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

[humanist rho]

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

Homework Statement

Prove that e^A.e^B = 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 ?

 

[CompuChip]

When you work out the brackets on the left hand side, you will get products like Aⁿ B^m, 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?

 

[humanist rho]

Yeah,Now I got it.
Thank you.

 

[LuisVela]

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

However, if I start with exp(a+b), I go like this:$$exp(a+b)=\sum_{n=0}^{\infty}{\frac{(a+b)^n}{n!}} = <br /> \sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{1}{n!}\binom{n}{k}a^k\cdot b^{n-k} =<br /> \sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{a^k}{k!}\frac{b^{n-k}}{(n-k)!}$$

-------------------------------------------------------------
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...

 

[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.

 

[berkeman]

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.

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