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On a compact spatial Cauchy hypersurface without boundary, the gravitational Hamiltonian consists entirely of constraints and vanishes on solutions. There is no asymptotic time-translation boundary charge to add. This is the canonical statement that a closed universe has zero total energy; it does not assert that local matter energy density or local curvature vanishes.

A regular phase-mixed distribution $F(H,p_{\varphi },I)$ in an axisymmetric Stäckel potential is even separately in $v_{\lambda }$ and $v_{\mu }$, because these integrals depend on those components only through squares. Odd integration then gives all three mixed second moments zero, even if the $p_{\varphi }$ dependence produces azimuthal streaming. Bare stationarity alone is insufficient in degenerate resonant potentials with additional integrals. For example the isotropic harmonic Hamiltonian has the extra conserved $K=v_x{v}_y+{\omega }^{2}xy$; a positive Gaussian $\exp (-aH+ϵK)$ with $a>|ϵ|>0$ is a steady tracer and has nonzero $⟨v_x{v}_y⟩$. Its harmonic gravitational potential is a Stäckel potential in these spheroidal coordinates. This tracer need not itself source that harmonic gravitational potential. The alignment claim thus needs the phase-mixed three-integral hypothesis, not merely the weak form of Jeans theorem.

For orbital angular momentum $\ell =1$ and spin $s=1/2$, considerThe quantum number $m_j=m_{\ell }+m_s$ is conserved. The stretched states with $m_j=\pm 3/2$ have energies $K\pm 2a$. The remaining eigenvalues follow from two $2\times 2$ blocks with off-diagonal entries $\sqrt{2}K$:The subscripts denote $m_j=\pm 1/2$. For $K>0$, the zero-field multiplets are $j=3/2$ at $K$ and $j=1/2$ at $-2K$. Their weak-field Zeeman splitting is $g_{j}a{m}_j$, with Landé g-factors $4/3$ and $2/3$, respectively. When $a\gg |K|$, the Paschen-Back effect gives approximate product states $(m_{\ell },m_s)$ and energies $a(m_{\ell }+2m_s)+2K{m}_{\ell }{m}_s$. Mixing within a fixed $m_j$ causes an avoided crossing; different $m_j$ sectors can cross because the Hamiltonian does not couple them.

For $H=\alpha a^{†}a+\Delta (a^2+a^{†2})/2$, a finite real squeeze requires $\alpha >|\Delta |$. At $\alpha =\Delta >0$, the Hamiltonian instead equals $\alpha Q^2-\alpha /2$, with $Q=(a+a^{†})/\sqrt{2}$. The formal oscillator frequency is zero, but this operator is not constant: $⟨2|H|0⟩=\alpha /\sqrt{2}$. Thus a normalizable oscillator ground state and finite canonical squeeze cannot be obtained by simply substituting zero frequency into the positive-frequency formula.