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For the sextic even Landau potentialput $q={\varphi }^2$. The stationary-point equation isso the nonzero stationary points haveAt such a point, $\epsilon =-q^2-gq$, and henceA first-order transition occurs when a nonzero minimum has the same free energy as the disordered minimum $G(0)=0$. ThereforeAt $g=0,\epsilon =0$, the first-order line meets the continuous line and forms the tricritical point. For $g>0$, the transition at $\epsilon =0$ is continuous.

For $g<0$, the disordered state $\varphi =0$ is a local minimum when $\epsilon >0$ and loses stability at the disordered spinodal point $\epsilon =0$. The ordered minima exist whileand merge with the intervening maxima at the ordered spinodal $\epsilon =g^2/4$. Consequently the two spinodals enclose a region of metastability. If increasing $\epsilon$ corresponds to heating, the ordered phase can persist above the coexistence curve throughoutwhile the disordered phase can persist below it throughout

Landau theory represents the appropriate thermodynamic potential as an analytic function of a small order parameter, retaining every low-order term allowed by the system's symmetry. At fixed control parameters, the equilibrium phase is the global minimum of the Landau free energy; local minima describe metastable phases. A continuous phase transition occurs when a minimum changes continuously and loses stability, whereas a first-order phase transition occurs when distinct minima exchange global stability.

Forthe stationary points satisfyIf $\epsilon >0$, $\varphi =0$ is the unique minimum. If $\epsilon <0$, $\varphi =0$ is a maximum and the two degenerate minima areThus the reflection symmetry $\varphi \mapsto -\varphi$ is spontaneously broken below $\epsilon =0$, and

Write $u=kr-l\pi /2$. Sincethe outgoing-to-incoming coefficient ratio is $-S_l$. Hence the Partial-wave S-matrix element and scattering phase shift areThe scattering length from a partial-wave S-matrix is

For the stated S-wave scattering solution, $\tanh (\lambda r)\to 1$ as $r\to \infty$, soUsing the same incoming/outgoing convention,Therefore

The analytic continuation of $S_0$ has a bound-state pole of the scattering amplitude at $k=i\lambda$ in the upper half-plane. Since $E={\hslash }^2{k}^2/(2m)$, this givesTo extract the bound-state wavefunction, multiply the scattering solution by $k-i\lambda$ and take the pole limit:Thus the unnormalized exterior solution isThese data are collected in the Hyperbolic-tangent S-wave exterior solution.

In the classical experiment, particles whose impact parameter lies between $b$ and $b+db$ occupy an incident annulus of areaFor a spherically symmetric potential, azimuthal symmetry scatters this annulus into the solid-angle stripThus the central-potential classical differential cross-section isIf the relation between $b$ and $\theta$ has several branches, one sums this expression over every branch. The corresponding diagram consists of an incoming line offset by $b$ from the centre, an outgoing asymptote making angle $\theta$ with the incident direction, and the annulus and solid-angle strip just described.

For quantum scattering by a central potential, the far-field wavefunction iswhere $f(\theta )$ is the scattering amplitude. The incident plane wave has probability currentTo leading order in $1/r$, the scattered spherical wave has radial currentThe scattered rate through area $r^{2}d\Omega$ divided by the incident flux is therefore the quantum differential cross-section from a scattering amplitude:

Write $r=e^x$, $ϵ=\hslash /\sqrt{2m}$, $L=l+1/2$, and $\lambda =-|\lambda |$. Since $V(r)=-V_0/r$, the transformed equation hasThe two turning radii $a<b$ are the positive roots ofsoInside the allowed interval,The semiclassical action integral is thereforeEquating this with $(n+1/2)\pi$ from part (a), and using $L=l+1/2$, givesSince ${ϵ}^2={\hslash }^2/(2m)$,Thus the Coulomb WKB quantization with the Langer correction yields

Between the classical turning points, $q<0$, while boundedness selects exponentially decaying WKB approximation solutions in the forbidden regions $x<a$ and $x>b$. The Airy turning-point connection formula at the left turning point matches the left decaying solution toin the allowed region. Matching the right decaying solution across $b$ givesFor these expressions to be proportional throughout $(a,b)$, their phases must be compatible. Each simple turning point contributes a phase shift $\pi /4$, giving the two-turning-point WKB quantization condition

Given the present state $(S_t^1,M_t)=(x,m)$, take an independent fresh copy of the stock path and denote its initial value, terminal value after $T-t$ steps, and running maximum by $(y,z,w)=(S_0^1,S_{T-t}^1,M_{T-t})$. Scaling the fresh path to start at $x$ makes its terminal value $xz/y$ and its future running maximum $xw/y$. Including the maximum already attained givesHence defineIndependent increments then give the Running-maximum state reduction in the Cox--Ross--Rubinstein modelwith

Let $N=T-t$ and use the discounted multipliers $u,d$. Under $Q$, the number $K$ of up moves in the remaining $N$ periods has the binomial distribution $\mathrm{Binomial}(N,q)$. Sincethe conditional expectation in part (b) becomesFor an undiscounted stock, replace $u,d$ by $U,D$ and discount the resulting payoff by the appropriate power of $R$.

The arbitrage-free value of the discounted claim is the conditional expectationConditional on the observed path $(S_0^1,\dots ,S_t^1)=(x_0,\dots ,x_t)$, the future multiplicative increments are independent of ${\mathcal{F}}_t$ and have the same law as a fresh stock path divided by its initial value. ThereforewhereThis is Risk-neutral valuation in the Cox--Ross--Rubinstein model for a path-dependent claim.

In the one-period building block of the Cox--Ross--Rubinstein model, the money-market account grows by a deterministic factor $R>0$, while the stock grows independently at each time byThus $S_t^0=R^t{S}_0^0$ and $S_t^1=S_0^1{\prod }_{j=1}^t{Y}_j$. Equivalently, the discounted stock has multipliers $u=U/R$ and $d=D/R$.

The model has no arbitrage precisely when the risk-free return lies strictly between the stock returns:Indeed, failure of either strict inequality gives a one-period arbitrage, while under the strict inequalities the probabilitylies in $(0,1)$. Taking the moves independent under $Q$ with up probability $q$ makes the discounted stock a martingale, since $qu+(1-q)d=1$. This is the unique Risk-neutral probability in the Cox--Ross--Rubinstein model, hence the unique equivalent martingale measure.

For observations $x=(x_1,\dots ,x_n)$, the likelihood function isA maximum-likelihood estimator is any measurable choiceFor one observation in a regular scalar model, the score function isand the Fisher information iswhere the second identity requires the usual differentiation-under-the-integral regularity. For $n$ independent observations, Tensorization of Fisher information gives $I_n(\theta )=nI(\theta )$.

Here $\Lambda$ is planar Lebesgue measure. For a Borel set $B\subseteq [0,\infty )$, polar coordinates giveThe pushforward is therefore locally finite and atomless. By the mapping theorem for Poisson point processes, $f(\Pi )$ is a homogeneous Poisson process on $[0,\infty )$ with constant intensity $\pi$. This is the two-dimensional radial volume transform of a homogeneous Poisson point process.

Put $S_k=R_k^2$. The variable $S_k$ is the $k$th arrival time of that rate-$\pi$ Poisson process, hence has the gamma distribution with shape $k$ and rate $\pi$:Since $s=r^2$ and $ds/dr=2r$, the change-of-variables formula for a probability density gives the Kth-nearest-neighbour distance in a homogeneous Poisson point process

The mapping theorem for Poisson point processes applies provided $f$ is a measurable function and its pushforward measureis locally finite, meaningfor every bounded Borel set $B\subseteq {\mathbb{R}}^s$. If “Poisson process” means a simple point process, one also requires $f_{\ast }\Lambda$ to be atomless, equivalentlyUnder these conditions, $f(\Pi )$ is a Poisson point process with mean measure

For every Borel set $A\subseteq {\mathbb{R}}^d$, letbe its random point count. A Non-homogeneous Poisson point process with intensity function of a point process $\lambda$ is characterized by:

Because $\Lambda$ has a density with respect to Lebesgue measure, it is atomless, so the process is simple: almost surely no location occurs more than once.