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Algebra of displacement operator

  1. Oct 23, 2016 #1
    1. The problem statement, all variables and given/known data

    Given an operator ##D(\alpha)=\exp\ (\alpha a^{\dagger}-\alpha^{*}a)## and a function ##g(a,a^{\dagger})##, where ##a## and ##a^{\dagger}## are operators and ##\alpha## and ##\alpha^{*}## are complex numbers, show that

    ##D^{-1}(\alpha)g(a,a^{\dagger})D(\alpha)=g(a+\alpha, a^{\dagger}+\alpha^{*})##

    2. Relevant equations

    3. The attempt at a solution

    ##D^{-1}(\alpha)g(a,a^{\dagger})D(\alpha)##

    ##=\exp\ (-\alpha a^{\dagger}+\alpha^{*}a)\ g(a,a^{\dagger})\ \exp\ (\alpha a^{\dagger}-\alpha^{*}a)##

    ##=(1-\alpha a^{\dagger}+\alpha^{*}a)\ g(a,a^{\dagger})\ (1+\alpha a^{\dagger}-\alpha^{*}a)##

    ##=g(a,a^{\dagger})+\alpha^{*}[a,g(a,a^{\dagger})]-\alpha[a^{\dagger},g(a,a^{\dagger})]##.

    On the other hand,

    ##g(a+\alpha, a^{\dagger}+\alpha^{*})##

    ##=g(a, a^{\dagger}) + \alpha \frac{\partial g(a,a^{\dagger})}{\partial \alpha} + \alpha^{*} \frac{\partial g(a,a^{\dagger})}{\partial \alpha^{*}}##.

    This seems to suggest that

    ##\frac{\partial g(a,a^{\dagger})}{\partial \alpha^{*}}=[a,g(a,a^{\dagger})]##

    and

    ##\frac{\partial g(a,a^{\dagger})}{\partial \alpha}=-[a^{\dagger},g(a,a^{\dagger})]##.

    Am I missing something here?
     
  2. jcsd
  3. Oct 23, 2016 #2
    You can't expand the operators like that - this assumes that ##\alpha## is infinitesimal, which is not true.
    First, establish how ##a## and ##a^{\dagger}## transform under the action of the displacement operator i.e. what are ##D(\alpha)^{-1} a D(\alpha)## and ##D(\alpha)^{-1} a^{\dagger} D(\alpha)##. Then insert factors of ##D(\alpha) D^{-1} (\alpha)## between each pair of creation/annihilation operators in ##g(a,a^{\dagger})##.
     
  4. Oct 23, 2016 #3
    But ##g(\alpha,\alpha^{\dagger})## is a generic function, and might not have pairings of ##a## and ##a^{\dagger}##.
     
  5. Oct 23, 2016 #4
    I think its fair to assume that any reasonably well behaved function of the operators can be expressed as a sum of operator products. After all, most elementary functions of operators (eg. exponential) are defined in terms of their power series expansions.
     
  6. Oct 23, 2016 #5
    But, the exponential ##\exp(a)## of an operator ##a## is

    ##\exp(a)=1+a+\frac{1}{2!}a^{2}+\frac{1}{3!}a^{3}+\dots##

    So, this is not a sum of operator products.
     
  7. Oct 23, 2016 #6
    What I meant by operator products is any term of the form ##a^{n}##, ##\left(a^{\dagger}\right)^{m}##, ##\left(a^{\dagger}\right)^{m}a^{n}##, ##\left(a^{\dagger}\right)^{m}a^{n} \left(a^{\dagger}\right)^{k} ##, and so on so forth.
     
  8. Oct 23, 2016 #7
    Okay, so I have

    ##D(\alpha)^{-1}aD(\alpha) = \exp(-\alpha a^{*}+\alpha^{*}a)a\exp(\alpha a^{*}-\alpha^{*}a)##

    Do I expand the exponential ##\exp(-\alpha a^{*}+\alpha^{*}a)## and use ##[a,a^{\dagger}]=1##?
     
  9. Oct 23, 2016 #8
    Yup, or you can just use the Baker-Hausdorff- Campbell formula (if you don't want to prove it)
     
  10. Oct 23, 2016 #9
    So, I get ##\exp(-\alpha a^{*}+\alpha^{*}a)=\exp(-\alpha a^{*})\ \exp(\alpha^{*}a)## using BCH, but I still have to expand ##\exp(-\alpha a^{*})## to find out how it commutes with ##a##, right?
     
    Last edited: Oct 23, 2016
  11. Oct 24, 2016 #10
    I was referring to this particular form of the BCH lemma:
    [tex]e^{iG\lambda}A e^{-iG\lambda} = A + i\lambda [G,A] + \frac{\left(i\lambda\right)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\underbrace{[G,[G,[G,\ldots[G}_{n\ times},A]]]\ldots]+\ldots[/tex]
    Also, your result isn't right; you're missing a factor of ##e^{-|\alpha|^2/2}##.
     
  12. Oct 30, 2016 #11
    Ok, so, using the Baker-Campbell-Hausdorff lemma

    $$e^{iG\lambda}A e^{-iG\lambda} = A + i\lambda [G,A] + \frac{\left(i\lambda\right)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\underbrace{[G,[G,[G,\ldots[G}_{n\ \text{times}},A]]]\ldots]+\ldots,$$

    we have

    ##D(\alpha)^{-1}aD(\alpha)##

    ##= \exp(-\alpha a^{\dagger}+\alpha^{*}a)a\exp(\alpha a^{\dagger}-\alpha^{*}a)##

    ##= a + [-\alpha a^{\dagger}+\alpha^{*}a,a]##

    ##= a- [\alpha a^{\dagger},a] + [\alpha^{*}a,a]##

    ##= a- \alpha[a^{\dagger},a] + \alpha^{*}[a,a]##

    ##= a- \alpha[a^{\dagger},a]##

    ##= a+ i\alpha##.

    Am I correct?
     
  13. Oct 30, 2016 #12
    Almost; there shouldn't be a factor of ##i##. The commutator is just ##-1##. Now just repeat the same for ##a^{\dagger}##.
     
  14. Oct 30, 2016 #13
    I did, and I got the desired answer.

    I was wondering if the formula you quoted is a different version of the BCH formula.
     
  15. Oct 31, 2016 #14
    It has the same origin really - its just convenient because this particular form (i.e. transformations) appears very frequently in quantum mechanics. I believe it's discussed in most standard texts such as Sakurai if you want more specific details.
     
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