Simplified Hausdorf-Campbell formula

  • Thread starter paweld
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  • #1
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The following formula is very useful in QM (it's simplifed version
of Hausdorf-Campbell formula):
[tex]
\exp(X+Y) = \exp(-[X,Y]/2)\exp(X) \exp(Y)
[/tex]
It holds for any operator X, Y which commute with their commutator (i.e.
[tex] [X,[X,Y]]= [Y,[X,Y]] = 0[/tex]).
I look for a simple proof of this fact. Do you have any idea.

I also wonder if this formula is correct (for X,Y as before
such that [tex] [X,[X,Y]]= [Y,[X,Y]] = 0[/tex]):
[tex]
\exp(X) \exp(Y) = \exp([X,Y]) \exp(Y) \exp(X)
[/tex]

Thanks for help.
 

Answers and Replies

  • #2
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4
The original Baker-Hausdorff-Campbell formula reads as

[tex]\exp(x)\exp(y)=\exp(x+y+BCH(x,y))[/tex]

where

[tex]BCH(x,y)=\frac12[x,y]+\frac{1}{12}[x,[x,y]]-\frac{1}{12}[y,[x,y]]-\frac{1}{24}[x,[y,[x,y]]]+\ldots[/tex]

Now, with your assumptions x,y and thus x+y commute with the commutator. Therefore BCH reduces to just the first term that commutes with x+y. Therefore you can move it to the LHS and you get what you are looking for.

Your second formula follows by writing [tex]\exp (y)\exp(x)[/tex] the same way and comparing the two expressions.
 
Last edited:
  • #3
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Thanks for answer. I hadn't noticed that the second formula follows directly
from the first one.

The proof of original Baker-Hausdorff-Campbell is quite difficult so I want to find
a proof of the weaker version (stated before). Do you have any idea?
 
  • #4
Are X and Y supposed to be matrices, this point has always confused me.
 
  • #5
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X and Y are bounded operators in some Banach space (i.e. they can be for
example matricies).
 
  • #6
1,444
4
Are X and Y supposed to be matrices, this point has always confused me.
They are supposed to be linear operators acting on a vector space, finite or infinite-dimensional. Of course in the latter case one has to worry about their domains of definition and convergence.
 
  • #7
1,444
4
The proof of original Baker-Hausdorff-Campbell is quite difficult so I want to find
a proof of the weaker version (stated before). Do you have any idea?
I think I have seen the proof. Will try to remember.

OK, I think it is standard. You start with

[tex]e^XYe^{-X}=Y+[X,Y]+0[/tex]

because [X,[X,Y]] and higher commutators vanish. From this

[tex]e^XY^ne^{-X}=(Y+[X,Y])^n[/tex]

Therefore

[tex]e^Xe^Ye^{-X}=\exp(e^XYe^{-X})=\exp(Y+[X,Y])=e^Ye^{[X,Y]}[/tex]

From there you drive home.
 
Last edited:

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