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Normalisation of harmonic oscillator classical action

  1. Apr 26, 2015 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    The transition amplitude for the harmonic oscillator may be written as ##\langle x_2, t_2 | x_1, t_1 \rangle = N_{\omega}(T) \exp(i/\hbar S_{cl})##, where ##T=t_2-t_1## and ##S_{cl}## is the classical action. Let the wave function at ##t=0## be ##\psi(x,o) = \left(\frac{mw}{\pi \hbar}\right)^{1/4} \exp(-mwx^2/2\hbar)##. Find ##\psi(x,t)##

    2. Relevant equations

    ##\langle 0, T | 0, 0 \rangle \equiv \langle 0, t_2|0, t_1 \rangle = N_{\omega}(T)##
    Classical action, ##S_cl = \frac{mw}{2 \sin w T} (-2x_1 x_2 + \cos w T(x_1^2 + x_2^2))##

    3. The attempt at a solution

    ##\psi(x,t) = \int dx' \langle x,t |x',t' \rangle \psi(x',t')##Evaluate for ##t' = 0## gives $$\psi(x,t) = \int dx' \langle x,t |x',0 \rangle \psi(x',0) $$ which is $$= \sqrt{\frac{mw}{2\pi i \hbar sin wt}} \left(\frac{mw}{\pi \hbar}\right)^{1/4} e^{\frac{i}{\hbar} \frac{mw}{2 \sin wt} x^2 \cos wt} \int dx' e^{\frac{i}{\hbar} \frac{mw}{2\sin wt} (x'^2 \cos wt) - 2xx')} e^{\frac{-mwx'^2}{2 \hbar}}$$
    I am just a bit confused on what Gaussian integral I should use to compute this? I was thinking $$\int e^{iax'^2 + bx'} dx' = ..$$ but the ##x'^2## term has both imaginary and real coefficients. I've tried completing the square too but no progress. Thanks!
     
  2. jcsd
  3. Apr 26, 2015 #2

    king vitamin

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  4. Apr 27, 2015 #3

    CAF123

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    Hi king vitamin,
    Hmm so I should write it like $$\int dx' e^{x'^2(A+if(t)) + ig(t)x'}?$$ I'm just not sure how to then apply the result $$\int dx e^{iax^2 + bx} = \sqrt{\frac{i\pi}{a}} \exp(-ib^2/4a)$$ Thanks.
     
  5. Apr 27, 2015 #4

    king vitamin

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    [tex]
    \int dx' e^{x'^2(A+if(t)) + ig(t)x'} = \int dx' e^{i (f(t) - iA) x'^2 + ig(t)x'} = \sqrt{\frac{i \pi}{f(t) - iA}}\exp\left(-ig(t)^2/4(f(t)-iA)\right)
    [/tex]

    Which can be simplified. But remember that A must be negative.
     
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