Periodic potential - energy bands

[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

 

[Jilang]

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

 

[Spinnor]

Interesting potential. For a=b and c = 3.4 Wolfram Alpha will plot your function,

http://www.wolframalpha.com/input/?i=plot+y+=+cos(theta)+++sin(3.4xtheta)

Change c.

 

[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?

 

[stevendaryl]

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.

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 $\theta$.

 

[houlahound]

Check out this baby;

http://demonstrations.wolfram.com/SchroedingerWavefunctionsInAContinuouslyVaryingPotential/

Or

http://demonstrations.wolfram.com/BandSpectrumInAPeriodicPotential/

 

[Konte]

Thanks everybody for all of your answers.

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?

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 $\hat{V}(\theta)\,=\, 3\,-\,sin(7\,\theta)$.

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 $\theta$.

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

 

[stevendaryl]

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

$x$ 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 $\theta = \frac{2\pi x}{L}$.

 

[Konte]

$x$ 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 $\theta = \frac{2\pi x}{L}$.

Ok, thanks a lot.