# Prove that exp[A].exp = exp[A+B] only if A and B commute.

1. Sep 19, 2011

### humanist rho

Prove that exp[A].exp = exp[A+B] only if A and B commute.

1. The problem statement, all variables and given/known data

Prove that eA.eB = e(A+B) only if A and B commute.

The attempt at a solution

I expanded both sides of the equation.

(1+A+A2/2!...)(1+B+B2/2!+..) = (1+(A+B)+(A+B)2/2!+..)

Now how to proceed ?

2. Sep 19, 2011

### CompuChip

Re: Prove that exp[A].exp = exp[A+B] only if A and B commute.

When you work out the brackets on the left hand side, you will get products like An Bm, where all the A's are to the left of the B's.

Now try to work out (A + B)n (e.g., start with (A + B)2). What do you need to get this in the same form?

3. Sep 20, 2011

### humanist rho

Re: Prove that exp[A].exp = exp[A+B] only if A and B commute.

Yeah,Now I got it.
Thank you.

4. Dec 9, 2011

### LuisVela

Re: Prove that exp[A].exp = exp[A+B] only if A and B commute.

Hi guys.
I was reading some QM and they mentioned exp(a+b)=exp(a)*exp(b) only if [a,b]=0; so I thought to go to the series definition of exp(x) and work it out by myself.

I do understand where the problem arises if suddenly a and b don't commute anymore, but thats not my problem. I want to write down the proof for exp(a+b)=exp(a)*exp(b) for "typical" a and b. iykwim

So, I came across your post and I would like some further details of how to preoceed in the proof.
this is what I have so far:

If:
$$exp(x)=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$

Then, for exp(a)*exp(b), we have:

$$exp(a)*exp(b)=\sum_{n=0}^{\infty} \sum_{m=0}^{\infty}{\frac{a^n}{n!}}{\frac{b^m}{m!}}$$

$$exp(a+b)=\sum_{n=0}^{\infty}{\frac{(a+b)^n}{n!}} = \sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{1}{n!}\binom{n}{k}a^k\cdot b^{n-k} = \sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{a^k}{k!}\frac{b^{n-k}}{(n-k)!}$$

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Summarizing, I get on one hand:
$$\sum_{n=0}^{\infty} \sum_{m=0}^{\infty}{\frac{a^n}{n!}}{\frac{b^m}{m!}}$$
While on the other hand I get:
$$\sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{a^k}{k!}\frac{b^{n-k}}{(n-k)!}$$

The two expressions look alike, but I can't put them in the very exact form. Could you help me here?

How should I proceed? I have a feeling that some index substitution is the answer, but I haven't figured out which one...

5. Sep 25, 2013

### jjalonsoc

You are quite there. If you define n-k=l, then you obtain only the sum over n of expressions of k and l with the condition that k+l=n for k,l>=0. This sum can then be written as two sums for k and l from 0 to infinity in both cases.

6. Sep 25, 2013

### Staff: Mentor

This homework thread is 2 years old, so the OP is probably not working on the problem any more...