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Work in units with $\hslash =1$. For unit mass and frequency, the two-dimensional isotropic harmonic oscillator hasThe product number states are therefore energy eigenvectors:For the level $E=N+1$, the possibilities areso its degeneracy isRestoring units multiplies every energy by $\hslash \omega$.

The addition energy has both charging and orbital terms,The larger diamonds indicate quantum dot shell closure after 2, 6 and 12 electrons. A two-dimensional isotropic harmonic oscillator confinement, $V(x,y)=m^{\ast }{\omega }_0^2(x^2+y^2)/2$, explains that sequence: shell $q=n_x+n_y$ has $q+1$ orbital states and capacity $2(q+1)$ after spin degeneracy. The cumulative capacities are $(q+1)(q+2)=2,6,12,20,\dots$. Within a degenerate shell, the extra orbital term vanishes to this approximation; crossing a shell boundary costs an extra $\hslash {\omega }_0$.

The gate capacitance lever arm is ${\alpha }_g=C_g/C_{\Sigma }=1/2$. Large and small gate widths differ by $7-5=2\mathrm{mV}$, so $\hslash {\omega }_0\simeq e{\alpha }_g(2\mathrm{mV})=1.0\mathrm{meV}$. Equivalently the tip energies differ by $3.5-2.5=1.0\mathrm{meV}$. Confinement anisotropy, exchange and disorder would split the ideal degeneracies; the diagram supports approximate shell symmetry rather than proving a perfectly parabolic potential.

For each mode in Cartesian coordinates, $[a_i,a_j^{†}]={\delta }_{ij}$ and $a_i^{†}{a}_i=H_i/(\hslash \omega )-1/2$. Consequently the two-dimensional isotropic harmonic oscillator hasThe normalized actions of the creation and annihilation operators areInvert their definitions as $x=\sqrt{\hslash /(2m\omega )}(a_x+a_x^{†})$ and $p_x=i\sqrt{m\hslash \omega /2}(a_x^{†}-a_x)$, and similarly for $y$. linear operators belonging to different coordinates commute, so substitution into $L_z=x{p}_y-y{p}_x$ givesEach term transfers one excitation between the two modes: $[N_x+N_y,a_x^{†}{a}_y]=(1-1)a_x^{†}{a}_y=0$, and likewise for its adjoint. Thus $[H,L_z]=0$ and, with no explicit time dependence, $d{L}_z/dt=(i/\hslash )[H,L_z]=0$. This is conservation of angular momentum under planar rotations.

Finally $L_z|1,0⟩=i\hslash |0,1⟩$ and $L_z|0,1⟩=-i\hslash |1,0⟩$. HenceThe angular momentum is $+\hslash$; division by $\sqrt{2}$ performs wavefunction normalization on this state without changing its eigenvalue. It is one excitation of the positive circular mode.

Use the canonical commutation relations $[\hat{x},{\hat{p}}_x]=[\hat{y},{\hat{p}}_y]=i\hslash$, with cross-coordinate commutators zero. For either coordinate,Thus $2\alpha \beta \hslash =1$. To obtain the required Hamiltonian, the ratio of its position operator and momentum operator coefficients must be preserved:where the positive root follows from the requested signs. ConsequentlyThe annihilation operators are dimensionless; $\alpha$ has units of inverse length and $\beta$ has units of inverse momentum. Operators belonging to different coordinates commute automatically. Writing the number operators as ${\hat{N}}_j={\hat{a}}_j^{†}{\hat{a}}_j$, substitution givesThe constant is the sum of the two zero-point energies of this two-dimensional isotropic harmonic oscillator.

Set $\hat{B}={\hat{a}}_x^{†}{\hat{a}}_y-{\hat{a}}_y^{†}{\hat{a}}_x$. Taking the Hermitian adjoint reverses the order of products, so ${\hat{B}}^{†}=-\hat{B}$. Complex conjugation also reverses the sign of $-i$, and thereforeThus $\hat{Q}$ is a self-adjoint operator on the usual oscillator states.

Using the coefficients in part (a), products containing two position operators or two momentum operators cancel in the difference. Cross-coordinate operators commute, leavingHenceIt is the component of orbital angular momentum perpendicular to the plane, with units $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ (equivalently $\mathrm{J}\mathrm{s}$). The rotational symmetry of the two-dimensional isotropic harmonic oscillator makes this a conserved observable; directly, each term in $\hat{B}$ preserves $N_x+N_y$, so $[\hat{H},\hat{Q}]=0$.

The Hamiltonian operator separates into two commuting one-dimensional quantum harmonic oscillators. WriteProducts of harmonic oscillator eigenfunctions giveHere $H_n$ is the physicists' Hermite polynomial, so $H_0=1$, $H_1(u)=2u$ and $H_2(u)=4u^2-2$. The first six energy eigenstates, ordered with multiplicity, areFor this two-dimensional oscillator level ordering, the excitation energies obeyEvery state with $n_1+n_2\ge 3$ costs at least $3\hslash {\omega }_1$, so none has been missed. Equalities within the list occur when ${\omega }_2={\omega }_1=\omega$. Then the two-dimensional isotropic harmonic oscillator hasThe quantum degeneracy counts $n_1=0,\dots ,N$ with $n_2=N-n_1$. The six listed states occupy three distinct energy levels, of quantum degeneracies one, two and three.

The orbital angular momentum $L_3=X_1{P}_2-X_2{P}_1$ generates rotations in the plane. The canonical commutation relations giveUse the commutator product rule $[A,B^2]=[A,B]B+B[A,B]$. Since different coordinates commute,Thus, for arbitrary positive angular frequencies,At ${\omega }_1={\omega }_2$, this becomes $\boxed{[L_3,H]=0}$. The rotational symmetry of the two-dimensional isotropic harmonic oscillator allows a basis of simultaneous energy eigenstates and orbital angular momentum eigenstates.

For a two-dimensional anisotropic quantum harmonic oscillator in the indicated range, the six lowest number states, counted with multiplicity, have occupations $(0,0),(1,0),(0,1),(2,0),(1,1),(0,2)$. Their excitation energies, divided by $\hslash$, are $0,{\omega }_1,{\omega }_2,2{\omega }_1,{\omega }_1+{\omega }_2,2{\omega }_2$. These are ordered because ${\omega }_2<2{\omega }_1$ and $2{\omega }_2<3{\omega }_1$; every omitted state has at least three quanta and costs at least $3\hslash {\omega }_1$. At equal angular frequencies, the two-dimensional isotropic harmonic oscillator has energy eigenvalue $\hslash \omega (N+1)$ and quantum degeneracy $N+1$, since there are $N+1$ ordered nonnegative pairs with sum $N$.

For the two-dimensional isotropic harmonic oscillator, $H=\hslash \omega (N+1)$ and the Schwinger boson representation gives $J^2={\hslash }^{2}N(N+2)/4$. Substitution yields the displayed relation, including its quantum constant term. The fixed-total-occupation sector is a spin-$j=N/2$ representation, so energy is $\hslash \omega (2j+1)$ with degeneracy $2j+1$.