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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.

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 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.

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.

For $V=m^2{\varphi }^2/2+\lambda ({\varphi }^2{)}^2/4$, $m^2<0$, the vacuum manifold is a sphere of radius $v=\sqrt{-m^2/\lambda }$. At a chosen vacuum the Hessian matrix is $2\lambda v^{2}n{n}^T$, where $n$ is its unit direction. The radial scalar has mass squared $2\lambda v^2$; the $n-1$ angular modes are physical Goldstone bosons when the symmetry is global.

For the two-component case, write $\mathbf{\varphi }=(v+\rho )(\cos \theta ,\sin \theta )$. The kinetic term becomes $\frac{1}{2}({\partial }_{\mu }\rho {)}^2+\frac{1}{2}(v+\rho {)}^2({\partial }_{\mu }\theta {)}^2$. At quadratic order the canonically normalized angular field is $\pi =v\theta$, with no mass term. Its Euler-Lagrange equation is the wave equation, so ${\omega }^2=|\mathbf{k}{|}^2$ in units where the propagation speed is one. The radial field instead obeys ${\omega }^2=|\mathbf{k}{|}^2+2\lambda v^2$. Thus a common rotation costs no energy, whereas a slowly varying rotation propagates as a collective wave. With the selecting source of continuous vacuum selection by an external source, use $\pi =w\theta$ and obtain ${\omega }^2=|\mathbf{k}{|}^2+h/w$. This elementary dispersion relation separates a Goldstone boson, a pseudo-Goldstone boson and a massive radial excitation within one example.

A particle constrained to a stationary smooth surface of revolution $z=f(r)$ under uniform gravity conserves $h=r^2\dot{\theta }$ and mechanical energy per unit mass. Substitute $\dot{z}=f^{\prime }(r)\dot{r}$ and $\dot{\theta }=h/r^2$ into that energy to obtainThis proves the radial reduction. Simple roots of $E-U_h$ are turning points; a stationary point $U_h^{\prime }(r_0)=0$ supports an exact circular orbit. At a local maximum of $U_h$, another trajectory at the same energy can approach the orbit asymptotically as a separatrix. Existence of roots elsewhere is insufficient for confinement: their dynamically accessible ordering relative to the launch radius matters.

For a central-force radial energy equation, choose a monotone radius $q=q(r)$ and clock $\tau$ such that $d\tau /dt=(q/r)dq/dr$. Then $q^{\prime }=(r/q)\dot{r}$, and multiplying the old energy equation by $r^2/q^2$ preserves the centrifugal form $L^2/(2q^2)$. A split of the remaining terms into constant energy and a new Newtonian gravitational potential produces a transformed radial orbit.

In static coordinates on anti-de Sitter spacetime with curvature radius $L$ and $c=1$, put $f=1+r^2/L^2$. The conserved energy per unit rest mass is $E=fdt/d\tau$. Timelike normalization for a radial geodesic gives $(dr/d\tau {)}^2=E^2-1-r^2/L^2$, hence $d^{2}r/d{\tau }^2=-r/L^2$. An outward launch from the origin at proper time ${\tau }_0$ gives the displayed harmonic solution until its first return, after proper time $\pi L$, independently of $E>1$. Beyond that return, use a signed diameter coordinate or change the radial direction to keep the areal radius nonnegative. This is a coordinate realization of Anti-de Sitter timelike geodesic refocusing.

Radiant flux is the power carried by electromagnetic radiation: radiation energy transferred per unit time, measured in watts, $\mathrm{W}=\mathrm{J}{\mathrm{s}}^{-1}$. Specify what is counted: all radiation emitted by a source, radiation within one beam, or radiation intercepted by a photodetector. Writing $Q(t)$ for the cumulative energy crossing the specified boundary givesThe first expression is an instantaneous derivative; the second is an average over the stated time interval. If $0.020\mathrm{J}$ arrives during $0.010\mathrm{s}$, the mean radiant flux is $2\mathrm{W}$. The photodetector's absorbed power equals this incident radiant flux only if it absorbs all of the incident radiation.

Keeping radiant flux fixed while widening a beam reduces its irradiance, the incident power per receiving area. Radiance additionally resolves the propagation direction and has units $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$; an optical camera can collect different radiant fluxes from sources with the same radiance but different apparent solid angles. Luminous flux, measured in lumens, weights the optical spectrum by the human eye's standardized response instead of counting energy equally at every wavelength. Equal radiant fluxes need not give equal luminous fluxes.^[ref]

Radiant power also differs from the optical power of an optical lens, which is reciprocal focal length in dioptres. Likewise, astronomical radiative flux is power per area. Stating the units prevents these shared words from hiding different physical quantities.

The energy entering and leaving a system through electromagnetic radiation. A top-of-atmosphere radiation budget compares absorbed sunlight with outgoing thermal radiation. A latitude-dependent imbalance can be balanced by atmospheric and ocean heat fluxes even when the global mean is zero.

For a hydrostatic monatomic ideal gas with negligible external pressure, $2U+{\Omega }_f=0$, where gravitational potential energy is negative. The final total energy is therefore ${\Omega }_f/2$, giving the displayed radiated energy when no mass or other energy carrier escapes. The common $-{\Omega }_f/2$ result additionally takes the initial total energy to be negligible. Initial rest does not remove initial self-gravitational binding energy.

The characteristic thickness over which a high-energy electron loses all but a fraction $1/e$ of its energy through bremsstrahlung. At fixed thickness $x$, smaller $X_0$ increases the radiative loss; the relevant dimensionless measure is $x/X_0$. Low-energy charged particles can instead be dominated by collisional stopping power.^[ref]

Radiation pressure is the normal momentum flux transferred by electromagnetic radiation to a surface. A photon travelling in vacuum carries momentum $p_{\gamma }=E_{\gamma }/c=h\nu /c$, where $E_{\gamma }$ is its energy, $\nu$ its frequency, $h$ the Planck constant and $c$ the speed of light. Photon absorption transfers that momentum; specular reflection reverses its normal component. This lets light exert a force even when almost no energy becomes heat in the receiver.

Two useful formulas describe different preparations. A directed beam normally incident on a stationary perfect absorber gives pressure $I/c$, and on a stationary perfect mirror gives $2I/c$. Here $I$ is light intensity, the incident power per beam cross-sectional area perpendicular to propagation, in $\mathrm{W}{\mathrm{m}}^{-2}$. An isotropic photon gas has pressure $u/3$, where $u=U/V$ is its radiation internal energy per volume, in $\mathrm{J}{\mathrm{m}}^{-3}$. Stored energy density and travelling energy flux must be distinguished: an isotropic field can have zero net energy flux and positive radiation pressure.

The physical appeal is that the same momentum conservation connects a single photon collision, a mirror's measurable force, and a radiation-filled cavity's equation of state. The directional calculation below complements radiation pressure on a perfectly absorbing surface; the equilibrium spectrum and its temperature dependence belong to blackbody radiation and the photon gas.