Consider 1-D Ising model of n lattice points with periodic boundary condition,
$\beta H(\sigma_1,\sigma_2,...,\sigma_N) = -\sum_{i=1}^nk(\sigma_i\sigma_{i+1})-\sum_{i=1}^n\sigma_i$
$k=\beta J$ and $h=\beta B$ where $J$ is the coupling constant and $B$ is the applied magnetic field.

Partition function, $Z=\sum_{\sigma_1,\sigma_2,...,\sigma_N}e^{-\beta H(\sigma_1,\sigma_2,...,\sigma_N)}$
where $\sum_k$ is the sum over all the possible microstates and $H_k$ is the hamiltonian of the $k-th$ microstate.
$Z = \sum_{\sigma_1,\sigma_2,...,\sigma_N}exp\Big(k(\sigma_1\sigma_2+\sigma_2\sigma_3+...+\sigma_N\sigma_1)\Big)exp\Big(h(\sigma_1+\sigma_2+...+\sigma_N)\Big)$
$\implies Z = \sum_{\sigma_1,\sigma_2,...,\sigma_N}exp\Big(k\sigma_1\sigma_2+\frac{(\sigma_1+\sigma_2)}{2}\Big)exp\Big(k\sigma_2\sigma_3+\frac{(\sigma_2+\sigma_3)}{2}\Big)...exp\Big(k\sigma_N\sigma_1+\frac{(\sigma_N+\sigma_1)}{2}\Big)$
The $Z$ can be written in terms of transfer matrix as

where $T=\begin{pmatrix}e^{k-h}&e^{-k}\\e^{-k}&e^{k+h}\end{pmatrix}$
$\sigma_1=-1$ corresponds to $|\sigma_1\rangle=\begin{pmatrix}1\\0\end{pmatrix}$ and $\sigma_1=+1$ corresponds to $|\sigma_1\rangle=\begin{pmatrix}0\\1\end{pmatrix}$

So, $Z = \sum_{\sigma_1}\langle\sigma_1|T^N|\sigma_1\rangle=\sum_{\sigma_1}\langle O\sigma_1|OT^NO^T|O\sigma_1\rangle\tag{1}$
where $O$ is the orthogonal matrix that diagonalize $T$ and thus $T^N$.
where $\lambda_{1,2}=e^k\cosh h\pm\sqrt{e^{2k}\sinh^2 h+e^{-2k}}$

So, $Z=Tr(T^N)=Tr(OT^NO^T)=\lambda_1^N+\lambda_2^N\tag{2}$

Now, I am trying to calculate $\langle\sigma_i\rangle$ when $h=0$ using transfer matrix.
We know that when $h=0$, $\langle\sigma_i\rangle=0$
As quantities inside $\sum$ are just numbers (matrix elements and $\sigma_i$ are numbers),
So, $\langle\sigma_i\rangle=\sum_{\sigma_1,\sigma_i}\sigma_i\langle\sigma_i|T^{N-i+1}|\sigma_1\rangle\langle\sigma_1|T^{i-1}|\sigma_i\rangle$
$\implies\langle\sigma_i\rangle=\sum_{\sigma_i}\sigma_i\langle O\sigma_i|OT^NO^T|O\sigma_i\rangle$
$\implies\langle\sigma_i\rangle=\sum_{\sigma_i}\sigma_i\langle O\sigma_i|\begin{pmatrix}\lambda_1^N&0\\0&\lambda_2^N\end{pmatrix}|O\sigma_i\rangle\tag{3}$

But now how to show that $\langle\sigma_i\rangle=0$?
One way is to calculate $O$ explicitly and plug this in $(3)$ and show that it is $0$. This I have verified.
But I have seen that in books, they introduce Pauli matrix to show that it is $0$. I am not able to understand how that follows from $(3)$.

  • $\begingroup$ Why use the transfer matrix at all? To show that $\langle\sigma_i\rangle=0$, it suffices to observe that the Gibbs measure gives the same probability (when $h=0$) to a configuration $(\sigma_i)_{i=1,\dots,N}$ and to the configuration $(-\sigma_i)_{i=1,\dots,N}$... $\endgroup$ Apr 16 at 7:58
  • $\begingroup$ @YvanVelenik, thanks a lot for the reply. Yeah we can simply show this as you have shown. But in the Goldenfield's book on Phase transition, he uses transfer matrix approach to calculate all the observables and correlations. I want to make myself more comfortable with this approach. Can you please help me how to proceed further from $(3)$? $\endgroup$
    – Iti
    Apr 16 at 8:33
  • $\begingroup$ My point is that, at each step of your derivation, it is obvious that the value is equal to $0$... For instance, in your last expression, the summand obviously changes sign when $\sigma_i$ is replaced by $-\sigma_i$. Therefore, the sum over $\sigma_i$ gives you $0$. So, for this particular problem, it is completely useless to introduce additional matrices. The introduction of such matrices is useful, however, if you are interested in higher-order correlation functions (say, the 2-point function); see, for instance, this earlier answer of mine. $\endgroup$ Apr 16 at 11:45
  • $\begingroup$ Can you please explain how Pauli matrix is being introduced. I am not able to see this. I have read your answer on the link given but not very clear about it. Can you please explain in the context of my question. It would be very helpful $\endgroup$
    – Iti
    Apr 16 at 12:58
  • $\begingroup$ OK. I wrote an answer detailing a step that was not spelled out in my other answer. I hope this helps. $\endgroup$ Apr 16 at 14:00

1 Answer 1


As I explained in the comments, you don't need to introduce any Pauli matrix to see that $\langle\sigma_i\rangle=0$. In fact, you don't even need to introduce the transfer matrix at all, but just observe that the Gibbs measure gives the same probability to the configurations $(\sigma_1,\dots,\sigma_N)$ and $(-\sigma_1,\dots,-\sigma_N)$.

So, I'll rather answer the underlying technical question that is spelled out in your comments, namely how to rewrite your final expression in terms of one of the Pauli matrices. (This is actually useful if you want to compute, for instance, the 2-point function; see this answer for such an application.)

The idea is very simple, just rewrite \begin{align*} \sum_{\sigma_i=\pm 1} \sigma_i (T^N)_{\sigma_i,\sigma_i} &= \sum_{\sigma_i=\pm 1} \sum_{\sigma'_i=\pm 1} \delta_{\sigma_i\sigma'_i} \sigma_i (T^N)_{\sigma_i,\sigma_i}\\ &= \sum_{\sigma_i=\pm 1} \sum_{\sigma'_i=\pm 1} S_{\sigma_i,\sigma'_i}(T^N)_{\sigma'_i,\sigma_i}\\ &= \sum_{\sigma_i=\pm 1} (ST^N)_{\sigma_i,\sigma_i}, \end{align*} where I have introduced the matrix $S=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, using the fact that $\delta_{\sigma_i,\sigma'_i}\sigma_i = S_{\sigma_i,\sigma'_i}$. In the second identity, I use the fact that the summand vanishes when $\sigma_i\neq\sigma'_i$ to replace the second instance of $\sigma_i$ by $\sigma'_i$.

  • $\begingroup$ Ok now I understood how to introduce Pauli Matrix. After that So, after that we have to diagonalize $ST^N$. $\endgroup$
    – Iti
    Apr 16 at 16:34
  • $\begingroup$ Well, as I said, in this particular case, you don't have to do anything, because it is immediately clear that the result will be zero (for instance because $T$ is invariant under the interchange $+\leftrightarrow-$, while $S$ gets multiplied by $-1$). For this particular case, all these computations are making the result more obscure, not clearer. However, if you wanted to compute $\langle\sigma_i\rangle$ with $+$ boundary condition, for instance, you could indeed proceed in this way. $\endgroup$ Apr 17 at 8:03
  • $\begingroup$ Thank you so much for the explanation. Now I have understood and able to solve two point correlation function using this approach. $\endgroup$
    – Iti
    Apr 17 at 13:52
  • $\begingroup$ I have one last doubt in the above proof we can also get $T^NS$ instead of $ST^N$ by interchanging $\sigma_i$ and $\sigma_i'$. Does that change the final result because matrix multiplication is not commutative. $\endgroup$
    – Iti
    Apr 17 at 13:54
  • $\begingroup$ In the particular setting above, it does not change anything, because you are actually computing the trace of $ST^N$, which is invariant under cyclic permutations. $\endgroup$ Apr 17 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.