Simplified Hausdorf-Campbell formula

[paweld]

The following formula is very useful in QM (it's simplifed version
of Hausdorf-Campbell formula):
<br /> \exp(X+Y) = \exp(-[X,Y]/2)\exp(X) \exp(Y)<br />
It holds for any operator X, Y which commute with their commutator (i.e.
[X,[X,Y]]= [Y,[X,Y]] = 0).
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 [X,[X,Y]]= [Y,[X,Y]] = 0):
<br /> \exp(X) \exp(Y) = \exp([X,Y]) \exp(Y) \exp(X)<br />

Thanks for help.

 

[arkajad]

The original Baker-Hausdorff-Campbell formula reads as

\exp(x)\exp(y)=\exp(x+y+BCH(x,y))

where

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

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 \exp (y)\exp(x) the same way and comparing the two expressions.

 

[paweld]

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?

 

[10001011011]

Are X and Y supposed to be matrices, this point has always confused me.

 

[paweld]

X and Y are bounded operators in some Banach space (i.e. they can be for
example matricies).

 

[arkajad]

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.

 

[arkajad]

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

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

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

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

Therefore

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

From there you drive home.