Past exam of the quantum theory course of the University of Oxford / 2017 / 2

On a smooth wavefunction $f$, the position operator and momentum operator giveThus the canonical commutation relation is $\boxed{[X,P]=i\hslash I}$. Since $X$ and $P$ are self-adjoint operators and $m,\omega$ are real, the adjoint operator of $a_{-}$ isExpanding the products in the stated order is essential:Subtracting gives $\boxed{[a_{-},a_{+}]=2m\hslash \omega I}$. These identities initially hold on smooth rapidly decaying functions, which contain the quantum harmonic oscillator energy eigenstates used below.

The factorization also proves a lower bound: for a normalized energy eigenstate $H\psi =E\psi$,Therefore $E\ge E_0:=\hslash \omega /2$. The commutator product rule, applied to $H=a_{+}{a}_{-}/(2m)+E_0$, givesSo a nonzero application of the creation operator $a_{+}$ raises the energy eigenvalue by $\hslash \omega$, while the annihilation operator $a_{-}$ lowers it by the same amount.

Begin with any energy eigenstate, whose existence is assumed. Repeated lowering must terminate: otherwise it would eventually produce a nonzero energy eigenstate below $E_0$, contradicting the lower bound. More explicitly, if $(E-E_0)/(\hslash \omega )=N+r$ with integer $N\ge 0$ and $0\le r<1$, the first $N$ lowering steps are nonzero. If $r>0$, one further step is also nonzero by its positive squared norm and has energy $E_0+(r-1)\hslash \omega <E_0$, an impossibility. Thus $r=0$ and the last state satisfies $a_{-}{\psi }_0=0$. This provesEvery level exists as well: $\Vert a_{+}{\psi }_n{\Vert }^2=2m(E_n+\hslash \omega /2)=2m\hslash \omega (n+1)>0$, so repeated raising from ${\psi }_0$ never vanishes. The same factorization bounds the energy expectation of any normalized state in the one-dimensional harmonic oscillator form domain below by $E_0$, proving that the terminated state really is a ground state.

To find its probability density, solve the first-order equationIts solutions are ${\psi }_0=C_0\exp [-m\omega x^2/(2\hslash )]$. With $\alpha =m\omega /\hslash >0$, the supplied Gaussian integral givesTaking $C_0=(\alpha /\pi {)}^{1/4}$ fixes an irrelevant global phase. ConsequentlyThe ground-state energy is positive even though the potential minimum is zero: it is the oscillator's zero-point energy.

Put $\gamma =2m\hslash \omega$. If ${\psi }_n$ is a normalized energy eigenstate at $E_n$, the factorization from part (a) givesIt follows inductively that $v_n=(a_{+}{)}^n{\psi }_0$ has squared norm ${\gamma }^{n}n!$. The commutator with $H$ establishes its energy eigenvalue at each step. Henceis normalized and has $E_n=(n+\frac{1}{2})\hslash \omega$. Distinct levels are orthogonal because $H$ is a self-adjoint operator. In particular,

To decompose the specified initial wavefunction, retain this exact phase convention for the creation and annihilation operators. Their difference givesSince $a_{-}{\psi }_0=0$ and $a_{-}{a}_{+}{\psi }_0=\gamma {\psi }_0$,This quadratic excitation of an oscillator ground state, $\Psi (x,0)=C_0^{-1}{X}^2{\psi }_0$, contains only the $n=0$ and $n=2$ energy eigenstates. The printed initial function has an unspecified overall normalization, so the Born rule requires division by its squared norm. After wavefunction normalization the state isThe minus sign is consistent with the paper's convention: its $a_{+}$ is $i$ times the usual real-coordinate raising operator multiplied by $\sqrt{\gamma }$, so its ${\psi }_2$ has the opposite sign to the usual real second harmonic oscillator eigenfunction. The energy measurement probabilities are independent of this choice of phase.

Therefore the complete answer isEvery other energy has probability zero. At any later time the two coefficients only acquire their separate dynamical phases, so these probabilities remain the same. A direct check of the normalization isThis also confirms the factor of three in the squared norm of the two-component expansion.