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 6
 Problem Statement

I want to calculate [tex]D(\beta)bD(\beta)[/tex]
where b is ladder operator in problem of LHO and ##\beta## is complex number.
[tex]D(\beta)=exp(\beta b^{\dagger}\beta^*b)[/tex]
 Relevant Equations

Commutator [tex][b,b^{\dagger}]=I[/tex]
BCH formula:
[tex]e^Ae^B=exp(A+B+\frac{1}{2}[A,B]+\frac{1}{12}[A,[A,B]]+\frac{1}{12}[B,[A,B]]+...)[/tex]
From
[tex][b,b^{\dagger}]=I[/tex] it is easy to see that
[tex][b,[b,b^{\dagger}]]=[b^{\dagger},[b,b^{\dagger}]]=0. [/tex]
And from that
[tex]e^be^{b^{\dagger}}=e^{b+b^{\dagger}}e^{\frac{1}{2}I}[/tex]
Using that I get
[tex]exp(\beta b^{\dagger}\beta^*b)=e^{\beta b^{\dagger}}e^{\beta^*b}e^{\frac{1}{2}\beta^2I}[/tex].
So
[tex]D(\beta)bD(\beta)=e^{\beta b^{\dagger}}e^{\beta^*b}e^{\frac{1}{2}\beta^2I}be^{\beta b^{\dagger}}e^{\beta^*b}e^{\frac{1}{2}\beta^2I}[/tex]
and I do not see what to do from here.
[tex][b,b^{\dagger}]=I[/tex] it is easy to see that
[tex][b,[b,b^{\dagger}]]=[b^{\dagger},[b,b^{\dagger}]]=0. [/tex]
And from that
[tex]e^be^{b^{\dagger}}=e^{b+b^{\dagger}}e^{\frac{1}{2}I}[/tex]
Using that I get
[tex]exp(\beta b^{\dagger}\beta^*b)=e^{\beta b^{\dagger}}e^{\beta^*b}e^{\frac{1}{2}\beta^2I}[/tex].
So
[tex]D(\beta)bD(\beta)=e^{\beta b^{\dagger}}e^{\beta^*b}e^{\frac{1}{2}\beta^2I}be^{\beta b^{\dagger}}e^{\beta^*b}e^{\frac{1}{2}\beta^2I}[/tex]
and I do not see what to do from here.