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Position representation of coherent state and time evolution

  1. Mar 16, 2016 #1
    1. The problem statement, all variables and given/known data
    I ended up solving the problem as I was typing it up, I am posting what I did anyway as it took so long to type and might be useful to someone else.

    I am trying to figure out the position representation of a coherent state and it's time evolution. I should be getting a Guassian wavefunction which is oscillating in position.

    2. Relevant equations
    Coherent state,
    [itex] | \alpha \rangle = \exp (- \frac{|\alpha |^2}{2} ) \sum\limits_{n = 0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} | n \rangle [/itex]
    Position representation of number state, with ## \xi = \sqrt{\frac{\omega}{ \hbar}} x ##
    ## \psi_n (\xi) = \langle x | n \rangle = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} (2^n n!)^{-1/2}\exp (-\frac{\xi^2}{2}) H_n(\xi)##
    where ## H_n## are Hermite Polinomials.
    Generating function for Hermite Polinomials is,
    ## \exp (2xt - t^2) = \sum\limits_{n = 0}^{\infty} H_n (x) \frac{t^n}{n!}##
    System Hamiltonian is,
    ## \mathcal{H} = \hbar \omega \left(\frac{1}{2} + a^{\dagger}a \right) = \hbar \omega \left(\frac{1}{2} + \hat{n} \right) ##
    3. The attempt at a solution

    The position representation of a coherent state should be,
    ## \psi (\alpha) = \langle x | \alpha \rangle = \exp (- \frac{|\alpha |^2}{2} ) \sum\limits_{n = 0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} \langle x | n \rangle \\ = \exp (- \frac{|\alpha |^2}{2} ) \sum\limits_{n = 0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} \left( \frac{\omega}{\pi \hbar}\right)^{1/4} (2^n n!)^{-1/2}\exp (-\frac{\xi^2}{2}) H_n(\xi) \\
    = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (-\frac{\xi^2}{2}) \sum\limits_{n = 0}^{\infty} \frac{(\alpha / \sqrt{2})^n}{n!} H_n (\xi) ##
    Using the generating function of the Hermite polinomials this reduces to,
    ## \psi (\alpha) = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (-\frac{\xi^2}{2}) \exp (\sqrt{2} \xi \alpha - \frac{\alpha^2}{2}) ##

    Completeing squares gives,
    ## \psi (\alpha) = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (\frac{\xi^2}{2}) \exp \left( - ( \xi - \frac{\alpha}{\sqrt{2}})^2 \right) ##


    Now the time evolved state will be,
    ## | \alpha (t) \rangle = \exp (-i\mathcal{H}t / \hbar) | \alpha \rangle ##
    ## = \exp(- \frac{i\omega t}{2})\exp (-i\omega \hat{n}t) \exp (- \frac{|\alpha |^2}{2} ) \sum\limits_{n = 0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} | n \rangle##
    ## = \exp(- \frac{i\omega t}{2}) \sum\limits_{n = 0}^{\infty} \frac{(\alpha e^{ -i\omega t})^n}{\sqrt{n!}} | n \rangle##
    ## = \exp(- \frac{i\omega t}{2}) | \alpha e^{-i \omega t} \rangle ##
    So the coherent state evolves to another coherent state with a phase ## \frac{-i\omega t}{2} ## added overall and ##\alpha(t) = \alpha e^{-i \omega t}##

    so,
    ## \psi (\alpha(t)) = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (\frac{\xi^2}{2}) \exp \left( - ( \xi - \frac{\alpha e^{i\omega t}}{\sqrt{2}})^2 \right)##

    If I seperate out the real and imaginary part of ## \alpha (t) ## as ## \alpha_r + i \alpha_i = | \alpha | \cos (\phi - \omega t) + i |\alpha | \sin (\phi - \omega t) ##, where ##\phi ## is the complex phase of ## \alpha##,
    ## \psi (\alpha(t)) = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (\frac{\xi^2}{2}) \exp \left( - ( (\xi - \frac{\alpha_r}{\sqrt{2}}) - i\frac{\alpha_i}{\sqrt{2}} )^2 \right)##
    ## = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (\frac{\xi^2}{2}) \exp \left( -(\xi - \frac{\alpha_r}{\sqrt{2}} )^2 + \frac{\alpha_i^2}{2} +2i \frac{\alpha_i}{\sqrt{2}}(\xi - \frac{\alpha_r}{\sqrt{2}} ) \right)##

    The probability distribution in x will be,
    ##| \psi (\alpha (t)) |^2 = \left( \frac{\omega}{\pi \hbar}\right)^{1/2} \exp (-|\alpha |^2) \exp \left( \xi^2 - 2(\xi - \frac{\alpha_r}{\sqrt{2}} )^2 + \alpha_i^2 \right)##
    ## = \left( \frac{\omega}{\pi \hbar}\right)^{1/2} \exp (-|\alpha |^2) \exp \left( -\xi^2 - \alpha_r^2 + 2\sqrt{2}\xi\alpha_r + \alpha_i^2 \right)##
    ## = \left( \frac{\omega}{\pi \hbar}\right)^{1/2} \exp (-|\alpha |^2) \exp \left( -(\xi - \sqrt{2}\alpha_r) ^2 + \alpha_r^2 + \alpha_i^2 \right)##
    ## = \left( \frac{\omega}{\pi \hbar}\right)^{1/2} \exp (-|\alpha |^2) \exp \left( -(\xi - \sqrt{2}\alpha_r) ^2 \right) \exp ( |\alpha|^2 ) ##

    ## | \psi_{\alpha}(x) |^2 = \left( \frac{\omega}{\pi \hbar}\right)^{1/2} \exp \left( -(\sqrt{\frac{\omega}{ \hbar}} x - \sqrt{2}|\alpha|\cos (\phi - \omega t)) ^2 \right) ##

    Which is clearly a normalized Guassian with center at ## \sqrt{\frac{2\hbar}{\omega}}| \alpha | \cos (\phi - \omega t) ##
     
  2. jcsd
  3. Mar 21, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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