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Annihilation of shifted gaussian vs. nonshifted gaussian

  1. Nov 11, 2015 #1
    Maybe a picture will help. http://chipreuben.com/annihilation-of-shifted-Gaussian.jpg [Broken]
    I get zero, both with the shifted and the non-shifted. Can anyone tell why the nonshifted should yield the ground state eigenfunction multiplied up by a constant? It seems the answer has to do with expanding in Fourier mode and then converting back to the distance domain...but I'm not sure.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 12, 2015 #2

    vanhees71

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    The most simple thing is to just check the equation. You've given the ground-state wave function (a Gaussian) and then you can simply apply the annihilation operator and check the given relation. That shows that the "shifted ground state" is a coherent state.
     
  4. Nov 12, 2015 #3

    samalkhaiat

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    Okay, I’m going to work in units where [itex]\hbar = m = \omega = 1[/itex]. In this units, the momentum operator is simply
    [tex]p = \frac{i}{\sqrt{2}} \ (a^{\dagger} - a) . \ \ \ \ \ \ \ \ \ \ \ \ \ (1)[/tex]
    Now, consider the translation operator, [itex]T(\epsilon) = \exp(- i \epsilon \ p)[/itex], whose action on coordinate state is given by [tex]T(\epsilon) | x \rangle = | x + \epsilon \rangle , \ \ \ \ \ \ (2)[/tex]
    and define a state [itex]|\Psi_{\epsilon} \rangle[/itex] by the action of [itex]T(\epsilon)[/itex] on the ground state [itex]|\Psi_{0}\rangle[/itex]:
    [tex]
    |\Psi_{\epsilon} \rangle = T(\epsilon) \ | \Psi_{0}\rangle . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)
    [/tex]
    By multiplying (3) from the left with [itex]\langle x|[/itex] and inserting the unit operator [itex]\int d \bar{x} |\bar{x}\rangle \langle \bar{x}| =1[/itex], we transfer (3) to a coordinate representation, i.e., equation between wave functions
    [tex]
    \langle x | \Psi_{\epsilon}\rangle = \int d \bar{x} \ \langle x | T(\epsilon) | \bar{x} \rangle \langle \bar{x}| \Psi_{0}\rangle . \ \ \ \ (4)
    [/tex]
    Using (2), we obtain
    [tex]
    \Psi_{\epsilon}(x) = \int d \bar{x} \ \langle x | \bar{x} + \epsilon \rangle \ \Psi_{0}(\bar{x}) . \ \ \ \ (5)
    [/tex]
    Now, change the integration variable [itex]\bar{x} = y - \epsilon[/itex], and use [itex]\langle x | y \rangle = \delta (x - y)[/itex] to obtain
    [tex]
    \Psi_{\epsilon}(x) = \int dy \ \langle x | y \rangle \ \Psi_{0}(y - \epsilon) = \Psi_{0}(x - \epsilon) . \ \ (6)
    [/tex]
    So, [itex]\Psi_{\epsilon}(x)[/itex] is the shifted ground state wavefunction [itex]\Psi_{0}(x - \epsilon)[/itex].


    Okay, now go back to (3) and let the operator [itex]a[/itex] acts on it
    [tex]
    a \ |\Psi_{\epsilon}\rangle = a \ e^{- i \epsilon \ p} | \Psi_{0} \rangle . \ \ \ \ \ \ (7)
    [/tex]
    Since, [itex]a \ |\Psi_{0}\rangle = 0[/itex], we can replace the right-hand-side of (7) by the following commutator
    [tex]
    a \ |\Psi_{\epsilon}\rangle = \left[a \ , \ e^{- i \epsilon \ p} \right] | \ \Psi_{0} \rangle . \ \ \ \ \ \ \ (8)
    [/tex]
    If you now expand the exponential and use the relation
    [tex]
    [a \ , -i \epsilon \ p ] = \frac{\epsilon}{\sqrt{2}}\ [a \ , a^{\dagger}] = \frac{\epsilon}{\sqrt{2}} ,
    [/tex]
    you get
    [tex]
    \left[ a \ , e^{- i \epsilon \ p} \right] = \frac{\epsilon}{\sqrt{2}} \ e^{- i \epsilon \ p} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)
    [/tex]
    Substituting (9) in (8), leads to
    [tex]
    a \ |\Psi_{\epsilon}\rangle = \frac{\epsilon}{\sqrt{2}} \ e^{- i \epsilon \ p} \ | \Psi_{0} \rangle = \frac{\epsilon}{\sqrt{2}} \ |\Psi_{\epsilon} \rangle . \ \ \ \ \ \ (10)
    [/tex]
    In wave function language, this becomes
    [tex]
    a \ \Psi_{\epsilon}(x) = \frac{\epsilon}{\sqrt{2}} \ \Psi_{\epsilon} (x) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (11)
    [/tex]
    And finally using (6), i.e., [itex]\Psi_{\epsilon}(x) = \Psi_{0}(x - \epsilon)[/itex], we obtain
    [tex]
    a \ \Psi_{0}( x - \epsilon ) = \frac{\epsilon}{\sqrt{2}} \ \Psi_{0}( x - \epsilon ) . \ \ \ \ \ \ \ \ \ \ (12)
    [/tex]
    Notice that setting [itex]\epsilon = 0[/itex] gives you back
    [tex]
    a \ \Psi_{0}(x) = 0 .
    [/tex]
     
    Last edited by a moderator: May 7, 2017
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