I apologize in advanced for how trivial this question will come to many of you. I am in fact not a Mathematician nor physicist, just a math friendly Biologist who needs to understand the transfer matrix method for the 1D Ising Model. I work in research and we have a biophysical model that requires my understanding of this model. I understand that there are interaction energies of nearest neighbor particle spin along the lattice, and the influence of the external field on each particle spin.

Every material I've come across has said to convince yourself of the following: The partition function $Z=\sum_{\{\sigma\}}\prod_{i}^NV(\sigma,\sigma')$ where $V(\sigma,\sigma')=e^{k\sigma\sigma'}e^{\frac{h}{2}(\sigma+\sigma')}$ and $k=\beta J$ (the interaction between particles) and $h=\beta B$ (influence from external field). Then the transfer matrices:

$Z=\sum_{\{\sigma\}}\prod_{i}^NV(\sigma,\sigma') = Tr\begin{pmatrix} V(+1,+1) & V(+1,-1)\\ V(-1,+1) & V(-1,-1) \end{pmatrix}^N$

Ok, so now let's say we only have 3 particles. To prove this to myself I wanted to show that the $2^3=8$ conformations of the system sum up to the same as the transfer matrix method.

So, firstly:

$V(+1,+1) = e^{k+h}\\ V(+1,-1) = e^{-k}\\ V(-1,+1) = e^{-k}\\ V(-1,-1) = e^{k-h}$

So for our system of 3 particles, I have the following 8 states:

$(+1,+1,+1) = e^{2(k+h)}\\ (+1,-1,+1) = e^{-2k}\\ (+1,+1,-1) = e^{h}\\ (-1,+1,+1) = e^h\\ (+1,-1,-1) = e^{-h}\\ (-1,-1,+1) = e^{-h}\\ (-1,+1,-1)=e^{-2k}\\ (-1,-1,-1) = e^{2(k-h)}$

Summing them together: $Z=e^{2(k+h)}+2e^h+2e^{-h}+2e^{-2k}+e^{2(k-h)}$

So for the transfer matrix, I have:

$\begin{pmatrix} V_{\sigma_1,\sigma_2}(+1,+1) & V_{\sigma_1,\sigma_2}(+1,-1)\\ V_{\sigma_1,\sigma_2}(-1,+1) & V_{\sigma_1,\sigma_2}(-1,-1) \end{pmatrix} \begin{pmatrix} V_{\sigma_2,\sigma_3}(+1,+1) & V_{\sigma_2,\sigma_3}(+1,-1)\\ V_{\sigma_2,\sigma_3}(-1,+1) & V_{\sigma_2,\sigma_3}(-1,-1) \end{pmatrix}$

This becomes:

\begin{pmatrix} e^{2(k+h)}+e^{-2k} & e^h+e^{-h}\\ e^h+e^{-h} & e^{-2k}+e^{2(k-h)} \end{pmatrix}

Taking the trace of this is clearly not $Z=e^{2(k+h)}+2e^h+2e^{-h}+2e^{-2k}+e^{2(k-h)}$ that I did earlier. But I did notice that the sum of ALL the matrix elements from that product is. I know I am doing something wrong, or not understanding something well. Please, any help would be greatly appreciated.


1 Answer 1


Let us write the partition function as $$\mathcal{Z} = \sum_{\text{MS}}\exp\left[\beta J\sum_{i<i'}s_is_{i'}+\beta h\sum_{i=1}^{N}s_i+Nf\right],$$ where MS denotes all possible microstates of $N$ spins. Here $f$ is the energy shift, $h$ is the external magnetic field and $J$ is the coupling constant. Now let us write it in more compact way, $$\mathcal{Z}=\sum_{\text{MS}}\exp\left[\sum_{i}^{N}\left\{Ks_is_{i+1}+\frac{b(s_i+s_{i+1})}{2}+f\right\}\right]$$ with $K=\beta J$, $b=\beta h$. The sum over all microstates is $$\sum_{\text{MS}}=\sum_{\{s_i\}},$$ which means that we sum over all spins and over all spin states ($\pm 1$). Using the property of exponential function, we can write $$\mathcal{Z}\propto\sum_{\lbrace s_i\rbrace}\left(e^{Ks_1s_2+b(s_1+s_2)/2}\right)...\left(e^{Ks_Ns_1+b(s_N+s_1)/2}\right),$$ where we write "$\propto$" instead of "$=$" because $e^{f}$ factor is omitted. Also, we imply periodic boundary conditions, $s_{N+1}=s_1$, which is reflected in the very last term in product. Then, we can see that each product contains two consecutive spins, for instance $s_1$ and $s_2$. Let us denote such products as $T_{s_1s_2}$, $$\mathcal{Z}=\sum_{\{s_i\}}T_{s_1s_2}T_{s_2s_3}T_{s_3s_4}...T_{s_Ns_1}.$$ The quantity $T_{s_1s_2}$ looks like $$T_{s_1s_2}=\exp\left\{Ks_1s_2+\frac{b(s_1+s_2)}{2}+f\right\}.$$ For $N=3$ spins, we have $$\mathcal{Z}=\sum_{s_1}\sum_{s_2}\sum_{s_3}T_{s_1s_2}T_{s_2s_3}T_{s_3s_1}.$$

Keypoint: as I remember, the periodic boundary conditions (PBC) are crucial, because without them, one cannot recognize the trace, isn't it? Only in $N\rightarrow\infty$ the difference between PBC and chain is negligible. See also this question.

  • 1
    $\begingroup$ Gosh thanks I didn't know the periodic boundary condition was a requirement for a small N! Thank you it resolved! $\endgroup$
    – Nick
    Oct 17, 2023 at 22:13
  • $\begingroup$ @Nick , let me emphasize that it is not a requirement to consider the case of finite $N$, but when you got to $N\rightarrow\infty$ limit, you impose PBC to make trace and simplify the problem. The argument to do it is that in such limit boundary conditions are negligible. So, if you compare computation via transfer matrix via explicit spin summation you have to use PBC. $\endgroup$ Oct 18, 2023 at 6:10

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