Past exam of the quantum mechanics and electromagnetism course of the University of Oxford / 2011 / A / 2 / Solution

Take $m,\omega >0$ and use the usual common invariant domain for the position operator, momentum operator and their products. Since $X$ and $P$ are self-adjoint,Expanding the commutator and using $[P,X]=-i\hslash I$ givesThe two products areConsequently the quantum harmonic oscillator Hamiltonian operator isThe identity $[AB,C]=A[B,C]+[A,C]B$ now givesThus $H(a_{\pm }\psi )=(E\pm \hslash \omega )a_{\pm }\psi$. Each ladder image is an eigenstate if it is nonzero; a zero vector is not an eigenstate. The distinction is needed for lowering the ground state. Their squared norms areThese follow by moving the operator to the other side of the inner product with its adjoint operator. The first inequality implies $E\ge \hslash \omega /2$ for every eigenstate.

For a lowest-energy quantum state, $a_{-}{\psi }_0$ must vanish, since otherwise it would be an eigenstate with lower energy. The norm formula then forces $E_0=\hslash \omega /2$. Raising never vanishes and producesTo see that there are no other eigenvalues, repeatedly lower any eigenstate. If $E/(\hslash \omega )-1/2$ were not a nonnegative integer, there would be a final nonzero lowered quantum state with energy strictly between $\hslash \omega /2$ and $3\hslash \omega /2$. Its lower image would have positive norm but energy below the lower bound, a contradiction. Thus every eigenstate reaches the ground eigenspace after an integer number of lowerings. On a positive level, lowering is injective by its strictly positive norm factor. The assumed one-dimensional ground eigenspace therefore makes every eigenspace one-dimensional, so the displayed raised quantum states give all eigenstates up to nonzero multiples. If ${\psi }_0$ is normalized, their norms are $\Vert {\psi }_n{\Vert }^2=n!$.

For $k\ge 1$, $(a_{+}{)}^k\psi$ has eigenvalue $E+k\hslash \omega$, distinct from $E$. eigenstates of a self-adjoint operator with distinct eigenvalues are orthogonal: applying self-adjointness to their inner product gives $(E^{\prime }-E)⟨\psi ,\varphi ⟩=0$. ThereforeThe expectation of $(a_{-}{)}^k$ also vanishes, either by lowering and orthogonality or by conjugating the adjoint relation.

Solving the ladder definitions for $X$ and $P$ givesso $T-V=\frac{1}{2}\hslash \omega ((a_{+}{)}^2+(a_{-}{)}^2)$. Both terms have zero expectation by the preceding orthogonality. Since $T+V=H$, the oscillator form of the quantum virial theorem isExpectations here are divided by $\Vert \psi {\Vert }^2$ if the eigenstate is not normalized; no normalization assumption was used in the ladder argument.