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I Periodic potential - energy bands

  1. Oct 30, 2016 #1
    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. jcsd
  3. Oct 31, 2016 #2
    I wouldn't think so. The potential is just a step function, the height of which depends on the direction.
     
  4. Oct 31, 2016 #3
  5. Oct 31, 2016 #4
    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?
     
  6. Oct 31, 2016 #5

    stevendaryl

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

    The potential [itex]V = a sin(\theta)[/itex] isn't periodic in this sense. Yes, [itex]V(\theta + 2 \pi) = V(\theta)[/itex], but the angle [itex]\theta + 2\pi[/itex] is the same angle as [itex]\theta[/itex].
     
  7. Oct 31, 2016 #6
  8. Oct 31, 2016 #7
    Thanks everybody for all of your answers.

    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)##.
    sin7x.png

    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
     
  9. Oct 31, 2016 #8

    stevendaryl

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    [itex]x[/itex] and [itex]\theta[/itex] 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 [itex]x[/itex] that runs from [itex]-\infty[/itex] to [itex]+\infty[/itex]. In contrast, the angular variable [itex]\theta[/itex] runs from [itex]-\pi[/itex] to [itex]+\pi[/itex], and the point [itex]\theta = -\pi[/itex] is identified with the point [itex]\theta = +\pi[/itex]. These differences lead to different constraints on the wave function. In terms of [itex]x[/itex], it must be the case that [itex]lim_{x \rightarrow \pm \infty} \psi(x) = 0[/itex]. In terms of [itex]\theta[/itex], it must be the case that [itex]\psi(\theta) = \psi(\theta + 2\pi)[/itex]. 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 [itex]x[/itex] with period [itex]L[/itex]---the wave function, the potential, everything---then the distinction disappears; In that case you can identify [itex]x[/itex] with the angular variable [itex]\theta = \frac{2\pi x}{L}[/itex].
     
  10. Oct 31, 2016 #9
    Ok, thanks a lot.
     
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