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

• LagrangeEuler
This results in$$\left[ e^A , b \right] = \sum_{n=0}^{\infty} \dfrac{A^n b - b A^n}{n!}$$Substituting this back into the original equation, we get$$D\left(\beta\right)bD\left(-\beta\right) = e^A b e^{-A} = \left( \sum_{n=0}^{\infty} \dfrac{A^n b - b A^n}{n!} + b e^A \right) e^{-A}$$Using the property that ##e^A e^{-A} = I##, we
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.

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) ##.

Last edited:

LagrangeEuler said:
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)##

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

LagrangeEuler said:
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*}

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?

LagrangeEuler said:
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.

LagrangeEuler said:
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.

## 1. What is the BCH formula?

The BCH (Baker-Campbell-Hausdorff) formula is a mathematical formula used to simplify the calculation of exponentials of operators in a Lie algebra. It is commonly used in quantum mechanics and other areas of physics.

## 2. How is the BCH formula used to solve with [b,b^{\dagger}]=I?

The BCH formula can be used to solve for the commutator [b,b^{\dagger}] in terms of the identity operator I. By substituting this solution into the original equation, [b,b^{\dagger}]=I, we can solve for the values of b and b^{\dagger}.

## 3. Can the BCH formula be used for any Lie algebra?

Yes, the BCH formula can be applied to any Lie algebra, as long as the operators involved satisfy certain conditions. These conditions include being linear, having a well-defined commutator, and being closed under multiplication.

## 4. Are there any limitations to using the BCH formula?

One limitation of the BCH formula is that it only provides an approximate solution for the exponential of an operator. This approximation becomes more accurate as the number of terms in the formula is increased, but it is not exact.

## 5. How is the BCH formula derived?

The BCH formula was first derived by mathematicians Einar Hille and Wilhelm Magnus in 1949. They used a series of mathematical techniques, including the Campbell-Baker-Hausdorff theorem, to simplify the calculation of exponentials of operators in a Lie algebra.

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