I am trying to show that
$$e^{A_n}e^{A_{n-1}}…e^{A_2}e^{A_1}=e^{\sum_iA_i}e^{\frac{1}{2}\sum_{i>j}[A_i,A_j]} \tag{1}$$
is true for a set of operators $A_1,A_2,…,A_n$ such that $[[A_i,A_j],A_k]=0$ for all $(i,j,k)$. I am having troubles obtaining the second exponent in the RHS of the equation above.
In my attempt am using the Baker Campbell Hausdorff (BCH) Theorem, which states that
$$ e^{A_i}e^{A_j}=e^{A_i+A_j}e^{\frac{1}{2}[A_i,A_j]}$$
is valid for operators satisftyng $[[A_i,A_j],A_j]=[[A_i,A_j],A_i]=0$.
We can then write
$$e^{A_n}e^{A_{n-1}}…e^{A_3}e^{A_2}e^{A_1}= e^{A_n}e^{A_{n-1}}…e^{A_3}e^{A_2+A_1}e^{\frac{1}{2}[A_2,A_1]}\\\hspace{60mm}= e^{A_n}e^{A_{n-1}}…e^{A_3+A_2+A_1}e^{\frac{1}{2}[A_3,A_2+A_1]}e^{\frac{1}{2}[A_2,A_1]}\\ \hspace{84mm}= e^{A_n}…e^{A_4+A_3+A_2+A_1}e^{\frac{1}{2}[A_4,A_3+A_2+A_1]}e^{\frac{1}{2}[A_3,A_2+A_1]}e^{\frac{1}{2}[A_2,A_1]}\tag{2}$$
Then, I want to show that
$$ e^{\frac{1}{2}[A_4,A_3+A_2+A_1]}e^{\frac{1}{2}[A_3,A_2+A_1]}e^{\frac{1}{2}[A_2,A_1]}=e^{\frac{1}{2}([A_4,A_3+A_2+A_1]+[A_3,A_2+A_1]+[A_2,A_1])} $$
and so on. Which am assuming is only true if terms like $[A_4,A_3]$ commute with $[A_2,A_1]$. Is this assumption correct?
Assuming is correct, then how do I show they commute by using our initial condition that $[[A_i,A_j],A_k]=0$ for all $(i,j,k)$? I have tried, but obtain the following
$$[[A_4,A_3],[A_2,A_1]]=A_4A_3A_2A_1-A_4A_3A_1A_2-A_3A_4A_2A_1+A_3A_4A_1A_2 \\ -A_2A_1A_4A_3+A_2A_1A_3A_4-A_1A_2A_4A_3+A_1A_2A_3A_4 \tag{3}$$
but I haven't found a way to regroup it and show this is $0$. Is there an easier way to show Eq. $(1)$ is true? Or to show that Eq. $(3)$ is zero?
Best Answer
You have that every commutator of As commutes with all As, (Lie algebra elements). You wish to probe whether $[[A_4,A_3],[A_2,A_1]]=0$.
Consider $[A_4 A_3, [A_2,A_1]]= A_4[A_3, [A_2,A_1]] + [A_4 , [A_2,A_1]] A_3 = 0+ 0=0$, by your postulate. So, subtracting $[A_3 A_4, [A_2,A_1]]=0$, likewise null, you derive $[[A_4,A_3],[A_2,A_1]]=0$.