Suppose we have a harmonic oscillator confined in an even infinite square well with width $2a$. The potential is given by $V(x) = \frac{1}{2} m \omega^2x^2, -a<x<a$ and $V(x) \to \infty, x<-a $ or $x>a$
Along with the harmonic oscillator, we have the boundary conditions:
- $\psi(a) = 0$
- $\psi(-a) = 0$
since it is confined in an infinite square well.
Suppose now that $a=\sqrt{\frac{3\hbar}{2m\omega}} \neq 0 $
The energy eigenfunctions for the harmonic oscillator are:
$$\psi_n(x) = \sqrt{\frac{n}{2^n\cdot n!}} (\frac{1}{\pi})^\frac{1}{4}H_n(bx)e^{\frac{-b^2x^2}{2}}$$
with $b = \sqrt(\frac{m\omega}{\hbar})$.
Then the only way for $\psi(a)$ and $\psi(-a)$ to be 0 is if the respective hermite polynomial is zero: $H_n(bx) = 0 $
So I guess that the ground state will be the first Hermite polynomial that is zero for this specific $a$.
It is:
- $H_1 = 1 \neq 0$
- $H_2 = 2ba \neq 0$
- $H_3 = 0$
So the ground state is for $n=3$.
But then, if I continue to examine Hermite polynomials, we find that no other is zero for $a=\sqrt{\frac{3\hbar}{2m\omega}}$.
So I am unable to find a first excited state!
Is it that the harmonic oscillator has only one energy level (ground state). If that's the case, what's the intuition behind it?
If not, what is happening here exactly?
