Simplified Hausdorf-Campbell formula

[paweld]

The following formula is very useful in QM (it's simplifed version
of Hausdorf-Campbell formula):
$$\exp(X+Y) = \exp(-[X,Y]/2)\exp(X) \exp(Y)$$
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 $$[X,[X,Y]]= [Y,[X,Y]] = 0$$):
$$\exp(X) \exp(Y) = \exp([X,Y]) \exp(Y) \exp(X)$$

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.