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The intrinsic parity $\eta$ of a particle is the parity eigenvalue of its one-particle state at rest:For the particles in this question, $\eta =\pm 1$. A two-particle state has multiparticle parity ${\eta }_1{\eta }_2(-1{)}^{\ell }$, where $\ell$ is its relative orbital angular momentum.

The pionic-deuterium ground state has atomic $\ell =0$. Since the pion has spin zero, its total angular momentum is therefore the deuteron spin, $J=1$, and its parity is ${\eta }_d{\eta }_{\pi }={\eta }_{\pi }$. The two neutrons are identical fermions. Their antisymmetric spin-one-half singlet state would require even orbital $\ell$, but $S=0$ and $J=1$ would instead require $\ell =1$. Hence they must occupy the symmetric spin-one-half triplet state, so the orbital state is antisymmetric and $\ell$ is odd. The addition of angular momentum with $S=J=1$ permits $\ell =0,1,2$, leaving $\ell =1$. Thus parity conservation givesso the pion has intrinsic parity $\boxed{{\eta }_{\pi }=-1}$.

In ${\Delta }^{-}\to {\pi }^{-}+n$, the final total spin is $S=1/2$. Coupling it to $\ell$ to obtain $J=3/2$ givesso the allowed orbital angular momenta are $\boxed{\ell =1,2}$. The observed angular probability is ${\cos }^2\theta =|P_1(\cos \theta ){|}^2$, where $P_1$ is a Legendre polynomial; it therefore selects the $\ell =1$ partial wave rather than $\ell =2$. The parity selection rule for a two-body decay now givesTherefore the delta particle has intrinsic parity $\boxed{{\eta }_{{\Delta }^{-}}=+1}$.

The equilibrium equations give $y=0$ and $x-x^3=0$, so the fixed points areThe Jacobian isAt the origin its determinant is $-1$, so the origin is a saddle equilibrium. At either of the other points the determinant is two and the trace is $\epsilon (1-\alpha )$. For small positive $\epsilon$, they are unstable foci when $\alpha <1$ and stable foci when $\alpha >1$; at $\alpha =1$ the linearization is nonhyperbolic. When $\epsilon =0$, each is a center equilibrium.

For $\epsilon =0$, the system is Hamiltonian withbecause $\dot{x}=H_y$ and $\dot{y}=-H_x$. Its conservative planar phase portrait has two wells centered at $(\pm 1,0)$, a saddle at the origin, periodic level curves around each center for $-1/4<H<0$, two homoclinic loops on $H=0$, and outer periodic curves surrounding both wells for $H>0$.

For positive small $\epsilon$,On the unperturbed level $H=H_0$,The upper and lower halves contribute equally because $dt=dx/y$. Hence the energy balance for the weakly perturbed double-well oscillator isso one may take

For the right homoclinic loop, $H_0=0$, $x_1=0$, and $x_2=\sqrt{2}$. Its persistence requiresThe substitution $u=1-x^2/2$ givesThus the homoclinic balance for the weakly perturbed double-well oscillator occurs atThe left loop gives the same condition by symmetry.

For an orbit surrounding one center, the corresponding balance ratio tends to $\alpha =1$ as the orbit shrinks to the center and to $\alpha =5/4$ as it approaches the homoclinic loop. The single-well periodic orbit range for the weakly perturbed double-well oscillator is therefore

Rotate the contour for the Gaussian integral through angle $-\pi /4$; a small positive damping factor justifies the deformation and can then be removed. After scaling by $t^{-1/2}$, the oscillatory Gaussian integral with negative phase givesSimilarly, first scale $u=t^{-1/3}v$ and then rotate the positive ray through angle $-\pi /6$. Since ${\int }_0^{\infty }{e}^{-v^3}dv=\Gamma (4/3)$, the cubic oscillatory Gamma integral gives

For fixed $\alpha$, consider the complex phaseIts unique interior stationary point iswhereThe endpoint contributions are $O(x^{-1})$, while the one-dimensional stationary-phase formula has order $x^{-1/2}$. Taking its real part proves the stationary-phase asymptotic for the generalized Bessel cosine integral:

For $Q_x(x)$ the phase is $x(\sin \theta -\theta )$. Its stationary point has merged with the endpoint $\theta =0$, whereUse the cubic endpoint stationary-phase scaling $\theta =(6/x{)}^{1/3}u$. The other endpoint and the region away from zero contribute only $O(x^{-1})$, soTherefore

For the finite state space $S$, the generator or Q-matrix $G=(g_{ij})$ is defined byIts off-diagonal entries are nonnegative jump rates and $g_{ii}=-{\sum }_{j\ne i}{g}_{ij}$. The Transition semigroup of a continuous-time Markov chain is $P(t)=e^{tG}$.

An invariant distribution of a continuous-time Markov chain is a probability row vector $\pi$ satisfying $\pi P(t)=\pi$ for every $t\ge 0$. Differentiation at zero gives $\pi G=0$; conversely, $\pi G=0$ implies $\pi e^{tG}=\pi$. ThusInvariant distributions need not be unique: each closed communicating class supports one, and their convex combinations are invariant. A finite irreducible chain has a unique invariant distribution.

Let $0=T_0<T_1<\cdots$ be the arrival times of the independent rate-$\lambda$ Poisson process. The increments $T_{n+1}-T_n$ are independent exponential variables of rate $\lambda$. Conditional on $Y_n=X_{T_n}=i$, the Markov property, time homogeneity, and independence of the next increment giveHence $(Y_n)$ is a discrete-time Markov chain. Its transition matrix is the Poisson-sampled continuous-time Markov chain resolvent

If $\pi G=0$, then $\pi e^{tG}=\pi$ and integration gives $\pi P_{\lambda }=\pi$. Conversely, if a probability row vector $\nu$ satisfies $\nu P_{\lambda }=\nu$, multiplyingon the right by $\lambda I-G$ yields $\nu G=0$. Therefore the sampled chain and the original chain have exactly the same invariant distributions.