Raising the Gaussian once multiplies it by $\xi$; raising again gives a quadratic times the same Gaussian; each rung adds one degree to the polynomial in front. The polynomials are Hermite’s, known long before quantum mechanics. They carry everything visible about the excited states: the $n$-th has $n$ nodes, and the states alternate between even and odd because the potential looks the same from both sides.

The modes of a plucked string are sines with one more crossing each; the modes of the oscillator are the same idea in a parabolic bowl, each with one more crossing than the last and all fading away at the edges.

The first four states of the oscillator, each drawn on its own level inside the potential, in units of $x_{0}$ and $\hbar\omega$. The $n$-th has $n$ nodes, the states are even and odd in turn, and each reaches a little beyond the dashed stretch a classical particle of its energy could cover.

Raising the ground state $n$ times with $\hat{a}^{\dagger}=\tfrac{1}{\sqrt2}\bigl(\xi-\tfrac{d}{d\xi}\bigr)$ gives a polynomial of degree times the same Gaussian. The polynomials are the Hermite polynomials .

• $H_{0}=1$, $H_{1}=2\xi$, , .

Apply $\hat{a}^{\dagger}$ in its position form to the Gaussian: differentiating brings down $-\xi$, and the minus sign in $\hat{a}^{\dagger}$ doubles the already there. Applying it again to gives the quadratic, and the factor of is divided out.

The adjoint of $\hat{a}=\frac{1}{\sqrt2}\bigl(\xi+\frac{d}{d\xi}\bigr)$: the derivative changes sign.

The derivative of the Gaussian is $-\xi$ times it.

Multiply by the normalisation $\pi^{-1/4}$; $\hat{a}^{\dagger}\ket{0}=\ket{1}$ needs no further factor.

The derivative of $\xi e^{-\xi^{2}/2}$ is $(1-\xi^{2})e^{-\xi^{2}/2}$.

Divide by $\sqrt2$, since $\hat{a}^{\dagger}\ket{1}=\sqrt2\ket{2}$. This is with its normalisation.

The $n$-th state is a polynomial of degree $n$ times a Gaussian, and the polynomial has $n$ real zeros: the state has $n$ nodes.

The potential is the same at $x$ and $-x$, and the states take turns: $\psi_{0}$, $\psi_{2}$, are even and , , are odd.

Each Hermite polynomial follows from the two before it, so any of them can be written down in a few lines.

States of different energy are orthogonal, because the Hamiltonian is Hermitian. For the oscillator this makes the Hermite functions orthonormal.

Great work! You now know how to:

• raise the ground state into Hermite functions
• use the Hermite recurrence and count nodes
• tell even oscillator states from odd ones