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JEOL's minimum-dose method focuses and corrects astigmatism away from the recording position before exposing it, reducing irradiation of the chosen field.^[ref] A downstream objective aperture cannot undo irradiation that has already reached the specimen: rejecting more outgoing electrons does not itself reduce specimen exposure.

For an original dose budget, define $D$ as incident electron count per specimen area, rather than absorbed energy per mass. A uniform beam of electric current magnitude $I$, illuminating area $A$ for time $T$, giveswhere $e$ is the elementary charge. Choosing $I=1\mathrm{pA}$, $T=1\mathrm{s}$ and $A=1\mu {\mathrm{m}}^2$ gives approximately $6.24\times 10^6$ incident electrons, or $6.24\mathrm{electrons}\mathrm{n}{\mathrm{m}}^{-2}$. This is an invented exposure calculation, not a safe-dose recommendation. Focusing and searching exposures must also enter the accumulated budget.

If detected counts follow a Poisson distribution, a mean count $\mu$ has standard deviation $\sqrt{\mu }$. Ignoring background, $100$ accepted counts therefore have $10\%$ relative counting uncertainty, while $25$ have $20\%$. Reducing accepted counts can reveal useful diffraction contrast while increasing counting uncertainty. It is useful to specify both contrast and count budget when discussing electron microscope resolution.

A proposed measurement calibration sequence records a beam-blocked detector background, an illuminated empty support region, and a specimen image with fixed voltage, exposure and detector gain. Test detector linearity with exposure time before comparing corrected intensities. Calibrate the image scale against a known spacing, as in microscope scale calibration, and independently calibrate electron diffraction spacing at its own lens settings; one calibration does not supply the other. Register compared regions, record objective aperture selection, specimen tilt and accumulated exposure, then repeat the first setting to test drift and irradiation changes. These are proposed controls, not undocumented steps attributed to JEOL's silicon example.

The point-dipole energy is $[{\mathbf{p}}_1\cdot {\mathbf{p}}_2-3({\mathbf{p}}_1\cdot \hat{r})({\mathbf{p}}_2\cdot \hat{r})]/(4\pi {ϵ}_0{r}^3)$. If the separation is perpendicular to the first dipole, the second term vanishes. Parallel dipoles then have positive energy $p_1{p}_2/(4\pi {ϵ}_0{r}^3)$, while mutually perpendicular dipoles give zero in this approximation. Real molecular interactions can include multipoles, induction and short-range contributions beyond this point-dipole term.

Choose the beam as the $z$ direction. A particle's transverse momentum is its momentum projected onto the perpendicular plane; $p_T=\sqrt{p_x^2+p_y^2}$. A Lorentz boost along the beam changes energy and longitudinal momentum, while leaving transverse momentum unchanged. Large transverse momentum identifies motion substantially across the beam rather than merely along it.

For an axially split $p$ multiplet with $p_z$ energy $E_0$ and equal $p_x,p_y$ energies $E_t$, a transverse $x$-directed magnetic field leaves $p_x$ unchanged and couples $p_y,p_z$ through the orbital angular momentum operator $L_x$. The matrix element has modulus ${\mu }_{B}B$, giving the two energy eigenvalues in the title. For ${\mu }_{B}B\ll |E_t-E_0|$, the orbital shifts are quadratic despite a linear Zeeman energy operator. A full quadratic-field calculation also includes the diamagnetic term; electron spin gives its own linear contribution.

Two particles with unequal rest energies $r,R$ and equal opposite velocities have total energy $\gamma (r+R)$ but nonzero total momentum. If they form a composite while emitting a photon transverse to their initial axis, the final composite rest energy $Q$ satisfies the displayed identity. It follows by squaring the residual four-momentum. Neglecting binding, put $Q=r+R$; the minimum incoming energy is $(r+R)[\chi +\sqrt{1+{\chi }^2}]$, where $\chi =\epsilon (r+R)/(4rR)$. An exact threshold requires the composite rest mass, not just the incoming constituent masses.

An internal qubit uses two distinguishable electronic or hyperfine structure quantum states, denoted $|g⟩,|e⟩$. The ion's centre-of-mass motion has a different Hilbert space. In one approximately harmonic direction, a number state $|n⟩$ has energy $E_n=\hslash \nu (n+1/2)$, where $\nu$ denotes a motional angular frequency, not ordinary frequency. The tensor product of quantum systems gives $|g,n⟩$. Cooling towards $n=0$ prepares motion; it does not automatically prepare the internal qubit. Nor does the ground state mean a stationary point particle: zero-point energy and position spread remain.

For several ions, Coulomb forces couple their displacements into collective normal modes. Each selected normal mode has its own quantum harmonic oscillator number states, while each ion retains internal qubit states. A pulse addressing one ion can change a shared mode. This supplies a route to correlations between distant internal quantum states; it also makes residual motional excitation and unwanted excitation of other normal modes relevant. A single-ion pulse calculation alone establishes no experimental multi-ion quantum gate performance.

A triple-axis spectrometer uses Bragg's law at an initial crystal monochromator to select ${\lambda }_i$ and at a second analyzing crystal to select ${\lambda }_f$. The sample sits between these selectors. The three axes refer to the monochromator, sample and analyzer rotations. The final counter measures accepted counts; the analyzer's wavelength selection makes those counts sensitive to outgoing energy even when the counter does not measure each neutron's energy directly.

For a constant-$\mathbf{Q}$ scan, change incident or outgoing energy while coordinating sample and analyzer settings so that ${\mathbf{k}}_i-{\mathbf{k}}_f$ stays fixed. Counts versus $E_i-E_f$ then probe excitations at one momentum transfer. Changing only the analyzer angle generally changes both transferred quantities. Incident-beam monitoring, analyzer response and the instrumental resolution must enter quantitative interpretation; the diagram's arrows specify information flow, not an absolute intensity calibration.

A trophic level describes feeding position relative to primary producers. Assign primary producers level 1; in the illustrated food chain, zooplankton occupy level 2, small fish level 3, and large fish level 4. Following a single feeding path adds one level at each step. The labels make dependence on the initial resource visible, but a branching food web can give an organism several feeding paths. See Clark, Douglas and Choi, Biology 2e, 46.1: feeding positions and branching food webs.

An omnivore can obtain food from more than one trophic level. As an explicitly assumed diet-weighted example, suppose an omnivore gets half its assimilated food energy from level-1 primary producers and half from level-2 consumers. Define its trophic level as one plus the weighted mean level of its food: $1+(1/2)1+(1/2)2=2.5$. This computed position depends on the specified diet and weighting convention; it is not an immutable property of the species. A different life stage or diet can change it.

A tunnel diode is a p-n junction with sufficiently heavy semiconductor doping on both sides to make a thin depletion region and permit substantial interband tunnelling. The neutral regions are degenerate semiconductors. Its distinctive forward current-voltage characteristic contains a peak followed by a valley: over part of the curve, increasing the applied electric potential difference decreases the electric current. The Esaki tunnel-diode experiment made this behavior experimentally concrete.

The useful quantity is the local slope, not the ratio of total voltage to total electric current. With conventional electric current $I$ entering the positive-voltage terminal, defineNegative differential resistance means $g_d<0$, hence $r_d<0$ where defined. A forward operating point can still have $V>0$, $I>0$ and absorbed power $VI>0$. A small voltage increase then gives a negative current increment; this does not make the total electric current negative or turn the device into an unlimited source of energy. The external supply sets the operating point and supplies the energy available for an electronic amplifier.

An original idealized construction explains the shape. Put the n-type semiconductor on the left and the p-type semiconductor on the right. Take $q>0$ as the elementary charge and $V=V_p-V_n$. Increasing forward bias lowers the right-hand electron energy levels relative to the left. Denote the two reservoir electrochemical potentials by ${\mu }_n,{\mu }_p$, so that ${\mu }_n-{\mu }_p=qV$. At zero temperature, occupied left-hand conduction band states extend up to ${\mu }_n$; empty right-hand valence band states start at ${\mu }_p$. For transfer conserving electron energy, the available interval isIt is nonempty only when its upper endpoint exceeds its lower endpoint. Both an occupied initial quantum state and an empty final quantum state must exist at the same energy. The Pauli exclusion principle prevents transfer into an already occupied final quantum state.

For the original diagram, choose arbitrary energy units with $E_{c,n}=-0.5$, ${\mu }_n=0$, ${\mu }_p=-u$ and $E_{v,p}=0.5-u$, where $u$ is $qV$ in those units. The available interval has widthThus $L(0.25)=0.25$, $L(0.50)=0.50$, $L(0.70)=0.30$, and $L(1.15)=0$. Initially forward bias opens a transfer window; further forward bias closes it by moving the right-hand valence band edge below the left-hand conduction band edge. This is a state-counting illustration, not a formula for a measured electric current.

For reverse bias, the transfer direction changes: occupied right-hand valence band states can feed empty left-hand conduction band states. In the same construction, $u=-0.70$ opens the interval $0<E<0.70$ for this reverse transfer. A narrow depletion region allows a large reverse quantum tunnelling electric current at small electric potential difference, whereas the ordinary Shockley diode equation predicts only a small reverse saturation electric current before diode reverse breakdown. At zero biasing, equal reservoir occupations give zero net electric current; quantum tunnelling does not require a net equilibrium flow.

A finite-temperature refinement replaces sharp occupation cutoffs with the Fermi-Dirac distribution. An illustrative weak-transfer expression isHere $\mathcal{W}\ge 0$ describes allowed initial/final quantum states and their coupling, and $f_n,f_p$ are Fermi-Dirac distributions with the respective electrochemical potentials. The occupation difference follows by subtracting the reverse transfer factor from the forward one: $f_n(1-f_p)-f_p(1-f_n)=f_n-f_p$. An overlap interval alone does not determine $\mathcal{W}$: transmission probability, density of states, crystal momentum selection and phonons can matter. A complete germanium calculation is more demanding than multiplying two constant density of states factors. At larger forward bias, ordinary carrier injection and diffusion supply a rising current branch even after direct band overlap is lost.

For a nonrotating uniform-stratification fluid under the Boussinesq approximation driven at $0<\omega <N$ by a localized vertical-wall velocity, positive vertical wavenumbers radiate down and right and negative ones up and right. The real velocity separates into functions of $Z_{+}$ and $Z_{-}$. Its in-phase component is half the imposed boundary profile on each beam; the quadrature component is its Hilbert transform partner. Thus compact localization at maximum boundary speed does not imply compact localization of all components at every phase. A finite-duration source contains several frequencies and spreads energy over their different ray angles and group speeds.

Consider one chosen model at a fixed total energy, with two open scattering channels, no absorption and no omitted outgoing modes. Channel 1 leaves the target unchanged; channel 2 excites it. Let $v_1=v>0$ and $v_2=v/2$. For the same projectile mass, conservation of energy then requires a target excitation $\Delta =3m{v}^2/8$. These are hypothetical parameters, not a fit to a particular atom. Each incoming or outgoing mode is normalized to unit probability current; the outgoing modes will have the angular profiles specified below.

Choose the S-matrixThis phenomenological unitary matrix specifies the channel conversion; it is not derived here from an interaction potential energy. For incidence in channel 1, its first column gives the outgoing flux amplitudesTo recover ordinary wave coefficients relative to an incident coefficient one, undo the velocity normalization:Thus $|c_2{|}^2=32/25>1$ is allowed: channel 2 travels more slowly. Its outgoing probability current fraction is $16/25$, not $32/25$. Renormalizing the unweighted squares would give $9/41$ and $32/41$, the wrong probabilities. More generally the ordinary coefficient map is $A=V^{-1/2}S{V}^{1/2}$, with $V=\mathrm{diag}(v,v/2)$; it obeys $A^{†}VA=V$. The conserved quantity is the squared coefficient norm weighted by velocity, which represents probability current.

Now specify an original angular realization of these two outgoing modes, with scattering angle $\theta$ and no azimuth dependence:The first normalization follows from $2\pi {\int }_{-1}^1(1+u^2)du=16\pi /3$, where $u=\cos \theta$. These angular profiles define the modes of this model; they are not asserted to follow from Born's approximation or an atomic potential energy. The joint directional probability densities are

Suppose a detector accepts the cone $D:0\le \theta \le \pi /3$, with full azimuth and unit efficiency. Its angular fractions within the channels areMultiplying the angular fractions by the outgoing flux weights givesFor $1600$ independent incident events, the expected values are therefore $171$ and $256$ accepted detections, compared with $576$ and $1024$ over all directions. The remaining expected $1173$ events fall outside the cone. The chosen incident number makes the expectations integers; actual counts fluctuate. Conditional on a detection inside the cone, the channel probabilities are $171/427$ and $256/427$, rather than $9/25$ and $16/25$. Angular acceptance changes the observed channel proportions even with equal detector efficiencies.

The final atomic states are orthogonal, so ignoring the atomic readout adds the two channel intensities; it does not produce $|b_1{g}_1+b_2{g}_2{|}^2$. Coherent paths that reach the same final state can instead interfere, as the existing probability amplitude and quantum superposition examples demonstrate. The distinction is physical distinguishability of final states, not whether the detector operator chooses to display their labels. Flux normalization and angular integration make the scattering interpretation useful: they turn complex outgoing waves into conserved event weights and then into the statistics of a specified detector.

For motion confined to a plane but retaining potential energy $-\kappa /r$, separation in polar coordinates gives angular wavefunctions $e^{im\varphi }$ with integer $m$. The regular radial factor behaves as $r^{|m|}$ at the origin. Removing this power and the decaying exponential function leaves an Generalized Laguerre polynomial differential equation. polynomial termination requires $N=n_r+|m|+1/2$, yielding the displayed energy. At fixed $q=n_r+|m|$, the spinless quantum degeneracy is $2q+1$. This is not the logarithmic potential of electrostatics formulated entirely in two spatial dimensions.

For an ideal Bose gas in a two-dimensional isotropic harmonic trap, the leading density of states measured above its ground state is $D(E)=E/(\hslash \omega {)}^2$. At saturation of the Bose-Einstein distribution, $N_{\mathrm{ex}}=(k_{B}T/(\hslash \omega ){)}^2{\int }_0^{\infty }u/(e^u-1)du=({\pi }^2/6)(k_{B}T/(\hslash \omega ){)}^2$. The finite excited-state capacity gives the displayed critical temperature and condensate fraction $N_0/N=1-(T/T_c{)}^2$ below it. The appropriate trapped-gas thermodynamic limit fixes $N{\omega }^2$ as $N\to \infty$ and $\omega \to 0$; a finite trap exhibits finite-size rounding. The number density is inhomogeneous, and the physical chemical potential approaches the oscillator ground state energy, not an unshifted zero by convention.

For a two-dimensional anisotropic quantum harmonic oscillator in the indicated range, the six lowest number states, counted with multiplicity, have occupations $(0,0),(1,0),(0,1),(2,0),(1,1),(0,2)$. Their excitation energies, divided by $\hslash$, are $0,{\omega }_1,{\omega }_2,2{\omega }_1,{\omega }_1+{\omega }_2,2{\omega }_2$. These are ordered because ${\omega }_2<2{\omega }_1$ and $2{\omega }_2<3{\omega }_1$; every omitted state has at least three quanta and costs at least $3\hslash {\omega }_1$. At equal angular frequencies, the two-dimensional isotropic harmonic oscillator has energy eigenvalue $\hslash \omega (N+1)$ and quantum degeneracy $N+1$, since there are $N+1$ ordered nonnegative pairs with sum $N$.

For $k_z=v_z=0$ and $k_y\ne 0$, define $Z=i(k_x{v}_y-k_y{v}_x)$. Uniform rotation does not enter its evolution: $\dot{Z}=-\nu k^{2}Z$, $v_x=i{k}_{y}Z/k^2$ and $v_y=-i{k}_{x}Z/k^2$. With $T=k_x/k_y$ and $\mathrm{Re}=A/(\nu k_y^2)$, the perturbation energy ratio is $E(T)/E(T_0)=[(1+T_0^2)/(1+T^2)]\exp [-(T+T^3/3-T_0-T_0^3/3)/\mathrm{Re}]$. Leading waves can gain energy as shear reduces their wavevector magnitude, before viscosity and subsequent winding damp them.

// Target: hydrodynamic-stability.bigb

For $Q,L\ge 2$, a periodic Potts chain has $Q$ uniform ground states with energy $-JL$. A single domain wall cannot close around the ring. Its first excited states have two domain walls and energy $-JL+2J$. Choose the wall bonds in $\left(\genfrac{}{}{0ex}{}{L}{2}\right)$ ways and assign distinct states to the two resulting arcs in $Q(Q-1)$ ways. This proves the displayed degeneracy factor without identifying globally relabelled configurations.

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.

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.