How Do Bloch Wavefunctions Constrain Real-Valued Potentials?

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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
-\frac{1}{2}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x)\psi(x)=E\psi(x)

According to Bloch theorem, the energy eigenstates are of form \psi(x)=exp(ikx)\phi(x)
where \phi(x) 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:

V(x)=\frac{\psi''(x)}{2\psi(x)}

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

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

From this equation one can easily see that the only way how a real-valued \phi(x) can correspond to a real-valued potential V(x) is that \phi(x) is the trivial constant function. Therefore, in most Bloch wavefunctions that correspond to a physically possible potential, \phi(x) 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 \phi(x) 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.
 
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Looks like no one's interested in this... I did some calculations myself. Let's say \phi(x)=f(x)+ig(x), where ##f## and ##g## are real functions. Now we have

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)

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

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?
 
If you put your Bloch function into your Schrodinger equation you will find the result is
\tfrac12(p + k)^2 \phi(x) + V(x) \phi(x) = E \phi(x)
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 v = \partial E / \partial k). 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.
 
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 \phi(x)=1, the full wavefunction becomes \psi(x)=e^{ikx}\times 1=e^{ikx} 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
 
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 \phi(x) is real or complex is generally unimportant, since the wavefunction \psi(x) 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 \phi(x).
 
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