( \sum_{i=1}^{u-1} \Bigl(\beta\sigma_i\sigma_{i+1} + \frac{h}{2}(\sigma_i + \sigma_{i+1}) \Bigr)\right) \sigma_u \\ &\phantom{=}&\phantom{\sum_{\sigma_1,\dots,\sigma_N} e^{\frac{h}{2}\sigma_1}}\!\!\!\!\!\!\times \exp \left( \sum_{i=u}^{v-1} \Bigl(\beta\sigma_i\sigma_{i+1} + \frac{h}{2}(\sigma_i + \sigma_{i+1}) \Bigr)\right) \sigma_v\\ &\phantom{=}&\phantom{\sum_{\sigma_1,\dots,\sigma_N} e^{\frac{h}{2}\sigma_1}}\!\!\!\!\!\!\times \exp \left( \sum_{i=v}^{N-1} \Bigl(\beta\sigma_i\sigma_{i+1} + \frac{h}{2}(\sigma_i + \sigma_{i+1}) \Bigr)\right) e^{\frac{h}{2}\sigma_N} . \end{eqnarray} It is thus natural to introduce the following matrices $$ T=\begin{pmatrix}e^{\beta+h} & e^{-\beta} \\ e^{-\beta} & e^{\beta-h} \end{pmatrix}, \quad S=\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}, \quad R=\begin{pmatrix}e^{\frac{h}{2}} & 0 \\ 0 & e^{-\frac{h}{2}} \end{pmatrix}, $$ that is, we set $T_{\sigma,\sigma'} = e^{\beta\sigma\sigma' + \frac{h}{2}(\sigma + \sigma')}$, $S_{\sigma,\sigma'} = \sigma\delta_{\sigma,\sigma'}$ and $R_{\sigma,\sigma'} = e^{\frac{h}{2}\sigma}\delta_{\sigma,\sigma'}$, where $\sigma,\sigma'\in\{1,-1\}$.

One can then write $$ Z_N \langle\sigma_u\sigma_v\rangle_N = \sum_{\sigma_1=\pm 1,\sigma_N=\pm 1}(RT^{u-1}ST^{v-u}ST^{N-v}R)_{\sigma_1,\sigma_N}. $$ The problem is thus reduced to the computation of the product of these matrices, which is straightforward (diagonalise $T$ to do that). Of course, you still have to compute the partition function. The same type of computations as above yield $$ Z_N = \sum_{\sigma_1=\pm 1,\sigma_N=\pm 1}(RT^{N-1}R)_{\sigma_1,\sigma_N}. $$

  • $\begingroup$ I wrote that quickly and did not check the computations, so I do not guarantee that there are no mistakes. But the procedure itself should be fine and, hopefully, clear. $\endgroup$ Commented May 6, 2022 at 17:49

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