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Possible Bloch wavefunctions

  1. Jul 17, 2013 #1

    hilbert2

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    Consider an electron in a periodic potential V(x) such that V(x+a) = V(x) for some real number a. The energy eigenstates are obtained from time-independent SE, which in atomic units is
    [itex]-\frac{1}{2}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x)\psi(x)=E\psi(x)[/itex]

    According to Bloch theorem, the energy eigenstates are of form [itex]\psi(x)=exp(ikx)\phi(x)[/itex]
    where [itex]\phi(x)[/itex] has the same periodicity as V(x).

    If at least one eigenfunction is known, the potential V(x) can be solved from the SE with an inverse formula:

    [itex]V(x)=\frac{\psi''(x)}{2\psi(x)}[/itex]

    Here the eigenvalue E has been arbitrarily chosen to be zero (changing its value only corresponds to adding a constant term to V(x). Plugging the expression for Bloch wavefunction in this equation and differentiating, we get

    [itex]V(x)=-k^{2}+ik\frac{\phi'(x)}{\phi(x)}+\frac{\phi''(x)}{2\phi(x)}[/itex]

    From this equation one can easily see that the only way how a real-valued [itex]\phi(x)[/itex] can correspond to a real-valued potential V(x) is that [itex]\phi(x)[/itex] is the trivial constant function. Therefore, in most Bloch wavefunctions that correspond to a physically possible potential, [itex]\phi(x)[/itex] is a complex-valued function.

    Questions: Why is the range of physically possible Bloch wavefunctions so limited? What's the simplest way to express the minimal condition for function [itex]\phi(x)[/itex] that guarantees real-valued V(x) ? Can anyone give even one nontrivial example of a (differentiable) Bloch-type wavefunction that corresponds to a real potential.
     
    Last edited: Jul 17, 2013
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  3. Jul 22, 2013 #2

    hilbert2

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    Looks like no one's interested in this... I did some calculations myself. Let's say [itex]\phi(x)=f(x)+ig(x)[/itex], where ##f## and ##g## are real functions. Now we have

    [itex]V(x)=-\frac{k^{2}}{2}+ik\frac{\phi'(x)}{\phi(x)}+\frac{\phi''(x)}{2\phi(x)}\\=-\frac{k^{2}}{2}+k\frac{f'(x)g(x)-g'(x)f(x)}{[f(x)]^{2}+[g(x)]^{2}}+\frac{1}{2}\frac{f''(x)f(x)-g''(x)f(x)}{[f(x)]^{2}+[g(x)]^{2}}+i\left(k\frac{f'(x)f(x)+g'(x)g(x)}{[f(x)]^{2}+[g(x)]^{2}}+\frac{1}{2}\frac{g''(x)f(x)-f''(x)g(x)}{[f(x)]^{2}+[g(x)]^{2}}\right)[/itex]

    and if we want [itex]Im\left(V(x)\right)=0[/itex], we must have [itex]f''(x)g(x)-g''(x)f(x)=2k(f'(x)f(x)+g'(x)g(x))[/itex] .

    Let's try ##f(x)=sin(x)##. Plugging this in the previous equation, we get a condition for ##g(x)##:

    ##sin(x)g''(x)+2kg'(x)g(x)+sin(x)g(x)+ksin(2x)=0## .

    This is a nonlinear ODE, and Wolfram gave me a terrifyingly complicated solution to it...

    Does anyone see a way how we could tell something more about the possible Bloch wavefunctions?
     
  4. Jul 22, 2013 #3
    If you put your Bloch function into your Schrodinger equation you will find the result is
    [tex]\tfrac12(p + k)^2 \phi(x) + V(x) \phi(x) = E \phi(x)[/tex]
    so you can see that the value of k is, in a sense, acting as some additional momentum. k is in fact called the pseudomomentum or crystal momentum because of its relationship to the momentum, and that it has a conservation law that is similar to conservation of momentum in that it arises from the translational symmetry of space.

    Anyway, physically, solutions where k /= 0 are traveling solutions. These solutions have a non-zero velocity (given by [itex]v = \partial E / \partial k[/itex]). It is fairly straightforward to prove that any wavefunction which is purely real, or can be made purely real by application of a constant phase factor, is stationary, and any wavefunction which is complex in a non-trivial way is not stationary. So these wavefunctions with k /= 0 have to be complex.
     
  5. Jul 22, 2013 #4

    hilbert2

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    Thanks for your reply. One can't a priori say that ##\phi(x)## has to be complex-valued for the full wave function to correspond to an unbound state. For example, if [itex]\phi(x)=1[/itex], the full wavefunction becomes [itex]\psi(x)=e^{ikx}\times 1=e^{ikx}[/itex] which is not a real function or a stationary state.

    What I'm after here, is an example of a function ##\phi(x)## that

    a) is periodic
    b) corresponds to a real valued potential V(x) and
    c) has a simple enough functional form that I can actually write it down and show to someone
     
  6. Jul 22, 2013 #5
    I sort of see what you're asking here. But first let me point out that your example satisfies all those conditions. Of course, the potential ends up being a constant.

    Except for when k = 0, whether [itex]\phi(x)[/itex] is real or complex is generally unimportant, since the wavefunction [itex]\psi(x)[/itex] will be complex.

    The thing that makes this unimportant is that you will have a single potential for all your electrons.
    Your good quantum numbers are the pseudomomentum k and a band index n, and you will generally have occupied states at every allowed value of k. All values of k from -pi/a to pi/a are relevant. While you may find, for some potential, that at some value of k /= 0 that phi(x) is real, phi(x) at another k won't be real. You have the differential equation which gives you the potential that satisfies phi(x) is real at some k, surely you can see that at a different k you will have a different potential. But this doesn't correspond to a real situation which is interesting in solid-state physics.

    AFAIK, there is no good example of a simple solution to the Bloch equation. I've never seen any example in a textbook. Even with the simplest periodic potential of V(x) = sin(x) you get a rather difficult to solve equation for [itex]\phi(x)[/itex].
     
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