# Displacement operator, squeeze operator - problem interpreting notation....

• Emil_M

## Homework Statement

Prove the following relations (for ##\zeta:=r e^{i\theta}##):
\begin{align} D(\alpha)^\dagger a D(\alpha)&=a+\alpha\\ S(\zeta)^\dagger a S(\zeta)&= a \cosh r- a^ \dagger e^{i\theta} \sinh r \end{align}

## Homework Equations

##|\alpha\rangle## is the coherent state. ##a## and ##a^\dagger## are the creation and annihilation operators. ##\alpha## is the eigenvalue of the annihilation operator.
Thus, ##a|\alpha\rangle=\alpha|\alpha\rangle##.

## The Attempt at a Solution

I really cannot attempt a solution here, because I don't seem understand expression (1). If ##a## is an operator, and ##\alpha## is a scalar, I really don't know how to interpret the expression ##a+\alpha##.

Thanks for your help!

## Answers and Replies

Identity operator times alpha?

Consider the relations
$$e^ABe^{-A} = B + [A,B] + \frac{1}{2!}[A,[A,B]] + \frac{1}{3!}[A,[A,[A,B]]] + \ldots$$
as well as ##[a,a^\dagger]=1##.

Consider the relations
$$e^ABe^{-A} = B + [A,B] + \frac{1}{2!}[A,[A,B]] + \frac{1}{3!}[A,[A,[A,B]]] + \ldots$$
as well as ##[a,a^\dagger]=1##.

Thank you for your reply! I know the Hadamard lemma, as well as the commutator relation (with ##1## being the unity operator). I am just not sure how to add a scalar to an operator. Could DrDu be right in assuming the equation was meant to read

$$D(\alpha)^\dagger a D(\alpha)=a+\mathbb{U}\alpha$$

with ##\mathbb{U}## being the unity operator?

Could DrDu be right in assuming the equation was meant to read