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Exp operator expansions

  1. Oct 19, 2006 #1
    Does anyone of an exp operator expansion identity that for operators of the form:

    [tex]
    \exp \left ( \hat{A} \hat{B} \right )
    [\tex]

    For example, I know of the BHC decomposition:

    [tex]
    e^{\hat{A} + \hat{B}} = e^{-\frac{ [ \hat{A}, \hat{B} ]}{2}} e^{\hat{A}} e^{\hat{B}}
    [\tex]

    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.

    Cheers.
     
    Last edited by a moderator: Oct 19, 2006
  2. jcsd
  3. Oct 19, 2006 #2
    If they commute, then yes, each term in the series is just [tex]A^n B^n[/tex]. Otherwise you have to just calculate a LOT of commutators and hope something nice works out. What specific problem were you trying to tackle?
     
  4. Oct 19, 2006 #3
    I am trying to decompose the Cross Kerr unitray transformation i.e. the unitary operator:

    [itex]\exp \left ( i \varphi \hat{n}_{1} \hat{n}_{2} \right )[/itex]

    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?
     
  5. Oct 20, 2006 #4
    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.
     
  6. Oct 20, 2006 #5

    selfAdjoint

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

    and knew that BCH didn't handle the case he was asking about.
     
  7. Oct 20, 2006 #6
    Could you use the operator product expansion, and then the BCH formula?
     
  8. Oct 24, 2006 #7
    I don't see how. The BHC formula states that:

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

    when [TEX][\hat{A},[\hat{A},\hat{B}]] = [\hat{B},[\hat{A},\hat{B}]] [/TEX]

    i.e. it refers to the sum of operators in an expotential. I want to decompose a product of (commuting operators).
     
  9. Oct 24, 2006 #8

    reilly

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    Science Advisor

    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
     
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