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A microstate is a complete microscopic specification within the chosen model. For two distinguishable, noninteracting two-level systems, label the units $A,B$ and their states $0,1$, with energies $0,\Delta$ and $\Delta >0$. The four microstates arewith total energies $0,\Delta ,\Delta ,2\Delta$. Distinguishability matters: $(0,1)$ and $(1,0)$ describe different assignments to the labeled units. Noninteraction matters too: their energies add without an extra coupling term.

For a classical particle model, a microstate can instead be a point in phase space, specifying every position and momentum. In the quantum canonical ensemble, the sum uses energy eigenstates; a general quantum state may also contain quantum coherence, which an energy label alone does not specify.

A mobility edge separates localized from extended energy eigenstates on the energy axis. Three-dimensional disordered models can have both. In the infinite one-dimensional tight-binding model with independent continuous random site energies, all states are localized for any nonzero disorder, with probability one. This statement excludes specially correlated disorder.^[ref]

A mobility edge is not a band gap: states can exist on both sides of $E_c$, but differ in spatial extent. For electrons, the location of the Fermi energy relative to a mobility edge matters for low-temperature electrical conductivity. A three-dimensional sample need not have an extended interval at every disorder strength. In particular, the finite one-dimensional calculations below neither locate a three-dimensional mobility edge nor establish a nonzero universal critical disorder.

An energy interval with no extended bulk energy eigenstates available for transport. It can contain localized eigenstates and have a nonzero density of states; a mobility gap need not be a spectral gap.

The Fourier transform in the title represents a one-dimensional state in the wavevector basis. Its probability density is $|\varphi (k){|}^2$, normalized by $\int |\varphi (k){|}^{2}dk=1$ when $\int |\psi (x){|}^{2}dx=1$. For physical momentum $p=\hslash k$, the density per unit momentum is $|\varphi (p/\hslash ){|}^2/\hslash$. A confined stationary energy eigenstate need not be a momentum eigenstate.

For the quantum harmonic oscillator with $a_{-}=P-im\omega X$, $a_{+}=a_{-}^{\ast }$ and $\gamma =2m\hslash \omega$, factorization gives $a_{+}{a}_{-}=2m(H-E_0)$, where $E_0=\hslash \omega /2$. For an energy eigenstate $\psi$ at $E_n=E_0+n\hslash \omega$,The first identity shows that lowering to the ground state is injective on the $E_n$ eigenspace; the second shows how to invert it by raising. If the ground eigenspace is one-dimensional, $(a_{-}{)}^n\psi =c{\psi }_0$ and therefore $\psi =c(a_{+}{)}^n{\psi }_0/({\gamma }^{n}n!)$. Thus every level is also nondegenerate. This proof explicitly rules out a second independent energy eigenstate at an excited level rather than merely constructing one ladder of states.

A number state is a normalized eigenstate of the number operator:For a single quantum harmonic oscillator, it is also an energy eigenstate with energy $\hslash \omega (n+1/2)$.

Choose normalized energy eigenstates $|n⟩$ of a time-independent Hamiltonian operator, with $H|n⟩=E_n|n⟩$, and assume first that they form a complete orthonormal basis. For a pure state and an observable $A$,The column index $n$ labels the input basis vector, and row $m$ labels the output component. Self-adjointness gives $A_{nm}=A_{mn}^{\ast }$, so a finite example is a Hermitian matrix. A matrix element can be negative or complex; by itself it is not generally a probability. The Born rule assigns measurement probabilities using the eigenstates of $A$, which need not be energy eigenstates. For an unbounded observable, these expressions require a suitable operator domain; a system with continuous spectrum also needs the corresponding spectral integrals.

Consider a noninteracting electron in a finite tight-binding model with open ends:The vectors $|n⟩$ form an orthonormal basis of site states. The hopping energy is $t$; it is not the time coordinate. Draw each ${ϵ}_n$ independently from the continuous uniform distribution on $[-W/2,W/2]$ and then hold it fixed. The disorder strength $W/t$ is dimensionless. Open ends mean no hopping bond connects sites $N$ and $1$; equivalently set ${\psi }_0={\psi }_{N+1}=0$. There are no electron interactions, phonons or wave absorption in this model.

The eigenvalue equation in the interior isAs a simple local calculation, if a state's peak is at $n_0$, a neighbouring site with $|E-{ϵ}_n|\gg t$ receives a small leading contribution ${\psi }_n\simeq -t{\psi }_{n_0}/(E-{ϵ}_n)$. Several successive large mismatches multiply small factors. This explains why large disorder can restrict spreading, but is not a proof of Anderson localization: near-resonant denominators and alternative interference paths prevent treating this expansion as uniformly valid.

For the first original calculation, take $N=80$, $t=1$, $W/t=6$ and the fixed random seed $20261015$. Diagonalizing the resulting real symmetric matrix with numpy.linalg.eigh and selecting the eigenvalue closest to zero gives $E/t\simeq -0.05962$, a probability peak at site $29$, and $P\simeq 6.0031$. The selected eigenvector has total probability one. This is one chosen realization, not a disorder average or a measured specimen.

The clean comparison needs care. With $W=0$, the open chain has standing wavesFor even $N$, the identity ${\sin }^{4}x=(3-4\cos 2x+\cos 4x)/8$ and finite trigonometric sums give ${\sum }_n|{\psi }_n^{(m)}{|}^4=3/[2(N+1)]$, hence $P=2(N+1)/3$. Thus the clean $80$-site comparison has $P=54$, although it extends across the chain. Extended does not mean uniform probability. For odd $N$, the special central mode $m=(N+1)/2$ has zeros on alternating sites and instead $P=(N+1)/2$; the even-size assumption avoids that exception.

The second original calculation compares $N=40$ and $80$ over $W/t=0,0.5,1,2,3,4,6,8$. For each size, draw $24$ independent site patterns using seed $20261016$, with the size included in the random-seed construction. Scale each fixed pattern across disorder strengths. At every point, average $P$ over the four energy eigenstates closest to $E=0$. The solid line is the median across these sample averages; shading spans their $10$th–$90$th percentiles. The selected states can change as $W/t$ varies: lines connect one sample's selections rather than tracking a single mode.

At $W/t=0$ the two clean values are $27\frac{1}{3}$ and $54$. At $W/t=8$, the computed median sample averages are approximately $3.17$ and $3.01$. In the latter regime, doubling $N$ hardly changes the typical occupied site count, so $P/N$ falls. At weak disorder the larger chain permits more spatial extent. Neither a single curve nor the overlap of finite-size bands proves the infinite-system statement under mobility edge; a long localization length can hide confinement in either size.

A normalized ground state initially supported on $(0,a)$ is $\sqrt{2/a}\sin (\pi x/a)$. Expanding the infinite square well to $(0,2a)$ leaves this wavefunction unchanged during an ideal sudden switch. Its overlap with the new ground state $a^{-1/2}\sin (\pi x/(2a))$ is $4\sqrt{2}/(3\pi )$, so the Born rule gives $32/(9{\pi }^2)$. This differs from ground-state overlap after sudden expansion of a square well with a fixed midpoint, because translation of the new well changes the overlap. Under a fixed final Hamiltonian operator, all final energy eigenstate probabilities remain constant during subsequent unitary time evolution.

For a spinless particle, minimal coupling gives $H=(\mathbf{p}-q\mathbf{A}{)}^2/(2m)+q\varphi$. Choose the vector potential $\mathbf{A}=(0,\alpha x^2/2,0)$ and $\varphi =0$. Its curl is $\mathbf{B}=\alpha x\hat{\mathbf{z}}$ and $[p_y,A_y]=0$, soThe translation separation in a magnetic field gradient follows from $[H,p_y]=0$: a simultaneous energy eigenstate can be written $\Psi =L_y^{-1/2}{e}^{iky}{\Phi }_k(x)$. The periodic boundary condition requires $e^{ik{L}_y}=1$, hence $k=2\pi n/L_y$, $n\in \mathbb{Z}$. The remaining Schrödinger equation isThis magnetic-gradient quartic confinement has a positive quartic coefficient for $g\ne 0$. If $k/g<0$, it is a single well at zero, with $V_k(0)={\hslash }^2{k}^2/(2m)$. Its harmonic curvature givesThus the two lowest states are respectively even and odd, with one central peak or two opposite-sign lobes, and approximately $E_j=V_k(0)+(j+1/2)\hslash {\omega }_s$. They are not a nearly degenerate pair.

If $k/g>0$, the two minima are $x=\pm x_0$, $x_0=\sqrt{2\hslash k/g}$, with zero potential there and a central barrier ${\hslash }^2{k}^2/(2m)$. For well-separated minima the lowest parity states of a symmetric double well areHere ${\varphi }_L,{\varphi }_R$ are the localized ground-state Gaussians. Their even combination is the lower state; the odd combination has a node at the origin. Weak quantum tunnelling gives an exponentially small double-well tunneling splitting.

To obtain their common leading energy, put $x=x_0+\delta x$. Since $\hslash k=g{x}_0^2/2$,The double-well oscillator scale in a linear magnetic field therefore givesThe asymptotic condition is $x_0\gg \sqrt{\hslash /(m{\omega }_d)}$, so the barrier separates narrow local oscillator states. It is this scale comparison that makes the source's “large” ratio meaningful.

The signed group velocity of a magnetic-gradient doublet follows by differentiating the band energy with respect to the conserved wavevector:For $q\alpha >0$ the double-well branch has $k>0$ and moves along $+y$; reversing both signs reverses its velocity. To this order the two parity states have the same velocity. Corrections include local anharmonicity and the exponentially small splitting. The spinless Hamiltonian specified here has no extra Zeeman effect term.

Centre the infinite square well at zero. Its ground state wavefunction is $\psi (x)=\sqrt{2/L}\cos (\pi x/L)$ inside the well and zero outside. The momentum representation in quantum mechanics gives the wavevector amplitudeConsequently the unnormalised wavevector distribution of an infinite-well ground state isThe apparent singularities at $k=\pm \pi /L$ are removable. The normalized probability density is $4\pi L$ times the boxed expression and has limit $L/(4\pi )$ there. Translating the well changes only the Fourier transform's phase. The wavevector is continuously distributed; a box energy eigenstate is not a free momentum eigenstate.

The quantum variational principle turns the ground-state problem into minimization of a Rayleigh quotient:The Hamiltonian must be a self-adjoint operator bounded below, and the nonzero trial state must have finite form energy. Expanding a normalized trial state in exact energy eigenstates gives $E[\psi ]={\sum }_n|c_n{|}^2{E}_n$, a weighted average no smaller than $E_0$. Equality requires support only in the ground-state eigenspace. The spectral version also accommodates continuous spectrum; a lowest normalizable eigenstate is not required merely to bound the spectral infimum.

The variational method chooses a trial family with the correct boundary conditions, fermionic exchange symmetry and other required quantum numbers, evaluates its energy and minimizes over its parameters. It gives an upper bound on ground-state energy. Enlarging a nested trial family can only improve or preserve its optimal bound. A low energy does not by itself establish a highly accurate wavefunction, especially near degeneracy; energy errors are second order in small admixtures orthogonal to a nondegenerate exact ground state.

For example, a normalized Gaussian trial for a one-dimensional quantum harmonic oscillator is ${\psi }_{\beta }=(\beta /\pi {)}^{1/4}{e}^{-\beta x^2/2}$, with $\beta >0$. Its energy isThe two terms balance kinetic localization cost against the potential cost of spreading out. This Gaussian variational energy for the harmonic oscillator happens to attain the exact answer because the family contains the exact ground state.

A trial expansion $\psi ={\sum }_j{c}_j{\chi }_j$ in a finite, possibly nonorthogonal basis gives the generalized eigenvalue problem $H\mathbf{c}=ES\mathbf{c}$, with $H_{ij}=⟨{\chi }_i|H|{\chi }_j⟩$ and $S_{ij}=⟨{\chi }_i|{\chi }_j⟩$. Its lowest Ritz value minimizes the quotient in that subspace. For excited states, the excited-state variational constraint imposes exact orthogonality to the lower eigenstates to obtain a corresponding upper bound; the Courant–Fischer min-max principle yields upper bounds for ordered Ritz eigenvalues from a common subspace. orthogonality only to an independently approximate ground state does not automatically provide an excited-state bound: residual true-ground-state contamination can lower the trial energy below the desired excited energy. Symmetry can impose exact orthogonality when the relevant states have different symmetry sectors.

Uniform system and dispersion. A plane wave reduces the differential Hamiltonian to a two-by-two matrix acting on its spinor:The Pauli anticommutation relations imply $H(\mathbf{k}{)}^2=|\mathbf{d}{|}^{2}I$. Choosing a spinor in one of its eigenspaces therefore givesAn arbitrary constant spinor multiplying a plane wave is not necessarily an energy eigenstate; it must be an eigenvector of this matrix. For example, for the upper branch and $\epsilon =|\mathbf{d}|$, one can useaway from the coordinate singularity $\epsilon +V=0$, where another normalized eigenvector chart is used.

For the usual $c>0$ convention, the massless branches are $E_{\pm }=\pm \hslash ck$. For nonzero $V$, the two branches have a gap $2|V|$ and are quadratic near $k=0$:Changing the sign of $V$ does not change these energies. It reverses the out-of-plane spin polarization of a given band at the origin. The plots for equal positive and negative mass magnitudes therefore have identical dispersion curves, despite their different spinors.

Spin direction in the massless case. For $k>0$ put $k_x+i{k}_y=k{e}^{i\phi }$. The normalized eigenspinors can be chosen asDirect multiplication gives $⟨{\sigma }_x⟩=s\cos \phi$, $⟨{\sigma }_y⟩=s\sin \phi$ and $⟨{\sigma }_z⟩=0$. Since the Pauli matrices represent physical spin in this question,The spin is parallel to momentum on the positive-energy branch and antiparallel on the negative-energy branch. Thus the branch-independent statement is collinearity; the printed parallel claim needs the positive-branch qualification. At $k=0,V=0$, the Hamiltonian is zero and spin is not fixed by a wavevector direction.

Mass-domain-wall eigenstate. Assume $c>0$ for the stated normalizable wall state and writeThen $F^{\prime }=-VF/(\hslash c)$, ${\sigma }_x{u}_{-}=-u_{-}$, and $(i{\sigma }_y+{\sigma }_z)u_{-}=0$. Applying the full operator givesThe transverse derivative and mass terms cancel pointwise. Hence the specified state is an eigenstate with $E=-\hslash c{k}_x$.

For a step mass $V(y)=V_{+}>0$ above the interface and $V(y)=V_{-}<0$ below it,The Localization length at a step Dirac mass wall sets the two exponential tails. The amplitude is continuous, peaks at the interface and decays on both sides, with generally unequal decay lengths. The probability density decays with lengths ${\xi }_{\pm }/2$. A normalized transverse function is $f(y)=\sqrt{2/({\xi }_{+}+{\xi }_{-})}F(y)$ when multiplying the unit spinor $u_{-}$. The longitudinal plane wave can independently be box-normalized. A step in $V$ allows a step in $F^{\prime }$; the first-order Dirac equation does not impose continuity of that derivative.

Propagation. The interface dispersion is the straight lineEvery component on this branch travels in the negative $x$ direction. A packet has the formThis dispersionless chiral interface wavepacket therefore translates rigidly with speed $c$ in that direction, without dispersion in this exactly linear model. It remains confined in $y$; there is no free transverse group velocity, because the bound-state family is not labelled by a transverse travelling-wave momentum.

The continuity equation gives the current components $j_x=c{\Psi }^{†}{\sigma }_x\Psi =-c|\Psi {|}^2$ and $j_y=c{\Psi }^{†}{\sigma }_y\Psi =0$. This explicitly confirms absence of probability transport away from the interface. For the step-mass case the profile is normalizable because its tails have positive decay constants and $c>0$. A general sign-changing profile also requires $F\in L^2(\mathbb{R})$; nonzero asymptotic masses suffice, whereas signs alone do not guarantee confinement if their integrated magnitudes grow too slowly. If $c<0$, the supplied exponential instead grows at infinity; a normalizable state requires the opposite spinor and envelope sign. Reversing the wall orientation also reverses the propagation chirality.

A coherent state is a normalized eigenstate of the annihilation operator, $a|\alpha ⟩=\alpha |\alpha ⟩$. Expanding in the Fock state eigenbasis gives a recurrence $c_{n+1}=\alpha c_n/\sqrt{n+1}$, and wavefunction normalization fixesThe displacement operator therefore shifts the oscillator ground state in phase space. Distinct coherent states are not orthogonal:

The occupation probabilities form a Poisson distribution,For the quantum harmonic oscillator, the expectation values of position and momentum areThe fluctuations equal those of the ground state,Thus the position wavefunction is a displaced Gaussian wave packet that saturates the Heisenberg uncertainty principle.

Time evolution under $H=\hslash \omega (a^{†}a+1/2)$ preserves this form:The center follows the classical oscillator trajectory while the packet width stays constant. Its mean energy is $\hslash \omega (N+1/2)$; it is generally a superposition of energy eigenstates, rather than a state of definite energy. For $N\gg 1$, the relative occupation fluctuation is $1/\sqrt{N}$, explaining the classical appearance of a highly excited coherent state.

For a bound energy eigenstate with a nondegenerate energy eigenvalue of an even one-dimensional potential, the wavefunction can be chosen with definite parity. A uniform electric field $\mathcal{E}$ gives the electric-dipole interaction $H^{\prime }=-q\mathcal{E}x$, which is odd under reflection. Hencesince $x|{\psi }_n(x){|}^2$ is odd. Second-order nondegenerate perturbation theory then gives the leading allowed Stark effect shift:Only opposite-parity intermediate states have nonzero position matrix elements. More generally, reflection maps $H(\mathcal{E})$ to $H(-\mathcal{E})$, so an analytic nondegenerate energy level is an even function of the field. This explains the absence of all odd orders and the usual quadratic Stark effect.

A level separation changes by the difference of the two quadratic coefficients. Symmetry alone does not guarantee a nonzero differential shift: in a displaced quantum harmonic oscillator every level moves by the same $-q^2{\mathcal{E}}^2/(2m{\omega }^2)$, leaving every spacing unchanged. Degenerate states would instead require degenerate perturbation theory and can split linearly; the nondegeneracy of one-dimensional bound states applies to ordinary isolated well levels.

For the quantum harmonic oscillator, the canonical commutation relation $[x,p]=i\hslash$ gives $[a,a^{†}]=1$ and $H_0=\hslash {\omega }_0(a^{†}a+1/2)$. Inverting the definition of the annihilation operator givesWith normalized energy eigenstates and consistent phases, $a|0⟩=0$ and $a^{†}|0⟩=|1⟩$. Their orthogonality therefore givesMore generally, $⟨n|x|0⟩=x_{01}{\delta }_{n1}$, the oscillator electric-dipole selection rule out of the ground state.

To derive the transition amplitude, insert the expansion in stationary energy eigenstates into the Schrödinger equation with $H=H_0+V_1(t)$. The derivatives of the explicit phases cancel the $H_0$ terms. Projection onto $⟨n|$ leaves the exact coefficient equationsThe initial conditions are $c_0(-\infty )=1$ and $c_{n\ne 0}(-\infty )=0$. To first order in the perturbation, replace the amplitudes on the right by these initial values. Since $E_n-E_0=n\hslash {\omega }_0$, integration gives time-dependent perturbation theory in the formThis neglects depletion and higher-order transitions; it is justified when the total transition probability is small. Here $⟨0|x|0⟩=0$, so there is also no first-order diagonal amplitude change from this particular perturbation.

For the exponential electric field, the electric-dipole interaction is $V_1=-qEx{e}^{t/\tau }$, with $\tau >0$. Only $n=1$ couples directly to the initial state, andThe Born rule givesThe real part $1/\tau >0$ makes the lower-limit contribution vanish. ThusSet $u={\omega }_0\tau$ and $\eta =qE{x}_{01}/(\hslash {\omega }_0)$. The dimensionless form is $p_1/{\eta }^2=u^2/(1+u^2)$, increasing monotonically from zero to one, with half the asymptotic value at $u=1$. A sufficient weak-field condition for the entire curve is $|\eta |\ll 1$.

For ${\omega }_0\tau \ll 1$, the force has acted for too little time to appreciably change the state at the instant of measurement. The short-time amplitude is approximately $iqE{x}_{01}\tau /\hslash$, soFor ${\omega }_0\tau \gg 1$, instead,The nonzero slow-switch limit is a probability in the original $H_0$ basis. It does not contradict the adiabatic theorem. At $t=0$ the instantaneous oscillator is displaced by $d=qE/(m{\omega }_0^2)$:Its ground state is $|0_E⟩=|0⟩+\eta |1⟩+O({\eta }^2)$ in the original basis. Adiabatic following therefore predicts exactly the finite original-basis limit found above.

The exponential switch-on of a driven quantum oscillator also permits an exact check. A linear force preserves a coherent state, with original-basis displacement $\alpha (0)=i\eta u/(1+iu)$. Thus the exact original-basis probability is $|\alpha {|}^2{e}^{-|\alpha {|}^2}$, whose weak-field limit is the expression above. In the instantaneous displaced basis the mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/(1+u^2)$ and does tend to zero for slow switching. The two bases measure different notions of excitation.

Take a normalized spin one-half state in the $S_z$ eigenbasis. Since $\mathbf{S}=(\hslash /2)\mathbf{\sigma }$, direct multiplication of the Pauli matrices givesSubstitution of the half-angle parametrization givesThus $\theta$ and $\varphi$ are precisely the polar and azimuthal angles of the mean spin. Two complex coefficients contain four real parameters; normalization removes one and an unobservable common quantum phase removes another. The resulting pure states form the Bloch sphere. One may choose $\alpha$ real and nonnegative away from the patch where $\alpha =0$. At the poles the azimuth is redundant. The length $|⟨\mathbf{S}⟩|=\hslash /2$ should not be confused with $\sqrt{⟨S^2⟩}=\sqrt{3}\hslash /2$.

For a spatial plane wave, $-i\hslash \nabla$ becomes $\hslash \mathbf{k}$, so the Hamiltonian operator reduces to the Weyl Hamiltonian matrixDefine $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and ${\chi }_{+}=(\cos (\theta /2),e^{i\varphi }\sin (\theta /2){)}^{\mathsf{T}}$. The matrixsatisfies $(\mathbf{\sigma }\cdot \mathbf{n}){\chi }_{+}={\chi }_{+}$ by the angle-addition identities. Consequently the displayed spinor is an energy eigenstate for either $\mathbf{k}=+k\mathbf{n}$ or $\mathbf{k}=-k\mathbf{n}$, with respective energies $+\hslash vk$ and $-\hslash vk$, for $k\ge 0$. Equivalently, at each fixed nonzero wavevector the two branches areIf $v<0$, the two energy labels exchange; the ordered eigenvalues are $\pm \hslash |v||\mathbf{k}|$. Taking $v>0$ sets the convention in the sketches. An orthogonal spinor ${\chi }_{-}=(-e^{-i\varphi }\sin (\theta /2),\cos (\theta /2){)}^{\mathsf{T}}$ obeys $(\mathbf{\sigma }\cdot \mathbf{n}){\chi }_{-}=-{\chi }_{-}$ and supplies the other branch at $+k\mathbf{n}$.

The Pauli matrix multiplication law independently gives $H(\mathbf{k}{)}^2=(\hslash v{)}^2|\mathbf{k}{|}^{2}I$. The dispersion relation is therefore a pair of cones touching at zero energy and momentum; a signed straight-line cut has upper and lower branches $\pm \hslash v|k_x|$. In the $k_z=0$ plane, $\theta =\pi /2$, and the upper-branch mean spin points radially outwards, the lower-branch spin radially inwards. This is spin-momentum locking, not a tangential spin texture. At the touching point the two states are degenerate and the spin direction is not selected by the Hamiltonian operator.

Now let a spatially constant perturbation have matrix $V=\left(\begin{array}{cc}u& w\\ w^{\ast }& z\end{array}\right)$, with $u,z$ real. The Hermitian two-by-two matrix in the Pauli basis decomposition follows explicitly fromwhere $d=(u+z)/2$, $c=(u-z)/2$, $a=\mathrm{Re}w$, and $b=-\mathrm{Im}w$ are real. Equivalently, $d=\mathrm{tr}V/2$ and $(a,b,c{)}_i=\mathrm{tr}(V{\sigma }_i)/2$. The four matrices are linearly independent under the trace inner product and span the four-dimensional real vector space of Hermitian two-by-two matrices.

Spatial constancy means the perturbation does not mix different wavevectors, so each momentum sector can still be diagonalized independently. Write $\mathbf{b}=(a,b,c)$. The full matrix and its exact eigenvalues areThis constant matrix perturbation of a Weyl Hamiltonian translates the cone toIn three-dimensional momentum space the energy bands still touch; the perturbation produces no full spectral gap. The new branch spin directions are parallel and antiparallel to $\hslash v\mathbf{k}+\mathbf{b}$, so their radial texture is centered on the shifted point.

A restricted cut must be distinguished from the full dispersion relation. In the fixed plane $k_z=0$, the energies are $d\pm \sqrt{(\hslash v{k}_x+a{)}^2+(\hslash v{k}_y+b{)}^2+c^2}$, with minimum branch separation $2|c|$. This plane misses the crossing if $c\ne 0$. Along any line through ${\mathbf{k}}_{\ast }$ the dispersion relation still crosses linearly. Thus a sketch of a gapped fixed-plane cut alone would not describe the gapless three-dimensional Hamiltonian operator. The cone after a general perturbation is also no longer a function of $|\mathbf{k}|$ about the old origin alone.

Expand the time-dependent quantum state in unperturbed energy eigenstates, factoring out free dynamical phases:The Schrödinger equation gives the exact interaction picture coefficient equationsInitially $c_n(-\infty )={\delta }_{ni}$. To first order in time-dependent perturbation theory, replace the coefficients on the right by these zeroth-order values and integrate. ThusFor a transition $j\ne i$, this is the desired leading probability amplitude, and $P_{ji}=|{\gamma }_{ji}^{(1)}{|}^2$ to leading order. The formula is understood with free asymptotic phases removed. For $j=i$, the full survival amplitude also contains the zeroth-order $1$; the displayed integral alone is its first-order correction, not the whole amplitude. This specifies the usual scattering operator meaning of the paper's notation at infinite times.

For the ion pulse, write $x_{ji}=⟨j|x|i⟩$ and $y_{ji}=⟨j|y|i⟩$. ThenConsider an excitation with ${\omega }_{ji}>0$ and put $\beta ={\omega }_{ji}b/v$, $u={\omega }_{ji}t$. The first real pulse is even in time and the second is odd. Their respective surviving Fourier components are therefore cosine and imaginary sine:Since $\beta \log (1/\beta )\to 0$, the longitudinal term is smaller for comparable position matrix elements. Hence the electric-dipole excitation by a passing ion hasThis leading limit extends to either sign of ${\omega }_{ji}$ using $|{\omega }_{ji}|b/v\ll 1$. If $x_{ji}=0$ by a selection rule, the retained leading term vanishes and the smaller longitudinal term may give the first nonzero probability. The conditions also include a small total transition probability, approximately straight constant-velocity motion, and $b$ large compared with atomic size for the electric-dipole approximation. The paper uses a Coulomb-coupling convention in which the factor is written $Z{e}^2$; in SI units replace $e^2$ throughout by $e^2/(4\pi {ϵ}_0)$.

The collision duration is of order $b/v$. Excitations with angular frequency ${\omega }_{ji}$ receive an unsuppressed impulse when this is short relative to $1/{\omega }_{ji}$. In contrast, for ${\omega }_{ji}b/v\gg 1$, cancellations in the Fourier transform suppress the transition: a slowly varying distant field approximately preserves the atomic energy eigenstate. The exact dipole-pulse transforms make this explicit:where $K_{\nu }$ is a Modified Bessel function of the second kind. Their large-argument exponential decrease explains the slow-passage cutoff. These transforms are optional refinements of the small-$\beta$ derivation above, not a separate assumption about the target.

To connect the microscopic transition probability to energy loss, suppose the atom begins in a low-energy state. Exciting bound electronic states and causing ionization transfer positive energies $E_j-E_i$ from the ion to the material. For target atomic number density $n_{\mathrm{at}}$ and path length $d$, a schematic independent-encounter sum iswith a continuum integral as well as a sum for ionized final states. The short-collision formula gives $P_{ji}\propto Z^2/(b^2{v}^2)$, so the impact-parameter integral has logarithmic dependence on its cutoffs. Its upper effective cutoff is of order $v/{\omega }_{ji}$, where the transition ceases to be sudden; the lower cutoff must exclude regions where the dipole or weak-collision approximation fails. Thus the leading high-speed charge and isotope scaling of electronic stopping isAlthough a faster ion can access more energetic transitions and a wider range of distant encounters, its shorter interaction time reduces each ordinary excitation probability. In the fast-ion regime shown, the $1/v^2$ decrease dominates the slower variation of the excitation factor. The energy loss consequently falls with increasing incident kinetic energy. The original figure has $\delta E$ increasing downward, so the curves' rise toward the top of the page means decreasing loss, not increasing loss. A low-speed extrapolation of the inverse-square law is invalid because excitation thresholds, slow-passage suppression, changing ion charge and nuclear recoil can then matter.