- #1

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- 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?

##\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?