Stationary states parity of an Anisotropic Harmonic Oscillator

Question

An anisotropic harmonic oscillator in three dimensions is given by the potential:

$$V(X,Y,Z)=\frac{1}{2}m\omega^2\bigg[\bigg(1+\frac{2\lambda}{3}\bigg)(X^2+Y^2)+\bigg(1-\frac{4\lambda}{3}Z^2\bigg)\bigg].$$

The energy eigenvalues of the stationary states are:

$$E_{n_x,n_y,n_z}=(n_x+n_y+1)\hbar\omega\sqrt{1+\frac{2\lambda}{3}}+(n_z+\frac{1}{2})\hbar\omega\sqrt{1-\frac{4\lambda}{3}}.$$

It is obvious that the potential has a defined parity, because under a change $\boldsymbol{r}\rightarrow-\boldsymbol{r}$ it remains the same function. How does this defined parity affect the eigenvalues of $H$?

Because there is no explicit dependence on the coordinates in $E_{n_x,n_y,n_z}$, it seems like there is nothing to say about the parity of the eigenvalues. Still, the problem I am posting is from Cohen-Tannoudji's book of quantum mechanics, and he asks to discuss the parity and the degree of degeneracy of the ground State for this harmonic oscillator.

Answer 1

Let us start from the one-dimensional case. The the parity operator $R$ is defined as $$(R\psi)(x) := \psi(-x)$$ for every wavefunction $\psi \in L^2(\mathbb R, dx)$.

$R$ is selfadjoint and unitary and, directly from its definition, we get $RX=-XR$ and $RP=-PR$ where $X$ and $P$ are the position and momentum operators.

Since the creation operator $a^\dagger$ and the annihilation operator $a$ of the harmonic oscillator are linear combinations of $X$ and $P$, the same commutation relations are valid for $a$ and $a^\dagger$. In particular, $$Ra^\dagger=-a^\dagger R\:.\tag{1}$$ Since $R|0\rangle =|0\rangle$ as it arises by direct inspection in position representation (the wavefunction of the ground state is proportional to $e^{-cx^2}$ which is an even function), we have from (1) that $$R|n\rangle = R \frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle = (-1)^n \frac{(a^\dagger)^n}{\sqrt{n!}}R|0\rangle = (-1)^n |n\rangle\:.\tag{2}$$

In the considered case $$|n_1 n_2 n_3\rangle = |n_1\rangle\otimes |n_2\rangle \otimes |n_3\rangle\tag{3}$$ and the parity operator $R$ is the tensor product of the corresponding three parity operators $$R = R_1 \otimes R_2 \otimes R_3$$ so that the parity of the state $|n_1 n_2 n_3\rangle$ is $(-1)^{n_1+ n_2 +n_3}$. Indeed, from (1),(2), and (3): $$R |n_1 n_2 n_3\rangle = (-1)^{n_1+ n_2 +n_3} |n_1 n_2 n_3\rangle\:.$$ This is the parity of the eigenvalue, unless there is accidental degeneracy: $$E_{n_1, n_2, n_3} = E_{m_1, m_2, m_3}\quad \mbox{for some $(n_1, n_2, n_3) \neq (m_1, m_2, m_3)$,}$$ in that case the eigenvalue may have not defined parity. Accidental degeneracy arises under swap of $n_1$ and $n_2$ in the considered case, however the parity is invariant under this operation.