- #1

LagrangeEuler

- 717

- 20

- Homework 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.