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With $y=x-x_0$ and oscillator length $a_0=\sqrt{\hslash /(m\omega )}$, the first two harmonic oscillator eigenfunctions and energies areThe ground state is even and nodeless; the first excited state is odd with one node of a wavefunction. Their rms widths are $a_0/\sqrt{2}$ and $\sqrt{3/2}{a}_0$. Also, $m{\omega }^2{a}_0^2\sim \hslash \omega$ gives the characteristic extent directly.

For the shaken harmonic trap, expanding about the unshifted Hamiltonian givesThe scalar term gives only a common phase and no transitions. In time-dependent perturbation theory, initially $c_0=1$, $c_{n\ne 0}=0$ and the oscillatory factor for $0\to 1$ is $e^{i(E_1-E_0)t/\hslash }=e^{i\omega t}$. ThusThe odd operator $x$ makes transitions to even $n>0$ vanish by parity. In fact the first-order selection rule for a linearly driven oscillator is stronger: since $x=a_0(a+a^{†})/\sqrt{2}$,All excited $n\ne 1$ amplitudes vanish at first order, including odd $n>1$. This assertion does not apply to the surviving $c_0$ or to higher perturbative orders.

Put $\delta =\omega -\Omega$, $\Sigma =\omega +\Omega$. The exact first-order time integral isNear resonance, the second term oscillates rapidly at approximately $2\omega$ and is bounded by $1/\Sigma$, whereas the first grows as $T/2$ on resonance. Thus $\omega T\gg 1$ justifies the rotating-wave approximation, $\cos \Omega t\mapsto e^{-i\Omega t}/2$. At zeros of the resonant term, the omitted small contribution can dominate the residual; the approximation has no uniform relative-error bound there.

Using the requested order-one approximation $⟨1|x|0⟩\simeq a_0$ and $m{\omega }^2{a}_0/\hslash =\omega /a_0$ gives the square-pulse resonance profileThe exact harmonic oscillator matrix element instead gives half this probability. The finite central maximum is at $\Omega =\omega$; first zeros are at $|\Omega -\omega |=2\pi /T$, with smaller side lobes and a $1/{\delta }^2$ envelope. Increasing $T$ narrows the peak as $1/T$ and increases its height as $T^2$, as long as weak resonant driving keeps $p_1\ll 1$. Small displacement alone does not ensure this for arbitrarily long $T$; near resonance one also needs $\epsilon \omega T/a_0\ll 1$.

The normalized quantum harmonic oscillator ground state has $|A{|}^2=\sqrt{m\omega /(\pi \hslash )}$ and energy $E_0=\hslash \omega /2$. Its classical turning points are $x=\pm \sqrt{\hslash /(m\omega )}$. With $y=x\sqrt{m\omega /\hslash }$, the ground-state oscillator probability beyond the turning points isThe stated rounded error function gives $0.16$; the more precise Gaussian integral is $0.15730$.

An impulse integrated across its full support produces the impulsive quantum kick $K=e^{-i\hat{U}/\hslash }$. Thus the leading transition amplitude is $-i{U}_{21}/\hslash$ andHere $U_{21}$ has units of action. The off-diagonal two-state impulse is exactly $P={\sin }^2(|U_{21}|/\hslash )$ for a Hermitian two-state impulse with zero diagonal. No energy denominator occurs: the impulse is instantaneous.

The central field approximation replaces the anisotropic many-electron environment by an average spherical potential, giving one-electron orbitals labelled by $n,l,m$. Screening changes the effective radial attraction and removes the pure Coulomb degeneracy between different $l$.

The Hartree method improves the potential self-consistently. For normalized occupied orbitals,Guess orbitals, construct their mean fields, solve the Hartree equations, and iterate to convergence. The $j\ne i$ sum excludes self-interaction in this product-state model. A spherical average can enforce the central-field specialization. A Hartree product state omits fermionic exchange and dynamical correlation; Hartree-Fock method adds antisymmetry and exchange through a Slater determinant. Converged orbital eigenvalues must not simply be added as the total energy, because that double counts pair interactions.^[ref]

At zero field, the spin-one-half triplet state has energy $\Delta /4$ and degeneracy three; the spin-one-half singlet state has energy $-3\Delta /4$ and degeneracy one:For $\Delta >0$ the singlet is lower; for $\Delta <0$ the triplet is lower. If $\Delta =0$, all four states coincide.

The quoted expansion is nondegenerate perturbation theory for a known isolated eigenstate of a time-independent Hamiltonian. A useful small-mixing condition is $|⟨k|H_1|i⟩|\ll |E_i-E_k|$ for coupled states. First order is a diagonal expectation value; second order describes virtual mixing. A degenerate or nearly degenerate coupled subspace must first be treated by diagonalization there. Symmetry can eliminate terms, and convergence still needs checking. Ground-state second-order shifts are nonpositive because all denominators are negative.^[ref]

For the rotational Stark effect, $H_1=-pE\cos \theta$, where $\theta$ is the polar angle to the field. The source calls it azimuthal, which is inconsistent with this coupling. Rotor states have parity $(-1{)}^J$, while $\cos \theta$ is odd, so every first-order matrix element within one fixed-$J$ manifold vanishes. The $m$ degeneracy therefore causes no first-order splitting.

Only $J\pm 1$ with the same $m$ contribute in second order. Using their energy differences $2BJ$ and $-2B(J+1)$,For $J\ge 1$ this simplifies to the quadratic Stark shift of a polar linear rotor,For $J=0$, only $J^{\prime }=1,m=0$ contributes, with squared angular matrix element $1/3$ and denominator $-2B$:Here $B$ is the rotational energy constant, unlike the magnetic field in Question3.

The unperturbed line gives $B/h=6.08\mathrm{GHz}$. The $J=1$ shifts are $(pE{)}^2/(10B)$ for $m=0$ and $-(pE{)}^2/(20B)$ for $m=\pm 1$. Subtracting the common lower-state shift gives transition shifts $4(pE{)}^2/(15Bh)$ and $7(pE{)}^2/(60Bh)$. Their separation determines the dipole moment from a rotational Stark doublet:The $\mathrm{D}$ unit is the debye. Using external $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{D}=3.33564\times 10^{-30}\mathrm{C}\mathrm{m}$ gives $pE/B\simeq 0.0593$, consistent with weak-field perturbation theory. The two shifted frequencies are about $12.165707$ and $12.162497\mathrm{GHz}$; polarization selects $\Delta m=0$ or $\pm 1$. The inferred value is a magnitude; these shifts do not determine the dipole's sign.^[ref][ref]

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.

For a particle of charge $q$ and mass $m$ in a uniform magnetic field $\mathbf{B}=B\hat{z}$, minimal coupling replaces momentum by $\mathbf{p}-q\mathbf{A}$. Choose Landau gauge $\mathbf{A}=(0,Bx,0)$ and first restrict motion to the $xy$ plane. Since $p_y=\hslash k_y$ is conserved,These Landau levels quantize the classical cyclotron motion. The wavefunctions are harmonic-oscillator functions centred at $x_c$, multiplied by $e^{i{k}_{y}y}$. Changing $k_y$ shifts the guiding centre without changing the energy. Gauge-dependent labels such as $k_y$ are not themselves observables; the level spacing and degeneracy are gauge independent. The magnetic length ${\ell }_B=\sqrt{\hslash /|qB|}$ is the transverse orbital scale.

For a large rectangle of sides $L_x,L_y$, periodicity along $y$ gives $\Delta k_y=2\pi /L_y$. Requiring centres inside the sample gives, for each orbital level and each specified spin state,Thus each magnetic flux quantum supplies one guiding-centre state. The continuum bulk count neglects boundary corrections; magnetic periodic conditions on a torus require integer flux count. A finite confining edge lifts the simple flat guiding-centre spectrum near that edge.

spin adds a Zeeman term. For electrons with gyromagnetic factor $g$, its contribution is $g{\mu }_{B}B{m}_s$, with $m_s=\pm 1/2$ and ${\mu }_B=e\hslash /(2m_e)$. If $g=2$ and the orbital mass is $m_e$, the branches become $\hslash {\omega }_{c}n$ and $\hslash {\omega }_c(n+1)$: the lowest combined level has one spin branch, while positive levels coincide in pairs. In a material, effective mass and effective $g$ need not give that coincidence.

In three dimensions, free motion parallel to $B$ adds ${\hslash }^2{k}_z^2/(2m)$, so each orbital index labels a one-dimensional continuum rather than an isolated three-dimensional energy. Resolved magnetic quantization requires the spacing to exceed substantial thermal and disorder broadening. The changing occupation of these levels underlies magnetic quantum oscillations; in two dimensions their degeneracy and filling are central to the integer quantum Hall effect. These statements assume nonrelativistic, approximately independent particles, rather than arbitrary interacting or disordered spectra.

A helium atom has two electrons bound to nuclear charge $+2e$. In the nonrelativistic fixed-nucleus approximation, using atomic units,The last term couples the electrons, so a product of independent hydrogenic orbitals is not an exact solution. The dominant ground configuration is $1s^2$. Its spatial wavefunction is symmetric, so Fermi antisymmetry requires the spin singletA triplet cannot put both electrons in the same $1s$ spatial orbital. Correlation changes the exact spatial function, including its dependence on $r_{12}$, while preserving this ground-state symmetry.

A useful helium variational effective charge takes normalized orbitals ${\varphi }_{\zeta }(\mathbf{r})=({\zeta }^3/\pi {)}^{1/2}{e}^{-\zeta r}$ for both electrons and multiplies their product by ${\chi }_{00}$. The energy in Hartree units isThe reduced effective charge expresses screening. Neglecting repulsion gives $-4E_h$, while its expectation in the unscreened $\zeta =2$ product gives $-2.75E_h$; optimizing the orbital improves that trial estimate. The exact ground energy is lower than the optimized product energy, since the product omits explicit electron correlation. This calculation keeps nuclear charge two in $H$; only the trial orbital parameter is varied.

For a singly excited helium state with distinct orthonormal orbitals $a=1s$ and $b=n\ell$, the spatial combinations areThe symmetric ${\Phi }_{+}$ pairs with the antisymmetric spin singlet, while antisymmetric ${\Phi }_{-}$ pairs with the symmetric spin triplet. With the same orbitals, their leading energies areHere $J_{ab}$ is the direct Coulomb integral and $K_{ab}$ is the exchange integral. The latter is positive for the Coulomb interaction, so the triplet is lower for this fixed-orbital comparison. singlet-triplet exchange splitting in helium arises from antisymmetry and spatial correlation in the Coulomb interaction, not from inserting a separate classical spin-spin force. Spin-singlet helium states are parahelium; spin-triplet helium states are orthohelium. The two spectra approximately form separate families because the leading electric dipole operator does not change total spin.^[ref]

The lowest $1s2s$ terms are ${}^3{S}_1$ and ${}^1{S}_0$, with the triplet lower. The $1s2p$ configuration gives ${}^3{P}_{0,1,2}^{\mathrm{o}}$ and ${}^1{P}_1^{\mathrm{o}}$, where the odd parity allows singlet $P$ to connect to the even-parity ground singlet by an electric dipole transition. Both $1s2s$ levels lack an allowed direct electric dipole decay to $1s^2$: the triplet is spin forbidden, and singlet $S\to S$ is forbidden by orbital angular momentum and parity. This explains helium metastability without implying an infinite lifetime once weaker radiative processes or collisions are allowed.

The observed first $1s2s$ excitation energies are approximately $19.82$ and $20.62\mathrm{eV}$ for the triplet and singlet, respectively. Higher singly excited Rydberg series approach the He II ground-state ionization threshold; doubly excited states above that threshold can autoionize and are not all stable bound excited levels.^[ref]

Introduce a bookkeeping parameter $\lambda$ in $H=H_0+\lambda H_1$ and expand $|{\varphi }_n⟩=|{\psi }_n⟩+\lambda |{\eta }_n⟩+O({\lambda }^2)$ and ${\mathcal{E}}_n=E_n+\lambda \Delta E_n+O({\lambda }^2)$. Fix phase and normalization at this order by $⟨{\psi }_n|{\eta }_n⟩=0$. The first-order Schrödinger equation isangular momentum projection onto $⟨{\psi }_n|$ gives $\Delta E_n=⟨{\psi }_n|H_1|{\psi }_n⟩$. angular momentum projection onto a different eigenstate givesCompleteness therefore yields the first-order nondegenerate perturbation theory resultNonzero energy denominators are essential. The displayed state has unit norm through first order; exact normalization would change terms of second order. A continuum, if present, contributes its corresponding integral as well as any discrete sum.

For the oscillator in a static electric field, electrostatic potential energy is $H_1=-qϵx$ and $\omega =\sqrt{k/m}$. parity gives $⟨0|x|0⟩=0$, and the supplied Gaussian wavefunctions giveSince $E_0-E_1=-\hslash \omega$, the two-state first-order ground state isThus $⟨x⟩=2\eta x_{\mathrm{zp}}+O({ϵ}^2)$, and the induced electric dipole isIt is aligned with the field for either sign of $q$. The harmonic-oscillator electric polarizability is $q^2/k$. The position operator connects $|0⟩$ only to $|1⟩$, so the requested two-state calculation contains the complete first-order dipole response, rather than dropping other first-order terms.

Complete the square to find the exact electric-field displacement of a quantum harmonic oscillator:The exact normalized ground wavefunction is ${\psi }_0(x-x_0)$, with energy $\hslash \omega /2-q^2{ϵ}^2/(2k)$. Its mean displacement is $x_0$, soThe first-order dipole equals the exact dipole, because the restoring force is exactly linear and translating a harmonic well does not change its width or frequency. This agreement of one expectation value does not make the truncated wavefunction exact. The Hellmann–Feynman theorem provides an independent check: $d=-\partial E_0/\partial ϵ$.

For perturbation theory to approximate the state, $|\eta |\ll 1$, or $|x_0|\ll 2x_{\mathrm{zp}}$. The exact translated ground state has the original-oscillator expansionIt is a coherent state with mean occupation ${\eta }^2$. Large fields require many oscillator levels, whereas the first-order state retains only two. The normalized two-state truncation of a displaced oscillator instead giveswhich incorrectly tends towards zero for large $|\eta |$ instead of following the displaced well. In position space the truncated polynomial times Gaussian also develops a node absent from the exact ground state. Taking a linearized formula outside its small-state-correction regime is therefore invalid even though its dipole coefficient happens to be exact.

Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem makes the potential energy of the electron constant inside the charged spherical shell. Relative to the point-proton Coulomb potential, the perturbation isIt is positive: the finite charge distribution weakens the attraction close to the origin.

The first-order energy correction is the expectation value of this perturbation in the unperturbed hydrogen atom ground state. Integrating its probability density over spherical volume givesFor a proton radius much smaller than the Bohr radius, expand the exponential uniformly over this short interval. The leading integral isSince $E_{1s}=-C/(2a_0)$, the signed fractional change isEquivalently, $\Delta E/|E_{1s}|=+4b^2/(3a_0^2)$ to leading order: the energy increases and the binding-energy magnitude decreases. This is the finite proton-shell correction to hydrogen ground energy.

If no small-radius expansion is made, the same time-independent perturbation theory integral gives, with $\beta =b/a_0$,Its expansion starts with $4{\beta }^2/3$, in agreement with the leading result. The integral is exact for the first-order calculation; it is not an exact solution of the perturbed Schrödinger equation.

A uniform magnetic field quantizes the transverse cyclotron motion of an electron. With electron charge $-e$ and the Landau gauge for a uniform magnetic field $\mathbf{A}=(0,Bx,0)$, minimal coupling givesFor an eigenstate of $p_y$ with eigenvalue $\hslash k_y$, this is a quantum harmonic oscillator centered at $x_c=-\hslash k_y/(eB)$, with cyclotron frequency ${\omega }_c=eB/m$. Its Landau levels areThe oscillator width is set by the magnetic length ${\ell }_B=\sqrt{\hslash /(eB)}$. In three dimensions, the free motion parallel to the magnetic field adds ${\hslash }^2{k}_z^2/(2m)$. The electron spin adds a Zeeman splitting $\pm g_s{\mu }_{B}B/2$. In a crystal the appropriate band effective mass and effective spin factor replace the free-electron values.

Changing $k_y$ moves the orbit center without changing the energy. In a rectangle $L_x{L}_y$, adjacent allowed centers differ by $2\pi {\ell }_B^2/L_y$. Requiring the center to lie in the sample gives the degeneracy of a Landau level, per spin,This large state degeneracy, together with the increasing level spacing, rearranges the occupied states as the magnetic field changes. To resolve the Landau levels, thermal broadening $k_{B}T$ and disorder broadening should both be small compared with $\hslash {\omega }_c$.

The de Haas-van Alphen effect is an oscillation of magnetization as successive Landau levels pass the Fermi energy. The Shubnikov-de Haas effect is the corresponding oscillation in longitudinal magnetoresistance. For a simple orbit, successive crossings occur periodically in $1/B$; the general relation to the extremal orbit area is Onsager quantization. These measurements reveal the geometry and effective masses of a Fermi surface.

In a two-dimensional electron system, filling an integer number $\nu$ of resolved levels leads to the integer quantum Hall effect, withand very small longitudinal resistivity on a plateau. Disorder localization helps produce finite plateau widths. Cyclotron resonance provides another direct signature: radiation is absorbed at the adjacent-level spacing $\hslash {\omega }_c$.

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 elastic scattering by the Yukawa potential, take $\kappa >0$ and use the first Born approximation in the plane-wave normalizationChoose the polar axis along the momentum transfer $\hslash \mathbf{q}$. The angular integral isThe radial volume element cancels the remaining powers of $r$, leavingFor real $A$, the differential scattering cross-section is thereforeThis also has a finite forward limit when $\kappa >0$. The Born amplitude for a Yukawa potential provides a screened route to the Coulomb potential by taking $\kappa \to 0^{+}$ at fixed $q>0$.

For an alpha particle of charge $2e$ scattering from a heavy nucleus of charge $Ze$, set $A=2Z{e}^2/(4\pi {ϵ}_0)$. Neglect recoil, so that the reduced mass is approximately the projectile mass, $E={\hslash }^2{k}^2/(2m)$, and elastic geometry givesThe Coulomb-limit scattering amplitude becomeswhich gives the Rutherford differential cross-sectionThe unscreened Coulomb potential has long range, so taking this limit does not justify the Born approximation for its full scattering state, particularly for a strongly charged target. The exact Rutherford scattering amplitude differs by a phase and has the same modulus. Thus the resulting differential scattering cross-section remains applicable to the ideal nonrelativistic Coulomb problem beyond the weak-potential regime.

To relate the detector counts to the atomic number, first compute the number of target atoms per illuminated area. With molar mass $M=0.197\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, the number density integrated through the foil isThe detector's transverse area is already its area normal to the scattered ray. Its small solid angle isThere is no extra $\cos 45^{\circ }$ factor: the foil is traversed normally and the detector area is transverse to its line of sight.

For a thin foil with negligible multiple scattering and energy loss, the detected fraction is $f=\mathcal{N}(d\sigma /d\Omega )\Delta \Omega$. Thus the measurements implyUsing $E=7.68\times 10^{6}e\mathrm{J}=1.23047\times 10^{-12}\mathrm{J}$ in the Rutherford differential cross-section gives

The stated measurements do not reproduce gold's atomic number 79 in this model. The values and units above agree with the original PDF. Substituting $Z=79$ instead would predict $f\simeq 3.17\times 10^{-7}$ with the other inputs unchanged, rather than $3.7\times 10^{-7}$. No stated correction accounts for that discrepancy; forcing the answer to 79 would require changing the data or adding an experimental correction not specified in the problem.

Neglect hyperfine structure and use the electronic weak-field Zeeman splitting regime, where ${\mu }_{B}B$ is small compared with the spin-orbit interaction splitting. The sodium levels are $3s{}^2{S}_{1/2}$, $3p{}^2{P}_{1/2}$ and $3p{}^2{P}_{3/2}$. The Landé g-factor formula givesFor example, inserting $L=1,S=1/2,J=1/2$ gives $3/2+(3/4-2)/(2\cdot 3/4)=2/3$. Each sublevel with magnetic quantum number $m_J$ shifts by $g_J{m}_J{\mu }_{B}B$. Thus the lower level has shifts $\pm {\mu }_{B}B$, the upper $J=1/2$ level has shifts $\pm {\mu }_{B}B/3$, and the upper $J=3/2$ level has shifts $\pm 2{\mu }_{B}B$ and $\pm 2{\mu }_{B}B/3$.

The electric-dipole selection rule allows $\Delta J=0,\pm 1$ except $0\leftrightarrow 0$, and $m_u-m_l=0,\pm 1$. Both fine structure branches are allowed. The emitted photon energy isAll the allowed components of the sodium D-line Zeeman pattern are listed below; the last column is the photon shift in units of ${\mu }_{B}B$:These are four components of D1 and six of D2. Their strengths need not be equal; the table and arrows specify allowed transitions, not intensities.

The $3p{}^2{P}_{3/2}$ level lies above $3p{}^2{P}_{1/2}$, so D2 has the larger zero-field photon energy. ^[ref] In the stated weak-field regime its most positive shift is $5{\mu }_{B}B/3$. Since $\lambda =hc/E_{\gamma }$, the shortest wavelength is therefore produced byThe tempting $m_u=3/2\to m_l=-1/2$ pair has a larger difference of level shifts but is forbidden because $|m_u-m_l|=2$.

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.