For the spin $1/2$ fermions we want to prove the ``Hubbard Stratonovich'' operator identity

$$e^{-a (n_{\uparrow}-1/2)(n_{\downarrow}-1/2) } = \frac{1}{2}... ...a = \pm 1} e^{\lambda \sigma (n_{\uparrow}- n_{\downarrow}) },$$ (1)

where $\cosh(\lambda ) = e^{a/2}$. We first note that this is a single site problem and so the Fock space is four dimensional. Next, we note that both the operators in the above equation are diagonal $4 \times 4$ matrices in the number basis (i.e., the basis in which number operators $n_{\uparrow}$ and $n_{\downarrow}$ are diagonal). We will show that these diagonal matrix elements are the same for the two operators.

In this basis we will represent the states by $\vert n_{\uparrow}, n_{\downarrow}\rangle$, i.e., $\vert, 0 \rangle$ is the empty state, $\vert 1, 0 \rangle$ is the single spin-up state, $\vert, 1 \rangle$ is the single spin-down state, and $\vert 1, 1 \rangle$ is the doubly occupied state. For brevity, we will call $\hat{L}= e^{-a (n_{\uparrow}-1/2)(n_{\downarrow}-1/2) }$, and $\hat{R}= 1/2e^{- a/4} \sum_{\sigma = \pm 1} e^{\lambda \sigma (n_{\uparrow}- n_{\downarrow}) }$.

We find that, $\langle 0, 0 \vert \hat{L}\vert 0, 0 \rangle = e^{- a/4}$, and $\langle 0, 0 \vert\hat{R} \vert 0, 0 \rangle = 1/2 e^{- a/4} \sum_{\sigma = \pm 1} 1 = e^{- a/4}$. Similarly, we have, $\langle 1, 1 \vert \hat{L}\vert 1, 1 \rangle = \langle 1, 1 \vert \hat{R}\vert 1, 1 \rangle = e^{- a/4}$.

Next, we find that $\langle 1, 0 \vert \hat{L}\vert 1, 0 \rangle = e^{-a(1/2)(- 1/2)} = e^{a/4}$, and $\langle 1, 0 \vert \hat{R}\vert 1, 0 \rangle = e^{- a/4} \cosh(\lambda )= e^{a/4}$. And, similarly, $\langle 0, 1 \vert \hat{L}\vert 0, 1 \rangle = \langle 0, 1 \vert \hat{R}\vert 0, 1 \rangle = e^{a/4}$. Thus, in general,

$$\langle n_{\uparrow}, n_{\downarrow}\vert \hat{L}\vert n_{\u... ...arrow}\vert \hat{R}\vert n_{\uparrow}, n_{\downarrow}\rangle .$$ (2)

Since the two operators have the same matrix elements, they must be identical.