# Harmonic Oscillator potential, proof that Gaussians remain Gaussians?

I read in several papers that for a Harmonic Oscillator Hamiltonian in the time dependent Schrödinger equation a Gaussian wave packet remains Gaussian.

Unfortunately I could not find any proof for this statement and trying to verify it myself I did not succeed. If I make a general ansatz with a spherically symmetric Gaussian wave packet with time dependent width and time and space dependent phase $$\psi(t,\vec x) = (\pi a(t)^2)^{-3/4} \exp\left(-\frac{x^2}{2 a(t)^2} + i \phi(t,\vec x)\right)$$ and insert it into the Schrödinger equation $$i \dot{\psi}(t,\vec x) = -\frac{1}{2m} \Delta\psi(t,\vec x) + \frac{k}{2} x^2 \psi(t,\vec x)$$ I get relatively complicated differential equations involving first time derivatives of a and $\phi$ as well as first and second order spatial derivatives of $\phi$. I failed to solve those equations or even show that a solution exists.

Is there an easy way to show this?

Is there any reference where this is shown?

Which are the solutions for $a(t)$ and $\phi(t,\vec x)$ for a Gaussian (given initial conditions $a(0) = \sigma$ and $\phi(0,\vec x)=0$)?

• Perhaps you could try to go to Fourier space, using the spatial Fourier decomposition?
– Danu
Commented Mar 17, 2014 at 11:42
• @Danu: Wouldn't that give me exactly the same equations? Since the Fourier transform of a Gaussian is again a Gaussian and in the Hamiltonian $x^2$ and $p^2$ appear symmetrically? Commented Mar 17, 2014 at 11:46
• I've seen a Fourier transform simplify some calculations involving Gaussians before, but it's up to you if you want to give it a shot.
– Danu
Commented Mar 17, 2014 at 12:09
• More on Gaussian wave packets: physics.stackexchange.com/search?q=Gaussian+wave+packet Commented Mar 17, 2014 at 19:02

There is a bit more structure to be had in the problem, and I would recommend that you take advantage of it. In particular, you know that $\phi$ is really a linear function of $x$, or otherwise it is not a gaussian; and that the real and imaginary parts of its coefficient have distinct physical meaning, which you can get by taking appropriate expectation values. If you apply this, then, you can make your Ansatz be $$\langle \vec x|\psi(t)\rangle= \psi(t,\vec x) = (\pi a(t)^2)^{-3/4} \exp\left(-\frac{(\vec x-\vec x_0(t))^2}{2 a(t)^2} + i \vec p_0(t)·\vec x/ħ\right).$$ Even better, you can easily calculate the expectation values of position and momentum to be $$\langle\psi(t)|\vec x|\psi(t)\rangle=\vec x_0(t)$$ and $$\langle\psi(t)|\vec p|\psi(t)\rangle=\vec p_0(t),$$ and you can apply Ehrenfest's theorem to get meaningful, easy-to-solve equations of motion for these quantities. This completely constrains your solution.
Of course, this should leave you slightly uncomfortable because you haven't proved that $\psi(t,\vec x)$ is a solution of the TDSE... but finding that out is a few differentiations away. Since you know it's a solution anyway, you're on pretty safe ground.