# Struggling with Baker-Campbell-Haussdorf-like formulas

1. Oct 17, 2009

### luisgml_2000

1. The problem statement, all variables and given/known data

I have been asked to prove $$e^{\alpha a}f(a,a^\dagger)e^{-\alpha a}=f(a,a^\dagger+\alpha)$$, where $$a$$ and $$a^\dagger$$ represent the annihilation and creation operators of the quantum harmonic oscillator, respectively.

2. Relevant equations

3. The attempt at a solution

The only thing I have done is to use the Hadamard Lemma (http://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula). Any ideas will be greatly appreciated!

2. Oct 17, 2009

### gabbagabbahey

3. Oct 17, 2009

### luisgml_2000

I get

$$e^{\alpha a}f(a,a^\dagger)e^{-\alpha a}=f(a,a^\dagger)+\frac{\alpha}{1!}[a,f(a,a^\dagger)]+\frac{\alpha^2}{2!}[a,[a,f(a,a^\dagger)]]+\ldots$$

but I don't know what to do next.

4. Oct 17, 2009

### luisgml_2000

It seems that I have to apply some kind of Taylor expansion but I'm not sure on how to do this.

5. Oct 17, 2009

### gabbagabbahey

Hint: Consider $f(a,a^{\dagger})=a^{\dagger}$ and $f(a,a^{\dagger})=a$....what do these considerations tell you about the unitary transformation $f(a,a^{\dagger})\to e^{\alpha a}f(a,a^{\dagger})e^{-\alpha a}$?

Last edited: Oct 17, 2009
6. Oct 17, 2009

### luisgml_2000

$$e^{\alpha a}a^\dagger e^{-\alpha a}=a^\dagger+\frac{\alpha}{1!}[a,a^\dagger]+\frac{\alpha^2}{2!}[a,[a,a^\dagger]]+\ldots=a^\dagger+\frac{\alpha}{1!}=\alpha+a^\dagger$$

$$e^{\alpha a}a e^{-\alpha a}=a+\frac{\alpha}{1!}[a,a]+\frac{\alpha^2}{2!}[a,[a,a]]+\ldots=a$$

In this sense the formula seems to be working fine.

Does it mean that the unitary transformation $f(a,a^{\dagger})\to e^{\alpha a}f(a,a^{\dagger})e^{-\alpha a}$ represents a displacement of $\alpha$ in the second argument of $f$?

7. Oct 17, 2009

### gabbagabbahey

Yes, exactly. Under the transformation $f(a,a^{\dagger})\to e^{\alpha a}f(a,a^{\dagger})e^{-\alpha a}$, you have $a\to a$ and $a^{\dagger}\to a^{\dagger}+\alpha$ and hence $f(a,a^{\dagger})\to f(a,a^{\dagger}+\alpha)$....nice and simple, no Taylor series needed

8. Oct 17, 2009

### luisgml_2000

Thanks a lot, I think I finally understand it!!