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For a dilute classical trapped gas, $\mu \simeq k_{B}T\ln (N/Z_1)$ with energy zero at the potential minimum. Thus $z=e^{\mu /(k_{B}T)}=N/Z_1$. Quantum occupation effects begin when $z$ is no longer small, giving the displayed order-of-magnitude particle threshold. Equivalently, the central density $n(0)=N/[8\pi (k_{B}T/a{)}^3]$ satisfies $n(0){\lambda }_T^3\sim 1$. This is a degeneracy crossover criterion rather than an exact Bose condensation threshold. In the continuum approximation, spinless bosons have excited-state saturation $N_c=\zeta (9/2)Z_1$, from the trap density of states proportional to $E^{7/2}$. A quantum ground energy should be subtracted consistently when discreteness matters.

Quantum escape of a current-biased Josephson junction is decay of a trapped collective quantum phase state through a barrier in the tilted Josephson washboard potential. This barrier lies in the potential energy of an electrical circuit coordinate. Transfer of Cooper pairs across the physical insulating layer already supplies the Josephson coupling energy; it does not by itself demonstrate this further collective escape. Single-electron superconducting tunneling is a third, distinct process.

At higher temperature $T$, thermal activation instead lifts the coordinate over the barrier. In the rare-escape regime its characteristic form is the Kramers escape ratewhere $k_B$ is the Boltzmann constant and the prefactor depends on damping and thermal equilibration. Cooling suppresses this route. For a quasistationary energy level at energy $E$, the leading conservative WKB approximation instead haswith the two turning points bracketing the forbidden barrier. This is an exponential estimate requiring a large action compared with $\hslash$, slow escape compared with intrawell motion, and weak environmental corrections. It is not a complete rate prediction near the barrier top or with substantial damping.

The familiar cubic-barrier exponent can be checked independently. Choose $V(x)=M_{\varphi }{\omega }_p^2{x}^2(1-x/a)/2$, with $a>0$. Its top is at $2a/3$ and $\Delta U=2M_{\varphi }{\omega }_p^2{a}^2/27$. At the well-bottom reference energy $E=0$,Substitution $u=x/a$ reduces the dimensionless integral to ${\int }_0^{1}u\sqrt{1-u}du=4/15$. This leading semiclassical action uses a cubic potential and a well-bottom reference; actual ground-state escape also needs the zero-point correction, prefactor and environment. In the chosen washboard example it gives $B\simeq 74.0$ after matching the exact barrier and curvature. This matching is an illustrative approximation, not an exact action integral for the cosine potential.

Escape need not yield a sustained voltage: the coordinate can become trapped again. The running-state readout assumes sufficient underdamping and bias to avoid such retrapping. The parallel model has quality factor $Q={\omega }_{p}RC$; large $Q$ makes energy loss per oscillation small. Low-temperature rate saturation alone cannot distinguish quantum tunnelling from unwanted heating or external noise.

A quantum fluid is a fluid whose collective behavior requires quantum mechanics, particularly when the thermal de Broglie wavelength becomes comparable with particle spacing. Exchange statistics and interactions then affect the state in ways that a dilute classical particle model cannot capture. Low temperature favors this regime, but temperature alone is insufficient without a density and particle-mass scale.

Superfluid helium-4, degenerate fermionic matter described by a Fermi liquid, and dilute ultracold atomic gases provide different realizations. A quantum fluid need not be a superfluid, and quantum degeneracy does not by itself remove dissipation. Measurements of heat capacity, collective waves and response to flow distinguish the relevant states. The useful organizing question is which low-energy excitations carry energy, momentum and conserved particle number, rather than whether individual particles merely have small kinetic energies.

Quantum ideal-gas statistics describes noninteracting identical particles while retaining the exchange rules of bosons or fermions. In thermal equilibrium, a one-particle state of energy $E$ has mean occupationThe minus sign is the Bose-Einstein distribution; the plus sign is the Fermi-Dirac distribution. A bosonic state permits arbitrary occupation, whereas Pauli exclusion principle restricts each fermionic state to zero or one particle. ^[ref]

The mean particle number and internal energy follow by summing $\overline{N}={\sum }_r{\overline{n}}_r$ and $U={\sum }_r{E}_r{\overline{n}}_r$. When every occupation is small, both denominators reduce to the same exponential and reproduce Maxwell–Boltzmann statistics. Thus absence of interactions does not make a gas classical.

A classical oscillator of amplitude $A$ has energy $m{\omega }^2{A}^2/2$. Comparing with the oscillator spacing $\hslash \omega$ gives the large-occupation scale displayed. A coherent state with this classical amplitude has exactly this mean excitation occupation and a Poisson number distribution; its total mean energy also includes the zero-point term. An observed macroscopic amplitude does not, by itself, establish a pure coherent state.

A quantum quench changes a quantum system's control parameters rapidly and studies the subsequent unitary time evolution. In the ideal sudden limit, write an initial energy eigenstate $|m_i⟩$ using a complete final orthonormal eigenbasis:The Born rule probabilities for a nondegenerate final energy measurement are $|c_n{|}^2$, with ${\sum }_n|c_n{|}^2=1$. For a degenerate energy eigenvalue, sum the probabilities over its orthonormal energy eigenstates. Subsequent relative quantum phases can alter other measurement probabilities even though these final energy populations stay fixed. A quench need not begin in an energy eigenstate; the same expansion applies to any initial pure state.

A quantum coordinate on a circle with kinetic Hamiltonian ${\ell }^2/(2I)$. Periodic wavefunctions have angular momentum $\hslash n$ and energy ${\hslash }^2{n}^2/(2I)$ for integer $n$. Its thermal trace has both a momentum-eigenstate sum and a topological winding-path representation. Twisted boundary conditions or flux would modify this spectrum and are separate models.

For a fixed Hilbert space, a bounded self-adjoint Hamiltonian operator with a finite step $H_i\to H_f$ at $t=0$ does not deliver an instantaneous rotation of the quantum state. Integrating the Schrödinger equation across a short interval givesFor a normalized quantum state and $\Vert H(t)\Vert \le M$ in operator norm, the difference has norm at most $2ϵM/\hslash$, which tends to zero. The time derivative can jump even though the quantum state is continuous. Here $\hslash$ is the reduced Planck constant. An instantaneous replacement of the eigenbasis changes coordinates, not the vector being represented.

For a real switch occupying $0\le t\le \tau$, remove any real scalar contribution $a(t)I$ that produces only a global phase, and write $|\tilde{\psi }(t)⟩=e^{i{\int }_0^{t}a(s)ds/\hslash }|\psi (t)⟩$. The same integral argument gives the sufficient bounded-model estimateThus $\epsilon \ll 1$ controls the sudden approximation up to a global phase. For any normalized final energy eigenstate, its probability amplitude differs from the instantaneous-switch value by at most $\epsilon$, and its Born rule probability differs by at most $2\epsilon$. These are sufficient error bounds, not necessary criteria. For the fixed-gap two-level sudden quench, any switching path $H(t)=-\Delta \mathbf{n}(t)\cdot \mathbf{\sigma }/2$ with a unit vector $\mathbf{n}(t)$ has operator norm $\Delta /2$, giving $\epsilon \le \Delta \tau /(2\hslash )$.

This estimate is deliberately for bounded Hamiltonian operators. An unbounded energy spectrum does not supply a finite global $M$; relevant quantum states, operator domains and excited energy eigenvalues require separate control. The adiabatic theorem also requires hypotheses about the path and its spectral gaps; there is no universal switch-duration criterion covering all spectra. The cold-atom collapse and revival after a lattice quench illustrates how a single switch can be sudden for one degree of freedom and slow for another.

A quantum time-reversal operator is an antiunitary operator reversing spatial momentum and angular momentum. For a time-reversal-invariant Hamiltonian operator, its conjugation of $i$ makes it intertwine forward and backward unitary time evolution without reversing the energy spectrum. A scalar momentum eigenstate may transform as $\hat{T}|\mathbf{p}⟩=e^{i\chi (\mathbf{p})}|-\mathbf{p}⟩$. With unit-modulus phases and a complete normalized basis, expansion of arbitrary states proves antiunitarity.

Quantum transmission is the passage of an incident quantum particle into an outgoing region through a specified interaction or potential. A wave function can have both reflected and transmitted components; the measurable transmission probability compares outgoing and incoming probability currents, rather than comparing wave amplitudes alone.

For a one-dimensional, constant-mass Schrödinger equation, write an incident amplitude of one and transmitted amplitude $t$, with positive asymptotic wavenumbers $k_L$ and $k_R$. The probability current givesOnly when the asymptotic wavenumbers agree does this become $|t{|}^2$. A real, static potential with no additional loss channels has reflection plus transmission probability equal to one. Quantum tunnelling is an important case in which the transmitted component crosses a region forbidden by the corresponding classical energy condition.

Quantum transport studies the flow of particles, electric charge or energy when quantum states, their occupation and their transmission probabilities determine the flow. In a short quantum wire, transverse quantum confinement permits only discrete transverse electronic modes, even though motion along the wire remains possible.

The Landauer formula treats two contacts as reservoirs supplying incident electrons. Their different occupations drive an electric current; a potential barrier changes the transmission probability of each incident mode. At low temperature, opening one perfectly transmitted spin-resolved mode increases the electrical conductance by the spin-resolved conductance quantum $e^2/h$. Quantum tunnelling, interference and Coulomb blockade explain departures from a simple classical resistor picture.

Quantum tunnelling is transmission through a potential barrier even when the particle energy is below the barrier height.

The wavefunction penetrates the classically inaccessible region; matching it to an outgoing scattering state on the far side gives a nonzero transmission probability. For a real, time-independent potential barrier, the transmitted energy equals the incident energy: quantum tunnelling does not require a temporary violation of conservation of energy. A tunnel junction turns this small transmission probability into a measurable electric current, whose sensitivity to separation makes the scanning tunnelling microscope possible.

For a normalized dilation ${\psi }_{\lambda }(x)={\lambda }^{d/2}\psi (\lambda x)$ and a homogeneous potential of degree $-n$, the energy is ${\lambda }^{2}T+{\lambda }^{n}V$. Stationarity at an eigenstate gives $2T+nV=0$.

A quantum vortex has integer winding number of a condensate's complex argument. For atomic mass $m$, its superfluid velocity gives circulation $2\pi \hslash q/m$ for integer $q$. A straight vortex has a density-depleted core and, outside it, tangential velocity $q\hslash /(mr)$. Its long-range flow energy is proportional to $q^2\log (R/a)$, while its angular momentum has the sign of $q$.

For parabolic bands confined by an infinite well of width $L$, a same-index transition has edge $E_{g,j}+{\pi }^2{\hslash }^2{n}^2(1/m_e^{\ast }+1/m_{h,j}^{\ast })/(2L^2)$. Above it the in-plane pair kinetic energy is ${\hslash }^2{k}_{\parallel }^2/(2{\mu }_j)$ with reduced mass ${\mu }_j^{-1}=(m_e^{\ast }{)}^{-1}+(m_{h,j}^{\ast }{)}^{-1}$.

For $\dot{x}=\alpha x-y+y^3$, $\dot{y}=-x$, the energy $H=x^2-y^2+y^4/2$ satisfies $\dot{H}=2\alpha x^2$. At $\alpha =0$, its compact level sets give two families of closed periodic orbits around $(0,\pm 1)$ and an outer family around both; $H=0$ is a figure-eight homoclinic orbit pair through the saddle point at the origin. The two minima of $H$ are center equilibria for zero damping, stable foci for small negative $\alpha$, and unstable foci for small positive $\alpha$. For $\alpha \ne 0$, strict monotonicity along nonconstant solutions excludes periodic orbits and homoclinic orbits.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

A reduction in fluorescence caused by a process competing with photon emission, for example a collision that removes excitation energy without emitting a photon. Measuring less light because fewer excitation photons reach the sample does not by itself establish fluorescence quenching.

The Q value of a reaction or particle decay is the initial total rest energy minus the final total rest energy. Positive $Q$ releases kinetic energy; a negative $Q$ requires energy supplied by incident motion or another source. Atomic and bare nuclear mass conventions must be distinguished when counting electrons in beta decay.

A Rabi oscillation is a coherent oscillation of populations in a driven two-level quantum system.

A resonant drive can transfer a quantum state from one energy eigenstate to another and then bring its population back. This is unitary time evolution: the drive exchanges energy with the selected two-level system, while total probability stays one. One complete Rabi cycle has population period $2\pi /\Omega$; halfway through, a Rabi pi pulse transfers the whole population in the ideal model. This makes Rabi oscillations useful for calibrating resonant quantum control, preparing quantum superpositions and supplying the pulses of Ramsey interferometry.