0
$\begingroup$

I am currently reading the second chapter of Quantum Phase Transitions by Subir Sachdev. I have the first edition of this textbook. The main idea of the second chapter is the following: quantum phase transitions in $d$ dimensions are intimately connected to finite-temperature classical transitions in $d + 1$ dimensions. This chapter considers the case where $d = 0$, in which we have single-site models that map to a 1 dimensional classical system. Section 2.1, therefore, starts off with a discussion of the Classical Ising Chain. The following is a physical and mathematical overview presented in this chapter. I fully understand the mathematics stated below.

The Hamiltonian of a Classical Ising Chain with $M$ spins is given by the following:

\begin{align*} H &= -K\sum_{i = 1}^M \sigma_i^z \sigma_{i+1}^z - h \sum_{i = 1}^M \sigma_i^z\\ &\text{Here we assume that K and h absorb the Boltzmann constant.}\\ &\text{Here we also assume periodic boundary conditions, i.e. $\sigma^z_{M+1} = \sigma^z_1$} \end{align*}

The canonical partition function is given by the following:

\begin{align} Z = \sum_{\{\sigma^z_i = \pm 1\}} e^{-H} \end{align}

From this we can see that the partition function evaluates to the following:

\begin{align} Z &= \sum_{\{\sigma^z_i\}} \Pi_{i = 1}^M e^{K \sigma_i^z \sigma^z_{i+1}}e^{h\sigma_i^z} \equiv \sum_{\{\sigma^z_i\}} \Pi_{i = 1}^M T_1(\sigma_i^z ,\sigma^z_{i+1}) T_2(\sigma_i^z) \end{align}

Here we see that $T_1(\sigma_i^z ,\sigma^z_{i+1}) = e^{K \sigma_i^z \sigma^z_{i+1}}$ and that $T_2(\sigma_i^z) = e^{h\sigma_i^z}$. We can treat the parameters as row-column index labels. From this, we see that $T_1$ is a $2 \times 2$ matrix and that $T_2$ is a $2 \times 2$ diagonal matrix. Here we also note that the summation is over all possible $\{\sigma^z_i\}$, i.e. over all possible pairwise combinations of $\pm 1$. Hence the summation is over the four pairs $(1,1); (-1,1), (1,-1), $ and $(-1,-1)$.

Hence we see the following:

\begin{align} T_1 = \begin{pmatrix} e^K && e^{-K}\\ e^{-K} && e^K \end{pmatrix} \end{align}

The first row consists of the $(1,1); (-1,1)$ terms, and the second row consists of the $(1,-1); (-1,-1)$ terms.

Now, we also clearly see that the matrix version of $T_2$ is the following:

\begin{align} T_2 = \begin{pmatrix} e^h && 0\\ 0 && e^h \end{pmatrix} \end{align}

From all of the above we see that the canonical partition function is given by the following:

\begin{align} Z = Trace([T_1 T_2][T_1 T_2]...[T_1 T_2] \text{ M times}) = [Trace(T_1 T_2)]^M = [Trace(T_2^{1/2}T_1 T_2^{1/2})]^M \end{align}

Since the trace of a matrix is the sum of the matrix's diagonal elements, we see that the trace of $T_1 T_2$ is $e^{K+h} + e^{K-h}$. Now, I interpreted $T_2^{1/2}$ as the following matrix:

\begin{align} T_2^{1/2} = \begin{pmatrix} e^{h/2} && 0\\ 0 && e^{h/2} \end{pmatrix} \end{align}

Essentially, $T_2^{1/2}$ seemed to me to be a "square root" of the matrix $T_2$, i.e $T_2^{1/2} T_2^{1/2} = T_2^{1/2 + 1/2} = T_2$. Using this intuitive guess I found that $T_2^{1/2} T_1 T_2^{1/2}$ is the matrix

\begin{pmatrix} e^{K + h} && e^{-K}\\ e^{-K} && e^{K - h} \end{pmatrix}

This equals the matrix in equation (2.7). We see immediately that the trace of $T_2^{1/2} T_1 T_2^{1/2}$ is also $e^{K+h} + e^{K-h}$

Finally, the chapter mentions that the partition function is $Z = \epsilon_1^M + \epsilon_2^M$, where $\epsilon_{1,2}$ are the eigenvalues of $T_2^{1/2} T_1 T_2^{1/2}$.

As mentioned before, I understand all of the mathematics above. The chapter then defines the two-point spin correlator as

\begin{align} \langle \sigma_i^z \sigma_j^z \rangle = \frac{1}{Z} \sum_{\{\sigma^z_i\}}e^{-H} \sigma^z_i \sigma^z_j \end{align}

The chapter then claims that if we set $h = 0$ in the Hamiltonian, then $\langle \sigma_i^z \sigma_j^z \rangle = \frac{1}{Z}Trace(T_1^i \hat{\sigma}^z T_1^{j - i}\hat{\sigma}^z T_1^{M-j})$ by the method I mentioned above. Now, I do not understand where the $T_1^i$, $T_1^{j - i}$, and $T_1^{M-j}$ come from. Can anyone show how I can prove that $\langle \sigma_i^z \sigma_j^z \rangle = \frac{1}{Z} \sum_{\{\sigma^z_i\}}e^{-H} \sigma^z_i \sigma^z_j$?

$\endgroup$

1 Answer 1

0
$\begingroup$

See the following link for a beautiful explanation:

https://stanford.edu/~jeffjar/statmech/lec4.html#solving

$\endgroup$

Your Answer

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