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I just got out of an exam today, and have been wrestling with one of the questions. I was given a simple (quantum) 1D harmonic oscillator, with potential given by: $${V(x)=\infty, \forall x<0}$$ $$V(x)=(\frac{1}{2}m\omega^{2}x^{2}), \forall x\ge 0$$

The question called for me to sketch the wave function $\psi(x)$ for the oscillator at its ground state, and compare the ground-state energy to that for the same oscillator without the potential spike.

I know what the basic graph sketches look like for the "normal" potential and wave function, but I'm completely puzzled on how $\psi(x)$ would look in this case, or even how one would derive a new expression for it.

Many thanks in advance for any advice you all can offer!

OldWorldBlues
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1 Answers1

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If we write down the Schrodinger equation, it has the same form as that for a harmonic oscillator for $x\geq 0$ (i.e., satisfied by the same solutions), but the boundary condition at $x=0$ is that at the infinite potential wall, $\psi(0)=0$. That is of all the solutions of HO we need to choose the ones, satisfying this bc.

Roger V.
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