# Confused about the B-C-H equation

• I
• dyn

#### 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

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.

Does this equation imply that eA+B ≠ eB+A ?
Thanks
No, the two are equal.

Demystifier
No, the two are equal.
My bad you are right, I misread that as eA eB ≠ eA+B.

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

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 ?
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}

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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 ?

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.

dyn
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...

dyn
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.

dyn