Expand $G(x,x^{\prime })$ in a complete orthonormal eigenbasis in its first argument. Applying the operator yields ${\sum }_n(E_n-E)c_n{\phi }_n(x)=\delta (x-x^{\prime })$. Completeness expands the Dirac delta distribution as ${\sum }_n{\phi }_n(x){\phi }_n^{\ast }(x^{\prime })$, soConsequently ${\lim }_{E\to E_n}(E_n-E)G={\phi }_n(x){\phi }_n^{\ast }(x^{\prime })$.

For the explicit formula, use $E_n-E\simeq -({\hslash }^2{k}_n/m)(k-k_n)$ and $\sin ka\simeq (-1{)}^{n}a(k-k_n)$. Also
$\sin [k_n(a/2-x)]=(-1{)}^{n+1}\sin [k_n(x+a/2)]$. The resulting limit isThe expression is symmetric in $x,x^{\prime }$ and holds on either side of the source. The usual complex-analysis residue with respect to $E-E_n$ has the opposite sign, since the source requests the limit with $E_n-E$.

Let $x_{<}=\min (x,x^{\prime })$ and $x_{>}=\max (x,x^{\prime })$. A homogeneous solution vanishing at the left wall is $u(x)=\sin [k(x+a/2)]$; one vanishing at the right wall is $v(x)=\sin [k(a/2-x)]$. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-k\sin ka$. Continuity and the source derivative jump $G_{+}^{\prime }-G_{-}^{\prime }=-2m/{\hslash }^2$ therefore giveThis is exactly the printed form because $x_{<}=(x+x^{\prime }-|x-x^{\prime }|)/2$ and $x_{>}=(x+x^{\prime }+|x-x^{\prime }|)/2$, with a minus sign from reversing the argument of the second sine.

The true poles occur at $k_n=n\pi /a$, $n=1,2,\dots$, corresponding toThe apparent zero at $k=0$ is removable: the numerator and $k\sin ka$ both vanish quadratically. Negative values of $k_n$ duplicate the same energies, rather than describing extra levels. The boundary conditions and source jump uniquely determine the inverse whenever $E$ is not an eigenvalue.

Write $L_0\psi =-U\psi$ and use the outgoing inverse of $L_0$. Adding the homogeneous incoming wave yieldsFor the attractive delta potential, let $\eta =m\alpha /(k{\hslash }^2)$. The equation reduces to $\psi (x)=e^{ikx}+i\eta e^{ik|x|}\psi (0)$ and $\psi (0)=1/(1-i\eta )$. Separating the two half-lines givesBoth sides have the same group velocity, soThe same amplitudes satisfy continuity and the exact delta-potential derivative jump ${\psi }^{\prime }(0^{+})-{\psi }^{\prime }(0^{-})=-2m\alpha \psi (0)/{\hslash }^2$, providing an independent local check on the sign.

Away from $x=x^{\prime }$, the free equation is $G^{\prime \prime }+k^{2}G=0$. The outgoing boundary condition selects $G=A{e}^{ik|x-x^{\prime }|}$, continuous at its source. Integrating the differential equation across the source gives the derivative jumpThe jump for the proposed form is $2ikA$, so $A=im/(k{\hslash }^2)$. ThusFor $x$ far to either side, the $x^{\prime }$-dependent phase is absorbed into the outgoing amplitude. The outgoing prescription is equivalently the boundary value of the resolvent with $E\to E+i0$; changing that prescription would give a different Green function.

Assume a real, time-independent potential with the same zero asymptote on both sides, so $k=\sqrt{2mE}/\hslash$ is the same incident and transmitted wave number. For unit incoming amplitudes,Multiplying the two Schrödinger equations crosswise and subtracting shows that the Wronskian $W={\psi }_L{\psi }_R^{\prime }-{\psi }_L^{\prime }{\psi }_R$ is constant. Its two asymptotic values are $-2ik{t}_R$ and $-2ik{t}_L$, hence $t_L=t_R$.

For a real potential the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ is conserved, giving $|r_L{|}^2+|t_L{|}^2=|r_R{|}^2+|t_R{|}^2=1$. ThereforeThe reflection amplitudes can have different phases. Reality or self-adjointness is essential for the reflection-probability conclusion: a complex absorbing potential need not have equal left and right reflectances. Equal asymptotic speeds are also used in identifying transmission with the squared amplitude.

Put $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The integral becomesThe poles are $-2i/3$ and $-3i/2$; only the first is inside the unit circle. The derivative of the denominator there is $12(-2i/3)+13i=5i$. The residue theorem therefore givesThere are no poles on the contour because the real denominator is at least one.

Integrate $e^{iz}/z$ around an upper semicircle, indenting the origin by a small upper semicircle traversed clockwise. The integrand is holomorphic inside the indented contour. The large arc vanishes by Jordan's lemma, and the small arc contributes $-i\pi$, soTaking imaginary and real parts yieldsThe first integrand has a removable singularity at zero and its tails converge conditionally by Dirichlet test for improper integrals. The second integral needs a principal value at zero; its zero also follows from oddness with symmetric exclusion. One must not assign an ordinary convergent integral to its singularity.

Axisymmetry supplies conserved $p_{\theta }$ in addition to conserved energy. Consequently $\boxed{f=F(H,p_{\theta })}$ is a more general stationary solution, since both arguments are constant on every trajectory. Additional independent integrals may permit still more general distributions. This is a sufficient construction, not a claim that every stationary distribution must have exactly these two arguments.

For $f=F(H)$ with no explicit time dependence, ${\partial }_{t}f=0$ and $[H,F(H)]=F^{\prime }(H)[H,H]=0$. It therefore solves the stationary transport equation. Normalizability and the accessible energy region still have to be checked for a proposed probability distribution.

Conservation of probability obeys ${\partial }_{t}f+{\nabla }_{q,p}\cdot (f\mathbf{v})=0$. Expand the divergence and use incompressibility to obtain ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{q,p}f=0$, or $\boxed{df/dt=0}$. This is fine-grained transport for Hamiltonian trajectories without sources or collisions. Coarse graining can obscure this constancy without contradicting it.

To match the source's order define $[a,b]={\sum }_i(a_{p_i}{b}_{q_i}-a_{q_i}{b}_{p_i})$. The chain rule and Hamilton's equations then giveAntisymmetry implies $[H,H]=0$, hence $dH/dt=H_t$, which vanishes for a time-independent Hamiltonian. With the alternative common convention $\{a,b\}={\sum }_i(a_{q_i}{b}_{p_i}-a_{p_i}{b}_{q_i})$, the same law is $dg/dt=g_t+\{g,H\}$; mixing the two orders would reverse the physical time evolution.

Hamilton's equations define phase-space velocity $({\dot{q}}_i,{\dot{p}}_i)=(H_{p_i},-H_{q_i})$. Its divergence isThus infinitesimal phase-space volumes are preserved, assuming sufficient differentiability. This is Liouville's theorem; spatial density alone need not be incompressible because the statement concerns the combined coordinate-momentum space.

Put $X={\omega }^2$ and $S={\omega }_1^2+{\omega }_2^2+4{\Omega }^2$. Its equation is $X^2-SX+{\omega }_1^2{\omega }_2^2=0$. If both effective stiffnesses are positive, the sum and product of roots are positive and $S^2-4{\omega }_1^2{\omega }_2^2\ge 0$, so both roots are positive. They can coincide in a degenerate zero-rotation isotropic case; “two” need not mean distinct.

If ${\omega }_1^2<0<{\omega }_2^2$, the product is negative, so one root is negative. Its imaginary frequency gives a growing exponential displacement: centrifugal softening has overwhelmed one elastic restoring direction, making the central configuration unstable. Coriolis coupling mixes the two coordinates but does not remove this negative-product instability.

Substitution of the exponential ansatz gives the homogeneous amplitude systemA nonzero mode requires zero determinant. Thus $\boxed{({\omega }_1^2-{\omega }^2)({\omega }_2^2-{\omega }^2)=4{\Omega }^2{\omega }^2}$ with the stated two effective stiffnesses. The off-diagonal product is positive, which determines the sign on the right.

Differentiate the quadratic Lagrangian with respect to coordinates and velocities. The equations areThe velocity-cross terms are Coriolis accelerations; moving $-{\Omega }^2{x}_i$ to the force side identifies the outward centrifugal acceleration. The remaining terms are elastic restoring forces. The sign pair is fixed by the rotating-basis convention in part a.

Use rotating basis vectors with ${\dot{\mathbf{e}}}_1=\Omega {\mathbf{e}}_2$, ${\dot{\mathbf{e}}}_2=-\Omega {\mathbf{e}}_1$. The inertial velocity is $({\dot{x}}_1-\Omega x_2){\mathbf{e}}_1+({\dot{x}}_2+\Omega x_1){\mathbf{e}}_2$. The two fixed rotating anchors are $(0,\pm a)$, so the spring lengths are ${\ell }_{\pm }=\sqrt{x_1^2+(x_2\mp a{)}^2}$. Combining kinetic and elastic energies givesA constant elastic offset has no dynamical effect.