Proving Exponential Rules for [A,B]=c

[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.

 

[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

 

[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?

 

[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 (*).

 

[Logarythmic]

I solved it. Thanks. =)