the Zassenhaus /Baker–Campbell–Hausdorff formula for cosine.

Question

This question concerns the expansion of non-commutative algebra $[X,Y] \neq 0$ for two operators $X,Y$. One can think of $X$ and $Y$ as some matrices.

If $[X,Y] = 0$, we have

$$e^{t(X+Y)}= e^{tX}~ e^{tY}$$

If $[X,Y] \neq 0$, We know the Zassenhaus formula or the Baker–Campbell–Hausdorff formula: $$e^{t(X+Y)}= e^{tX}~ e^{tY} ~e^{-\frac{t^2}{2} [X,Y]} ~ e^{\frac{t^3}{6}(2[Y,[X,Y]]+ [X,[X,Y]] )} ~ e^{\frac{-t^4}{24}([[[X,Y],X],X] + 3[[[X,Y],X],Y] + 3[[[X,Y],Y],Y]) } \cdots$$

This expresses $e^{t(X+Y)}$ in terms of $e^{tX}$ and $e^{tY}$, and their further commutators $[X,Y]$.

Question: Do we have a similar form for $\cos(A+B)$ when $[A,B]=C \neq 0$? (We may take $[C,A]=[C,B]=0$ for the simplest case to extract the first order term.)

If $[A,B]=0$, we have $$\cos(A+B)=\cos(A)\cos(B) -\sin(A)\sin(B).$$

If $[A,B]=C \neq 0$, do we have some similar expression like the Zassenhaus formula or the Baker–Campbell–Hausdorff formula: $$\cos(A+B)=\cos(A)\cos(B) \dots-\sin(A)\sin(B) \dots+ \dots$$

Can we express $\cos(A+B)$ in terms of $\cos(A)$,$\cos(B)$,$\sin(A)$,$\sin(B)$ and some function of $C$?

Answer 1

It seems to me that if $[A,B]=C$ with $[C,A]=[C,B]=0$, it will be a simple case.

We have $$e^{A+B}=e^A e^B e^{-\frac{1}{2}[A,B]}$$

and $$\cos(A+B)=\frac{e^{i(A+B)}+e^{-i(A+B)}}{2}=\frac{e^{iA}e^{iB}e^{\frac{1}{2}[A,B]}+e^{-iA}e^{-iB}e^{\frac{1}{2}[A,B]}}{2}$$ $$=e^{\frac{1}{2}[A,B]}\frac{e^{iA}e^{iB}+e^{-iA}e^{-iB}}{2}=e^{\frac{1}{2}[A,B]}\big(\cos(A)\cos(B) -\sin(A)\sin(B)\big).$$

Notice that $\frac{e^{iA}e^{iB}+e^{-iA}e^{-iB}}{2}=\cos(A)\cos(B) -\sin(A)\sin(B)$ is true. and notice that we use the condition $[A,B]=C$ with $[C,A]=[C,B]=0$.

Thus my answer is $$\cos(A+B)=e^{\frac{1}{2}[A,B]}\big(\cos(A)\cos(B) -\sin(A)\sin(B)\big)$$ up to higher order terms when $[A,B]=C$ with $[C,A]\neq0, [C,B]\neq 0$.

Maybe someone else can fill in the higher order term computation?

Answer 2

Let us write the Zassenhaus formula in the form $$ e^{X+Y}=e^Xe^Ye^{C_2(X,Y)}e^{C_3(X,Y)}\ldots, $$ where $C_n(X,Y)$ satisfies $C_n(\alpha X,\alpha Y)=\alpha^nC_n(X,Y)$. The first couple of these are $$ C_2(X,Y)=-\frac12[X,Y],\qquad C_3(X,Y)=\frac13[Y,[X,Y]]+\frac16[X,[X,Y]]. $$ Applying this to the definition of $\cos(A+B)$, we get $$ 2\cos(A+B)=e^{i(A+B)}+e^{-i(A+B)}\\=e^{iA}e^{iB}e^{-C_2}e^{-iC_3}e^{C_4}e^{iC_5}e^{-C_6}\ldots+e^{-iA}e^{-iB}e^{-C_2}e^{iC_3}e^{C_4}e^{-iC_5}e^{-C_6}\ldots, $$ where we have used the abbreviation $C_n=C_n(X,Y)$. To get a sense of what is going on, let us assume $C_n=0$ for all $n>3$. Thus we have $$ e^{iA}e^{iB}e^{-C_2}e^{-iC_3}+e^{-iA}e^{-iB}e^{-C_2}e^{iC_3} \\= (\cos A+i\sin A)(\cos B+i\sin B)e^{-C_2}(\cos C_3-i\sin C_3)\\ +(\cos A-i\sin A)(\cos B-i\sin B)e^{-C_2}(\cos C_3+i\sin C_3)\\ =2(\cos A\cos B-\sin A\sin B)e^{-C_2}\cos C_3+2(\cos A\sin B+\sin A\cos B)e^{-C_2}\sin C_3, $$ yielding $$ \cos(A+B)=(\cos A\cos B-\sin A\sin B)e^{-C_2}\cos C_3\\+(\cos A\sin B+\sin A\cos B)e^{-C_2}\sin C_3. $$ It is now easy to see that the general formula is given by $$ \cos(A+B)=\big[\cos(x_0+x_1-x_3+x_5-x_7+\ldots)e^{-x_2+x_4-x_6+\ldots}\big]\Big|_{x_0=A,x_1=B,x_2=C_2,x_3=C_3,\ldots}, $$ where it is understood that we expand $\cos(a+b-x_3+x_5-x_7+\ldots)e^{-x_2+x_4-x_6+\ldots}$ as if all the variables were simply numbers, with the terms ordered in such a way that the terms with $x_n$ come after all $x_1,\ldots,x_{n-1}$, and then substitute the matrices for the variables.