Struggling with Baker-Campbell-Haussdorf-like formulas

[luisgml_2000]

Homework Statement

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.

Homework Equations

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!

 

[gabbagabbahey]

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!

That seems like a good place to start...what do you get when you do that?

 

[luisgml_2000]

That seems like a good place to start...what do you get when you do that?

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.

 

[luisgml_2000]

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

 

[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 $f(a,a^{\dagger})\to e^{\alpha a}f(a,a^{\dagger})e^{-\alpha a}$?

 

[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 $f$?

 

[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

 

[luisgml_2000]

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

Thanks a lot, I think I finally understand it!