Mapping a 1D quantum Ising chain to a 2-dimensional classical Ising system

Question

Going through Ref. 1 (I'll stick with the book's equation numbering), I'm learning about the mapping of quantum systems into classical systems. First of all let me briefly recap notation and some details:

Recap

The Hamiltonian of the 1D quantum Ising chain is $$\hat{H}_I=-Jg\sum_i\hat{\sigma}_i^x-J\sum_i\hat{\sigma}_i^z\hat{\sigma}_{i+1}^z\tag{5.1},$$ where the $\sigma$'s are the Pauli matrices, as we're dealing with spin $\frac{1}{2}$ particles The procedure is pretty standard; we want divide imaginary time in small steps $a$ and write the partition function as that of a classical system via the transfer matrix method, so I'm not describing that part specifically. In what follows I will denote with $\lvert m_i\rangle$, $m=\pm1$ the eigenvectors of $\hat{\sigma}_i^z$: $$\hat{\sigma}^z_i\lvert m_i\rangle=m_i\lvert m_i\rangle \qquad m_i=\pm1$$ and naturally, with $M$ spins $$\lvert\boldsymbol{m}\rangle:=\bigotimes_{i=1}^M \lvert m_i\rangle$$

The problem

A crucial part of the construction of the mapping is to evaluate matrix elements of the transfer matrix in the eigenbasis above. Since $\hat{\sigma}^z$'s are diagonal in this basis, it all boils to down to evaluating matrix elements of $\exp\{Jga\sum_i\hat{\sigma}_i^x\}$, for which we can focus on the single spin and evaluate (let me drop the $i$ for simplicity) $\langle m\lvert\exp\{Jga\hat{\sigma}^x\}\lvert m'\rangle$. Now, the book claims that the following identity may easily be derived $$\langle m\lvert\exp\{Jga\hat{\sigma}^x\}\lvert m'\rangle\tag{5.63}=A\exp(Bmm') \qquad\begin{cases}A=\frac{1}{2}\cosh(2Jga) \\ \exp(-2B)=\tanh(Jga)\end{cases}$$ on the other hand, by direct evaluation of the matrix element I find$^{(\star)}$ $$\langle m \lvert \exp[gJa\hat{\sigma}_x]\lvert m'\rangle=\frac{1}{2}(\mathrm{e}^{Jga}+m'm\mathrm{e}^{-Jga})$$ which is in clear disagreement with $(5.63)$. Nevertheless, in the construction, it turns out that the $A$ constant is irrelevant as it only contributes with an inconsequential normalization. Furthermore, the relevant part is the exponential: even if it doesn't have the same coefficient, it is crucial that an exponential appears for the mapping to be successful. Now, in the limit $Jga\gg 1$, we may write: $$\frac{1}{2}(\mathrm{e}^{Jga}+m'm\mathrm{e}^{-Jga})=\frac{1}{2}\mathrm{e}^{Jga}(1+\underbrace{m'm\mathrm{e}^{-2Jga}}_{\ll 1})\simeq\underbrace{\frac{1}{2} \mathrm{e}^{Jga}}_{A'}\exp\{\underbrace{\mathrm{e}^{-2Jga}}_{B'}m'm\}=A'\exp(B'mm').$$ This approximation can't be correct, though: we are working in the $a\to 0$ limit (small imaginary time steps) and even if the condition $Jga\gg1$ were attained tuning the magnetic field, I wouldn't be happy with it: the mapping shouldn't depend on the regime.

So, wrapping it all up: Is $(5.63)$ wrong? In case it is, how do I get the correct mapping?

References

1. Quantum Phase Transitions, Subir Sachdev, second edition, 2011. Chapter 5, section 5.6.

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$^{(\star)}$ _{Maybe this result is more familiar written in terms of hyperbolic functions, using $\exp(\alpha\hat{\sigma}^x)=\cosh(\alpha)+\sinh(\alpha)\hat{\sigma}^x$, but nonetheless it is correct}

Answer 1

I got just stuck at the same point, so here my thoughts on it:

$ \langle m ' | e^{\alpha \sigma^x} | m \rangle = \cosh\alpha \delta_{m m'} + \sinh \alpha (1-\delta_{m m'})$ and we claim this to be equal to $A e^{B m m'}$. If this is true, then

$m=m'$: $A e^B = \cosh \alpha$

$m=-m'$: $A e^{-B} = \sinh \alpha$

we can solve this to $A^2= \frac{1}{2}\sinh2\alpha$ and $e^{-2B} = \tanh \alpha$.

So, we found the correct expression for B, but I guess the given expression for $A$ is a typo in the book - it is not relevant for the further discussion, since it is hidden in the normalization of the partition sum.

Answer 2

YoungLee has already posted an answer, which is the one I have accepted. However, I want to make some considerations as to why my argument breaks. As I wrote in my original post and as YoungLee points out, the value of $A$ itself is not relevant as it will be discarded for reasons of normalization; although YoungLee's answer shows that Sachdev is wrong about $A$, it won't affect the argument at all. What matters is $B$ and Sachdev got it right.

Then, why is my argument broken? Well, first of all I overcomplicated things trying to work with $m$ and $m'$, while working by cases would have shown the solution immediately. The main problem is the factorization $$\frac{1}{2}(\mathrm{e}^{Jga}+m'm\mathrm{e}^{-Jga})=\frac{1}{2}\mathrm{e}^{Jga}(1+m'm\mathrm{e}^{-Jga})$$ and the subsequent assumption that the first factor will yield $A'$ and the second factor $B'$ (I put primes just to avoid confusion with Sachdev's equations): $$A'=\mathrm{e}^{Jga}/2 \qquad B'=\mathrm{e}^{-2Jga}.\tag{A}\label{A}$$ Such splitting was unjustified and indeed led to the assumption $Jga\gg 1$ to make things work and we easily understand why it works in this limit, as it obviously has the same behaviour of the correct solution, posted by YoungLee: $$A=\sqrt{\cosh(Jga)\sinh(Jga)}\qquad B=-\frac{1}{2}\ln\tanh(Jga).$$