Discover the Exp Operator Expansion Identity for Form-Based Operators

[MaverickMenzies]

Does anyone of an exp operator expansion identity that for operators of the form:

[tex] \exp \left ( \hat{A} \hat{B} \right )<br /> [\tex]<br /> <br /> For example, I know of the BHC decomposition:<br /> <br /> $$e^{\hat{A} + \hat{B}} = e^{-\frac{ [ \hat{A}, \hat{B} ]}{2}} e^{\hat{A}} e^{\hat{B}}<br /> [\tex]<br /> <br /> I've heard that there exists such an expansion for the first case but i can't find it anywhere. I'd be grateful if anyone could supply me a reference or at least point me to it.<br /> <br /> Cheers.$$[/tex]

 

[StatMechGuy]

If they commute, then yes, each term in the series is just $$A^n B^n$$. Otherwise you have to just calculate a LOT of commutators and hope something nice works out. What specific problem were you trying to tackle?

 

[MaverickMenzies]

I am trying to decompose the Cross Kerr unitray transformation i.e. the unitary operator:

$\exp \left ( i \varphi \hat{n}_{1} \hat{n}_{2} \right )$

Where the operators in question are photon number operators in two different radiation modes.

I understand that there exists a decomposition (besides the taylor series expansion) of this operator using the Hubbard-Stratonovich decomposition. However, this decomposition requires that the creation and annihilation operators are fermomic. I, however, need a corresponding decomposition for bosonic operators.

Any suggestions?

 

[Epicurus]

The formula for exponential operator expansions of this sort is called the Baker-Campbell-Haussdorf formula which holds regradless if the operators are bosonic or fermionic and is a result of the mentioned taylor series expansion.

 

[selfAdjoint]

The formula for exponential operator expansions of this sort is called the Baker-Campbell-Haussdorf formula which holds regradless if the operators are bosonic or fermionic and is a result of the mentioned taylor series expansion.

Too quick off the mark, Epicurus. The OP already knew that:

For example, I know of the BHC decomposition:

and knew that BCH didn't handle the case he was asking about.

 

[Severian]

Could you use the operator product expansion, and then the BCH formula?

 

[MaverickMenzies]

Could you use the operator product expansion, and then the BCH formula?

I don't see how. The BHC formula states that:

$$e^{\hat{A} + \hat{B}} = e^{- [ \hat{A}, \hat{B} ]/2} e^{\hat{A}} e^{\hat{B}}$$

when $$[\hat{A},[\hat{A},\hat{B}]] = [\hat{B},[\hat{A},\hat{B}]]$$

i.e. it refers to the sum of operators in an expotential. I want to decompose a product of (commuting operators).

 

[reilly]

Yes, BCH, or CBH or whatever you want to call it, states that

exp(A +B) = exp(A)exp(B)exp([A,B]/2),

and this hold only when [A,[A,B]] = [B,[A,B]] = 0.

There's nothing in this theorem that precludes fermion operators.
(See Mandel and Wolf, Optical Coherence and Quantum Optics, p 319-320 for a detailed discussion.)

Regards,
Reilly Atkinson