3.3 Energy Eigenstates of the QHO

There is one final step to writing out the eigenstates, which is to return to physical units and remove the dimensionless length $y$ . Once we do this, the energy eigenstates (including time dependence) have the form:

\psi_{n}(x,t)=N_{n}h_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-\frac{m% \omega}{2\hbar}x^{2}}e^{-i\omega t\left(n+\frac{1}{2}\right)}.

(3.43)

The normalisation constant $N_{n}$ turns out to be

N_{n}=\frac{1}{\sqrt{2^{n}n!}}\sqrt[4]{\frac{m\omega}{\pi\hbar}}.

(3.44)

The spatial part of the first few eigenstates is

$\displaystyle u_{0}(x)$	$\displaystyle=N_{0}e^{-\frac{m\omega}{2\hbar}x^{2}}$	(3.45)
$\displaystyle u_{1}(x)$	$\displaystyle=N_{1}xe^{-\frac{m\omega}{2\hbar}x^{2}}$	(3.46)
$\displaystyle u_{2}(x)$	$\displaystyle=N_{2}\left(\frac{2m\omega}{\hbar}x^{2}-1\right)e^{-\frac{m\omega% }{2\hbar}x^{2}}$	(3.47)
$\displaystyle u_{3}(x)$	$\displaystyle=N_{3}\left(8\sqrt{\left(\frac{m\omega}{\hbar}\right)^{3}}y^{3}-1% 2\sqrt{\frac{m\omega}{\hbar}}y\right)e^{-\frac{m\omega}{2\hbar}x^{2}}$	(3.48)

3.3.1 Comparison of the Quantum to Classical Harmonic Oscillator

Here are the probability density of the first few eigenstates, with the classical probability density of a simple harmonic oscillator with energy $E=\hbar\omega\left(n+\frac{1}{2}\right)$ overlayed. We can see that as $n$ increases, the spatial probability density of the oscillator is more spread out, just like how the amplitude of them classical turning points increase for higher energy. The quantum probability density is actually nonzero beyond the classical turning point, so we have some tunnelling, but it decays exponentially in this region. Also note that the wavelength of $\psi_{n}$ is shortest at the center of the well, which is where a classical particle would be travelling the fastest.

3.3.2 Parity

Recall that we saw in section 3.2.3 that the Hermite polynomials are either even or odd. This property is passed on to the eigenstates as well. We can investigate this symmetry by defining an operator which will detect whether a given wavefunction is odd or even.

Definition 3.1.

The parity operator switches $x$ to $-x$ , and can be defined by its action on a wavefunction as

\hat{P}\psi(x,t)=\psi(-x,t).

(3.50)

What are the possible eigenvalues of parity? If we apply parity twice, we get

\hat{P}^{2}\psi(x)=\hat{P}\hat{P}\psi(x)=\hat{P}\psi(-x)=\psi(x),

(3.51)

so all wavefunctions are eigenfunctions of $\hat{P}^{2}$ with eigenvalue $+1$ . Using this fact, we can show that the eigenvalues of $\hat{P}$ are $\pm 1$ . Suppose that $\psi(x)$ is an eigenfunction of $\hat{P}$ with eigenvalue $P$ , then

\hat{P}^{2}\psi(x)=\hat{P}P\psi(x)=P^{2}\psi(x),

(3.52)

implies that $P^{2}=1$ , hence $P=\pm 1$ . If a wavefunction is an eigenfunction of parity with $\hat{P}\psi(x)=\psi(x)$ , then this means that $\psi(x)$ is an even function (symmetric about $x=0$ ) since this implies $\psi(-x)=\psi(x)$ . Likewise, if we have $\hat{P}\psi(x)=-\psi(x)$ , then $\psi(x)$ must be odd (antisymmetric about $x=0$ ) since this implies $\psi(x)=-\psi(-x)$ . Not all wavefunctions are even or odd, therefore not all wavefunctions are eigenfunctions of parity.

Is parity Hermitian? We can test this by calculating the adjoint, note that the integral has to be from $-\infty$ to $\infty$ for the harmonic oscillator.

	$\displaystyle\langle\hat{P}^{\dagger}\rangle$	$\displaystyle=\int_{-\infty}^{\infty}(\hat{P}\psi(x))^{\ast}\psi(x)% \operatorname{\mathrm{d}}\!x$		(3.53)
		$\displaystyle=\int_{-\infty}^{\infty}\psi^{\ast}(-x)\psi(x)\operatorname{% \mathrm{d}}\!x,$		(3.54)

now we make a substitution $y=-x$ to get

$\displaystyle\langle\hat{P}^{\dagger}\rangle$	$\displaystyle=-\int_{\infty}^{-\infty}\psi^{\ast}(y)\psi(-y)\operatorname{% \mathrm{d}}\!y$	(3.55)
	$\displaystyle=\int_{-\infty}^{\infty}\psi^{\ast}(y)\hat{P}\psi(y)\operatorname% {\mathrm{d}}\!y$	(3.56)
	$\displaystyle=\langle\hat{P}\rangle,$	(3.57)

hence $\hat{P}$ is Hermitian and therefore it is observable.

3.3.3 Superposition States

We can write superposition states just like we can for the infinite square well, using an expansion of energy eigenfunctions.

\psi(x,t)=\sum_{n=0}^{\infty}c_{n}u_{n}(x)e^{-\frac{iE_{n}t}{\hbar}}.

(3.58)

The only difference between this formula and the one for the infinite square well is that the index $n$ starts from 0 instead of 1.