I Periodic potential - energy bands

1. Oct 30, 2016

Konte

Hello everybody,

I have some questions about treatment of Schrodinger equation where $\hat{V}(\theta)$, the potential energy part of Hamiltonian $\hat{H}=\hat{T}(\theta)+\hat{V}(\theta)$ is a trigonometric function like:
$\hat{V}(\theta) = a sin(\theta)$
or
$\hat{V}(\theta) = a cos(\theta)+ b sin(c\theta)$ where $\theta$ is an angular variable.
I read something in solid-state physics that a system which evolve inside a periodical potential ends up with energy bands as eigenvalues solutions.

Do I have the same case here, with those two examples of potential energy?
In other words, will I obtain energy band too, even here I have nothing to do with lattice nor crystals?

Thank you everybody.

Konte

Last edited: Oct 30, 2016
2. Oct 31, 2016

Jilang

I wouldn't think so. The potential is just a step function, the height of which depends on the direction.

3. Oct 31, 2016

Spinnor

4. Oct 31, 2016

houlahound

The potential would be easier to follow with vectors. Is that from a point or repeats or?

Sorry I need more spatial information. What am I missing?

5. Oct 31, 2016

stevendaryl

Staff Emeritus
That's not a periodic potential. When people call a potential periodic, what they mean is that there is some vector $\vec{\delta r}$ such that $V(\vec{r} + \vec{\delta r}) = V(\vec{r})$. For example, in one dimension, if the potential looks like this: $V(x) = a cos(kx)$, then $V(x+\frac{2\pi}{k}) = V(x)$

The potential $V = a sin(\theta)$ isn't periodic in this sense. Yes, $V(\theta + 2 \pi) = V(\theta)$, but the angle $\theta + 2\pi$ is the same angle as $\theta$.

6. Oct 31, 2016

houlahound

7. Oct 31, 2016

Konte

To be clear, it's about an angular (thus periodic) potential that can hinder a quantum rotor.

As an example, I show here the case of $\hat{V}(\theta)\,=\, 3\,-\,sin(7\,\theta)$.

So the word "periodic" has two different senses, that I want to understand. However, it stays really blurred for me. In looking at the case of one dimension as you shown, mathematically, we cannot make a difference between:

$V(x) = a cos(kx)$ and $V(\theta) = a cos(n \theta)$, since $\theta$ and $x$ are dummies variables.

So I am troubled when the first gives band structure and the second doesn't.

Thank you much.

Konte

8. Oct 31, 2016

stevendaryl

Staff Emeritus
$x$ and $\theta$ are not interchangeable, though, so they aren't completely "dummy variables". In 1-D quantum mechanics, it is assumed that space is described by a coordinate $x$ that runs from $-\infty$ to $+\infty$. In contrast, the angular variable $\theta$ runs from $-\pi$ to $+\pi$, and the point $\theta = -\pi$ is identified with the point $\theta = +\pi$. These differences lead to different constraints on the wave function. In terms of $x$, it must be the case that $lim_{x \rightarrow \pm \infty} \psi(x) = 0$. In terms of $\theta$, it must be the case that $\psi(\theta) = \psi(\theta + 2\pi)$. So they are very different types of variables, and they have different consequences.

On the other hand, if you demand that EVERYTHING be periodic in $x$ with period $L$---the wave function, the potential, everything---then the distinction disappears; In that case you can identify $x$ with the angular variable $\theta = \frac{2\pi x}{L}$.

9. Oct 31, 2016

Konte

Ok, thanks a lot.