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Question:

If a particle is attached to a semi-harmonic oscillator (that is, for example, the spring is stretchable but not compressible) such that the potential $V(x)$ is infinity for $x\leq0$ and $V(x)=\frac{1}{2}m\omega^2x^2$ for $x>0$. What is the allowed energies for the particle?

My intuition was that this doesn't really make a change in the energies, since the values of the energy won't be related to the region $x\leq0$ and for elsewhere it's just the positive part of the simple harmonic oscillator. So the energies are still given by $E_n=\hbar\omega(n+\frac{1}{2})$.

But it concerns me that since the oscillator can only be stretched, the particle will be bounced back when it reaches the equilibrium position $x=0$. I don't know if this will change the energies of the particle. Could someone tell me whether my thinking is right, and if not, what am I missing?

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  • $\begingroup$ @dmckee you are correct I forgt to mention that only a subset of the solutions will be valid. I guess I was too tired yesterday. I'll erased my answer. $\endgroup$ – Wolphram jonny Dec 12 '14 at 14:03
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This seems like a reasonable homework-like question, so I'll provide a hint.

  1. Realize that, for $x>0$, both the simple and semi oscillators have the same potential $V(x)$.
  2. For the simple oscillator, draw out the first few of the energy eigenfunctions $\psi_n(x)$.
  3. Now for the semi-harmonic oscillator, think about what the boundary condition on any wavefunction $\psi(x)$ at the equilibrium point $x=0$ ought to be. If you don't know, perhaps think about the infinite square well.
  4. Think about how (1)--(3) above can help you determine the allowed energy eigenstates for the semi oscillator.
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