In the Heisenberg picture the oscillator is solved in one line: the lowering operator just turns with a phase, and position and momentum then move exactly like the classical coordinates. So every average swings at the frequency $\omega$, whatever the state. A number state has nothing to swing — its averages are zero and stay zero — but a superposition of neighbouring levels sways from side to side at exactly the classical frequency.

A swing pushed off at any height comes back at the same rate: that is what a harmonic force does. The quantum oscillator keeps the rule for its averages, whatever its state.

The averages of position and momentum in the state $(\ket{0}+\ket{1})/\sqrt2$ over two periods, in units of $x_{0}$ and $p_{0}$ against $\omega t$: $\langle x\rangle=\frac{x_{0}}{\sqrt2}\cos\omega t$ and $\langle p\rangle=-\frac{p_{0}}{\sqrt2}\sin\omega t$, the motion of a classical oscillator released from rest.

Heisenberg’s equation for the lowering operator is solved at once, and position and momentum then move exactly as classical ones do. Every average oscillates at $\omega$, whatever the state.

• In every state, $\langle x\rangle(t)=\langle x\rangle_{0}\cos\omega t+\frac{\langle p\rangle_{0}}{m\omega}\sin\omega t$: the classical motion, as Ehrenfest’s theorem promises for a quadratic potential.

Heisenberg’s equation for $\hat{a}$ uses its commutator with the Hamiltonian, which is $-\hbar\omega\,\hat{a}$, so $\hat{a}$ only acquires a phase. Rebuilding from and its adjoint turns the phases into a cosine and a sine.

Heisenberg’s equation of motion.

The commutator from the lesson on ladder operators.

Solve, and take the adjoint.

$\hat{x}=\sqrt{\hbar/2m\omega}\,(\hat{a}+\hat{a}^{\dagger})$ at each time.

Substitute $\hat{a}$ and $\hat{a}^{\dagger}$ through $\hat{x}$ and : the phases combine into and .

Every state of the oscillator returns to itself, up to an overall phase, after one classical period, because all its energies differ by whole quanta.

Heisenberg’s equation for the lowering operator gives $d\hat{a}/dt=-i\omega\hat{a}$, so it only acquires a phase as time passes.

In a number state every average is constant in time, as in any stationary state: the phase of the state cancels in every expectation value.

The superposition of the two lowest states has a mean position that swings at the classical frequency, with amplitude $x_{0}/\sqrt{2}$.

Great work! You now know how to:

• solve Heisenberg’s equations for $\hat{a}$, $x$ and $p$
• explain why averages follow the classical motion exactly
• tell which superpositions sway and which stand still