Algebra of displacement operator

[spaghetti3451]

Homework Statement

Given an operator $D(\alpha)=\exp\ (\alpha a^{\dagger}-\alpha^{*}a)$ and a function $g(a,a^{\dagger})$, where $a$ and $a^{\dagger}$ are operators and $\alpha$ and $\alpha^{*}$ are complex numbers, show that

$D^{-1}(\alpha)g(a,a^{\dagger})D(\alpha)=g(a+\alpha, a^{\dagger}+\alpha^{*})$

Homework Equations

The Attempt at a Solution

$D^{-1}(\alpha)g(a,a^{\dagger})D(\alpha)$

$=\exp\ (-\alpha a^{\dagger}+\alpha^{*}a)\ g(a,a^{\dagger})\ \exp\ (\alpha a^{\dagger}-\alpha^{*}a)$

$=(1-\alpha a^{\dagger}+\alpha^{*}a)\ g(a,a^{\dagger})\ (1+\alpha a^{\dagger}-\alpha^{*}a)$

$=g(a,a^{\dagger})+\alpha^{*}[a,g(a,a^{\dagger})]-\alpha[a^{\dagger},g(a,a^{\dagger})]$.

On the other hand,

$g(a+\alpha, a^{\dagger}+\alpha^{*})$

$=g(a, a^{\dagger}) + \alpha \frac{\partial g(a,a^{\dagger})}{\partial \alpha} + \alpha^{*} \frac{\partial g(a,a^{\dagger})}{\partial \alpha^{*}}$.

This seems to suggest that

$\frac{\partial g(a,a^{\dagger})}{\partial \alpha^{*}}=[a,g(a,a^{\dagger})]$

and

$\frac{\partial g(a,a^{\dagger})}{\partial \alpha}=-[a^{\dagger},g(a,a^{\dagger})]$.

Am I missing something here?

 

[Fightfish]

You can't expand the operators like that - this assumes that $\alpha$ is infinitesimal, which is not true.
First, establish how $a$ and $a^{\dagger}$ transform under the action of the displacement operator i.e. what are $D(\alpha)^{-1} a D(\alpha)$ and $D(\alpha)^{-1} a^{\dagger} D(\alpha)$. Then insert factors of $D(\alpha) D^{-1} (\alpha)$ between each pair of creation/annihilation operators in $g(a,a^{\dagger})$.

 

[spaghetti3451]

Then insert factors of $D(\alpha) D^{-1} (\alpha)$ between each pair of creation/annihilation operators in $g(a,a^{\dagger})$.

But $g(\alpha,\alpha^{\dagger})$ is a generic function, and might not have pairings of $a$ and $a^{\dagger}$.

 

[Fightfish]

But $g(\alpha,\alpha^{\dagger})$ is a generic function, and might not have pairings of $a$ and $a^{\dagger}$.

I think its fair to assume that any reasonably well behaved function of the operators can be expressed as a sum of operator products. After all, most elementary functions of operators (eg. exponential) are defined in terms of their power series expansions.

 

[spaghetti3451]

But, the exponential $\exp(a)$ of an operator $a$ is

$\exp(a)=1+a+\frac{1}{2!}a^{2}+\frac{1}{3!}a^{3}+\dots$

So, this is not a sum of operator products.

 

[Fightfish]

But, the exponential $\exp(a)$ of an operator $a$ is

$\exp(a)=1+a+\frac{1}{2!}a^{2}+\frac{1}{3!}a^{3}+\dots$

So, this is not a sum of operator products.

What I meant by operator products is any term of the form $a^{n}$, $\left(a^{\dagger}\right)^{m}$, $\left(a^{\dagger}\right)^{m}a^{n}$, $\left(a^{\dagger}\right)^{m}a^{n} \left(a^{\dagger}\right)^{k}$, and so on so forth.

 

[spaghetti3451]

Okay, so I have

$D(\alpha)^{-1}aD(\alpha) = \exp(-\alpha a^{*}+\alpha^{*}a)a\exp(\alpha a^{*}-\alpha^{*}a)$

Do I expand the exponential $\exp(-\alpha a^{*}+\alpha^{*}a)$ and use $[a,a^{\dagger}]=1$?

 

[Fightfish]

Do I expand the exponential $\exp(-\alpha a^{*}+\alpha^{*}a)$ and use $[a,a^{\dagger}]=1$?

Yup, or you can just use the Baker-Hausdorff- Campbell formula (if you don't want to prove it)

 

[spaghetti3451]

So, I get $\exp(-\alpha a^{*}+\alpha^{*}a)=\exp(-\alpha a^{*})\ \exp(\alpha^{*}a)$ using BCH, but I still have to expand $\exp(-\alpha a^{*})$ to find out how it commutes with $a$, right?

 

[Fightfish]

So, I get $\exp(-\alpha a^{*}+\alpha^{*}a)=\exp(-\alpha a^{*})\ \exp(\alpha^{*}a)$ using BCH, but I still have to expand $\exp(-\alpha a^{*})$ to find out how it commutes with $a$, right?

I was referring to this particular form of the BCH lemma:
$$e^{iG\lambda}A e^{-iG\lambda} = A + i\lambda [G,A] + \frac{\left(i\lambda\right)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\underbrace{[G,[G,[G,\ldots[G}_{n\ times},A]]]\ldots]+\ldots$$
Also, your result isn't right; you're missing a factor of $e^{-|\alpha|^2/2}$.

 

[spaghetti3451]

I was referring to this particular form of the BCH lemma:
$$e^{iG\lambda}A e^{-iG\lambda} = A + i\lambda [G,A] + \frac{\left(i\lambda\right)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\underbrace{[G,[G,[G,\ldots[G}_{n\ times},A]]]\ldots]+\ldots$$
Also, your result isn't right; you're missing a factor of $e^{-|\alpha|^2/2}$.

Ok, so, using the Baker-Campbell-Hausdorff lemma

$$e^{iG\lambda}A e^{-iG\lambda} = A + i\lambda [G,A] + \frac{\left(i\lambda\right)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\underbrace{[G,[G,[G,\ldots[G}_{n\ \text{times}},A]]]\ldots]+\ldots,$$

we have

$D(\alpha)^{-1}aD(\alpha)$

$= \exp(-\alpha a^{\dagger}+\alpha^{*}a)a\exp(\alpha a^{\dagger}-\alpha^{*}a)$

$= a + [-\alpha a^{\dagger}+\alpha^{*}a,a]$

$= a- [\alpha a^{\dagger},a] + [\alpha^{*}a,a]$

$= a- \alpha[a^{\dagger},a] + \alpha^{*}[a,a]$

$= a- \alpha[a^{\dagger},a]$

$= a+ i\alpha$.

Am I correct?

 

[Fightfish]

Almost; there shouldn't be a factor of $i$. The commutator is just $-1$. Now just repeat the same for $a^{\dagger}$.

 

[spaghetti3451]

I did, and I got the desired answer.

I was wondering if the formula you quoted is a different version of the BCH formula.

 

[Fightfish]

It has the same origin really - its just convenient because this particular form (i.e. transformations) appears very frequently in quantum mechanics. I believe it's discussed in most standard texts such as Sakurai if you want more specific details.