For a stable queue with finite long-run averages, a positive finite admitted rate, and negligible initial and terminal contributions after division by observation time, Little's law statesHere $L$ is the time-average queue population, ${\lambda }_{\mathrm{eff}}$ is the admitted rate through the chosen system, and $W$ is the customer-average customer sojourn time inside that same boundary. Without abandonment, admitted rate equals completed-service queue throughput in this stable regime. It may differ from the offered arrival rate because of blocking. When customers abandon, count their admissions and exits too, and average their actual time inside; using only service completions would mismatch the population being measured.

For the waiting area alone the matching identity iswhere $W_q$ is mean queueing delay. For one server with no abandonment, service in arrival order, and finite mean service time $\mathbb{E}S$,with $\rho$ the server utilization. Mixing total population $L$ with waiting-only duration $W_q$ would count different portions of each customer's visit. Customers who start service immediately contribute zero to $W_q$, and must still be included in that customer average.

The strength of Little's law is that it does not require a Poisson process, an exponential distribution of service times, or service in arrival order. The sample-path area identity for Little's law explains why: counting customer-time across time slices or across customer visits gives the same area. A sufficient probabilistic setting is repeated independent and identically distributed empty-to-empty cycles with finite positive mean cycle duration, finite positive mean customer count, and finite mean population area; the renewal-reward theorem then supplies the long-run averages. Stability alone does not guarantee finite mean queueing delay: the Pollaczek-Khinchine mean formula shows why a finite second moment of service time matters for that mean in an M/G/1 queue.

For example, suppose a service desk receives six requests per hour but admits only four; measured mean customer sojourn time of admitted requests is half an hour. Its mean queue population is $4\cdot (1/2)=2$ customers. Multiplying the offered rate by that duration would incorrectly give three. Conversely, counting two customers on average and measuring admitted queue throughput of four per hour gives $W=2/4=1/2$ hour. This lets one estimate a hard-to-track duration from occupancy and completed-job counts.

John D. C. Little published a rigorous proof in 1961 under finite-mean, strict stationarity, and ergodicity assumptions on the arrival process. Those were assumptions of that proof, rather than a requirement that every useful form of the identity have random arrivals. ^[ref]. ^[ref].

A service discipline specifies which customers receive the server's capacity and where newly arriving customers join the queue. First come first served and processor sharing give different sample paths even when their equilibrium populations agree.

An ordered state $x=(c_1,\dots ,c_n)$ records the classes and positions of customers at a single server. Class-$r$ arrivals have rate ${\alpha }_r$ and exponential service parameter ${\mu }_r$. Under a symmetric service discipline, its stationary distribution is $(1-\rho ){\prod }_i({\alpha }_{c_i}/{\mu }_{c_i})$, where $\rho ={\sum }_r{\alpha }_r/{\mu }_r<1$. If every ${\mu }_r=\mu$, class-independent insertion and service position rules need not be equal: time reversal of a continuous-time Markov chain interchanges them and proves the same law. Aggregating words gives $(1-\rho )n!{\prod }_r[({\alpha }_r/{\mu }_r{)}^{n_r}/n_r!]$ by the multinomial coefficient. With class-dependent service rates, first come first served need not have this product law.

A busy period is the interval from an arrival that finds the system empty until the next departure leaving it empty. It excludes the preceding idle waiting time. In a stable M/G/1 queue with arrival rate $\lambda$ and mean service time $m$, its mean is $m/(1-\lambda m)$.

A queue with a Poisson process of arrivals and two independent service times with exponential distributions of different rates must distinguish the two possible single-customer states. If an arrival chooses either idle server with probability $1/2$, the continuous-time Markov chain is reversible. Write $a,b$ for the states with only server one or two busy, and $n\ge 2$ for total population. Its detailed balance equations give ${\pi }_a={\pi }_0\nu /(2{\mu }_1)$, ${\pi }_b={\pi }_0\nu /(2{\mu }_2)$ and ${\pi }_n={\pi }_0{\nu }^2(\nu /\mu {)}^{n-2}/(2{\mu }_1{\mu }_2)$. These weights normalize exactly when $\nu <\mu$. In equilibrium, reversed upward transitions have total rate $\nu$ in every state, so time reversal of a continuous-time Markov chain makes the departure stream a Poisson process of rate $\nu$. This output conclusion requires stationarity.

A queue consists of customers or jobs waiting for or receiving service. An arrival process adds customers and a service rule specifies which waiting customer is served next. A stochastic model of these events is studied in queueing theory; its queue length is the number of customers waiting or in service, according to the stated convention.

For positive $\lambda ,\mu ,\alpha ,\beta$, a continuous-time Markov chain may alternate between an arrival phase $C$, with upward transitions at rate $\lambda$, and a service phase $W$, with downward transitions at rate $\mu$ when the level is positive. Switching rates $C\to W$ and $W\to C$ are $\alpha ,\beta$. Its stationary balance equations give a geometric invariant tail with ratio $\theta =\lambda (\mu +\beta )/[\mu (\lambda +\alpha )]$. A stationary probability exists exactly when $\mu \alpha >\lambda \beta$, consistent with the stationary average service rate $\mu \alpha /(\alpha +\beta )$ exceeding the arrival rate $\lambda \beta /(\alpha +\beta )$.

Poisson arrivals of rate $\lambda$ receive independent service times distributed as $S$, with infinitely many servers so service starts on arrival. Starting empty, the occupancy at time $t$ is Poisson with mean $\lambda {\int }_0^t\mathbb{P}(S>u)du$, by independent thinning of the marked arrivals. If $\mathbb{E}S<\infty$, the equilibrium count is Poisson with mean $\lambda \mathbb{E}S$. Occupancy need not be a Markov process for a general service law; remaining service times carry additional state information.

Let the independent arrivals satisfy $\mathrm{Pr}(G_n=j)=q{p}^j$, with $0<p<1$ and $q=1-p$. During a busy period, each served customer produces $G_n$ further customers, so the total number served has the total-progeny law of a Galton-Watson process with mean $p/q$. Its extinction probability solves $s=q/(1-ps)$ and is the smaller of $1$ and $q/p$. The queue is positive recurrent for $p<1/2$, null recurrent for $p=1/2$, and transient for $p>1/2$. In the positive recurrent case its stationary distribution is ${\pi }_j=(1-p/q)(p/q{)}^j$.

An asymptotic description obtained by increasing queue resources or input size, or by approaching critical load while rescaling time and workload. Large-buffer queue scaling, many-source queue scaling, moderate-deviation queue scaling and the heavy-traffic limit of a queue preserve different relations between mean load, random fluctuations and storage.

For $\delta \ge 0$, a geodesic metric space is $\delta$-hyperbolic when each side of every metric geodesic triangle lies in the closed $\delta$-neighbourhood of the other two sides.

For a quasi-isometry $f:X\to Y$ with multiplicative constant $\lambda$ and additive/coarse-surjectivity constant $ϵ$, choose $f^{\prime }(y)$ with $d_Y(f(f^{\prime }(y)),y)\le ϵ$. The triangle inequality and the two distortion inequalities giveAlso $d_X(f^{\prime }(f(x)),x)\le 2\lambda ϵ$. Hence $f^{\prime }$ is a quasi-isometry and both composites stay uniformly close to their respective identities. No continuity is required.

The interquartile range measures the width between the lower and upper quartiles, in the same units as the observations. It describes central spread rather than the distance between the observed extremes. In our toy box plot, $\mathrm{IQR}=6.25-2.75=3.5$; moving the largest value from $20$ to $200$ would leave the quartiles and interquartile range unchanged under the chosen rule, because the order statistics used in their interpolation would not change.

A tunnel junction consists of two conducting electrodes separated by a thin region through which electrons can undergo quantum tunnelling. The separating region may be an insulating layer or a narrow vacuum gap. A difference in electrochemical potential supplies occupied incident states on one side and accessible states on the other.

For a simple barrier, the transmission probability decreases approximately exponentially with thickness, making the junction very sensitive to separation. This is the basis of scanning tunneling microscopy. The measured electric current also depends on the electronic density of states and the applied voltage. When a junction connects a sufficiently small conducting island, its capacitance can make individual charging energies large enough to produce Coulomb blockade.

Tunneling through a controllable vacuum gap by Gerd Binnig and others made barrier width experimentally adjustable. Their apparatus used a pointed electrical conductor opposite a conducting plate, with fine positioning by piezoelectricity. Rubber-tube support and magnetic levitation above a cooled bowl exhibiting superconductivity suppressed vibrations; the junction itself remained near room temperature.^[ref]

The readout was electrical resistance versus displacement at a fixed electric potential difference of 60 mV of either polarity. They observed an exponential function dependence spanning four orders of magnitude. Cleaning and better vacuum increased the inferred barrier from approximately 0.6-0.7 eV to 3.2 eV. Reproducible curves and the cleaning response argued against deforming contamination layers; an appropriate slope also helped exclude an ordinary conducting contact. The paper reports residual contamination, electronics-limited range and a jump attributed to release of thermal stress; its curve origins generally do not determine absolute gap widths.^[ref]

The ideal tunnel junction model explains the slope. At fixed small electric potential difference, fixed tip geometry and approximately unchanged density of states, take an effective barrier energy $\Phi$ above the relevant electron states and approximate the current magnitude byThe coefficient $C$ includes the applied electric potential difference, available occupied and empty quantum states, and tip geometry. A single transmission probability is not an electric current: a junction combines contributions from many electron states in both directions, with the electric potential difference producing a net imbalance. Its tilted, material-dependent barrier is only approximated by the equal-exterior rectangular potential barrier; the exact single-particle formula above is not a complete junction model.

For the chosen model value $\Phi =4$ electronvolts, $\kappa \simeq 10.25{\mathrm{nm}}^{-1}$. Increasing separation by $0.1\mathrm{nm}$ multiplies $I$ by about $0.129$, a reduction by about $7.8$. The following diagram chooses $I(0.4\mathrm{nm})=1\mathrm{nA}$ solely for illustration; it contains no measured data and is not a fit to the 1982 curves.

Distance calibration is essential. In Scanning Tunneling Microscopy: From Birth to Adolescence by Gerd Binnig and others, the authors describe an initial inferred barrier near 1 eV. Recalibration found only half the quoted piezo displacement sensitivity and raised the inferred barrier above 4 eV. Thus an exponential trace alone was insufficient evidence for tunnelling through vacuum. The model makes the sensitivity explicit: the inferred $\Phi ={\hslash }^2(d\ln I/ds{)}^2/(8m_e)$ depends quadratically on the distance scale.^[ref]

For scanning tunnelling microscopy, feedback control adjusts the vertical position to hold electric current fixed during a lateral scan; the recorded height follows separation only when the surface is electronically homogeneous. Changes in the local density of states can also change apparent height. This limitation is already explicit in the authors' account of constant-current imaging.^[ref]

A nonzero probability density inside a barrier does not by itself imply transport. In one dimension, the probability current of a wavefunction is $J=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$. A single evanescent wave $C{e}^{-\kappa x}$ has ${\psi }^{\ast }{\psi }^{\prime }=-\kappa |C{|}^2{e}^{-2\kappa x}$, a real quantity, hence $J=0$. This is what happens on the inaccessible side of a potential barrier extending to infinity with $E<V_0$.^[ref]

For the finite-barrier wavefunction $A{e}^{\kappa x}+B{e}^{-\kappa x}$, the cross terms instead giveThe relative quantum phase of the two probability amplitudes permits a nonzero, spatially constant probability current, even though neither exponential function separately carries it. The probability continuity equation gives ${\partial }_{x}J=0$ for this stationary state. Matching the wavefunction and its derivative also makes $J$ continuous at both faces, soHere $R=|r{|}^2$ is the quantum reflection probability, and $T=J_{\mathrm{tr}}/J_{\mathrm{inc}}$ is the transmission probability. The net exterior probability current equals the interior probability current; it is not lost as the wavefunction decreases. A physical incident wave packet supplies finite total probability, while the stationary state describes the energy component used to compute its transmitted flux.

Take a nonrelativistic particle of constant mass $m$ with energy $E>0$, and choose the same zero of potential energy on both sides of a rectangular potential barrier:The equality of the exterior potential energies matters: it gives the same exterior wavenumber $k=\sqrt{2mE}/\hslash$ and makes $T=|t{|}^2$. Here $\hslash =h/(2\pi )$ is the reduced Planck constant. This model has a finite jump in potential energy, no delta potential, and no absorbing term.

For $0<E<V_0$, the time-independent Schrödinger equation gives the spatial wavefunctionThe full stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. We set the incident probability amplitude to one and impose no incoming wave from the right; this scattering wavefunction is not a normalizable wavefunction. The tunnelling decay constant $\kappa$ describes exponential decay, rather than an oscillatory wavenumber in the barrier. Both exponential functions are needed on the finite interval: the growing term is not excluded by a condition at infinity.

The wavefunction and its derivative are continuous at each face, giving four linear matching conditions:Their solution supplies the transmission amplitudeA short derivation is to use the two right-face equations to express $A{e}^{\kappa a}$ and $B{e}^{-\kappa a}$ in terms of $t$, then substitute into the left-face equations. Taking the squared modulus gives the rectangular barrier transmission probability.

To identify the physical channel $\rho \mapsto V\rho V^{†}$ of a single-qubit unitary gate, prepare each of the three positive Pauli matrix eigenstates and estimate each output Pauli expectation. Their nine means areThey determine the Bloch vector rotation $\mathbf{r}\mapsto R\mathbf{r}$, hence the channel on all states. If two gates have the same channel, their relative unitary commutes with all three Pauli matrices: commuting with $Z$ forces it diagonal and commuting with $X$ forces equal diagonal entries. Thus they differ by a scalar global phase, proving uniqueness up to that phase. This procedure needs no controlled gate or eigenstate of the unknown gate. Estimating its bounded measurement means to error $ϵ$ with fixed confidence uses order ${ϵ}^{-2}$ repetitions. It determines the relative eigenphases but cannot determine an overall eigenvalue phase from uncontrolled gate access alone.

A polarizing beam splitter converts a photon polarization qubit to two paths. Rotating one arm's photon polarization to match the other and recombining at a balanced beam splitter gives $P_1(\theta )=1/2+\mathrm{Re}({\alpha }^{\ast }\beta e^{i\theta })$. Measurements at phases zero and $\pi /2$ recover the transverse Bloch vector components; direct counts in the two separated arms recover the population difference. All three components, rather than a fringe alone, determine an arbitrary photon polarization density matrix.

// Target: quantum-theory.bigb

A convex roof extension extends a function $\mu$ on pure states to arbitrary density operators by minimizing its ensemble average over all pure state decompositions. It agrees with $\mu$ on pure states: every positive-weight state in a decomposition of a rank-one density operator has the same one-dimensional support. Concatenating nearly minimizing ensembles proves that $F$ is a convex function. Moreover, every convex function that is no larger than $\mu$ on pure states is no larger than $F$, by applying convexity to each decomposition. Infima need not be attained; an arbitrarily small additive error suffices when $F$ is finite.