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Show that wave packet is an eigenstate to operator

  1. Jun 28, 2009 #1
    Show that wave packet is an eigenstate to operator [SOLVED]

    1. The problem statement, all variables and given/known data

    For a harmonic oscillator we can define the step up and down operators [tex]\hat{a}[/tex] and [tex]\hat{a}^{\dagger}[/tex] and their action as

    [tex]\hat{a}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}+\frac{\imath}{m\omega}\hat{p}) \quad \hat{a}|n\rangle = \sqrt{n}|n-1\rangle[/tex]

    [tex]\hat{a}^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}-\frac{\imath}{m\omega}\hat{p}) \quad \hat{a}^{\dagger}|n\rangle = \sqrt{n+1}|n+1\rangle[/tex]

    Show that the Gaussian wave-packet

    [tex]\Psi(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}(x-x_t)^2+\frac{\imath}{\hbar}p_t(x-x_t)}[/tex]

    is an eigenstate to [tex]\hat{a}[/tex] with eigenvalue [tex]\alpha[/tex].
    [tex]\hat{a}|a\rangle = \alpha | a\rangle[/tex]

    Express the eigenvalue in terms of [tex]x_t[/tex] and [tex]p_t[/tex].

    2. Relevant equations

    See above.

    3. The attempt at a solution

    [tex]\hat{p} \rightarrow -i\hbar\frac{\partial}{\partial x}[/tex]

    gives the expression for

    [tex]\hat{a}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}+\frac{\hbar}{m\omega}\frac{\partial}{\partial x})[/tex]

    [tex]\hat{a}|\Psi\rangle = \sqrt{\frac{m\omega}{2\hbar}} \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(e^{-\frac{m\omega}{2\hbar}(x-x_t)^2+\frac{\imath}{\hbar}p_t(x-x_t)} + \frac{\hbar}{m\omega}(-\frac{m\omega}{\hbar}(x-x_t)+\frac{\imath}{\hbar}p_t(e^{-\frac{m\omega}{2\hbar}(x-x_t)^2+\frac{\imath}{\hbar}p_t(x-x_t)})\right) =
    \Psi\sqrt{\frac{m\omega}{2\hbar}}\left(1+(x-x_t)+\frac{\imath}{m\omega}p_t\right)[/tex]

    which is obviously not a scalar times [tex]\Psi[/tex].

    What am I doing wrong here?
     
    Last edited: Jun 28, 2009
  2. jcsd
  3. Jun 28, 2009 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You forgot the x in the first term in parentheses (which comes from the x operator multiplied by Psi).

    Here, cut out the clutter (h is supposed to be hbar in what follows):

    C = (mω/h)1/2

    A = (h/mω)

    Then:

    a|Ψ> = C(x + A ∂/∂x)Ψ

    = C(xΨ + A∂Ψ/∂x)

    = C(xΨ + A[(-mω/h)(x-xt) + (i/h)pt]Ψ)

    =C(xΨ + A[-A-1xΨ + A-1xtΨ + (i/h)ptΨ])

    See anything that might cancel?
     
  4. Jun 28, 2009 #3
    Yes. Thank you!
     
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