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

  1. Apr 11, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove the following relations (for ##\zeta:=r e^{i\theta}##):
    [tex]
    \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}
    [/tex]
    2. Relevant 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##.
    3. 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!
     
  2. jcsd
  3. Apr 11, 2016 #2

    DrDu

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    Science Advisor

    Identity operator times alpha?
     
  4. Apr 11, 2016 #3

    blue_leaf77

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    Homework Helper

    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##.
     
  5. Apr 11, 2016 #4
    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

    [tex]
    D(\alpha)^\dagger a D(\alpha)=a+\mathbb{U}\alpha
    [/tex]

    with ##\mathbb{U}## being the unity operator?
     
  6. Apr 11, 2016 #5

    blue_leaf77

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    Yes he's completely right.
     
  7. Apr 11, 2016 #6
    Thanks,!
     
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