Time Evolution for particle with potential suddenly removed

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1. Mar 19, 2015

Xyius

1. The problem statement, all variables and given/known data

This is a problem from my Statistical Mechanics book by Pathria.

At $t=0$, the ground state wavefunction of a one-dimensional quantum harmonic oscillator with potential $V(x)=\frac{1}{2}\omega_0^2 x^2$ is given by,

$$\psi(x,0)=\frac{1}{\pi^{1/4} \sqrt{a}}e^{-\frac{x^2}{2a^2}}$$

where $a=\sqrt{\frac{\hbar}{m \omega_0}}$. At $t=0$, the harmonic potential is abruptly removed. Use the momentum representation of the wavefunction at $t=0$ and the time-dependent Schrodinger equation to determine the spatial wavefunction and density at time $t>0$.
2. Relevant equations
$$i\hbar \frac{\partial}{\partial t}\psi=\hat{H} \psi$$

3. The attempt at a solution

So first I took the Fourier transform of the wave function given at $t=0$ to get,

$$\Phi(k,0)=-\frac{\sqrt{a}}{\pi^{1/4}}e^{\frac{1}{2}i a ^2 k^2}$$

So now that at $t>0$ the potential is removed, the new Hamiltonian is that of a free particle. I want to see how the original wave function evolves with time with this new potential. So I will multiply it by the time evolution operator.

$$\Phi(k,t)=\Phi(k,0)e^{-i \frac{\hat{H}t}{\hbar}}=\Phi(k,0)e^{-i \frac{\hat{p}^2 t}{2m\hbar}}$$

Generally, we can write this as,

$$|\psi(t)>=e^{-i \frac{\hat{p}^2 t}{2m\hbar}}|\psi_0>$$

The whole goal is to find the spatial wave function so lets project it onto $x$.

$$<x|\psi(t)>=<x|e^{-i \frac{\hat{p}^2 t}{2m\hbar}}|\psi_0>$$

What is given in the beginning of the problem is $<x|\psi_0>$ so I do the following.

$$<x|\psi(t)>=\int <x|e^{-i \frac{\hat{p}^2 t}{2m\hbar}}|x'><x'|\psi_0>dx'$$

Now, the Hamiltonian in the exponential purely depends on the momentum operator, so it would be easiest to deal with momentum eigenkets. So I will insert two more identities in there.

$$<x|\psi(t)>=\int <x|p'><p'|e^{-i \frac{\hat{p}^2 t}{2m\hbar}}|p><p|x'><x'|\psi_0>dx'dpdp'$$

$$<x|\psi(t)>=\int <x|p'>e^{-i \frac{p^2 t}{2m\hbar}}\delta(p'-p)<p|x'><x'|\psi_0>dx'dpdp'$$

$$<x|\psi(t)>=\int <x|p>e^{-i \frac{p^2 t}{2m\hbar}}<p|x'><x'|\psi_0>dx'dp$$

My problem is, I do not know what $<x|p>$ and $<p|x'>$ are. I remember back when I took quantum mechanics dealing with these types of terms but for the life of me I cannot remember what their value is or why! If anyone can help I would greatly appreciate it!

EDIT:
A ha! I just realized that these terms are simply the plane wave solution to the free particle! I will go though it now to make sure I have no other issues.

Last edited: Mar 19, 2015
2. Mar 20, 2015

Goddar

Hi. I think you're taking a detour here...
You obtained the full wave function in momentum space:
Φ(k,t ) = Φ(k,0)⋅exp(–ik2t/2m)
(remember in momentum space p-hat is just p, not a differential operator)
So to get the spatial wave function, just take the inverse Fourier transform.