BCH Formula Example: Solving with [b,b^{\dagger}]=I

[LagrangeEuler]

Homework Statement

I want to calculate $$D(\beta)bD(-\beta)$$
where b is ladder operator in problem of LHO and $\beta$ is complex number.
$$D(\beta)=exp(\beta b^{\dagger}-\beta^*b)$$

Relevant Equations

Commutator $$[b,b^{\dagger}]=I$$
BCH formula:
$$e^Ae^B=exp(A+B+\frac{1}{2}[A,B]+\frac{1}{12}[A,[A,B]]+\frac{1}{12}[B,[A,B]]+...)$$

From
$$[b,b^{\dagger}]=I$$ it is easy to see that
$$[b,[b,b^{\dagger}]]=[b^{\dagger},[b,b^{\dagger}]]=0.$$
And from that
$$e^be^{b^{\dagger}}=e^{b+b^{\dagger}}e^{\frac{1}{2}I}$$
Using that I get
$$exp(\beta b^{\dagger}-\beta^*b)=e^{\beta b^{\dagger}}e^{-\beta^*b}e^{-\frac{1}{2}|\beta|^2I}$$.
So
$$D(\beta)bD(-\beta)=e^{\beta b^{\dagger}}e^{-\beta^*b}e^{-\frac{1}{2}|\beta|^2I}be^{-\beta b^{\dagger}}e^{\beta^*b}e^{-\frac{1}{2}|\beta|^2I}$$
and I do not see what to do from here.

 

[fresh_42]

For my taste these notations are far too sloppy, esp. the fact that you do not distinguish $b$ and $\mathfrak{ad}b$. So assuming you made no mistake, you have
Correction:
\begin{align*}
D(\beta)bD(-\beta) &= (\exp(\beta b^\dagger)(\exp(-\beta^*b)b(\exp(\beta b)(\exp(-\beta b^\dagger)\cdot e^{-|\beta|^2}\\ &= e^{-|\beta|^2} \cdot \operatorname{Ad}(\exp(\beta b^\dagger)\cdot (\exp(-\beta^*b) ) (b)
\end{align*}
since $e^I$ commutes with everything and the rest is the conjugation by $\exp( \beta b^\dagger)$ and $\exp (- \beta^* b)$.

 

[LagrangeEuler]

What is $Ad()$?

 

[fresh_42]

What is $Ad()$?

The adjoint representation of the Lie group on its tangent space. It is a conjugation: $\exp(-\beta b^*) \stackrel{ \operatorname{Ad}}{\longmapsto} \left( b \longmapsto \exp(-\beta b^*)b\exp(\beta b^*)\right)$

 

[LagrangeEuler]

Correct result should be
$$D(\beta)bD(-\beta)=b-\beta I$$. And I am still not sure how to get this.

 

[fresh_42]

Correct result should be
$$D(\beta)bD(-\beta)=b-\beta I$$. And I am still not sure how to get this.

Why not calculate it directly without BCH?
\begin{align*}
D(\beta)bD(-\beta)&= \exp(\beta b^\dagger - \beta^* b)b \exp(-(\beta b^\dagger - \beta^* b)) \\
&= \operatorname{Ad}(\exp(\beta b^\dagger - \beta^* b))(b) \\
&= \exp(\operatorname{ad}(\beta b^\dagger - \beta^* b))(b) \\
&= \left( 1 + \operatorname{ad}(\beta b^\dagger - \beta^* b) + \dfrac{1}{2!} \operatorname{ad}^2(\beta b^\dagger - \beta^* b)+\ldots \right)(b)\\
&= b+ [\beta b^\dagger - \beta^* b,b] + \dfrac{1}{2} [\beta b^\dagger - \beta^* b,[\beta b^\dagger - \beta^* b,b]] + \dfrac{1}{3!} [\beta b^\dagger - \beta^* bv,[\beta b^\dagger - \beta^* b,[\beta b^\dagger - \beta^* b,b]]]+\ldots
\end{align*}

 

[LagrangeEuler]

Thanks. You will obviously get good result in this way. However I am not familiar with adjoint representation. Is it possible to solve it without it?

 

[fresh_42]

Thanks. You will obviously get good result in this way. However I am not familiar with adjoint representation. Is it possible to solve it without it?

Without it in physics means nothing else than drop the notation. Physicists tend to use the same words regardless whether it takes places on the manifold (Lie group) or its tangent spaces (Lie algebra). If you're lucky they call the tangents generators. So without it means nothing else as to jump from line 1 directly to line 5.

Here is what formally happens (at the example of $SU(2)$):
https://www.physicsforums.com/insights/representations-precision-important/
Btw. you can still use BCH. Just use it for the last line.

 

[George Jones]

Problem Statement: I want to calculate $$D(\beta)bD(-\beta)$$
where b is ladder operator in problem of LHO and $\beta$ is complex number.
$$D(\beta)=exp(\beta b^{\dagger}-\beta^*b)$$
Relevant Equations: Commutator $$[b,b^{\dagger}]=I$$

Writing $A = \beta b^\dagger - \beta^* b$ gives
$$\begin{align}
D\left(\beta\right)bD\left(-\beta\right) &= e^A b e^{-A} \nonumber \\
&= \left( \left[ e^A , b \right] + b e^A \right) e^{-A} \nonumber
\end{align} $$
Now, Calculate $\left[ e^A , b \right]$ using
$$\begin{align}
\left[ A^n , b \right] &= \left[ A A^{n-1} , b \right] \nonumber \\
&= A \left[ A^{n-1} , b \right] + \left[ A , b \right] A^{n-1} \nonumber
\end{align} $$
and mathematical induction.