Past exam of the quantum theory course of the University of Oxford / 2015 / 2 / c

If $a_{-}\psi \ne 0$, it has energy $E-\hslash \omega$, distinct from the energy of $\psi$. Eigenstates of a self-adjoint operator with distinct eigenvalues are orthogonal. Explicitly,so $⟨\psi ,a_{-}\psi ⟩=0$. If $a_{-}\psi =0$, the same conclusion is immediate.

Dividing by $\Vert \psi {\Vert }^2$ and using the definition of $a_{-}$ givesBoth expectation values are real because $P$ and $X$ are self-adjoint operators. Its real and imaginary parts therefore vanish separately:An oscillator energy eigenstate is centered in both position and momentum, despite having nonzero quantum uncertainties.

Twice applying the creation operator or twice applying the annihilation operator changes the energy eigenvalue by $2\hslash \omega$, unless they give zero. The same orthogonality argument as in part (i) therefore givesAlso the factorization in part (a) givesNow $P=(a_{+}+a_{-})/2$ and $X=(a_{+}-a_{-})/(2im\omega )$. Squaring in the correct operator order yieldsBy part (i), the means are zero, so these are the variances. Taking positive square roots gives the requested standard deviations:These harmonic-oscillator number-state second moments also show equal mean kinetic energy and potential energy, each $E/2$. Their uncertainty product is $E/\omega =(n+\frac{1}{2})\hslash$, so the ground state saturates the Heisenberg uncertainty relation.