Past exam of the quantum theory course of the University of Oxford / 2023 / 2 / e

Set ${\omega }_1={\omega }_2=\omega$. Inverting the creation and annihilation operators givesSubstitute these expressions in $L_3=X_1{P}_2-X_2{P}_1$. Terms with two creation operators or two annihilation operators cancel because different modes commute. The remaining orbital angular momentum in oscillator ladder operators isEquivalently it is $-i\hslash (a_{1,+}{a}_{2,-}-a_{2,+}{a}_{1,-})$, consistent with the circular ladder operators of a two-dimensional oscillator.

Choose the first three energy eigenstates consistently as ${\psi }_{0,0}$, ${\psi }_{1,0}=a_{1,+}{\psi }_{0,0}$ and ${\psi }_{0,1}=a_{2,+}{\psi }_{0,0}$. The Fock vacuum is annihilated by both lowering operators, soFor example $a_{1,-}{a}_{1,+}{\psi }_{0,0}={\psi }_{0,0}$, while $a_{2,-}{\psi }_{1,0}=0$, giving the middle formula. Thus the simultaneous eigenstates can be chosen asTheir pairs of eigenvalues $(H,L_3)$ are respectively $(\hslash \omega ,0)$, $(2\hslash \omega ,+\hslash )$ and $(2\hslash \omega ,-\hslash )$. The factor $1/\sqrt{2}$ normalizes the combinations when the original three states are normalized. Mixing only inside the degenerate first-excited energy eigenspace preserves the energy eigenvalue while diagonalizing $L_3$.

Write $\alpha =m\omega /\hslash$. The first Hermite polynomials give ${\psi }_{0,0}\propto e^{-\alpha (x_1^2+x_2^2)/2}$ and ${\psi }_{1,0},{\psi }_{0,1}$ proportional, with the same constant, to $x_1{e}^{-\alpha (x_1^2+x_2^2)/2}$ and $x_2{e}^{-\alpha (x_1^2+x_2^2)/2}$. Thus in Cartesian coordinates the new wavefunctions areUsing polar coordinates $x_1=r\cos \phi$, $x_2=r\sin \phi$ givesThe orbital angular momentum is $L_3=-i\hslash {\partial }_{\phi }$, which immediately verifies its eigenvalues $0,\pm \hslash$. Multiplying these spatial wavefunctions by $e^{-i\omega t}$ for ${\chi }_0$ and $e^{-2i\omega t}$ for ${\chi }_{\pm }$ gives the corresponding full stationary states.