Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 17B (Quantum Mechanics)

Retain the order of the noncommuting position operator and momentum operator. Using their commutator $[\hat{x},\hat{p}]=i\hslash$, expansion givesThus the two factorizations of the quantum harmonic oscillator are

Part (a) implies $[\hat{A},\hat{B}]=2m\hslash \omega$. Since $\hat{H}=\hat{B}\hat{A}/(2m)+\hslash \omega /2$, it follows thatFor an eigenfunction $\varphi$ of the Hamiltonian with eigenvalue $E$,Thus $\hat{A}$ is a ladder operator lowering the energy by $\hslash \omega$, provided $\hat{A}\varphi \ne 0$. The zero function is not an eigenfunction; in particular the ground state is annihilated by this operator. This qualification is essential for the later argument.

With the inner product $⟨\chi ,\psi ⟩=\int {\chi }^{\ast }\psi dx$, the assumed Hermiticity of $\hat{x}$ and $\hat{p}$ givesConsequently the adjoint relation isAs with the stated Hermiticity assumption, these identities concern wavefunctions in the appropriate operator domains. Square integrability alone does not guarantee that derivatives or multiplication by $x$ are square integrable; one may first work with smooth rapidly decaying wavefunctions, where integration by parts has no boundary term, and then use the operator domains.

Apply the adjoint operator identity of part (c) with $\chi =\hat{A}\varphi$ and use part (a):For a nonzero eigenfunction, positivity of the norm in particular forces $E\ge \hslash \omega /2$. Hence the requested norm isIt vanishes exactly at the ground-state energy.

The Hamiltonian of the quantum harmonic oscillator is the sum of two positive quadratic forms. For a nonzero eigenfunction in its domain,Thus $\boxed{E\ge 0}$. Part (d) supplies the stronger bound $E\ge \hslash \omega /2$.

Start with a nonzero eigenfunction of energy $E$ and apply the lowering ladder operator repeatedly. Whenever the resulting wavefunction is nonzero, part (b) makes it an eigenfunction of energy $E-k\hslash \omega$. This cannot remain nonzero for every $k$, since part (e) rules out negative energies. Let $\psi ={\hat{A}}^n\varphi$ be the last nonzero member. Then $\hat{A}\psi =0$, and part (d) forces its energy to be $\hslash \omega /2$. Therefore every energy eigenvalue has the formThese levels actually occur: solving $\hat{A}{\psi }_0=0$ with $\hat{p}=-i\hslash d/dx$ gives the normalizable Gaussian ${\psi }_0\propto \exp (-m\omega x^2/(2\hslash ))$. The raising operator $\hat{B}$ increases the energy by $\hslash \omega$, and $\Vert \hat{B}\psi {\Vert }^2=2m(E+\hslash \omega /2)\Vert \psi {\Vert }^2>0$. Repeated raising therefore produces a nonzero normalizable state at every listed level.