Exponentials and other functions of sums of anti-commuting operators I know that if $A$ and $B$ are commuting operators, then $\exp(A+B) = \exp(A) \exp(B)$.  Is there a similar formula if $A$ and $B$ are anti-commuting (that is, $AB+BA = 0$)?
I have developed a formula for $f(A+B)$ when $f$ is analytic and $A$ and $B$ anti-commute, and I want to know if this is already in the literature.
We can formally write $f(z) = g(z^2) + z h(z^2)$ where $g(z) = \dfrac{f(\sqrt z) + f(-\sqrt z)}2$ and $h(z) = \dfrac{f(\sqrt z) - f(-\sqrt z)}{2\sqrt z}$.  Then
$$ f(A+B) = g(A^2 + B^2) + (A+B) h(A^2 + B^2) .$$
Since $A^2$ and $B^2$ commute, $g$ and $h$ can be expanded as power series in the traditional manner.
It is not hard to prove, and I think it is quite cute.
 A: Suppose that $x,y$ are anti-commuting so that $xy=-yx$. Suppose furthermore that $f(x)=\sum_{j=0}^{\infty}a_{j}x^{j},g(y)=\sum_{k=0}^{\infty}b_{k}y^{k}$ are power series.
Then $$g(y)f(x)=\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}b_{k}a_{j}y^{k}x^{j}=\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}b_{k}a_{j}x^{j}y^{k}(-1)^{jk}$$
$$=\sum_{j=0}^{\infty}a_{j}x^{j}\sum_{k=0}^{\infty}b_{k}y^{k}(-1)^{jk}=\sum_{j=0}^{\infty}a_{j}x^{j}\sum_{k=0}^{\infty}b_{k}((-1)^{j}y)^{k}$$
$$=\sum_{j=0}^{\infty}a_{j}x^{j}g((-1)^{j}y)=\sum_{j=0}^{\infty}a_{2j}x^{2j}g((-1)^{2j}y)+\sum_{j}a_{2j+1}x^{2j+1}g((-1)^{2j+1}y)$$
$$=\sum_{j=0}^{\infty}a_{2j}x^{2j}g(y)+\sum_{j=0}^{\infty}a_{2j+1}x^{2j+1}g(-y)
=\sum_{j}^{\infty}a_{2j}x^{2j}g(y)+\sum_{j=0}^{\infty}a_{2j+1}x^{2j+1}g(-y)$$
$$=\frac{1}{2}(f(x)+f(-x))g(y)+\frac{1}{2}(f(x)-f(-x))g(-y).$$
$$=\frac{1}{2}(f(x)g(y)+f(-x)g(y)+f(x)g(-y)-f(-x)g(-y)).$$
In particular, if $r=e^{x},s=e^{y}$, then $r,s$ satisfy the following version of commutativity that I have not seen before:
$$sr=\frac{1}{2}\cdot(rs+r^{-1}s+rs^{-1}-r^{-1}s^{-1}).$$
A generalization
Observe the similarity between the above formula and the $2\times 2$-Fourier transform matrix/Hadamard matrix. Let $\omega_{r}=e^{2\pi i/r}$.
Suppose that $x,y$ satisfy the skew commutativity relation $yx=\omega_{r}xy$, and $f(z)=\sum_{j=0}^{\infty}a_{j}z^{j},
g(z)=\sum_{j=0}^{\infty}b_{j}z^{j}$ are power series. Then
$$g(y)f(x)=\frac{1}{r}\sum_{p=0}^{r-1}\sum_{q=0}^{r-1}\omega_{r}^{-pq}f(\omega_{r}^{p}x)g(\omega_{r}^{q}y).$$
