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!

 

[DrDu]

Identity operator times alpha?

 

[blue_leaf77]

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 $[a,a^\dagger]=1$.

 

[Emil_M]

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?

 

[blue_leaf77]

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

D(α)†aD(α)=a+UαD(α)†aD(α)=a+Uα​

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

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

Yes he's completely right.

 

[Emil_M]

Yes he's completely right.

Thanks,!