Past exam of the quantum theory course of the University of Oxford / 2021 / 2

The position operators and momentum operators are self-adjoint operators. Using conjugate linearity of the adjoint operator, $(cA+dB{)}^{\ast }=\overline{c}{A}^{\ast }+\overline{d}{B}^{\ast }$, givesWrite $D=\sqrt{2m\omega \hslash }$. Bilinearity and antisymmetry of the commutator, together with the canonical commutation relations, give $[P_j,X_k]=-i\hslash {\delta }_{jk}I$. HenceThe terms involving two position operators or two momentum operators vanish by the canonical commutation relations. Finally, reversing products under the adjoint operator, $(AB{)}^{\ast }=B^{\ast }{A}^{\ast }$, gives $[a_j,a_k{]}^{\ast }=-[a_j^{\ast },a_k^{\ast }]$. ThereforeHere $I$ is the identity operator. These are the relations for two independent pairs of creation and annihilation operators. The operator algebra in this question is understood on a common invariant operator domain, such as finite linear combinations of the number states constructed in part (c).

First apply the commutator product rule in the form $[AB,C]=A[B,C]+[A,C]B$:Repeated use of the commutator product rule gives, for $n\ge 1$,This is the commutator with powers of an eigenoperator. For $n=0$, the commutator with the identity operator is zero, agreeing with the formula.

Each $a_k^{\ast }{a}_k$ is unchanged by taking the adjoint operator, so $N^{\ast }=N$ in this algebra. Taking the adjoint of a commutator yields $[N,(a_j^{\ast }{)}^n{]}^{\ast }=[a_j^n,N]=n{a}_j^n$. By antisymmetry,In particular, each creation operator raises the number operator eigenvalue by one, and each annihilation operator lowers it by one.

Inverting the definitions of the creation and annihilation operators givesSubstituting in the orbital angular momentum $L$ givesOperators belonging to distinct modes commute. Therefore the terms with two annihilation operators and those with two creation operators cancel, and $a_1^{\ast }{a}_2=a_2{a}_1^{\ast }$, $a_2^{\ast }{a}_1=a_1{a}_2^{\ast }$. The surviving terms give the orbital angular momentum in oscillator ladder operators:

From part (a), $[N,a_j]=-a_j$ and $[N,a_j^{\ast }]=a_j^{\ast }$. The commutator product rule givesConsequentlyEach term in $L$ transfers one excitation between the two modes, preserving the total number operator eigenvalue. This is an instance of the oscillator bilinear commutator; every number operator eigenspace is an invariant subspace of $L$.

The common Fock vacuum obeys $N\psi ={\sum }_j{a}_j^{\ast }(a_j\psi )=0$. Put $B=(a_1^{\ast }{)}^{n_1}(a_2^{\ast }{)}^{n_2}$. By the commutator product rule and part (a),Moving $N$ through $B$ therefore givesDividing by $\sqrt{n_1!n_2!}$ provesThe constructed number state is nonzero: the relation $a_j(a_j^{\ast }{)}^r\psi =r(a_j^{\ast }{)}^{r-1}\psi$ follows from $[a_j,a_j^{\ast }]=I$ and $a_j\psi =0$. Commuting different modes and successively moving each creation operator across the inner product yieldsThus the given factorial denominator also gives unit norm, as expected for a number state.

Write $u={\psi }_{1,0}=a_1^{\ast }\psi$ and $v={\psi }_{0,1}=a_2^{\ast }\psi$. The canonical commutation relations for the creation and annihilation operators giveFor example, $a_1{a}_1^{\ast }\psi =(I+a_1^{\ast }{a}_1)\psi =\psi$. The expression for $L$ from part (b), with distinct modes reordered, then yieldsThus $\mathrm{span}\{u,v\}$ is an invariant subspace, and the matrix of $L$ in the ordered orthonormal basis $(u,v)$ isIts characteristic polynomial is ${\ell }^2-{\hslash }^2$, giving the eigenvalues and normalized eigenstatesIndeed, $L(u\pm iv)=\pm \hslash (u\pm iv)$ and $\Vert u\pm iv{\Vert }^2=2$ by the assumed orthonormality. Both eigenstates remain in the $N=1$ eigenspace.