Path Integrals Harmonic Oscillator

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SUMMARY

The discussion centers on solving problem 3-12 from "Quantum Mechanics and Path Integrals" by Feynman and Hibbs, specifically regarding the wavefunction evolution of a harmonic oscillator. The initial wavefunction is given as ψ(x,0) = exp[-(mω/2h)(x-a)²]. The solution involves using the kernel K for a harmonic oscillator, defined as K=(mω/(2πihsin(ωτ)))^(1/2) * exp[(imω/(2h sin(ωτ)))((x_b² + x_a²)cos(ωτ) - 2x_bx_a)]. The integral form of the wavefunction evolution is confirmed to be ψ(x,t) = ∫K(x,t;x',0)ψ(x',0)dx', leading to a derived expression for ψ(x,τ).

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Wislan
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Hi,

I am reading through the book "Quantum Mechanics and Path Integrals" by Feynman and Hibbs and am having a bit of trouble with problem 3-12. The question is (all Planck constants are the reduced Planck constant and all integrals are from -infinity to infinity):
The wavefunction for a harmonic oscillator is (at t=0): ψ(x,0) = exp[-\frac{mω}{2h} (x-a)^{2}]
then, using Eq.(3.42) (ψ(x_{b},t_{b})=∫K(x_{b},t_{b};x_{c},t_{c}) * ψ(x_{c},t_{c}) dx_{c}) and the results of problem 3-8 (the Kernel for a harmonic oscillator K=(\frac{mω}{2πihsin(ωτ)})^{1/2} * exp[\frac{imω}{2hsin(ωτ)}((x_{b}^{2} + x_{a}^{2})cos(ωτ)-2x_{b}x_{a})]
show that
ψ(x,τ)=exp[-\frac{iωτ}{2}-\frac{mω}{2h}(x^{2}-2axe^{-iωτ}+a^{2}cos(ωτ)e^{-iωτ})]

Now so far as I can tell, a solution is to multiply the wavefunction at time 0 by the kernel from x at t=0 to x at t=τ and then integrate over all x, however (unless I've made an algebra mistake which is always possible) this doesn't give the required answer. Any ideas?

Thanks,
Will
 
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Yes, the evolution of the wavefunction is given by

$$ \psi(x,t) = \int_{-\infty}^\infty K(x,t;x',0) \psi(x',0) dx'.$$

This integral is of the form

$$ \begin{split}
\int_{-\infty}^\infty \exp{\left[ -A (x')^2 + B x' + C\right]} dx' &= \int_{-\infty}^\infty\exp{\left[ -A \left(x'-\frac{B}{2A}\right)^2 + C + \frac{B^2}{4A} \right]} dx'\\
& = \sqrt{\frac{\pi}{A}} \exp{\left[ C + \frac{B^2}{4A} \right]}. \end{split}$$

Hopefully some of this is familiar and you can figure out the algebra.
 

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