# I Confused about the B-C-H equation

1. May 11, 2016

### dyn

Hi. I have come across the following equation

eA eB = eA+B e [A , B]/2

where A and B are operators.
Is this the B-C-H formula or is it a version that only applies under specific conditions on the operators ?

Does this equation imply that eA+B ≠ eB+A ?

Thanks

2. May 11, 2016

### blue_leaf77

It only applies when $[A,[A,B]] = [B,[A,B]]=0$.
Generally yes, they are not equal.

3. May 12, 2016

### DrDu

No, the two are equal.

4. May 12, 2016

### blue_leaf77

My bad you are right, I misread that as eA eB ≠ eA+B.

5. May 12, 2016

### dyn

So this condition implies [ A , B ] = complex constant ?

This means eA+B = e-[A,B]/2 eA eB and eB+A = e-[B,A]/2 eB eA

How do we know these 2 expressions are the same as A and B do not necessarily commute ?

6. May 12, 2016

### blue_leaf77

Without parameter they indeed may seem different, but once you give a parameter, say $x$, and expand into Taylor series, their identicality will unveil.
Let $A \to Ax$ and $B \to Bx$. Then we have
1) For $e^{Ax+Bx}$:
\begin{aligned} e^{Ax+Bx} &= e^{-[A,B]x^2/2}e^{Ax}e^{Bx} = 1 + x(A+B) + x^2\left( AB -\frac{1}{2}[A,B] + \frac{A^2+B^2}{2}\right) + \ldots \\ &= 1 + x(A+B) + x^2\frac{ (A + B)^2}{2} + \ldots \end{aligned}
2) For $e^{Bx+Ax}$:
\begin{aligned} e^{Bx+Ax} &= e^{-[B,A]x^2/2}e^{Bx}e^{Ax} = 1 + x(B+A) + x^2\left( BA -\frac{1}{2}[B,A] + \frac{A^2+B^2}{2} \right) + \ldots \\ &= 1 + x(A+B) + x^2\frac{ (A + B)^2}{2} + \ldots \end{aligned}

Last edited: May 14, 2016
7. May 13, 2016

### vanhees71

8. May 14, 2016

### dyn

Thanks for your reply. Even though it wouldn't make any difference between A+B and B+A should the above expressions include in the powers of the x2 term the factors A2 + B2 ?

9. May 14, 2016

### blue_leaf77

You are right, I have corrected it.

10. May 15, 2016

### ChrisVer

No it does not...
It implies exactly what you wrote; that you can't combine exponentials with operators/matrices in their arguments in the same way you do with just numbers [which commute].

Not necessarily; you can have [A,B]= K which K commutes with both A and B....One such K could be the equivalent to a complex constant for a matrix, that is a scaled unit-matrix $[A,B]= \alpha I_{n\times n}$, or it may be something else...

11. May 15, 2016

### A. Neumaier

It is the BCH formula for the special case where $[A,B]$ commutes with both $A$ and $B$. In general, the BCH formula has the form $e^Ae^B= e^{A\oplus B}$ where $A\oplus B=A+B+[A , B]/2+\ldots$ is an infinite series where the dots stand for multiple commutators. (Notation in wikipedia: $X\oplus Y=Z(X,Y)$.)

Under the condition given, all terms but the ones given vanish, and one gets $e^Ae^B= e^{A+B+[A , B]/2}$, which gives the right hand side in post #1 since $A+B$ and $[A , B]/2$ commute.