1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.