Confused about the B-C-H equation

[dyn]

Hi. I have come across the following equation

e^A e^B = e^A+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 e^A+B ≠ e^B+A ?

Thanks

 

[blue_leaf77]

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

It only applies when $[A,[A,B]] = [B,[A,B]]=0$.

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

Generally yes, they are not equal.

 

[DrDu]

Does this equation imply that e^A+B ≠ e^B+A ?
Thanks

No, the two are equal.

 

[blue_leaf77]

No, the two are equal.

My bad you are right, I misread that as e^A e^B ≠ e^A+B.

 

[dyn]

It only applies when $[A,[A,B]] = [B,[A,B]]=0$.

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

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

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

 

[blue_leaf77]

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

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

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

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 $e^{Bx+Ax}$:
$$
\begin{aligned}
e^{Bx+Ax} &= e^{-[B,A]x^2/2}e^{Bx}e^{Ax} = 1 + x(B+A) + x^2\left( BA -\frac{1}{2}[B,A] + \frac{A^2+B^2}{2} \right) + \ldots \\
&= 1 + x(A+B) + x^2\frac{ (A + B)^2}{2} + \ldots
\end{aligned}
$$

 

[vanhees71]

For the derivation of the BCH formula and also the here discussed application, see

http://theory.gsi.de/~vanhees/faq/quant/node99.html

It's in German, but the formula density is so high, that it shouldn't be a problem to follow the arguments.

 

[dyn]

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 $e^{Ax+Bx}$:
$$
\begin{aligned}
e^{Ax+Bx} &= e^{-[A,B]x^2/2}e^{Ax}e^{Bx} = 1 + x(A+B) + x^2\left( AB -\frac{1}{2}[A,B] \right) + \ldots \\
&= 1 + x(A+B) + x^2\frac{ AB + BA}{2} + \ldots
\end{aligned}
$$
2) For $e^{Bx+Ax}$:
$$
\begin{aligned}
e^{Bx+Ax} &= e^{-[B,A]x^2/2}e^{Bx}e^{Ax} = 1 + x(B+A) + x^2\left( BA -\frac{1}{2}[B,A] \right) + \ldots \\
&= 1 + x(A+B) + x^2\frac{ AB + BA}{2} + \ldots
\end{aligned}
$$

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 ?

 

[blue_leaf77]

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 ?

You are right, I have corrected it.

 

[ChrisVer]

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

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].

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

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...

 

[A. Neumaier]

Is this the B-C-H formula

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.