# Exponential rules

1. Jul 24, 2006

### Logarythmic

Can someone give me a hint on how to prove that

exp[A+B]=exp[A]expexp[-c/2]

where A and B are two operators such that [A,B]=c, where c is a complex number.

This is not homework or something, I'm just curious when reading the rules.

2. Jul 25, 2006

### nazzard

Hello Logarythmic,

the identity would also be true if [A,B] commutes with both A and B. c doesn't necessarily have to be a complex number in the first place.

A common proof starts with defining:

$$f(x)=e^{Ax}\,e^{Bx}$$

Calculating $\frac{df}{dx}$ (*) will lead to a first-order differential equation. After finding the solution, f(1) will give the identity.

(*)hint: if [A,B] commutes with A and B then:

$$e^{-Bx}\,A\,e^{Bx}=A+x[A,B]$$

Regards,

nazzard

3. Jul 25, 2006

### Logarythmic

Thanks for your answer but I don't get it at all. When calculating the derivative of f(x), should I keep in mind that the operators could depend on x? If A and B commutes, then I could prove it just like with ordinary numbers, right? So suppose that A and B do not commute. Calculating df/dx and solving the differential equation just gives me the originally defined f(x)...? Isn't the easiest way to prove it by using the series expansion of e^A?

4. Jul 25, 2006

### nazzard

Treat the operators A and B as if they would not depend on x:

$$\frac{d}{dx}(e^{Ax}\,e^{Bx})=e^{Ax}\,A\,e^{Bx}+e^{Ax}\,e^{Bx}\,B$$

$$=e^{Ax}\,e^{Bx}\,e^{-Bx}\,A\,e^{Bx}+e^{Ax}\,e^{Bx}\,B$$

Now you can factor out f(x) and apply (*).

Last edited: Jul 25, 2006
5. Jul 29, 2006

### Logarythmic

I solved it. Thanks. =)