# Why is this identity an if, rather than if and only if?

A recent question (Product of exponential of operators) asked who to proved that the exponentials of operators multiply in same manner as those of scalars if and only if the commutator of the operators is zero. As nervxxx's answer there points out, the only if part of that theorem isn't true. This is clearly correct, as nervxxx gives an example of two operators where $[A, B] \neq 0$, but where $e^{A+B} = e^A e^B = I$.

I remain confused by one point. nervxxx gives a proof of this theorem using a differential equation approach. I understand that approach. I can see clearly that it relies on the operators having well-defined inverses. The counterexample works because neither of those operators is invertible. What confuses me is that I have also worked out this derivation by expanding the exponentials in their Taylor series, and matching terms of equal order in each operator. When you do this, the $I, A,$ and $B$ terms all cancel out regardless. Equating the terms containing $AB$ and $BA$ gives the condition for the commutator. When I try to equate higher order terms, I don't get any new information. As far as I can tell, every step in that derivation can be reversed, so I should be getting an if and only if for the original proposition. But that's clearly false. So what's going wrong?

I can think of two things. First, there might be an implicit assumption that the operators are invertible. I don't see where that would be, but I might be overlooking it. Second, it could be the expansion and term-by-term equating of the infinite series. Just because they are equal term-by-term doesn't mean they are equal. That doesn't make a whole lot of sense to me either, unless I am missing something really basic about infinite series.

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I do kind of wonder if this wouldn't be more appropriate on Mathematics? –  David Z Mar 25 '13 at 23:44
@DavidZaslavsky I think this would be more appropriate in math. –  joshphysics Mar 26 '13 at 15:40
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## 3 Answers

Comments to the question (v1):

1. OP wrote: The counterexample works because neither of those operators are invertible. No, invertibility of $A$ and $B$ is irrelevant. Consider e.g. $2\times 2$ Pauli matrices $\sigma^i$ and define $$\sigma^{\pm}~:=~\frac{\sigma^1\pm i \sigma^2}{2}.$$ Then it is easy to check that $$A~=~\alpha {\bf 1}_{2 \times 2} +2\pi i \sigma^3~=~\begin{bmatrix}\alpha+2\pi i& 0\\ 0&\alpha-2\pi i \end{bmatrix}$$ and $$B~=~ \beta{\bf 1}_{2 \times 2} +\gamma \sigma^+ ~=~\begin{bmatrix}\beta& \gamma\\ 0&\beta \end{bmatrix},$$ satisfy $e^Ae^B=e^{A+B}$ for arbitrary complex numbers $\alpha, \beta, \gamma \in \mathbb{C}$. Moreover $0\neq[A,B]= 4\pi i \gamma \sigma^+$ iff $\gamma\neq 0$.

2. Let us now assume $e^Ae^B=e^{A+B}$, and ponder if it is possible to deduce that $[A,B]=0$? We of course already know that the answer is No. We expand $e^{t(A+B)}e^{-tA}e^{-tB}$ as a Taylor series in $t$. The lowest obstruction in $t$ is precisely the commutator $[A,B]$, and OP essentially wonders why this doesn't imply $[A,B]=0$? The problem is that we only have information about that series for one single point $t=1$. There is no perturbative sense in which we can say that some conditions are of lower order than other conditions, and that they should be solved first. We only have one big non-perturbative condition $e^{A+B}e^{-A}e^{-B}={\bf 1}$. So there is no apparent contradiction.

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Let $A,B\in M_k(\mathbb C)$ be given. You want to show that that if $$e^{A+B} = e^Ae^B$$ then $AB = BA$ (or rather that this cannot be shown). Now, if for example you knew instead that $$e^{\alpha (A + B)} = e^{\alpha A}e^{\alpha B}$$ for all $\alpha\in\mathbb R$, then it would follow that you could equate coefficients order-by-order in $\alpha$ on each side, which would give the erroneous "only if." Otherwise, its not clear to me how you are matching terms "of the same order."

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Doesn't $e^{\alpha A + \alpha B} = e^{\alpha A} e^{\alpha B}$ for all $\alpha \in \mathbb{R}$ also imply that $A$ and $B$ commute, by matching terms of the same order? –  Peter Shor Mar 25 '13 at 21:16
@PeterShor You're right I don't know why I made it more complicated than it needed to be; I switched it for simplicity; thanks. –  joshphysics Mar 25 '13 at 21:22
@PeterShor I guess it's one thing to say for all $a$ vs. for the specific case $a=1$ which was nervxxx's example. –  santa claus Mar 25 '13 at 21:23
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I believe that you made the (wrong) implicit assumption that \begin{align} e^{A+B} = e^A e^B \implies e^{t(A+B)} = e^{tA} e^{tB}, \end{align} $\forall t \in \mathbb{C}$, for any complex matrices $A,B$. (I haven't been able to disprove this statement). But if this statement was true, then you could, as Alec S and Qmechanic have pointed out, just look at the second order term in the expansion of $e^{t(A+B)}$ and set it to 0, i.e. set $[A,B] = 0$, because each coefficient of $t^n$ in the expansion must be 0.

I conjecture that this statement is false, hence we only have the $t = 1$ point and thus cannot just look at it 'term by term'. Someone prove/disprove it...? :)

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