- #1

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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

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- #1

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e

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

Thanks

- #2

blue_leaf77

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- #3

- #4

blue_leaf77

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- #5

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So this condition implies [ A , B ] = complex constant ?It only applies when ##[A,[A,B]] = [B,[A,B]]=0##.

This means e

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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Without parameter they indeed may seem different, but once you give a parameter, say ##x##, and expand into Taylor series, their identicality will unveil.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 ?

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:

- #7

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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.

- #8

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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 x

- #9

blue_leaf77

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You are right, I have corrected it.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 x^{2}term the factors A^{2}+ B^{2}?

- #10

ChrisVer

Gold Member

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No it does not...Does this equation imply that eA+B ≠ eB+A ?

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 [itex][A,B]= \alpha I_{n\times n}[/itex], or it may be something else...So this condition implies [ A , B ] = complex constant ?

- #11

A. Neumaier

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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)##.)Is this the B-C-H formula

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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