# An identity about exponential of operators

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.

• 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 $1/2$) Nov 19, 2021 at 16:19

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).