Normalisation of harmonic oscillator classical action

[CAF123]

Homework Statement

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 $\psi(x,t)$

Homework Equations

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

The Attempt at a Solution

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$\psi(x,t) = \int dx' \langle x,t |x',t' \rangle \psi(x',t')$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!

 

[king vitamin]

The formula for the Gaussian integral you give (see http://en.wikipedia.org/wiki/Common...grals_with_a_complex_argument_of_the_exponent for example) should work for any complex value a in the exponent as long as the imaginary part of a is positive.

 

[CAF123]

Hi king vitamin,

The formula for the Gaussian integral you give (see http://en.wikipedia.org/wiki/Common...grals_with_a_complex_argument_of_the_exponent for example) should work for any complex value a in the exponent as long as the imaginary part of a is positive.

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.

 

[king vitamin]

<br /> \int dx&#039; e^{x&#039;^2(A+if(t)) + ig(t)x&#039;} = \int dx&#039; e^{i (f(t) - iA) x&#039;^2 + ig(t)x&#039;} = \sqrt{\frac{i \pi}{f(t) - iA}}\exp\left(-ig(t)^2/4(f(t)-iA)\right)<br />

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