The interior time-independent Schrödinger equation for a free particle isConfinement by the rectangular infinite well imposes Dirichlet boundary conditions: $\psi =0$ at $x=0,b$ and $y=0,\sqrt{5}b$. The wave function is square-integrable and may be normalized to unit probability on the rectangle.

Separation of variables gives products of the one-dimensional well states,The Hamiltonian is the sum of the independent $x$ and $y$ kinetic-energy operators, so the one-dimensional energies add. Put $C={\pi }^2{\hslash }^2/(2m{b}^2)$. ThenFor $n=1$, increasing $\ell$ gives $6C/5,9C/5,14C/5,21C/5$. The smallest energy with $n\ge 2$ is also $21C/5$, attained at $(2,1)$; all $n\ge 3$ give larger energies. Thus the ground and first three distinct excited energy levels areThe largest listed energy therefore has energy-level degeneracy $\boxed{2}$, since the two product eigenfunctions are linearly independent and no other pair solves $5n^2+{\ell }^2=21$.

Away from a collision, let $r=q_1-q_2$ and $D=|r|>0$. The Lagrangian isDifferentiating $D^2=r\cdot r$ with respect to $x_1$ and $x_2$ gives $2D{\partial }_{x_1}D=2(x_1-x_2)$ and $2D{\partial }_{x_2}D=-2(x_1-x_2)$. ThereforeThe other coordinates give ${\nabla }_{q_1}D=r/D$ and ${\nabla }_{q_2}D=-r/D$. Substituting these into the Euler-Lagrange equations yields equal and opposite central forces:Adding the equations shows that the derivative of the total linear momentum is zero, so $P=m_1{\dot{q}}_1+m_2{\dot{q}}_2$ is constant. For the total angular momentum, the terms ${\dot{q}}_i\times {\dot{q}}_i$ vanish, givingThus both stated vectors are conserved quantities.

The total energy is also constant. Define $E=\frac{1}{2}{m}_1|{\dot{q}}_1{|}^2+\frac{1}{2}{m}_2|{\dot{q}}_2{|}^2+V(D)$. Since $\dot{D}=r\cdot ({\dot{q}}_1-{\dot{q}}_2)/D$,Hence $\boxed{E=T+V(D)\text{ is conserved}}$. The explicit cancellation establishes energy conservation for an arbitrary smooth separation-dependent potential, throughout any collision-free trajectory.

For variations vanishing at the endpoints, the Euler-Lagrange equations areDifferentiate the stated energy expression using the chain rule. The acceleration terms cancel, leavingAlong a stationary trajectory the two brackets vanish. Thus $dH/dt=-F_t$, and in the absence of explicit time dependence $\boxed{H\text{ is constant}}$.

In plane polar coordinates, the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2$, so the kinetic energy is $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$. An inverse-square central gravitational force has potential energy proportional to $1/r$. With the sign convention printed in this paper, write $V=km/r$, givingFor attraction to a star of mass $M_{\star }$, this convention has $k=-G{M}_{\star }<0$. Keeping $k$ real and signed is essential; no sign is changed in differentiating the printed functional.

Since $F_r=mr{\dot{\theta }}^2+km/r^2$, $F_{\dot{r}}=m\dot{r}$, $F_{\theta }=0$ and $F_{\dot{\theta }}=m{r}^2\dot{\theta }$, the equations of motion areThe second equation gives the conserved angular momentum $J=m{r}^2\dot{\theta }$; equivalently $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$. The autonomous-functional identity gives a second conserved quantity,This is the total mechanical energy; for attractive gravity the potential contribution is negative because $k<0$.