Path Integrals Harmonic Oscillator

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The discussion revolves around a problem from "Quantum Mechanics and Path Integrals" regarding the wavefunction of a harmonic oscillator and its evolution over time. The user is struggling to apply the kernel to the initial wavefunction and integrate to derive the correct expression for the wavefunction at time τ. A response clarifies that the evolution of the wavefunction involves integrating the product of the kernel and the initial wavefunction, suggesting that the user may need to revisit the algebra involved in this integral. The response also provides a formula for evaluating the integral, indicating that it may help in resolving the user's confusion. The conversation emphasizes the importance of careful algebraic manipulation in quantum mechanics problems.
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.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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