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Freeze the magnitude of an order parameter and retain just one common orientation $\theta$ for $N$ rigidly aligned cells. Give this collective coordinate energy $E(\theta )=-Nh\cos \theta$ at positive temperature $T$. The Boltzmann distribution is the displayed normalized probability density function, where $k_B$ is the Boltzmann constant. Here $h$ is energy per cell, a different normalization from the source density in continuous vacuum selection by an external source. This is a deliberately restricted model of selection among already ordered orientations: it does not derive alignment or an ordering critical temperature from a spatial interaction.

Its mean orientation is an elementary ratio of integrals:At $h=0$ every angle has equal probability, and the numerator vanishes exactly. For small $x$, expansion of the exponential gives $m_N(h)=x/2+O(x^3)$ because the angular mean of ${\cos }^2\theta$ is $1/2$. For $x=1$ and $5$ the means are approximately $0.446390$ and $0.893383$. At fixed $h>0$, increasing $N$ concentrates the probability density function near $\theta =0$; expanding $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$ gives angular variance approximately $1/x$. ConsequentlyThe uniform angular average remains a symmetric mixture if one sets $h=0$ first. The other order selects a definite direction. This is the continuous-orientation version of the symmetry-breaking order of limits; it illustrates selection without replacing the separate problem of stability against spatial fluctuations.

Consider an original one-dimensional effective mass model with $V(x)=0$ for $|x|<a$ and $V(x)=V_0>0$ outside. Take the same positive effective mass $m^{\ast }$ everywhere, an abrupt interface, and no interactions. The wavefunction represents a transverse envelope; free motion parallel to the layer is outside this calculation. A double heterostructure laser need not be thin enough for this quantization to dominate its operation.

Set $s=x/a$, $ϵ={\hslash }^2/(2m^{\ast }{a}^2)$, $z_0^2=V_0/ϵ$, $z^2=E/ϵ$ and $w=\sqrt{z_0^2-z^2}$. The time-independent Schrödinger equation becomesHere $u(s)$ denotes the dimensionless potential; a bound state has $0<z<z_0$. Matching the wavefunction and its derivative at $s=1$ givesFor the even bound state, the interior solution is $A\cos (zs)$ and the exterior solution is $A\cos z{e}^{-w(|s|-1)}$. For the odd bound state, replace $\cos (zs)$ by $\sin (zs)$ and $\cos z$ by $\mathrm{sgn}(s)\sin z$. The physical wavefunction is $\psi (x)=\varphi (x/a)/\sqrt{a}$.

Wavefunction normalization includes both exterior tails, not just the active layer. Direct integration givesThe first two terms are the interior integral; the last is the total exterior integral. Thus the interior probability is $P_{\mathrm{in}}=A^2[1\pm \sin (2z)/(2z)]$, with the upper sign for even parity.

Choose $z_0=2$. There is one even root in $(0,\pi /2)$ and one odd root in $(\pi /2,2)$; higher branches start beyond $z_0$. Each left-hand side increases on its stated interval while $\sqrt{4-z^2}$ decreases, proving uniqueness there. Numerically,These are calculated values of the chosen finite square well. The excited bound state lies closer to the barrier and has a longer exterior tail: being bound does not mean having unit probability inside the well.

This model demonstrates electronic confinement, not optical refractive index confinement or a complete laser. Unequal effective masses, band bending, interface disorder and interactions require a more detailed model. In particular, continuity of the ordinary derivative above uses the chosen constant effective mass assumption.

For first cousins in an otherwise outbred pedigree, compare the copies transmitted from their related parents, who are full siblings. These transmissions trace to the same shared grandparent with probability $1/2$ and then to the same homologous copy of that grandparent with probability $1/2$, giving $k_1=1/4$. The unrelated parental contributions cannot match, so two shared copies have probability zero. Selecting one random copy from each cousin chooses those two potentially shared copies with probability $1/4$, giving kinship coefficient $k_1/4=1/16$.

The first moment method uses the expected value of a count to control whether any counted occurrence exists. For a random variable $N$ taking nonnegative integer values, Markov's inequality givesConsequently, $\mathbb{E}N<1$ implies $\mathbb{P}(N=0)\ge 1-\mathbb{E}N>0$, proving that an outcome with no counted occurrence exists. For a sequence of such counts, $\mathbb{E}N\to 0$ shows that $N=0$ with probability tending to one. The first moment method uses information about the average count, while the complementary second moment method also uses its variance.

Draw balls sequentially without replacement from $g$ green and $r$ red balls. Let $T$ be the first positive draw count at which the numbers of drawn green and red balls agree. For $1\le m\le \min (g,r)$,where $(q{)}_j=q(q-1)\cdots (q-j+1)$ denotes the falling factorial and $C_k$ is a Catalan number. Odd first return times are impossible because equality requires an even total number of draws.

To count the successful colour orders, encode green by a step $+1$ and red by a step $-1$. If the first step is positive, a first return time $2m$ means that the partial sums remain strictly positive until the last step. Delete the first positive step and the last negative step, and shift the remaining heights down by one. The result is a Dyck path of length $2m-2$, of which there are $C_{m-1}$. Interchanging the colours doubles this count. Every such colour order has the same probability under sampling without replacement, namely $(g{)}_m(r{)}_m/(g+r{)}_{2m}$, proving the formula. In particular, the numbers of successful colour orders at lengths two, four and six are two, two and four; balanced orders with an earlier return must not be counted.

In a uniform linearized chord diagram, $\mathrm{Pr}(\sqrt{n}{R}_1\ge x)=(1-x^2/n{)}^n\to e^{-x^2}$. Conditional on the right endpoints, $d_1(n)=2+{\sum }_{i=2}^n{1}_{\{L_i\le R_1\}}$, with independent success probabilities $R_1/R_i$. A Chernoff bound gives $R_i=(1+o(1))\sqrt{i/n}$ uniformly for $i\ge n^{1/10}$. The early indicators are $o(\sqrt{n})$, and the remaining conditional mean is $(2+o(1))n{R}_1$. Their conditional variance is $O_{\mathbb{P}}(\sqrt{n})$. Hence $d_1(n)/\sqrt{n}-2\sqrt{n}{R}_1\to 0$ in probability, proving the displayed tail limit. A self-loop's contribution is included in the constant two.

Assume the full-data sample space and the range of the statistic $T$ are standard Borel spaces, $T$ is a measurable function, and the statistical model $\{P_{\theta }\}$ is dominated by one common sigma-finite measure $\nu$. Write $f_{\theta }=d{P}_{\theta }/d\nu$ for its densities. Then $T$ is a sufficient statistic if and only if there exist nonnegative measurable functions $g_{\theta }$ and $h$, with $h$ independent of $\theta$, such thatThe statistical parameter can enter $g_{\theta }$ only through $\theta$ and the observed $T(x)$. ^[ref]

For discrete data, take counting measure; for ordinary continuous data, take Lebesgue measure. Independence, identical sampling laws, smooth dependence on $\theta$, and a fixed positive support of a probability distribution are not requirements of this theorem. Parameter-dependent support must nevertheless be included in $f_{\theta }$, usually through an indicator function, and must fit the displayed factorization. The sufficient statistic for a symmetric uniform sample illustrates exactly this: its indicator function depends on the unknown scale through the observed absolute maximum. The null sets in the theorem concern versions of densities; the factors are not generally themselves normalized probability distributions.

For a discrete sample space, the cancellation behind the theorem is particularly transparent. On a group with positive probability, factorization givesConversely, multiply a common conditional probability mass function by $P_{\theta }(T=T(x))$ to obtain the required factorization. For continuous data, a sum over a group is not a density integral over the original Lebesgue measure: the group may have measure zero. Use the regular conditional distribution formulation and the Fisher-Neyman factorization theorem instead of that discrete calculation.

Assume independent physical-error probability $p$, correction of zero or one errors, and failure on every block with at least two errors. The failure map is $f(p)=1-(1-p{)}^5-5p(1-p{)}^4$. Factoring $f(p)-p$ gives $p(1-p)(4p^3-11p^2+9p-1)$. The unique interior threshold is the root of $4p^3-11p^2+9p-1=0$. Below it concatenation decreases the failure probability toward zero. Two levels give $f(f(p))=1000p^4+O(p^5)$, compared with $10p^2+O(p^3)$ at one level.

// Target: quantum-information-theory.bigb

Consider the chosen finite minimization problemOnly adjacent integers are neighbors. At each interior state propose left or right with probability $1/2$ each; at an endpoint propose its sole neighbor with probability $1/2$ and itself with probability $1/2$. This proposal distribution is symmetric, including at the endpoints. The local minimum at $1$ is relative to these allowed moves; $4$ is the unique global minimum.

At $1$, improving-only local search rejects both neighbors because $E(0)=2$ and $E(2)=4$ exceed $E(1)=1$. For simulated annealing, choose $T_{\ast }=1/\ln 2$, using the natural logarithm. Conditional on proposing $2$ from $1$, the acceptance probability isThe one-step transition probability is instead $P_{T_{\ast }}(1,2)=(1/2)(1/8)=1/16$. Both subsequent rightward proposals are downhill, so the conditional probability of the precise next three states being $2,3,4$, given $X_0=1$ and fixed $T_{\ast }$, isThis is one possible escape route, not the probability of eventually reaching $4$. Rejections, excursions toward $0$, and repeated attempts give many other sample paths. The calculation establishes possibility, without claiming a general performance advantage from a selected sample path.

For comparison, halving $T_{\ast }$ makes the uphill acceptance probability $1/64$; doubling it makes that probability $1/(2\sqrt{2})$. Cooling makes it harder to leave this local minimum, even though the stationary distribution increasingly favors the global minimum. That tension is the central scheduling problem.

Take a birth-death chain on $\{0,1,2,3,4\}$. At each interior state, move up one state with probability $p=1/2$ and down one state with probability $q=1/2$. The endpoints are absorbing states: $P_{00}=P_{44}=1$. This is a small gambler's ruin model in which the state can represent capital, and reaching $4$ is success.

Let $T_a=\inf \{n\ge 0:X_n=a\}$ be the hitting time of endpoint $a$, and let $T=\min (T_0,T_4)$ be the first endpoint hitting time. The desired hitting probability is $h_i={\mathbb{P}}_i(T_4<T_0)$. First-step analysis and the Markov property giveThe interior equation says adjacent differences are equal. Four equal increments must bridge the boundary values $0$ and $1$, soFor example, from state $2$ the hitting probability is $1/2$, and its one-step average is $\frac{1}{2}(1/4)+\frac{1}{2}(3/4)=1/2$. Individual sample paths can wander back and forth, but the same averaging rule already accounts for all of them.

Now ask how long it takes to reach either endpoint. The expected hitting time $t_i={\mathbb{E}}_{i}T$ counts steps, including the transition that first reaches an endpoint. First-step analysis givesThis is a discrete Poisson equation, $(I-P)t=1$ in the interior. The extra $1$ counts the step just taken. The quadratic valuessatisfy the equations; from state $2$, $t_2=1+\frac{1}{2}(3)+\frac{1}{2}(3)=4$ steps. The hitting probability is harmonic, whereas the expected hitting time has a nonzero source term. These are special cases of expected duration of symmetric gambler's ruin; see Hao Wu's MIT lecture 16, gambler's ruin, slide 9.

For a biased random walk with $p=0.6$ upward and $q=0.4$ downward, keep the same absorbing states. The interior harmonic equation becomes $h_i=p{h}_{i+1}+q{h}_{i-1}$. Using biased gambler's ruin probability with these boundary values givesThe three interior hitting probabilities are approximately $0.415385$, $0.692308$, and $0.876923$. Thus starting at $2$ now succeeds about $69.23\%$ of the time. A direct check is $0.6(57/65)+0.4(27/65)=9/13$. The curved profile reflects the unequal transition weights: its successive increments shrink by the factor $q/p=2/3$. The formula, with the endpoint denoted $T$ there, is verified in Karger and Lynch's MIT lecture 26, sections 2.3-2.4; the five-state numbers here are direct substitutions.

Why does harmonic averaging produce the correct hitting probability? With the natural filtration ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$, the Markov property gives the conditional expectation identityFor a function harmonic everywhere, this makes $h(X_n)$ a martingale whenever it is integrable. For a function harmonic only in the interior, $M_n=h(X_{n\wedge T})$ is a stopped martingale. In either five-state example, $0\le M_n\le 1$. Also, from any interior state at most three consecutive upward steps reach $4$, with probability at least $p^3>0$. Repeated three-step blocks give ${\mathbb{P}}_i(T>3k)\le (1-p^3{)}^k$, so $T$ is almost surely finite and has finite expected value; compare finite-interval exit times have geometric tails. Applying optional stopping at $n\wedge T$ and then the bounded convergence theorem givesThe boundedness and finite hitting time justify this passage to the limit. See Steven Lalley's notes, harmonic functions and stopped expectations, theorem 1, and stopped martingale in discrete time for this bounded version of optional stopping.

Uniqueness has a short maximum principle explanation. On a finite state space, the difference of two harmonic extensions with the same boundary conditions is zero on the boundary. If its positive maximum occurs inside, the weighted average forces every successor with positive transition probability to have the same maximum. A path to the boundary propagates that maximum to zero, a contradiction. Applying the argument to the negative of the difference proves uniqueness. Reachability of the boundary is essential: an interior closed communicating class can retain an arbitrary constant.

// Target: markov-process.bigb

For a uniform random permutation $\pi$ of $n\ge 1$ objects, let $M_n$ count its fixed points. Write $M_n={\sum }_{i=1}^n{1}_{\{\pi (i)=i\}}$. Each indicator random variable has expected value $1/n$, so linearity of expectation gives $\mathbb{E}[M_n]=1$ for every $n\ge 1$. This average does not require independence: for $n\ge 2$, fixing two specified objects has probability $1/[n(n-1)]$, greater than the product $1/n^2$ of their individual probabilities.

For four objects, exact enumeration gives counts $9,8,6,0,1$ for $0,1,2,3,4$ fixed points among the $24$ equally likely permutations. Their weighted arithmetic mean is $(0\cdot 9+1\cdot 8+2\cdot 6+4\cdot 1)/24=1$. Exactly three fixed points are impossible: the last object must then also be fixed.

Choose the $m$ fixed objects and derange the remaining $n-m$. Their count is $\left(\genfrac{}{}{0ex}{}{n}{m}\right)D_{n-m}$ among $n!$ equally likely permutations. The inclusion-exclusion principle for $D_{n-m}$ proves the displayed law, valid for $0\le m\le n$ and zero otherwise. With $D_0=1$, it includes the all-fixed case and the impossibility of exactly $n-1$ fixed objects. For fixed $m$, the law tends to $e^{-1}/m!$, the Poisson distribution of mean one.

On a finite state space $F$ with finite objective function values, the fixed-$T$ target is a Boltzmann distribution:The partition function $Z_T$ normalizes the weights but cancels from the Metropolis–Hastings acceptance probability. A short detailed balance proof explains the target. With symmetric $q$ and $x\ne y$,Summing over $x$ gives ${\pi }_T{P}_T={\pi }_T$, so this is a stationary distribution. On a finite state space, a connected graph of positive symmetric proposals makes the accepted Markov chain an irreducible Markov chain. Convergence from any starting state additionally requires an aperiodic Markov chain; one positive self-transition suffices under irreducibility. Replacing $q$ by $(I+q)/2$ ensures this. For asymmetric $q$, require connectivity through bidirectional proposals: one-way proposals are rejected.^[ref]

In the five-state barrier example for simulated annealing, $T_{\ast }=1/\ln 2$ gives exact Boltzmann weights $2^{-E(x)}$. Multiplying by $16$ givesFor the uphill edge, detailed balance is visible numerically:Thus the global minimum has the largest individual stationary probability, $16/33$, but is far from certain at this positive $T_{\ast }$. At $T_{\ast }/2$ the exact stationary distribution is $(16,64,1,16,256)/353$.

For any finite $F$, let $M$ be the set of global minima and $E_{\min }$ the smallest objective function value. Cancelling $e^{-E_{\min }/T}$ from numerator and denominator shows that each nonminimal state's weight tends to zero as $T\downarrow 0$, while every state in $M$ has equal weight. Therefore ${\pi }_T$ tends to the uniform distribution on $M$. This is a statement about a family of equilibrium probability distributions; it does not prove that a cooling Markov chain follows that family.

A binary erasure channel marks its failed transmissions with a special output symbol, whereas a binary symmetric channel changes a bit without announcing whether it changed. At the same nominal failure probability $p=0.1$, the two communication channels therefore leave different uncertainty. With a fair input, an unerased output determines the input exactly, while the erasure flag leaves one bit of conditional entropy. Thus $H(X\mid Y)=p$ and $I(X;Y)=1-p=0.9$ for the binary erasure channel. This is the binary case of capacity of a finite-alphabet erasure channel.

For the binary symmetric channel at $p=0.1$, neither possible received symbol rules out either input, and $H(X\mid Y)=h_2(0.1)\approx 0.46899559$. Its channel capacity is only $1-h_2(0.1)\approx 0.53100441$. A receiver benefits from knowing where information was lost, even when the fraction of affected transmissions is the same.

This comparison concerns the specified transition probabilities, rather than a general ordering of every error mechanism. In particular, at $p=1$, an erasure flag conveys no input information, while deterministic inversion still conveys a full bit. Reporting only an error percentage loses the distinction between hidden corruption, announced loss, and a predictable reversible transformation.

A fluctuation theorem compares the probability of a fluctuating sample trajectory with that of its reverse. Positive mean entropy production permits negative individual realizations: what is constrained is their relative probability, not a prohibition on every reversal. The finite statement below compares specified forward and reverse probability laws; replacing the reverse law by the forward law requires an additional symmetry. The biased-step entropy production model exhibits that symmetry exactly, while the Crooks fluctuation theorem compares two different driving protocols.

For the two-choice Achlioptas process, fix $a>0$ and $K\ge 1$. Conditional on any current graph with $N_{\ge K}\ge an$, put $W$ equal to the vertices in its large graph components. There are at most $n/K$ such initial graph components. If after $s$ more steps no graph component contains $an/3$ vertices of $W$, greedily assign final graph components to two groups so that each meets at least $|W|/3\ge an/3$ vertices of $W$. This induces a set partition of the initial large graph components, of which there are at most $2^{n/K}$ possibilities.

For any fixed such split $W=A\cup B$, a uniform candidate edge joins $A$ to $B$ with probability at least $2a^2/9$. If both candidates do so, the selected edge cannot avoid the crossing. Thus the probability that the split survives all $s$ steps is at most $(1-a^4/81{)}^s$. The deliberately weaker constant also covers distinct candidate sampling for sufficiently large $n$. In the absent-edge version, as long as the split has survived, every $A$--$B$ edge is absent, so conditioning on absence can only increase this crossing probability. A union bound with $s=\lceil A(a)n/K\rceil$ and $A(a)=81(\log 2+2)/a^4$ bounds failure by $e^{-2n/K}$. For fixed $K$, a further union bound makes the conclusion simultaneous over all starting steps $m\le 3n$. This proof is independent of how the rule chooses between the two candidates.

Take a cycle graph with four labeled graph vertices and four distinct edges $\{1,2\},\{2,3\},\{3,4\},\{4,1\}$. Set $h=0$, $J>0$, and impose periodic boundary conditions by writing ${\sigma }_5={\sigma }_1$. The square drawing represents a one-dimensional ring: it has no diagonal interactions and is not a two-dimensional lattice.

Call an edge between opposite Ising spins a domain wall. If $d$ is their number, the four bond products sum to $4-2d$, so the energy isEach domain wall costs $2J$ relative to an aligned bond. Returning around the ring must return to the starting Ising spin, so $d$ is even: equivalently, ${\prod }_{i=1}^4({\sigma }_i{\sigma }_{i+1})=1$. Choosing an even set of $d$ wall edges and then choosing ${\sigma }_1$ determines the entire pattern. The state degeneracy is therefore $g_d=2\left(\genfrac{}{}{0ex}{}{4}{d}\right)$, using a binomial coefficient.

The twelve middle microstates consist of eight patterns with one exceptional Ising spin and four patterns with two adjacent $+1$ Ising spins and two adjacent $-1$ Ising spins. The two alternating patterns have four domain walls. All $2+12+2=16$ microstates are shown below.

Writing $K=\beta J$, the partition function and the probability of each energy class areFor an original exact example, choose $K=(\ln 2)/4$. Then $e^{4K}=2$, $Z=17$, and $P(d=0)=4/17$, $P(d=2)=12/17$, $P(d=4)=1/17$. Each aligned microstate has probability $2/17$, each middle microstate $1/17$, and each alternating microstate $1/34$: multiply by state degeneracy before comparing whole classes. As $K\to 0$, all sixteen microstates become equally likely, but the class probabilities approach $1/8,3/4,1/8$. As $K\to \infty$, the two all-aligned microstates each have limiting probability $1/2$.

A useful Franck-Hertz energy-loss model separates three operations: accelerating an electron, losing a discrete energy $\Delta$ in an inelastic collision, and selecting the surviving electrons with a retarding collector. The collector measures an accepted fraction of an outgoing energy distribution. This makes the experiment a small inverse problem: the electric current depends on both the collision physics and the readout.

For a deliberately simple one-channel model, let $u\ge 0$ be the incoming kinetic energy, and let $p(u)$ be the probability that one encounter excites the chemical atom. Set $p(u)=0$ for $u<\Delta$ and $0\le p(u)\le 1$ otherwise. The conditional probability distribution of outgoing kinetic energy $E$ is the discrete loss kernelHere $\delta$ is the Dirac delta function. The kernel has unit integral, no negative outgoing energy, and mean outgoing kinetic energy $u-p(u)\Delta$. It neglects the small recoil loss in an elastic collision, other excitation channels and ionization. The threshold permits excitation; it does not require $p(u)=1$ above threshold. In particular, choosing a single $\Delta =4.9\mathrm{eV}$ here isolates the characteristic channel discussed above without claiming it is mercury's lowest possible excitation.

For electrons directed normally toward an ideal planar collector, a retarding electric potential difference $V_r>0$ imposes the barrier $e{V}_r$. Write its transfer as $T(E)=H(E-e{V}_r)$, where $H$ is the Heaviside step function and $e$ the positive elementary charge. Thus a perfectly sharp loss need not produce a sharp drop in electric current: the drop also depends on where the outgoing energy lies relative to the barrier. For an oblique incident electron at angle $\theta$ to the normal, the same planar model instead tests $E{\cos }^2\theta >e{V}_r$; angular scattering can change collection even at unchanged total kinetic energy.

For example, choose a monoenergetic incoming beam with $u=5.6\mathrm{eV}$, $p=0.75$ and $\Delta =4.9\mathrm{eV}$. After one encounter, $25\%$ have $5.6\mathrm{eV}$ and $75\%$ have $0.7\mathrm{eV}$. A $1.0\mathrm{V}$ retarder collects only $0.25I_0$, whereas a $0.5\mathrm{V}$ retarder collects $I_0$; here $I_0$ is the incident electric current with fixed geometrical acceptance and no background. The mean outgoing kinetic energy is $1.925\mathrm{eV}$, independently checked by $5.6-0.75(4.9)$. These are chosen inputs and calculated fractions, not historical measurements.

More generally, let $f(E;V)$ be the normalized outgoing probability density function at a fixed accelerating electric potential difference $V$, for the normally incident idealization. A retarding scan givesThe second equation follows by differentiating the lower integration limit, with Dirac delta functions understood for sharp beams. It explains why an integrated collector readout can reveal an energy distribution. With angles present it probes the distribution of normal kinetic energy; finite instrumental resolution smooths it, and numerical differentiation amplifies noise. Keep the incoming beam fixed during this scan. A scan of $V$ changes $f$ itself and cannot be differentiated by the same inversion formula.

In 1972, Stuart J. Freedman and John F. Clauser excited a calcium beam from an oven with a filtered deuterium arc lamp. A calcium emission cascade supplied paired photons; an optical lens system, optical filters, removable glass-plate polarizers and cooled photomultipliers selected their linear polarizations. They compared coincidence circuit rates at relative angles $22.5^{\circ }$ and $67.5^{\circ }$, normalized with polarizers removed. Alternating configurations controlled drift; delayed counts estimated accidental coincidences. The violation agreed with quantum mechanics, conditional on their explicit assumption that passage through a polarizer did not change a photon's detection probability. This was a static-analyzer test with a detection loophole, rather than the ideal four-output CHSH inequality statistic.^[ref]

FRET efficiency is the probability that an excitation of the FRET donor ends in transfer to the FRET acceptor. Write $k_D=k_{\mathrm{r}}+k_{\mathrm{nr}}$ for the total decay rate without a FRET acceptor, including radiative fluorescence and other decay, and $k_T$ for the transfer rate. For constant competing rates,The integral counts transfer events among all excitations in a population with exponential decay. Substitution gives the separation law above. This short rate argument assumes one FRET donor, one FRET acceptor, weak coupling, an available FRET acceptor, and fixed geometry during the decay; it is not a model for arbitrary interacting labels.^[ref]