# Wavefunction for shifted harmonic oscillator potential

1. Sep 24, 2016

### spaghetti3451

1. The problem statement, all variables and given/known data

Consider the following potential, which is symmetric about the origin at $x=0$:

$V(x) = \begin{cases} x^{2}+(x+\frac{d}{2}) &\text{for}\ x < -d/2\\ x^{2} &\text{for}\ -d/2 < x < d/2\\ x^{2}-(x-\frac{d}{2}) &\text{for}\ x > d/2 \end{cases}$

Find the ground state energy and wavefunction for this potential.

2. Relevant equations

3. The attempt at a solution

For $x < -d/2$, $V(x) = (x+\frac{1}{2})^{2}+\frac{2d-1}{4} = (x+\frac{1}{2})^{2}$,and

for $x > d/2$, $V(x) = (x-\frac{1}{2})^{2}+\frac{2d-1}{4} = (x-\frac{1}{2})^{2}$.

So, the wavefunctions in these two sectors are the shifted-in-position harmonic oscillator wavefunctions.

So, for $x < -d/2$, $\psi_{n}(x) \sim \text{exp}\Big(-\frac{m\omega}{2\hbar}(x-\frac{1}{2})^{2}\Big)\ H_{n}\Big(\sqrt{\frac{m\omega}{\hbar}}(x-\frac{1}{2})\Big)$,

for $-d/2 < x < d/2$, $\psi_{n}(x) \sim \text{exp}\Big(-\frac{m\omega}{2\hbar}x^{2}\Big)\ H_{n}\Big(\sqrt{\frac{m\omega}{\hbar}}x\Big)$, and

for $x > d/2$, $\psi_{n}(x) \sim \text{exp}\Big(-\frac{m\omega}{2\hbar}(x+\frac{1}{2})^{2}\Big)\ H_{n}\Big(\sqrt{\frac{m\omega}{\hbar}}(x+\frac{1}{2})\Big)$.

But, the wavefunctions at the potential kinks at $x=-d/2$ and $x=d/2$ do not match.

Is there some sorcery of the kink in the potential - perhaps the discontinuity at $V'(-d/2)$ and at $V(d/2)$ - that causing this behaviour of the wavefunction?

2. Sep 24, 2016

### vela

Staff Emeritus
I haven't looked into this problem much, but the thought that occurs to me is that you can't assume the solution for the middle region is the regular SHO solution. Those solutions were derived assuming the wave function had to be normalizable when integrating over all x. You no longer have that constraint.