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Normalize the incident plane wave to unit amplitude. The scattering asymptotics are $\psi (\mathbf{r})\sim e^{ikz}+f(\theta )e^{ikr}/r$; the outgoing coefficient $f$ is the scattering amplitude. The general axisymmetric free solution for $r>0$ isusing Spherical Bessel functions, spherical Neumann functions and Legendre polynomials.

For a real short-range central potential, the exterior radial solution in each partial wave is proportional to $\cos {\delta }_{\ell }{j}_{\ell }-\sin {\delta }_{\ell }{n}_{\ell }$, whose large-$r$ form is $\sin (kr-\ell \pi /2+{\delta }_{\ell })/(kr)$. This defines the scattering phase shift. Multiplying this combination by $(2\ell +1)i^{\ell }{e}^{i{\delta }_{\ell }}$ matches the incoming spherical component of the incident plane wave: the factor $e^{i{\delta }_{\ell }}$ cancels the shifted incoming phase. The outgoing coefficient instead acquires $e^{2i{\delta }_{\ell }}$. Subtracting the plane wave's outgoing component therefore givesFor the barrier's $s$-wave, the regular reduced radial function inside is $u(r)=A\sinh (\kappa r)$, with $\kappa =\sqrt{{\gamma }^2-k^2}$. Outside take $u(r)=B\sin (kr+{\delta }_0)$. Continuity of $u$ and its derivative at $a$ equates logarithmic derivatives:Use the branch with ${\delta }_0\to 0$ as $k\to 0$. The matching equation gives ${\delta }_0=k(\tanh (\gamma a)/\gamma -a)+o(k)$, so the $s$-wave scattering amplitude tends toIts negative is the repulsive spherical barrier scattering length, positive for this repulsive barrier.

For spin $1/2$, $\mathbf{S}=\hslash \mathbf{\sigma }/2$. The Pauli matrices satisfy ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$. The antisymmetric term vanishes on contraction with $n_i{n}_j$, and $|\mathbf{n}|=1$, so $(\mathbf{n}\cdot \mathbf{\sigma }{)}^2=I$. Separating the even and odd terms of the exponential series givesFor the single-particle Hamiltonian, unitary time evolution consequently acts asFor two spins the joint eigenstates of total angular momentum are the spin-one-half triplet statewith $J^2=2{\hslash }^2$ and $J_3=m\hslash$, and the spin-one-half singlet stateThese follow by adding the two spin-$1/2$ angular momenta, or by direct action of the Pauli matrices.

The two terms of the specified Hamiltonian commute, so its evolution factorizes into rotations with opposite angles on the two tensor factors. The unique initial $J_3=\hslash$ state is $|\uparrow \uparrow ⟩$. With $c=\cos (\lambda t/2)$, $s=\sin (\lambda t/2)$, its evolved state isThe singlet amplitude is $i\sqrt{2}cs$, giving the requested probability

Write $({\varphi }_{\epsilon }\ast f)(x)=\int \varphi (y)f(x-\epsilon y)dy$. Since $\int \varphi =1$,A Schwartz function is bounded and uniformly continuous. Given a tolerance, first choose $R$ so the tail integral of $|\varphi |$ is small. On $|y|\le R$, uniform continuity makes the difference uniformly small for every $x$ when $\epsilon R$ is small. On the tail it is bounded by $2\Vert f{\Vert }_{\infty }$. This proves uniform convergence, without assuming $\varphi \ge 0$, and is the approximate identity argument.

For the radial regularization of the three-dimensional fundamental solution of the Laplace equation, direct differentiation givesThe radial integral is $\int \varphi =3{\int }_0^{\infty }{r}^2(1+r^2{)}^{-5/2}dr=1$. This even approximate identity therefore gives $\int (-\Delta N_{\epsilon })f\to f(0)$ for $f\in \mathcal{S}({\mathbb{R}}^3)$.

The function $N=1/(4\pi |x|)$ is locally integrable and determines a tempered distribution. We have $0<N_{\epsilon }\le N$, and $N|\Delta f|$ is integrable near zero and infinity for a Schwartz test function. Integration by parts and the dominated convergence theorem giveThus $-\Delta N={\delta }_0$, proving that $N$ is a fundamental solution of a linear differential operator. In this last part the test-function space is $\mathcal{S}({\mathbb{R}}^3)$; the printed ${\mathbb{R}}^n$ in the three-dimensional integral is understood with $n=3$.

Write $V(x)=F(\pi ,x)$. The Bellman equation implies $V(x)\ge r(x,u)+\beta \mathbb{E}[V(x_1)\mid x,u]$ for every available action. Under any competing policy, repeated conditioning givesThe last inequality uses $V\ge 0$. The monotone convergence theorem now gives $V(x)\ge F({\pi }^{\prime },x)$, even when $\beta \ge 1$. If $V(x)=\infty$ the comparison is immediate; otherwise the nonnegative conditional bounds justify each finite induction step. This proves Bellman comparison with nonnegative rewards and superunit discount without a vanishing-terminal-term assumption.

For the unit-bet policy in this gambler's ruin problem, put $\beta =\sqrt{9/8}=3/(2\sqrt{2})$. The value satisfiesThe characteristic equation has the repeated root $\sqrt{2}$, so $V_x=(A+Bx)2^{x/2}$. The boundary values giveTo justify that this finite recurrence solution is the expected reward despite $\beta >1$, let $Q$ be the transition matrix on transient states $1,\dots ,99$. The matrix $\beta Q$ is diagonally similar, with diagonal factors $2^{x/2}$, to the symmetric tridiagonal matrix with off-diagonal entries $1/2$. Its eigenvalues are $\cos (j\pi /100)$, $1\le j\le 99$, with absolute value below one. Thus its Neumann series converges and yields exactly the absorbed-payoff expectation and the unique recurrence solution.

When $p_2=1/4$, at wealth $2$ bet $2$ once and then follow the unit-bet policy. Its expected value is $\beta V_4/4=\beta V_2>V_2$, since $V_4=4V_2$. Therefore the unit-bet policy is not optimal.