# Can Bloch Waves Reveal Periodic Potentials in Quantum Mechanics?

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• hilbert2
In summary: The function ##\psi## is not periodic, so the potential energy corresponding to it is not periodic either.
hilbert2
Gold Member
TL;DR Summary
Finding a 1d quantum system with potential energy V(x) producing a closed form wave function.
I was thinking about a problem I had considered a long time ago in some thread, finding an example of a wave function ##\displaystyle \psi (x) =e^{iax}\phi (x)## with ##\displaystyle\phi (x)## being periodic with period ##\displaystyle L## and the corresponding Schrödinger equation

##\displaystyle -\frac{\hbar^2}{2m}\frac{d^2 \psi (x)}{dx^2} + V(x)\psi (x) = E\psi (x)##,

such that the

##\displaystyle V(x) = E+\frac{\hbar^2}{2m}\frac{\psi'' (x)}{\psi (x)}##

is a periodic potential with period ##L##, not singular anywhere and has a closed form expression.

Now, I was able to guess a wave function ##\displaystyle \psi (x) = \sin (x) - \frac{1}{4}\sin (2x) + \frac{1}{16}\sin (3x)##, for which the potential is

##\displaystyle V(x) = -\frac{\hbar^2}{2m}\frac{\sin(x) - \sin (2x) + \frac{9}{16}\sin (3x)}{\sin (x) - \frac{1}{4}\sin (2x) + \frac{1}{16}\sin (3x)}##

(when the energy eigenvalue is arbitrarily set to ##E=0##)

and for which the graphs of ##\displaystyle\psi (x)## (red line) and ##\displaystyle V(x)## (blue line) are like in the image (where it is set that ##\hbar^2 /2m = 1##)

The plane wave part of the Bloch wave function, ##\displaystyle e^{iax}##, is just a constant ##1## in this case, which means that ##a=0##.

Had I chosen ##\displaystyle \psi (x) = \sin (x) + \sin (2x) + \sin (3x)##, the potential ##V(x)## would have singular points and not look like a physically possible one.

So, anyone have clues on how it could be seen from the coefficients ##C_k## of ##\displaystyle\psi (x) = \sum\limits_{k=1}^{\infty}C_k \sin (k x)## whether the potential energy producing that eigenstate is also a 'nice' function?

The only thing that really jumps to mind is trying to figure out a relation between ##\psi## and ##\psi''## such that their zeroes are coincident. The singularities must come from ##\psi## passing through zero. Also, for the taylor series around a zero of ##\psi''## the lowest order term can't have a power of ##x## less than that of the ##\psi##. Maybe you can play around with the series and some of these facts to get a relation on the ##C_k##?

Yes, the position of zeroes also came to my mind. If the series contains terms like ##\cos kx## instead of ##\sin kx##, it's more difficult to make them coincide because a zero of ##\cos (x)## is not necessarily a zero of ##\cos (kx)## with ##k\in\mathbb{N}##.

I couldn't find a nonzero value of ##a\in\mathbb{R}## such that the potential energy corresponding to the function ##\psi (x) = e^{iax}\left(\sin (x) - \frac{1}{4}\sin (2x) + \frac{1}{16}\sin (3x)\right)## would be a real valued function, so this probably only happens at individual values of ##a##. The idea of the Bloch theorem is that a periodic ##V(x)## should produce a wave function that is periodic except possibly for the phase, as it is in the case of that ##\psi (x)## with period ##L=\pi##. Maybe I should plot functions ##\psi'' /\psi## for many different values of ##a## and then see where the imaginary part stays small.

hilbert2 said:
##\displaystyle V(x) = -\frac{\hbar^2}{2m}\frac{\sin(x) - \sin (2x) + \frac{9}{16}\sin (3x)}{\sin (x) - \frac{1}{4}\sin (2x) + \frac{1}{16}\sin (3x)}##
has the wrong dimensions.

Yeah, the multipliers of ##x## in the ##\sin (kx)## terms should have dimensions of reciprocal length, but the system will be made dimensionless anyway by setting ##\hbar^2 /2m = 1## and this calculation is just to see what form of the function ##\psi## produces an acceptable function ##V##.

hilbert2 said:
Yeah, the multipliers of ##x## in the ##\sin (kx)## terms should have dimensions of reciprocal length, but the system will be made dimensionless anyway by setting ##\hbar^2 /2m = 1## and this calculation is just to see what form of the function ##\psi## produces an acceptable function ##V##.
But it should work with dimensions. You can have a normalization constant with dimensions of ##L^{-1/2}## to take care of the dimensions of ##\psi##, but ##\psi''## will not have the right dimensions. It makes me think that the ##\psi## you used is not a proper wave function.

If the ##\psi## has terms like ##\sin (kx)## and the ##k## is of dimensions ##1/L##, then the 2nd derivative gets multipliers like ##1/L^2## in front of the terms. I'd usually handle something like this by choosing atomic units by setting ##\hbar = m = 1## and then taking for granted that the ##V(x)## is in units of hartree and ##x## in units of bohr. A more complicated way would be to form a dimensionless length variable ##s## of form ##s = \hbar^\alpha m^\beta e^\gamma \epsilon_{0}^{\delta} x## with ##\alpha ,\beta ,\gamma ,\delta## some rational numbers and a similar dimensionless energy variable. The ##e## and ##\epsilon_{0}## there are the elementary charge and vacuum permittivity. Then the ##\psi (x)## could be written in terms of ##s##.

Haborix said:
The only thing that really jumps to mind is trying to figure out a relation between ##\psi## and ##\psi''## such that their zeroes are coincident.
This must be it. We all know that we can't divide by zero.

Looking at your graph of ##V(x)##, your choice of prefactors gives you -1 at ##x=0##. But inputting 0 into ##V(x)## gives you ##-0/0##. I applied L'Hopitals rule to $$V(x)=-\frac{C_1\sin{x}+4C_2\sin{2x}+9C_3\sin{3x}}{C_1\sin{x}+C_2\sin{2x}+C_3\sin{3x}}$$as ##x\rightarrow 0## and set it equal to -1. ##C_1## subtracts out and leaves ##C_2=-4C_3##. That is consistent with your prefactors. Looking at your graph for the next value and repeating that same method with $$\lim_{x \rightarrow \pi} V(x) = -11/4$$ gives me a value for ##C_1## when I choose ##C_3=\frac{1}{16}##. However, it does not reproduce your graph and, in fact, gives me poles when plotted.

It's late here. You should probably try the calculation yourself to see if I made a mistake.

Dr_Nate said:
Looking at your graph for the next value and repeating that same method with $$\lim_{x \rightarrow \pi} V(x) = -11/4$$ gives me a value for ##C_1## when I choose ##C_3=\frac{1}{16}##. However, it does not reproduce your graph and, in fact, gives me poles when plotted.

If I calculate the ##\lim\limits_{x\rightarrow \pi}V(x)## with L'Hopital, it gets the value

##\displaystyle\lim\limits_{x\rightarrow \pi}\left(-\frac{\sin x - \sin 2x + \frac{9}{16}\sin 3x}{\sin x - \frac{1}{4}\sin 2x + \frac{1}{16}\sin 3x}\right)\\\displaystyle =\lim\limits_{x\rightarrow \pi}\left(-\frac{\cos x - 2\cos 2x + \frac{27}{16}\cos 3x}{\cos x - \frac{1}{2}\cos 2x + \frac{3}{16}\cos 3x}\right)\\\displaystyle =-\frac{\cos \pi .-2\cos 2\pi + \frac{27}{16}\cos 3\pi}{\cos \pi - \frac{1}{2}\cos 2\pi + \frac{3}{16}\cos 3\pi}\\\displaystyle =-\frac{1+2+\frac{27}{16}}{1+\frac{1}{2}+\frac{3}{16}}\\\displaystyle =-\frac{25}{9} \approx -2.78##.

This is clearly the same value as in the graph.

I now get ##C_1 = 16 C_3, C_2=-4C_3## given your values at ##x=0## and ##\pi##. This seems like a method that works.

Another wavefunction I thought of is an approximation of the cosine function made from alternating downward and upward opening parabolas:

##\displaystyle\psi (x) = \sum\limits_{n\in 2\mathbb{Z}}(-1)^{n/2}(1-(x-n)^2 )\theta (1-|x-n|)##

where ##\theta (x)## is the Heaviside theta function and the graph of which is like

Then, because for ##\displaystyle\psi (x) = 1-(x-n)^2##

##\displaystyle\frac{\psi '' (x)}{\psi (x)} \propto \frac{1}{1-(x-n)^2}##

the potential energy ##V(x)## is like a 1D "lattice" of inverse square potential energies

at least if the "lattice spacing" is made large enough for the ##V(x)## to be very close to zero at the midpoint between two singularities.

Didn't immediately find any applications of this lattice model with a Google search, but that piecewise defined parabola function was a quite curious object.

## 1. What is an exactly solved Bloch wave function?

An exactly solved Bloch wave function is a mathematical expression that describes the behavior of a quantum particle in a periodic potential. It takes into account the periodicity of the potential and allows for the calculation of the particle's energy and momentum.

## 2. How is an exactly solved Bloch wave function different from other wave functions?

An exactly solved Bloch wave function is unique because it takes into account the periodicity of the potential, whereas other wave functions may not. This makes it particularly useful for studying particles in periodic structures, such as crystals.

## 3. What are some applications of exactly solved Bloch wave functions?

Exactly solved Bloch wave functions have a wide range of applications in physics, particularly in the study of solid state systems. They are used to understand the electronic properties of materials, such as their conductivity and band structure.

## 4. How is an exactly solved Bloch wave function calculated?

An exactly solved Bloch wave function is typically calculated using the Schrödinger equation and boundary conditions that take into account the periodicity of the potential. This can be a complex process and often requires the use of advanced mathematical techniques.

## 5. What are some limitations of exactly solved Bloch wave functions?

While exactly solved Bloch wave functions are useful for studying particles in periodic structures, they may not accurately describe particles in non-periodic systems. Additionally, they may not take into account certain effects, such as electron-electron interactions, which can play a significant role in some systems.

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