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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.

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    $\begingroup$ The Baker–Campbell–Hausdorff formula can be used for $\exp(A + B)$. $\endgroup$
    – LSpice
    Feb 3, 2022 at 0:15
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    $\begingroup$ Thank you @LSpice for fixing my typos. $\endgroup$
    – Stephen Montgomery-Smith
    Feb 3, 2022 at 5:21

1 Answer 1

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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

$$g(y)f(x)=\frac{1}{r}\sum_{p=0}^{r-1}\sum_{q=0}^{r-1}\omega_{r}^{-pq}f(\omega_{r}^{p}x)g(\omega_{r}^{q}y).$$

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    $\begingroup$ Your formula gives the commutator as a sort of Fourier inversion on the Klein Vierergruppe: $r s - s r = \frac1 2(r s - r^{-1}s - r s^{-1} + r^{-1}s^{-1})$. \\ Applying your formula for $s r$ again to $r s$ yields $3s r = 2r^{-1}s + s^{-1}r + 2r s^{-1} + s r^{-1} - 2r^{-1}s^{-1} - s^{-1}r^{-1}$, and hence $3[r, s] = [s^{-1}, r] + [r, s^{-1}] + [r^{-1}, s^{-1}]$. That's strange! $\endgroup$
    – LSpice
    Feb 3, 2022 at 4:59
  • $\begingroup$ My last commutator formula above should be $3[r, s] = [s^{-1}, r] + [s, r^{-1}] - [s^{-1}, r^{-1}]$ (unless I've made more sign errors); or maybe I could write $4[s, r] = [s, r] - [s^{-1}, r] - [s, r^{-1}] + [s^{-1}, r^{-1}]$. $\endgroup$
    – LSpice
    Feb 3, 2022 at 5:08
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    $\begingroup$ I think it looks like the Fourier inversion on the group $\mathbb{Z}_{2}$ instead of the Klein four group (so that the Fourier matrix is just the Hadamard gate). But the formula will probably make more sense if I generalized it at least to the Fourier transform of a finite cyclic group. $\endgroup$
    – Joseph Van Name
    Feb 3, 2022 at 6:00
  • $\begingroup$ I was referring to the two commuting actions of $\mathbb Z/2\mathbb Z$ on pairs $(r, s)$ of invertible elements by $(r, s) \mapsto (r^{-1}, s)$ and $(r, s) \mapsto (r, s^{-1})$. But of course one can count either of these, or others, as actions just of $\mathbb Z/2\mathbb Z$ alone. $\endgroup$
    – LSpice
    Feb 3, 2022 at 16:58
  • $\begingroup$ Another variation: writing $f=f_0+f_1$, $g=g_0+g_1$ where $f_0$, $g_0$ are even and $f_1$, $g_1$ are odd, we obtain that $f_0$ commutes with $g_0$, $f_0$ commutes with $g_1$, $f_1$ commutes with $g_0$ and $f_1$ anticommutes with $g_1$. $\endgroup$
    – მამუკა ჯიბლაძე
    Feb 4, 2022 at 7:12

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