How does a sine wave evolve under the Schrodinger equation?

Question

On pg 72 of "Something Deeply Hidden," Sean Carroll discusses how the uncertainty principle is just a consequence of the Schrodinger equation. He writes:

Consider a simple sine wave, oscillating up and down in a regular pattern throughout space. Plug such a wave function into the Schrodinger equation and ask how it will evolve. We find that a sine wave has a definite momentum, with shorter wavelengths corresponding to faster velocity. But a sine wave has no definite position; on the contrary, it's spread out everywhere.

I'm interested in how you describe. Carroll doesn't describe exactly what happens to a wave function that looks like a sine wave when you let it evolve under the Schrodinger equation. It seems like he might be saying it stays a sine wave, but I can't tell. What happens to a sine wave when you let it evolve under the Schrodinger equation?

If anyone knows of any kind of animation of this, that would be awesome.

Answer 1

I would suggest to see this analogy in your book as a visualization of the fact that momentum eigenstates ( $\hat{p}\Psi=\hbar k\Psi$ ) are (infinitely)extended objects in real space and vice versa for the position space eigenstates. Therefore momentum eigenstates can never be eigenstates of the position operator which will lead to the commutator relation used in the uncertaity principle.

A time evolution of a unnormalizable mometum state $\sin(kx)$ characterised by the quantum number $k\in\mathbb{R}$ for a free particle Hamiltonian (only kinetic energy) can be done in the following way.

$$i\hbar\partial_t\sin(kx)\psi(t)=-\frac{\hbar^2\nabla^2}{2m}\sin(kx)\psi(t)$$ $$i\hbar(\partial_t\psi(t))\sin(kx)=-\frac{\hbar^2k^2}{2m}\sin(kx)\psi(t)$$

This differential equation is solved by $\psi(t)=e^{-i\omega(k) t/\hbar}$ ans $\omega(k)=-\frac{\hbar k^2}{2m}$ (dispersion of a free particle).

Therefore the time dependent $\sin(kx)$ state under following the dynamic of a free particle $\hat{H}=\frac{-\hbar^2\nabla}{2m}$ results in $\Psi(x,t)=\sin(kx)\exp(-i\omega(k)t)$

This is nothing special you will just get a sine wave with the dispersion relation $\omega(k)$ and therefore some phase velocity (https://en.wikipedia.org/wiki/Phase_velocity).

Maybe this video can help you :

• https://www.youtube.com/watch?v=KKr91v7yLcM
• at 4:00 this sould visualize the statement in your book.