Past exam of the physics A course of the University of Cambridge / 2021 / Paper 1 / Section B / B7

Write $a=a_{-}$ and $a^{†}=a_{+}$. Inverting the annihilation operator and creation operator definitions givesThe ladder operators take $|n⟩$ to orthogonal quantum harmonic oscillator energy eigenstates $|n-1⟩$ and $|n+1⟩$. Thus $⟨n|a|n⟩=⟨n|a^{†}|n⟩=0$, andFor $n=0$, the lowering term is simply zero. The conclusion concerns an individual energy eigenstate; interference between different eigenstates can produce nonzero expectation values.

For the harmonic-oscillator number-state second moments, the number operator gives $⟨a^{†}a{⟩}_n=n$, while the canonical commutation relation gives $⟨a{a}^{†}{⟩}_n=n+1$. Terms with $a^2$ or $(a^{†}{)}^2$ have zero diagonal matrix elements, since they change the occupation number by two. ThereforeConsequently the potential energy and kinetic energy means areTheir sum is the oscillator energy eigenvalue $E_n=\hslash \omega (n+1/2)$. Their equality is also the virial theorem result for a quadratic potential, here derived directly from the ladder operators.

The standard deviations defining the position and momentum uncertainties satisfy $(\Delta x{)}^2=⟨x^2⟩-⟨x{⟩}^2$ and $(\Delta p{)}^2=⟨p^2⟩-⟨p{⟩}^2$. Part (a) makes both means zero. Using the second moments from part (b),Multiplication givesThe Heisenberg uncertainty relation requires only $\Delta x\Delta p\ge \hslash /2$. Among the oscillator energy eigenstates, equality occurs solely for the ground state $n=0$; the excited states have larger uncertainty products.

Each energy eigenstate picks up its own phase under unitary time evolution. The relative phase therefore changes at $(E_1-E_0)/\hslash =\omega$. After removing an irrelevant global phase, the state isThe diagonal position matrix elements vanish, but the annihilation operator and creation operator giveThe two off-diagonal cross terms in the expectation value yieldThus the angular frequency is $\omega$, and the magnitude of the oscillation, meaning its amplitude, isThe peak-to-peak displacement is $2A$. The phase $\varphi$ fixes where the oscillation starts but does not change $A$. A pure $|0⟩$ or $|1⟩$ preparation has no mean oscillation, whereas an equal-weight quantum superposition maximizes its amplitude within this two-state family. The motion is produced by interference between energy eigenstates, rather than by a changing occupation of either one.