Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / Section B / B7

The classical minimum is $E_{\mathrm{cl}}=0$, attained at $x=p=0$. The quantum oscillator has $E_0=\hslash \omega /2$, its zero-point energy; simultaneous zero position and momentum uncertainties are forbidden.

Let $\ell =\sqrt{\hslash /(m\omega )}$ and $\xi =x/\ell$. The first three normalized harmonic oscillator eigenfunctions areThey have respectively zero, one and two nodes, with even, odd and even parity. Their phases may be reversed without changing the state.^[ref]

Invert the given ladder operators:Because the Energy eigenstate are orthonormal, $⟨n|{\hat{a}}_{\pm }|n⟩=0$: each operator changes the state to one orthogonal to $|n⟩$. HenceThe result also follows from the definite parity of every oscillator energy eigenfunction, but here it follows directly from the specified operators.

Orthogonality gives $⟨{\hat{a}}_{+}^2⟩=⟨{\hat{a}}_{-}^2⟩=0$, while the ladder coefficients give $⟨{\hat{a}}_{+}{\hat{a}}_{-}⟩=n$ and $⟨{\hat{a}}_{-}{\hat{a}}_{+}⟩=n+1$. Squaring the expressions for $\hat{x}$ and $\hat{p}$ therefore givesThusTheir equality is the harmonic-potential case of the quantum virial theorem, and their sum is the oscillator energy $E_n=\hslash \omega (n+1/2)$.

For the uncertainty product of oscillator number states, the two means vanish, so the standard deviations are the square roots of the second moments:Consequently,Only the ground state $n=0$ saturates the Heisenberg uncertainty principle in this sequence of energy eigenstates; the excited number states have a larger uncertainty product.