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For a deliberately simplified neutrino oscillation calculation, take two neutrino flavors related to two mass eigenstates by a real rotation matrix with angle $\theta$. A quantum superposition acquires different phases during propagation; projecting back onto the initial neutrino flavor gives the probabilityHere $L$ is path length and $E$ is neutrino energy; the numerical coefficient incorporates unit conversion.

The paper states the corresponding conversion formula.^[ref]

The path dependence has a geometrical origin. In an original spherical-Earth model, let $R$ be the radius and $h$ a fixed production height, neglecting detector depth. Let $\Theta$ be the zenith distance of the direction in the sky from which the neutrino arrives, so $c=\cos \Theta =+1$ means downward propagation. The backward ray from detector to production obeys $R^2+L^2+2RLc=(R+h{)}^2$. Its positive solution isThus $L(+1)=h$ and $L(-1)=2R+h$. The calculation explains why opposite directions test very different baselines, rather than attributing every observed direction to the same travel distance. Actual production heights and the angle between the outgoing charged lepton and incoming neutrino smear that relationship.

For an original chosen example, set $\theta =\pi /4$ and $\Delta m^2=2.5\times 10^{-3}{\mathrm{eV}}^2$. At $E=1\mathrm{GeV}$, $L=20\mathrm{km}$ gives $P_{\mu \mu }\simeq 0.9960$, while $L=500\mathrm{km}$ gives $P_{\mu \mu }\simeq 0.000279$. The first minimum is at $L/E\simeq 494.7km/GeV$. These are stipulated inputs, not a fit to the 1998 data. A broad distribution of true $L/E$, together with energy and directional uncertainty, smooths the sharp minima.

A direct amplitude check makes the mechanism explicit. DefineThen $S_{11}={\cos }^2\theta +{\sin }^2\theta e^{-2i\phi }$. Taking its squared absolute value yields the displayed survival formula because $1-\cos (2\phi )=2{\sin }^2\phi$. Unitarity gives $|S_{11}{|}^2+|S_{21}{|}^2=1$: disappearance from one channel becomes conversion to the other within this closed model.

This illustration does not describe all solar neutrino propagation. Coherent forward scattering in matter changes the effective propagation operator, and three active neutrino flavors require a larger mixing description. Even in the simplified model, neutrino oscillation measures a squared-mass difference; it does not determine the absolute neutrino mass.

For inviscid unforced two-layer QG dynamics, $H_1{F}_1=H_2{F}_2=f_0^2/g\prime$ makes the coupling symmetric in the layer-depth inner product. Multiplication by $H_i{\psi }_i$ gives the positive energy $E=\frac{1}{2}[{\sum }_i{H}_i|\nabla {\psi }_i{|}^2+(f_0^2/g\prime )({\psi }_1-{\psi }_2{)}^2]$. It is conserved after integration when the lateral energy flux vanishes. The interface term is available potential energy; the gradient terms are kinetic energy.

A nondegenerate donor-bound state of energy zero and an excited representative conduction level of energy $\Delta >0$ and degeneracy $g$ have the displayed bound probability by canonical state counting. The inferred degeneracy from one measurement is $g=(P_b^{-1}-1)e^{\Delta /(k_{B}T)}$. At finite fixed $g$, the probability decreases from1 toward $1/(1+g)$ as temperature grows. A temperature-independent effective degeneracy is an assumed two-level approximation; a full conduction-band density of states can behave differently.

Use the independent spin-one-half two-level system with fixed $N$ distinguishable sites and Hamiltonian levels $E_{\pm }=\pm \mu B$, where $\mu >0$ is the magnitude of the field-axis magnetic moment and $B>0$ is a uniform externally controlled magnetic field. Assume positive temperature, no interactions between sites, and no appreciable lattice or load entropy. The two levels differ by $2\mu B$. With inverse temperature $\beta$, the canonical ensemble gives the one-site and total partition functionsHere the minus subscript denotes the lower energy $-\mu B$. Factorization of the partition function requires additive energy and independent sites; there is no particle-exchange factor $N!$ for these fixed distinguishable sites.

Differentiate the partition function, holding $B$ fixed when differentiating with respect to $\beta$. The internal energy, Helmholtz free energy, total field-axis magnetic moment $m$, and magnetic entropy areThe extensive moment $m$ is distinct from magnetization, the moment per unit volume. Direct substitution of $p_{\pm }$ into the Gibbs entropy gives the same $s(x)$ independently of the free-energy derivative.

At fixed $B$, the magnetic heat capacity isStrict decrease makes $s$ invertible on $x>0$. Constant magnetic entropy therefore fixes $x$, not $B$ or $T$ separately:The partition function may be used to label these equilibrium states even though the isolated cooling leg has no external heat bath. Actual thermalization of the lattice and load additionally requires a sufficiently fast spin-lattice relaxation; perfectly uncoupled spins by themselves cannot refrigerate an attached object.

The limiting cases are instructive. For $x\ll 1$, $s(x)=\ln 2-x^2/2+O(x^4)$, $m\simeq N{\mu }^{2}B/(k_{B}T)$, and $C_{B,\mathrm{mag}}\simeq N{k}_B{x}^2$. The moment response reproduces Curie's law. For $x\gg 1$, $s(x)\simeq (1+2x)e^{-2x}$ and the upper-state occupation is exponentially small. Taking $B\to 0$ at fixed $T>0$ gives $s\to \ln 2$; taking $T\to 0$ at fixed $B>0$ gives $s\to 0$. The origin has a path-dependent limit. Extrapolating a line of constant $x$ to $B=T=0$ does not construct a usable zero-temperature state or prove that a real material retains independent degenerate spins there.

Take an isolated two-level system, or a spin one-half qubit, with Pauli matrices ${\sigma }_x,{\sigma }_z$. Let ${\sigma }_z|0⟩=|0⟩$ and ${\sigma }_z|1⟩=-|1⟩$. The initial and final Hamiltonian operators areThe energy gaps are ${\Delta }_i,{\Delta }_f$. Prepare the initial ground state $|0⟩$, with energy eigenvalue $-{\Delta }_i/2$. A final orthonormal eigenbasis isDirect matrix multiplication gives $H_f|g_f⟩=-{\Delta }_f|g_f⟩/2$ and $H_f|e_f⟩=+{\Delta }_f|e_f⟩/2$. The unchanged initial quantum state is $|0⟩=c|g_f⟩-s|e_f⟩$. Its new probability amplitudes are the inner products $c,-s$; the Born rule gives $p_g=c^2,p_e=s^2$. The angle between the real amplitude vectors is $\theta /2$, whereas the Hamiltonian axis on the Bloch sphere rotates through $\theta$.

At $\theta =\pi /3$, $c=\sqrt{3}/2$ and $s=1/2$, so the final energy measurement gives $-{\Delta }_f/2$ with probability $3/4$ and $+{\Delta }_f/2$ with probability $1/4$. At $\theta =0$ the initial ground state remains the final ground state; at $\theta =\pi$ it is the final excited energy eigenstate. These statements are about the new energy labels, not an instantaneous physical rotation of the quantum state.

A two-level system is a model with two accessible states or two groups of states separated by an energy gap. Choosing the lower energy as zero gives $E_0=0$ and $E_1=\Delta >0$. The approximation is useful when all other levels are inaccessible on the temperature and driving scales being studied. A spin in an applied magnetic field can provide a two-state example through Zeeman splitting.

For two nondegenerate levels in a canonical ensemble, $Z=1+e^{-\beta \Delta }$ and the upper-state probability is $p_1=1/(1+e^{\beta \Delta })$. At low positive temperature the upper level is rarely occupied; at high temperature the two probabilities approach one half. The mean energy is $\Delta p_1$, while the heat capacity is largest at an intermediate scale because occupation is then most sensitive to warming. This produces a Schottky anomaly. A coherent superposition is also possible in a quantum model, but the equilibrium occupation calculation specifies probabilities in the energy basis rather than a chosen coherent drive.

If a trial quantum state is dominated by two energy eigenstates of distinct energies $E_a<E_b$, then its Rayleigh quotient is approximately their probability-weighted mean. Rearranging that mean gives the displayed relative admixture. Neglected states must make small contributions to the energy numerator as well as to wavefunction normalization; small probabilities alone need not suffice if their energies are large.

Project a quantized field onto two low-energy orbitals. A sufficient regime has temperature, hopping and interaction shifts small compared with the energy gap to discarded orbitals. Small interwell contact interaction integrals can then be neglected while retaining a small quantum tunnelling matrix element. The projected field's operator commutator is the two-mode projection operator kernel, rather than a complete spatial Dirac delta function.

The normalized vacuum state annihilated by $c=ua-v{b}^{†}$ and $d=v{a}^{†}+ub$, for positive $u,v$ with $u^2+v^2=1$. Since $(a^{†}{b}^{†}{)}^2=0$, it can also be written $u\exp [(v/u)a^{†}{b}^{†}]|0⟩$. Acting with either annihilation operator makes the two terms cancel. Its original occupation numbers each have expected value $v^2$ and its pair amplitude is $⟨ab⟩=-uv$. It has even fermion parity although the original total number operator is not sharp. For the two-orbital fermionic pairing Hamiltonian, it is the unique ground state of energy $\epsilon -\sqrt{{\epsilon }^2+{\Delta }^2}$.

For the two-dimensional isotropic harmonic oscillator, $H=\hslash \omega (N+1)$ and the Schwinger boson representation gives $J^2={\hslash }^{2}N(N+2)/4$. Substitution yields the displayed relation, including its quantum constant term. The fixed-total-occupation sector is a spin-$j=N/2$ representation, so energy is $\hslash \omega (2j+1)$ with degeneracy $2j+1$.

absorption of two photons whose summed energy matches an atomic transition. In the weak-field regime its rate scales with the product of the two intensities. Two electric-dipole interactions preserve the initial parity.

Two-photon excitation microscopy excites fluorescence through the nearly simultaneous absorption of two photons. In the weak unsaturated regime, the excitation rate is proportional to the square of the local illumination intensity. Focusing a pulsed laser therefore strongly favors excitation near its focus rather than throughout the illuminated path.

Scanning that focus provides optical sectioning without relying on a detection pinhole to reject all out-of-focus emission. The photons can have lower individual energies than a single excitation photon, but their combined energy must match an allowed transition. High peak intensity, bleaching and phototoxicity remain relevant; longer excitation wavelength alone does not guarantee less damage.^[ref]

A coherent transition involving two atom-field interactions. two-photon absorption uses a sum of photon energies; stimulated absorption-emission uses their difference. A second-order intermediate contribution need not represent real population in a stationary atomic state.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.

Two localized orbitals coupled by quantum tunnelling and on-site interactions give the smallest Bose-Hubbard model. A symmetric double-well potential with lowest one-particle energies $E_0,E_1$ has $t=(E_1-E_0)/2$ and common diagonal energy $(E_0+E_1)/2$. At fixed total particle number, that common diagonal term is constant. A fixed-number two-mode condensate gives a simple variational method for this model, while stronger interactions can require occupation-correlated trial quantum states.

For free fermions on a ring, each one-particle orbital has integer angular momentum label $l$ and two spin states. Two particles occupy the $l=0$ orbital in a unique singlet state at zero kinetic energy. At the next distinct energy ${\hslash }^2/(2m{R}^2)$, orbitals $0,+1$ or $0,-1$ can be combined with one singlet state or three triplet state spin states, giving total quantum degeneracy eight. The lowest triplet state has six states and its antisymmetric spatial factor vanishes when the particles coincide.

The two-site quantum version of an XY model couples the transverse components of two spin one-half degrees of freedom. In the ordered basis $|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩$, its only off-diagonal entries are $H_{23}=H_{32}=-2J$. The parallel states have energies $-2h$ and $2h$; the symmetric and antisymmetric opposite-spin combinations have energies $-2J$ and $2J$ respectively. The latter is the singlet state. The formulas remain valid at accidental degeneracies.

Here $s(m)=\frac{1+m}{2}\ln \frac{1+m}{2}+\frac{1-m}{2}\ln \frac{1-m}{2}$. This free-energy density is the energy minus temperature times entropy of independent sublattice Ising spins. Since $s^{\prime }(m)=\mathrm{atanh}m$, stationarity yields the two-sublattice Ising mean-field approximation. Writing $M_{\pm }=(M_B\pm M_W)/2$, the quadratic coefficients are $(k_{B}T+qJ)/2$ for $M_{+}^2$ and $(k_{B}T-qJ)/2$ for $M_{-}^2$. The positive quartic coefficient $k_{B}T/12$ in the staggered direction stabilizes the continuous transition. Unlike a log-cosh auxiliary potential, this variational expression can be minimized away from self-consistency.

For two same-height throats carrying the same light-layer volume flux and reduced gravity, the narrower throat has the larger critical minimum energy. A narrower downstream throat can back up and submerge an upstream throat. If the first throat is narrower, it can control discharge and feed a supercritical current through the second; a hydraulic jump followed by a second control is also possible when the available energy drop and intermediate geometry permit it. Equal throat widths are a marginal lossless case. Relative width alone does not fix jump position or mixing in the intervening reservoir.

A single recoil endpoint leaves a tradeoff between incident speed and projectile mass. Two targets can remove the unknown incident speed. Assume both runs sample the same maximum projectile speed $v$ and that single elastic collisions reach the endpoint. For initially stationary target masses $M_1<M_2$, let their maximum recoil speeds be $u_1,u_2$. The maximum elastic recoil energy fraction calculation givesIf energies rather than speeds are inferred, use $R=\sqrt{T_{1,\max }{M}_2/(T_{2,\max }{M}_1)}$. The energy ratio itself is not the speed ratio.

For the stipulated $4\mathrm{MeV}$ example, hydrogen and nitrogen giveThe inferred mass then determines $v=u_1(m+M_1)/(2m)$. Changing the assumed mass to improve one recoil endpoint changes the prediction for the other: this is the constraining power of the comparison.

James Chadwick's June paper instead used approximate endpoints $u_p=3.3\times 10^{7}m/s$ and $u_N=4.7\times 10^{6}m/s$, obtaining $m/m_p\simeq 1.15$; it allowed roughly ten percent uncertainty in $u_N$. This was a rough mass comparison.^[ref]

The inverse calculation has useful built-in checks. A positive finite mass requires $1<R<M_2/M_1$. As $R$ approaches $1$, distinguishing large projectile masses becomes difficult. Differentiation givesUncertain recoil endpoints therefore produce an uncertain inferred mass; the collision model cannot turn imprecise range measurements into exact spectroscopy. If different target runs select different incident energy groups, the common-$v$ cancellation also fails.

For comparison, Compton scattering of a photon of energy $E_{\gamma }$ from a stationary free target of mass $M$ has the exact relativistic endpointThis follows by combining momentum conservation with the target's relativistic energy, maximizing transfer when the photon reverses direction. Unlike the massive-projectile calculation, this equation does not approximate the incident photon as a nonrelativistic particle.

As a separate, explicitly hypothetical test, take $m_p{c}^2=938\mathrm{MeV}$ and target rest energies $938\mathrm{MeV}$ and $14\times 938\mathrm{MeV}$. To reproduce the stipulated hydrogen recoil endpoint $T=4\mathrm{MeV}$, the photon would need $E_{\gamma }\simeq 45.36\mathrm{MeV}$. The same photon would give the heavier target at most $0.311\mathrm{MeV}$, whereas the stipulated massive projectile gave $0.996\mathrm{MeV}$. This is the structure of a two-target discrimination, not a transcription of a 1932 measurement. Actual radiation can contain more than one energy group or both photons and neutrons; a useful model must account for the observed endpoints and their origin together.