In a number state the particle is on average at the centre and at rest, yet its position and momentum spreads grow with every quantum added, both as $\sqrt{n+\tfrac12}$. The energy is shared equally between kinetic and potential, as it is for a classical oscillator averaged over a swing. And for large $n$ the density itself begins to look classical: it piles up near the ends of the swing, where a classical particle slows down and lingers.

A pendulum photographed at random moments is most often caught near the ends of its swing, where it moves slowly, and least often in the middle, where it rushes through. A highly excited oscillator’s density has the same shape, with ripples.

The density $|\psi_{4}|^{2}$ of the state with four quanta, against the density $1/\pi\sqrt{A^{2}-\xi^{2}}$ of a classical oscillator of the same energy, which swings to in units of . Averaged over its ripples the quantum density follows the classical one, and both are largest near the ends of the swing.

Writing $\hat{x}$ and $\hat{p}$ through the ladder operators turns every average in a number state into a count of quanta. Averages of $\hat{x}$ and vanish; averages of their squares grow with .

• $\langle x\rangle=\langle p\rangle=0$ in every $\ket{n}$: $\hat{x}$ and change by one, and different number states are orthogonal.

Write $\hat{x}$ through the ladder operators and square it. The terms $\hat{a}^{2}$ and $\hat{a}^{\dagger2}$ change by two and average to zero; the mixed terms count quanta, and .

Invert the definitions of the ladder operators.

Square, keeping the order of the mixed terms.

They change $n$ by two, and $\ket{n\pm2}$ is orthogonal to $\ket{n}$.

Use $\hat{a}^{\dagger}\hat{a}=\hat{N}$ and $$.

Collect. Since $\langle x\rangle=0$, this is $(\Delta x)^{2}$.

In the $n$-th state the mean square position is $x_{0}^{2}(n+\tfrac12)$, while the mean position is zero.

Position and momentum each change the number of quanta by one, so their averages in any single number state vanish.

The two spreads grow together, and their product is $n+\tfrac12$ times $\hbar$.

In every number state the average kinetic and potential energies are equal, each half the total.

Great work! You now know how to:

• compute $\langle x^{2}\rangle$ and $\langle p^{2}\rangle$ in number states with ladder operators
• find the uncertainty product $(n+\tfrac12)\hbar$