Confused about the B-C-H equation

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  • #1
dyn
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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
 

Answers and Replies

  • #2
blue_leaf77
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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.
 
  • #3
DrDu
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Does this equation imply that eA+B ≠ eB+A ?
Thanks
No, the two are equal.
 
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  • #4
blue_leaf77
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No, the two are equal.
My bad you are right, I misread that as eA eB ≠ eA+B.
 
  • #5
dyn
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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 ?
 
  • #6
blue_leaf77
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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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  • #8
dyn
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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 ?
 
  • #9
blue_leaf77
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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.
 
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  • #10
ChrisVer
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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 [itex][A,B]= \alpha I_{n\times n}[/itex], or it may be something else...
 
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  • #11
A. Neumaier
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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.
 
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