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For three odd fermionic operators, the identityconverts the commutator with a quadratic Hamiltonian into the given canonical anticommutation relations. Writing a mode label as $i=(\mathbf{p},s)$, this yieldsThe opposite-species number terms commute with each fermionic creation operator or fermionic annihilation operator, because two anticommutations give a positive sign. Restoring the continuous normalization, the Dirac delta distribution in the mode brackets cancels $d^3\mathbf{q}/(2\pi {)}^3$. ConsequentlySubstitute these into the mode expansion of a Dirac field. The result isSince $i{\partial }_t{e}^{-ip\cdot x}=E_{\mathbf{p}}{e}^{-ip\cdot x}$ and $i{\partial }_t{e}^{ip\cdot x}=-E_{\mathbf{p}}{e}^{ip\cdot x}$, this is exactlyThe Heisenberg equation of motion uses an ordinary commutator with the even Hamiltonian, despite the odd fermionic operators obeying anticommutators among themselves.

Use the computational basis in the stated order and let the rightmost operation act first. The controlled phase gate is $|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes Z$. From the rotation definition,The control-zero block remains the identity, and the control-one block becomes $X$. ThusThis proves the relation for arbitrary superpositions, rather than only for computational-basis probabilities.

For the second relation write $S=U_{\mathrm{SW}}^{1/2}$. Its central block is $\left(\begin{array}{cc}a& b\\ b& a\end{array}\right)$, with $a=(1+i)/2$, $b=(1-i)/2$. Because $a^2=i/2$, $b^2=-i/2$ and $ab=1/2$, direct multiplication givesMeanwhile $R_z^{(1)}(\pi /2)R_z^{(2)}(-\pi /2)=\mathrm{diag}(1,-i,i,1)$. Multiplying these diagonal matrices and the overall factor $i$ givesThe supplied $S$ is indeed a square root of the SWAP gate: $S^2$ exchanges $|01⟩$ and $|10⟩$ while fixing the other basis states. It acts as one on the triplet subspace and as $i$ on the singlet subspace.

For an isotropic exchange interaction with Hamiltonian $H=J(t){\mathbf{\sigma }}_1\cdot {\mathbf{\sigma }}_2/4$, define $\phi =\int J(t)dt/\hslash$. Triplet and singlet energies differ by $J$, soAn exchange pulse with $\phi =\pi /2$ therefore implements $e^{-i\pi /8}S$. Two such pulses, interspersed with the specified local rotations, implement the controlled phase gate; the initial and final target rotations then turn it into a CNOT gate. Their common overall phase has no effect on an isolated gate's outcome probabilities; a subsequently controlled version must keep track of conditional phases.

The construction converts a physically available two-spin interaction into a standard entangling gate, which together with arbitrary one-qubit rotations supplies universal quantum computation. Practical operation needs calibrated exchange pulse area and timing, individually controllable rotations, and reproducible phase references. Charge noise can change $J$; magnetic-field gradients add unwanted single-spin evolution, and anisotropic exchange modifies the assumed Hamiltonian. Pulses must suppress leakage into unwanted orbital states while finishing before excessive decoherence. Readout, preparation, cross-talk and the extra operations in this decomposition all affect the achieved quantum gate fidelity. The algebraic identity alone does not establish a fault-tolerant quantum computation implementation.

The Hartree-Fock method approximates the ground state wavefunction by a single Slater determinant, minimizing the expectation value of the Hamiltonian over its occupied orbitals. It retains fermionic exchange symmetry exactly within that determinant but omits electron correlation beyond a single Slater determinant.

For $V\gg |t|$, the leading objective is to avoid adjacent electrons of the same spin. The unpolarized Fermi sea, although optimal within the preceding restricted family, has interaction energy per site $V(1/2-2/{\pi }^2)>0$. Its cost grows linearly with $V$, while states with no forbidden neighbouring pairs have zero interaction energy.

There are two independent spinless fermion t-V chains, one for each spin. On an even ring, avoiding all neighbouring pairs allows at most $M/2$ particles in each chain. Since the total is $M$, both bounds must be saturated, and each chain must alternate occupied and empty sites. Their two independent choices give four limiting occupation patterns:The first two are Néel states with one electron per site; the last two are charge density waves alternating doubly occupied and empty sites. Opposite-spin double occupation costs nothing in this particular Hamiltonian. Each species has alternating occupation in all four patterns.

These are the exact zero-interaction ground states at $t=0$, and the limiting ordered configurations as $V/|t|\to \infty$. At large finite $V/|t|$, virtual hopping admixes configurations with forbidden pairs, producing an energy correction of order $-M|t{|}^2/V$. A finite ring can have superpositions of the ordered configurations, with possible quantum tunnelling splittings; four exactly degenerate product ground states are not implied at finite hopping. The four ordered choices describe the strong-coupling thermodynamic limit, with spontaneous symmetry breaking understood.

For a normalized orbital $\int d^{2}r|\varphi {|}^2=1$, the symmetric Hartree product state and its equivalent Fock space form areThere are exactly $N(N-1)/2$ interacting pairs, so the expectation value of the grand Hamiltonian is the functionalAt fixed $N$ the term $-\mu N$ is constant. Enforce wavefunction normalization separately by varying ${\mathcal{E}}_{\mu }-N\eta (\int |\varphi {|}^2-1)$ with respect to ${\varphi }^{\ast }$. The resulting stationary Gross–Pitaevskii equation for the normalized orbital isThe Lagrange multiplier $\eta$ enforces the norm; fixed-$N$ variation cannot identify it with the independently inserted chemical potential merely by differentiating the constant $-\mu N$.

For the standard large-$N$ convention put $\Psi =\sqrt{N}\varphi$, so $n=|\Psi {|}^2$ and $\int n=N$, and replace $N-1$ by $N$. Varying the resulting grand energy functionalwith unrestricted $\Psi$ givesEquivalently, at fixed $N$ this same $\mu$ is chosen as the normalization Lagrange multiplier. This is the chemical potential convention used below. The contact interaction coupling $g$ has dimensions of energy times area in two dimensions.

The orthogonal coordinate change preserves the kinetic energy and takes the particle exchange operator to $(x_{+},x_{-})\mapsto (x_{+},-x_{-})$. Thus the spatial wave function is even in $x_{-}$ for bosons and odd for fermions.

The total potential energy and Hamiltonian separate asAssume $1+2\kappa >0$ so the relative oscillator is confining. Its spectrum is $E_{n_{+},n_{-}}=\hslash \omega (n_{+}+1/2)+\hslash \Omega (n_{-}+1/2)$, and its parity in the two-particle relative coordinate is $(-1{)}^{n_{-}}$.

The lowest boson state uses $n_{+}=n_{-}=0$, whereas the lowest fermion state uses $n_{+}=0$, $n_{-}=1$. ThereforeFor an oscillator of frequency $\Omega$, $⟨x_{-}^2{⟩}_n=(n+1/2)\hslash /(m\Omega )$. Since $(x_1-x_2{)}^2=2x_{-}^2$,The factor three is the extra relative-coordinate excitation required by antisymmetry. The supplied oscillator wave functions correspond to kinetic energy $P^2/(2m)$; the extra factor $1/2$ printed before that term in the hint is inconsistent with those functions and the stated energies. The calculation uses the conventional kinetic energy dictated by particle mass $m$.

Write $b=(\ell +1)a$, $R(r)=r^{\ell }{e}^{-r/b}$ and $Y(\theta ,\varphi )={\sin }^{\ell }\theta e^{i\ell \varphi }$, for nonnegative integer $\ell$. The angular Laplacian satisfies ${\Delta }_{S^2}Y=-\ell (\ell +1)Y$. Direct radial differentiation givesBecause $(\ell +1)/b=1/a$, the $1/r$ contribution cancels the Coulomb potential in the Hamiltonian. ThereforeThe volume element contributes $r^2\sin \theta$. The squared state factors into radial, polar and uniform azimuthal densities, so the angular normalization cancels in every radial expectation. With $I_j={\int }_0^{\infty }{r}^j{e}^{-2r/b}dr$,The supplied integration-by-parts recurrence gives $I_{j+1}=(j+1)b{I}_j/2$. HenceIn particular $\mathrm{Var}(r)/{\mu }^2=(2\ell +3{)}^{-1}$. Put $h=(2\ell +3{)}^{-1/3}$. Chebyshev's inequality gives

For the polar variable set $J_j={\int }_0^{\pi }{\sin }^j\theta d\theta$. The polar density is proportional to ${\sin }^{2\ell +1}\theta$, soAn exact expression for the requested tail, as well as its useful estimate, isThe inequality is Markov's inequality applied to ${\cos }^2\theta$. If a finite expression is desired, expand $(1-z^2{)}^{\ell }$: the numerator is ${\sum }_{j=0}^{\ell }(-1{)}^j(\genfrac{}{}{0ex}{}{\ell }{j})(1-h^{2j+1})/(2j+1)$ and the denominator is the same sum with $h=0$.

By the union bound, with probability at least $1-2h$ both $|r/\mu -1|\le h$ and $|\cos \theta |\le h$. The height above the equatorial plane is then at most $(1+h)\mu h$. The cylindrical radius differs from $\mu$ by at most $\mu h+(1+h)\mu h^2$. Thus the distance from the circle of radius $\mu$ in the equatorial plane is $O(\mu h)$ with probability tending to one. This is relative concentration around a growing circle; it does not assert an absolutely shrinking thickness. The probability concentration of a circular Coulomb eigenstate is a precise version of the large-angular-momentum orbit picture.

Let $U(t)=e^{-iHt/\hslash }$ and write ${\psi }_S(t)=U(t){\psi }_0$ in the Schrödinger picture, with observables $A_S$ constant unless they have an explicit time dependence. In the Heisenberg picture the state is ${\psi }_H={\psi }_0$ andDifferentiate the two unitary factors to obtain the equation. Unitarity gives $⟨{\psi }_S,A_S{\psi }_S⟩=⟨{\psi }_0,A_H{\psi }_0⟩$.

For the interaction picture choose a reference split $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and define ${\psi }_I=U_0^{\ast }{\psi }_S$, $A_I=U_0^{\ast }{A}_S{U}_0$. Then $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$, $V_I=U_0^{\ast }V{U}_0$, while $A_I$ evolves with $H_0$ (and its explicit time dependence). AgainThese are different unitary allocations of the same dynamics between states and observables, so measured expectations agree.

The canonical commutation relation now givesMultiplication of the last equation by $e^{i\omega t}$ and integration givesSince $A_t^{\ast }-A_t=2im\omega X_t$ and $A_0{\psi }_0=0$, both initial expectations of $A_0,A_0^{\ast }$ vanish. ThereforeThe printed positive displacement endpoint has the wrong sign for the Hamiltonian's positive $\mathcal{E}X$ term. The two turning values are $0$ and $-2\mathcal{E}/(m{\omega }^2)$; their ordering depends on the sign of $\mathcal{E}$. The result oscillates about the displaced equilibrium $-\mathcal{E}/(m{\omega }^2)$, exactly as the force $-\mathcal{E}$ predicts.

Inside the infinite well, the time-independent Schrödinger equation is $-{\hslash }^2{\Psi }^{\prime \prime }/(2m)=E\Psi$ with Dirichlet endpoints. Negative or zero energy gives no nonzero solution satisfying both endpoints. For $E>0$, the sine solution requires $ka=n\pi$. ThusThe normalization follows from ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$. These functions form an orthonormal basis. Substitute the expansion into $i\hslash {\partial }_t\psi =H\psi$ and take the inner product with each basis vector: orthonormality gives $i\hslash {\dot{c}}_n=E_n{c}_n$, equivalentlyFor a general infinite expansion this is interpreted in the usual Hamiltonian domain or its unitary evolution; the particular state below has only two terms.

Put $\theta =\pi x/a$. The identity $\sin \theta (16{\cos }^2\theta -1)=4\sin 3\theta +3\sin \theta$ gives $c_1=3/5$, $c_3=4/5$, all others zero. Their squared magnitudes sum to one. HenceThe evolved state is $(3e^{-i{E}_{1}t/\hslash }{\Psi }_1+4e^{-i{E}_{3}t/\hslash }{\Psi }_3)/5$. Its position probability includes an interference term with frequency $\Omega =(E_3-E_1)/\hslash =4{\pi }^2\hslash /(m{a}^2)$. Specifically,The three required integrals are $\pi /8-1/4$, $\pi /8+1/12$ and $1/4$, respectively. ThereforeThe energy expectation is constant while the spatial probability oscillates, because relative energy phases affect interference but not the energy probabilities.

For a nonzero trial vector use the Rayleigh quotient $\mathcal{E}(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its first variation in any allowed direction $\eta$ isVanishing for all complex directions implies $(H-\mathcal{E})\psi =0$; an eigenvector conversely makes it vanish. The ground energy is the infimum of this quotient for a lower-bounded Hamiltonian, so minimizing over a trial family provides an upper bound to it, exact when that family contains a ground eigenvector.

Set $a=⟨\varphi ,H\varphi ⟩$, $b=⟨\varphi ,I\varphi ⟩$ and $c=⟨\varphi ,HI\varphi ⟩$. All three are real: $H,I$ are self-adjoint and commute. Also $I^{†}I=I^2=1$. Expansion of the numerator and norm givesDifferentiate the quotient, retaining only nonzero trial vectors:Thus the stationarity equation is $\boxed{(t^2-1)(c-ab)=0}$. If $c=ab$, the quotient is identically $a$ wherever defined. Otherwise $t=\pm 1$ are the two extrema, withIn this nonconstant case $|b|<1$, so both denominators are positive. If $c-ab>0$, the plus value is the maximum and the minus value the minimum; the assignments reverse when $c-ab<0$. One can also see global extremality by resolving the trial vector into the orthogonal $I=\pm 1$ sectors: its energy is a weighted mean of their two fixed projected energies.

The null-vector exception matters. If $|b|=1$, then $I\varphi =\pm \varphi$, the covariance $c-ab$ vanishes and one of $\varphi \pm I\varphi$ is zero. That value has no Rayleigh quotient and is not an extra stationary physical state.

For particle exchange, change of integration variables shows that $I\varphi (x_1,x_2)=\varphi (x_2,x_1)$ is unitary and $I^2=1$, hence self-adjoint. Exchanging coordinates interchanges the two kinetic derivatives and leaves the symmetric potential unchanged, so $[I,H]=0$. The vectors $\varphi \pm I\varphi$ lie in its symmetric and antisymmetric sectors. Their normalized nonzero versions are respectively bosonic and fermionic spatial trial wavefunctions.

Because the energy of a general trial state is the weighted average of its sector energies, at least one nonzero symmetry projection has energy no greater than the original trial. Thus minimizing can be performed in a definite exchange-symmetry sector. This is a variational approximation within the trial subspace; it does not assert that projecting an arbitrary trial vector produces an exact eigenvector of $H$. For particles with a prescribed statistics sector one minimizes within that physical sector.

For bound hydrogen atom states, the radial wave function is regular and finite at the origin, and decays at infinity so that $4\pi {\int }_0^{\infty }{r}^2|\Psi (r){|}^{2}dr<\infty$. Equivalently the reduced radial function $w(r)=r\Psi (r)$ has $w(0)=0$ and is square integrable on the half-line. A singular $1/r$ branch is not an admissible regular Coulomb eigenfunction; its derivative energy diverges and it introduces an origin singularity absent from the Hamiltonian domain.

Write $\kappa =e^2/(4\pi {ϵ}_0)$, so $a={\hslash }^2/(m\kappa )$. Multiplication by $r$ turns the radial equation intoSince $w=a{e}^{-\lambda R}f(R)$ and $r=aR$, substitution givesChoose $\boxed{\lambda =\mu a>0}$ to remove the constant coefficient and factor out the decaying exponential. For $f={\sum }_{k\ge 0}{a}_k{R}^{k+c}$, the lowest power has coefficient $c(c-1)a_0$, so $\boxed{c(c-1)=0}$. The $c=0$ behavior would give $\Psi \sim a_0/R$, violating regularity; choose $c=1$.

Matching successive powers then yieldsA nonterminating tail eventually has one sign, and its coefficient ratio is asymptotic to $2\lambda /k$. More precisely, for any $0<\delta <\lambda$, the ratio is eventually at least $(2\lambda -\delta )/(k+1)$. Its same-sign tail consequently dominates a positive constant times the exponential series of rate $2\lambda -\delta$, up to finitely many terms. The factor $e^{-\lambda R}$ cannot then make $\Psi$ decay. Thus square integrability forces termination.

If the last nonzero term is $a_{n-1}{R}^n$, termination requires $\lambda n=1$. ThereforeFor $n=1$, $\lambda =1$, $f$ is proportional to $R$, andThe radial volume element is $4\pi r^{2}dr$, so the mean radius is the ratio of the radius-weighted norm integral to the norm integral. With $r=aR$, the common angular and normalization factors cancel:In the ground state $\boxed{⟨r⟩=3a/2}$.

With $\mathbf{B}=B_0\hat{\mathbf{z}}$ and $S_z=(\hslash /2){\sigma }_z$,ThusFor $B_0>0$, spin down is the lower energy eigenstate. The negative electron magnetic moment is already accounted for in the positive coefficient multiplying $\mathbf{S}\cdot \mathbf{B}$ in the given Hamiltonian.

Let $U(t)=e^{-iHt/\hslash }$ and write ${\psi }_S(t)=U(t){\psi }_0$ in the Schrödinger picture, with observables $A_S$ constant unless they have an explicit time dependence. In the Heisenberg picture the state is ${\psi }_H={\psi }_0$ andDifferentiate the two unitary factors to obtain the equation. Unitarity gives $⟨{\psi }_S,A_S{\psi }_S⟩=⟨{\psi }_0,A_H{\psi }_0⟩$.

For the interaction picture choose a reference split $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and define ${\psi }_I=U_0^{\ast }{\psi }_S$, $A_I=U_0^{\ast }{A}_S{U}_0$. Then $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$, $V_I=U_0^{\ast }V{U}_0$, while $A_I$ evolves with $H_0$ (and its explicit time dependence). AgainThese are different unitary allocations of the same dynamics between states and observables, so measured expectations agree.

The canonical commutation relation now givesMultiplication of the last equation by $e^{i\omega t}$ and integration givesSince $A_t^{\ast }-A_t=2im\omega X_t$ and $A_0{\psi }_0=0$, both initial expectations of $A_0,A_0^{\ast }$ vanish. ThereforeThe printed positive displacement endpoint has the wrong sign for the Hamiltonian's positive $\mathcal{E}X$ term. The two turning values are $0$ and $-2\mathcal{E}/(m{\omega }^2)$; their ordering depends on the sign of $\mathcal{E}$. The result oscillates about the displaced equilibrium $-\mathcal{E}/(m{\omega }^2)$, exactly as the force $-\mathcal{E}$ predicts.

For a nonzero trial vector use the Rayleigh quotient $\mathcal{E}(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its first variation in any allowed direction $\eta$ isVanishing for all complex directions implies $(H-\mathcal{E})\psi =0$; an eigenvector conversely makes it vanish. The ground energy is the infimum of this quotient for a lower-bounded Hamiltonian, so minimizing over a trial family provides an upper bound to it, exact when that family contains a ground eigenvector.

Set $a=⟨\varphi ,H\varphi ⟩$, $b=⟨\varphi ,I\varphi ⟩$ and $c=⟨\varphi ,HI\varphi ⟩$. All three are real: $H,I$ are self-adjoint and commute. Also $I^{†}I=I^2=1$. Expansion of the numerator and norm givesDifferentiate the quotient, retaining only nonzero trial vectors:Thus the stationarity equation is $\boxed{(t^2-1)(c-ab)=0}$. If $c=ab$, the quotient is identically $a$ wherever defined. Otherwise $t=\pm 1$ are the two extrema, withIn this nonconstant case $|b|<1$, so both denominators are positive. If $c-ab>0$, the plus value is the maximum and the minus value the minimum; the assignments reverse when $c-ab<0$. One can also see global extremality by resolving the trial vector into the orthogonal $I=\pm 1$ sectors: its energy is a weighted mean of their two fixed projected energies.

The null-vector exception matters. If $|b|=1$, then $I\varphi =\pm \varphi$, the covariance $c-ab$ vanishes and one of $\varphi \pm I\varphi$ is zero. That value has no Rayleigh quotient and is not an extra stationary physical state.

For particle exchange, change of integration variables shows that $I\varphi (x_1,x_2)=\varphi (x_2,x_1)$ is unitary and $I^2=1$, hence self-adjoint. Exchanging coordinates interchanges the two kinetic derivatives and leaves the symmetric potential unchanged, so $[I,H]=0$. The vectors $\varphi \pm I\varphi$ lie in its symmetric and antisymmetric sectors. Their normalized nonzero versions are respectively bosonic and fermionic spatial trial wavefunctions.

Because the energy of a general trial state is the weighted average of its sector energies, at least one nonzero symmetry projection has energy no greater than the original trial. Thus minimizing can be performed in a definite exchange-symmetry sector. This is a variational approximation within the trial subspace; it does not assert that projecting an arbitrary trial vector produces an exact eigenvector of $H$. For particles with a prescribed statistics sector one minimizes within that physical sector.

The Time-dependent Schrödinger equation, with the Hamiltonian for a particle of mass $m$ in a real potential energy $V$, isA stationary state is an energy eigenstate: its wavefunction evolves by a single global phase,Its probability density is independent of time. For a normalized wavefunction, the probability density and probability current areSubtracting the complex conjugate of the Time-dependent Schrödinger equation, multiplied by $\Psi$, from the original equation multiplied by ${\Psi }^{\ast }$ cancels the real potential energy terms and gives the continuity equation ${\partial }_t\rho +{\partial }_{x}j=0$. Thus $j$ measures signed probability flow in the positive $x$ direction.

Assume the given wavefunction is zero outside $[0,L]$. Wavefunction normalization requiresAn arbitrary overall quantum phase is unobservable, so choose the positive real representative $N=\sqrt{3}/L^{3/2}$.

The linear ramp state in an infinite square well has a boundary subtlety: the limiting value at $L$ is $\sqrt{3/L}$, rather than zero. The function is still a normalized element of the well's Hilbert space $L^2(0,L)$, so position measurement and energy-projection probabilities are meaningful. It is not in the domain of the Dirichlet Hamiltonian, nor is it a finite-energy box state. Altering the value at the single endpoint does not cure its nonzero boundary trace; a physical finite-energy preparation must smooth the function to zero near the wall. The following probabilities refer to the supplied idealized state.

To match the printed order $df/dt=\{H,f\}$, define the Poisson bracket with the sign conventionHamilton's equations imply $\dot{f}={\sum }_i(f_{q_i}{H}_{p_i}-f_{p_i}{H}_{q_i})=\{H,f{\}}_{\mathrm{paper}}$. With the other common Poisson bracket convention the correct statement is instead $\dot{f}=\{f,H{\}}_{\mathrm{standard}}$. For example $H=p^2/(2m)$, $f=q$ gives $\{H,q{\}}_{\mathrm{standard}}=-p/m$, contradicting the printed order unless its sign convention is adopted. Since $H_t=0$ and $\{H,H\}=0$, the Hamiltonian is conserved. Also ${\dot{p}}_{\varphi }=-H_{\varphi }=0$: this canonical momentum is the axial angular momentum.

For $|x|\ll s_0$ the quadratic Hamiltonian describes a simple harmonic oscillator withAn exactly circular orbit closes for every nonzero $l$. A nonzero infinitesimal radial oscillation closes at leading order precisely when $\sqrt{3}\sin \theta$ is rational. This condition concerns the cone angle: changing $s_0$ or $l$ cannot change the leading-order ratio. Thus there are no special nonzero $l$ values selecting closure for a fixed cone at this order.

At finite radial amplitude the system is equivalent, after unrolling the cone with planar angle $\chi =\varphi \sin \theta$, to a planar central force with potential proportional to $s$. Its angle advance per radial period changes continuously from $2\pi /\sqrt{3}$ near a circle towards $\pi$ at very large radial energy. Hence the physical advance changes from $2\pi /(\sqrt{3}\sin \theta )$ towards $\pi /\sin \theta$. This nonzero interval contains rational multiples of $2\pi$, so tuning a larger initial $|p_s|$ can produce closed orbits even when the small-amplitude ratio is irrational. Generic amplitudes still produce a precessing orbit; neither energy conservation nor increasing $|p_s|$ guarantees closure.

Put $\ell =p_{\varphi }$. The Hamiltonian has no explicit time or azimuthal dependence, so $\dot{H}=\{H,H\}=0$ and $\dot{\ell }=-H_{\varphi }=0$. Its equations of motion areWrite $C=p_R\ell \sin \varphi +({\ell }^2/R+A)\cos \varphi$. Differentiating along these equations givesEquivalently, the same cancellation evaluates $\{C,H\}$. Hence $\boxed{H,p_{\varphi },C\text{ are all constants of motion}.}$ The extra conserved component is associated with the Runge-Lenz symmetry of this inverse-distance central potential, rather than with conservation of radial momentum.

For equal masses $m$, the newly conserved relative $y$ energy isIt is kinetic energy relative to the free $y$ center of mass plus the two interactions. Adding its constant center kinetic energy and the already conserved $x,z$ energies gives the total Hamiltonian. The equal-mass oscillator rotations simplify to $J_{ij}=x_i{p}_{x_j}-x_j{p}_{x_i}$, but are not new: for unequal masses their conserved predecessors contain the mass-weight factors. As a concrete check, independent transformations of $X_i=\sqrt{m_i}{x}_i$ rotate an isotropic oscillator at any positive masses. Complete mass equality does not magically make generic $U_{12}(y_1-y_2)+U_{13}(y_1-y_3)+V(z_1-z_2)$ invariant under rotations of physical space.