Question regarding the half Harmonic Oscillator

Question

In the normal Quantum Harmonic Oscillator (QHO), we normally use the operator method (because it's to elegant), but I recently discovered the problem in Griffiths (prob 2.42) where they ask the same question of QHO, but where the harmonic potential is there only for $x>0$, while for $x<0$, it is $\infty$. Everywhere I looked for the possible solutions, I found the same reasoning that the even parity state solutions cannot exist due to the boundary condition $\psi(0)=0$.

I wanted to prove this more rigorously, so I again applied the operator method for the region of $x>0$, since for $x<0$ the Hamiltonian is just $\infty$. The part which made me wonder the most was that the spectrum is still formally the same, $$E_n=\left(n+\frac{1}{2}\right)\hbar\omega,$$ where there is no constraint on n for the full oscillator, whereas for the half oscillator, n can only take odd values ($n=1,3,5,...$). My question is: what is going on?

The operator approach fails miserably. What is wrong with this? I thought that it might be that n=0 for my case corresponds to $n=1$ solution, but nope: it just fails.

My Questions are:

1. what happens to the operator approach in half harmonic oscillator? Am I doing something wrong?
   

2. Is there a change in the Ladder operators?
   

3. What about the boundary conditions in the operator approach?

Answer 1

Your vacuum breaks parity, and @JosBergervoet takes you somewhere close, but not quite. 1. Probably, but you never showed your work. 2. In a messy way, see below and his point. 3. The BCs best encapsulate parity.

Let's look at the new "vacuum", and "translate" from the wave function approach, since students learn best by example. Ignore normalization distractions for the time being. Background: 1, 2, 3.

The wavefunctions of your half-oscillator are the truncated odd-parity states of the full oscillator, so your new parity breaking ground state ("vacuum") is $(\hat x +|\hat x|)/2$ acting on $$ \psi_1(x)- \psi_1(-x)= \langle x| a^\dagger|0\rangle - \langle -x| a^\dagger|0\rangle= \langle x|(1-\hat \Pi)a^\dagger|0\rangle , $$ since the parity operator projects out even powers of the full oscillator creation operator, $$ \hat \Pi ^2={\mathbb I}, \qquad \hat \Pi |x\rangle =|-x\rangle,\\ \hat \Pi a \hat \Pi = -a, \qquad \hat \Pi a^\dagger \hat \Pi = -a^\dagger \\ \hat \Pi = \hat \Pi \int\!\! dx ~~|x\rangle\langle x|=\int\!\! dx ~~|-x\rangle\langle x|= e^{i\pi a^\dagger a }. $$

So then, e.g., $$ (1-\hat \Pi)(a^\dagger)^n|0\rangle =(1-e^{i\pi\hat N})(a^\dagger)^n|0\rangle $$ vanishes for n even, including 0, but not odd.

Your first excited state is then the $(\hat x +|\hat x|)/2$ truncation of $$ (1-e^{i\pi\hat N})(a^\dagger)^3|0\rangle, $$ etc. Your new hamiltonian does not commute with parity, of course...

It is quite problematic to set up a consistent ladder operator recursion, then, but I should not presume to discourage you...

_{PS One might well use a conventional approximation for $|\hat x|$...}