= Solution
The <anisotropic quantum harmonic oscillator> is a sum of three commuting one-coordinate oscillator <Hamiltonians>. Substituting $\psi=X(x)Y(y)Z(z)$ and dividing by $XYZ$ gives a sum of three functions of independent coordinates, so each equals a separation constant. Consequently
$$
\psi_{n_xn_yn_z}=X_{n_x}Y_{n_y}Z_{n_z},\qquad
E=\hbar\sum_{i=x,y,z}\omega_i(n_i+1/2).
$$
For a time-independent operator, conservation for arbitrary states follows from $[H,Q]=0$, since $d\langle Q\rangle/dt=(i/\hbar)\langle[H,Q]\rangle$. The first labelled operator is $iL_z/\hbar$; the second is $-L^2/\hbar^2$. The <kinetic energy> is rotationally invariant. The potential is invariant under <rotations> about $z$ precisely when \b[(i) $\omega_x^2=\omega_y^2$]; full rotational invariance requires \b[(ii) $\omega_x^2=\omega_y^2=\omega_z^2$]. For positive oscillator <frequencies> these are equality of the <frequencies> themselves. Acting with the <commutators> on arbitrary test functions excludes anisotropic potentials in case (ii).
In the isotropic case let $b=m\omega/\hbar$. Then
$$
\boxed{E_N=\hbar\omega(N+3/2),\quad N=n_x+n_y+n_z},\qquad
\psi_{100}=\sqrt2\,\frac{b^{5/4}}{\pi^{3/4}}\,x e^{-br^2/2},\qquad
\boxed{\alpha=\frac{m\omega}{2\hbar}}.
$$
The polynomial $x$ is the first excited oscillator factor and the other two factors are ground-state <Gaussians>. The first excited spatial level has three independent choices, proportional to $x,y,z$. Odd <parity> gives $\langle x\rangle=0$, and <Gaussian> integration yields
$$
\langle x^2\rangle=\frac{3}{2b},\qquad
\boxed{\Delta x=\sqrt{\frac{3\hbar}{2m\omega}}}.
$$
For <oscillator uncertainty and allocation of classical energy>, the $x$ mode has <energy> $3\hbar\omega/2$, whose classical amplitude is $A_x=\sqrt{3\hbar/(m\omega)}$, giving $\Delta x=A_x/\sqrt2$. If “same <energy>” instead means assigning the entire three-dimensional <energy> $5\hbar\omega/2$ to one classical coordinate, its amplitude is $\sqrt{5\hbar/(m\omega)}$ and the ratio is $\sqrt{3/10}$. A general three-dimensional classical orbit's $x$ amplitude is not fixed by total <energy> without specifying its allocation among modes.
The <capacity of an isotropic oscillator shell>, including the two <electron> <spin> states, is $(N+1)(N+2)$ <electrons>. Nine non-interacting <electrons> occupy shells as $2,6,1$ for $N=0,1,2$. Thus
$$
\boxed{E_g=\left[2\frac32+6\frac52+\frac72\right]\hbar\omega=\frac{43}{2}\hbar\omega,\qquad
E_{\rm first\ higher}=\frac{45}{2}\hbar\omega}.
$$
The least positive excitation promotes an <electron> by one oscillator quantum. Rearrangements within the partly filled degenerate shell remain <ground states> and cost no <energy>.
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