Suppose we have a finite set $\mathcal{S}$ and operators (actually, matrices): $$b_{x-\frac{1}{2},x} = a_{x} + a_{x}^{\dagger} \quad \mbox{and} \quad b_{x,x+\frac{1}{2}} = i(a_{x}-a^{\dagger}_{x})$$ The $b$'s operators satisfy $\{b_{\alpha},b_{\beta}\} = 2\delta_{\alpha\beta}$. I'm trying to prove the following identity: $$e^{K\sum_{x}(a^{\dagger}_{x}a_{x}-a_{x}a^{\dagger}_{x})} = \prod_{x}\bigg{(}e^{-K}+2\operatorname{sinh}Ka^{\dagger}_{x}a_{x}\bigg{)}.$$

My work so far is the following. Because of the anti-commutation relations of the $b$'s, we get (if my calculations are correct!): $$[a^{\dagger}_{x},a_{x}] = a^{\dagger}_{x}a_{x}-a_{x}a^{\dagger}_{x}= 1$$ I replaced this into the exponential, to get: $$e^{K\sum_{x}(a^{\dagger}_{x}a_{x}-a_{x}a^{\dagger}_{x})} = \prod_{x}e^{-K+2Ka^{\dagger}_{x}a_{x}}$$ Then I tried using the formula: $$e^{2\alpha} = 2\operatorname{sinh}(\alpha)e^{\alpha}+1$$ but then I get: $$\prod_{x}e^{-K+2Ka^{\dagger}_{x}a_{x}} = \prod_{x}\bigg{(}e^{-K}+\operatorname{sinh}2Ka^{\dagger}_{x}a_{x}e^{Ka^{\dagger}_{x}a_{x}-K}\bigg{)}$$

What am I doing wrong?

ADD: The $b$ matrices are degined as follows:

$$b_{-\frac{1}{2},0} = \sigma_{0}^{1} \quad \mbox{and} \quad b_{x-\frac{1}{2},x} = \bigg{(}\prod_{x'=1}^{x-1}\sigma_{x'}^{3}\bigg{)}\sigma_{x}^{1} \quad \mbox{for $x \in \{1,...,N-1\}$} $$

$$b_{0, \frac{1}{2}} = \sigma_{0}^{2} \quad \mbox{and} \quad b_{x, x+\frac{1}{2}} = \bigg{(}\prod_{x'=1}^{x-1}\sigma_{x'}^{3}\bigg{)}\sigma_{x}^{2} \quad \mbox{for $x \in \{1,...,N-1\}$} $$

These are Jordan-Wigner transformations. Here, $\sigma_{i}^{k}$ is the operator on $\mathbb{C}^{2}\otimes \cdots \otimes \mathbb{C}^{2}$ ($N$ factors) given by: $$\sigma_{i}^{k} = I\otimes \cdots I \otimes \sigma^{k}\otimes \cdots \otimes I$$ where $I$ is the identity matrix and $\sigma^{k}$, $k=1,2,3$ are Pauli matrices. Also, the $\sigma^{k}$ above goes in the $i$-th entry.

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    $\begingroup$ Where do these $\hat{b}_{x-\tfrac{1}{2},x}$ and $\hat{b}_{x,x+\tfrac{1}{2}}$ operators come from? I am confused about what the indices mean (with the factors of $1/2$) $\endgroup$ Nov 19, 2021 at 16:19

1 Answer 1


Firstly, your commutation relation is wrong; it should be a fermionic relation $\left\{a_{x}^{\dagger}, a_{x}\right\}=a_{x}^{\dagger} a_{x}+a_{x} a_{x}^{\dagger}=1$. I guess you just have a typo since later your calculation is fine.

Second, the last line should read $\prod_{x} e^{-K+2 K a_{x}^{\dagger} a_{x}}=\prod_{x}\left(e^{-K}+2 \sinh K a_{x}^{\dagger} a_{x} e^{K a_{x}^{\dagger} a_{x}-K}\right)$ since your $\alpha=Ka_{x}^{\dagger} a_{x}$.

Then, since $a_{x}^{\dagger} a_{x}=0,1$ as eigenvalues due to the fermionic nature, and the whole expression commutes with $a_{x}^{\dagger} a_{x}$ which means it is diagonal in the eigenbasis, you only have to check these two cases and you can found the form I wrote is equivalent to what you want (if $a_{x}^{\dagger} a_{x}=0$, then the second term vanishes; if $a_{x}^{\dagger} a_{x}=1$, then the second term's exponent is 0).


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