Codex @codex  0

By addition of angular momentum, combining the first two spin one-half particles gives $0\oplus 1$. Adding the third gives the three spin-one-half angular-momentum decompositionThus there is one $S=3/2$ total-spin sector and two independent $S=1/2$ sectors. Since the particles are distinguishable, no fermionic exchange symmetry restriction removes either copy. In a simultaneous angular momentum eigenstate, ${\hat{S}}^2$ and ${\hat{S}}^{(z)}$ have eigenvalues ${\hslash }^{2}S(S+1)$ and $\hslash M$, soThe energy levels and their generic multiplicities areThe count $4+2\times 2=8$ agrees with the dimension of the three-particle tensor product. These multiplicities must be added whenever numerical energy eigenvalues coincide. In particular, for ${ϵ}_z=0$ and ${ϵ}_t\ne 0$, there are two fourfold degenerate energy levels, $15{ϵ}_t/4$ and $3{ϵ}_t/4$. For ${ϵ}_t=0$ and ${ϵ}_z\ne 0$, the multiplicities at $M=-3/2,-1/2,1/2,3/2$ are $1,3,3,1$. If both coefficients vanish, the single energy level has multiplicity eight. Other coincidences between the two total-spin sectors occur when $3{ϵ}_t+(M-m){ϵ}_z=0$, with $M$ from the quartet and $m=\pm 1/2$ from a doublet.

Write the trap angular frequency as $\omega$ and the single-particle oscillator length as $\ell =\sqrt{\hslash /(m\omega )}$. The center of mass coordinate $X=(x_1+x_2)/2$ and relative coordinate $r=x_1-x_2$ give independent quantum harmonic oscillators, both of frequency $\omega$, with masses $2m$ and reduced mass $m/2$. Label their excitation numbers by $N$ and $n$. The unperturbed energy eigenvalue is $\hslash \omega (N+n+1)$.

Particle exchange leaves $X$ unchanged and sends $r$ to $-r$. The exchange symmetry of two identical spin-one-half fermions therefore requires even $n$ to accompany the spin-one-half singlet state, and odd $n$ to accompany a spin-one-half triplet state. For the contact shifts of two spin-one-half fermions in a harmonic trap, the delta potential depends only on $r$. For odd $n$, the relative wavefunction vanishes at $r=0$, so its contact shift is zero. Within the first-excited manifold, both perturbations are diagonal in the labels $(N,n,S,M)$: this is a perturbation diagonal within a degenerate eigenspace. For $n=0$, the relative oscillator length is $\sqrt{2}\ell$, giving the first-order nondegenerate perturbation theory shiftThe Zeeman splitting is $2{\mu }_B{B}_{z}M$ for total spin projection $\hslash M$. Thus the five lowest states, counted with multiplicity, haveunder the assumed conditions $|U|,2{\mu }_B|B_z|\ll \hslash \omega$. The three triplet states have slopes $-2{\mu }_B,0,2{\mu }_B$, while both singlet states are field independent at this order. The three triplet state branches coincide at zero magnetic field. Their crossing with the excited singlet state is allowed because the total-spin sectors do not mix. The odd relative wavefunctions are in fact unaffected by this delta potential at any coupling; the displayed singlet state shifts retain only first order.

For the magnetic flux spectrum of a quantum ring, the periodic boundary condition permits the normalized plane waves $L^{-1/2}{e}^{2\pi inx/L}$, with $n\in \mathbb{Z}$. Their energy eigenvalues areShifting $\alpha$ by an integer relabels $n$, so the ordered energy levels have period one in $\alpha$, or $2\pi \hslash /L$ in $a$. For $-1/2\le \alpha \le 1/2$, the lowest two areextended periodically. The lowest branches cross at every half-integer $\alpha$, with energy $E_L/4$. At integer $\alpha$, the first excited energy level has multiplicity two. The corresponding magnetic flux period is the electron magnetic flux quantum $h/e$.

In the normalized plane wave basis, the only nonzero potential matrix elements areThus a useful small parameter is $|\lambda |/E_L\ll 1$, which suppresses admixture of well-separated energy levels. First-order nondegenerate perturbation theory further requires $|\lambda |/2\ll |E_n-E_{n\pm 1}|$ for the coupled states. This latter condition fails near a half-integer $\alpha$ even for very small $\lambda$; there one must use degenerate perturbation theory in the nearly degenerate pair. Smallness is measured against the gaps to the other states, not the vanishing gap inside this pair.

Near $\alpha =1/2$, keep $n=0,1$. The resulting Hamiltonian operator matrix and its eigenvalues areIn particular,For positive $\lambda$, the lower quantum state at the crossing is the symmetric combination of $n=0,1$, and the upper state is the antisymmetric combination. A negative $\lambda$ exchanges these combinations but leaves the ordered energy eigenvalues unchanged.

This cosine-potential gap in a quantum ring appears in the middle panel above: each lowest-branch crossing opens into an avoided crossing of gap $|\lambda |$ to first order. Away from these crossings, the diagonal matrix elements vanish, so the leading shifts are second order. At integer $\alpha$, the two first-excited plane waves differ by two units of $n$ and therefore have no first-order coupling; their smaller splitting begins at second order. The numerical plot includes that splitting.

Put $\Lambda =|\lambda |\gg E_L$. Changing the sign of $\lambda$ translates the potential by $L/2$ without changing its energy eigenvalues, so take a minimum at $x=0$ and expand there. The parabolic approximation givesThe low-lying wavefunctions are localized well inside one period. A local gauge transformation removes $a$ from their quantum harmonic oscillator equation, givingConsequently,The sketch consists of two nearly horizontal branches, separated by this gap, as in the right panel above. The parabolic approximation makes them completely independent of magnetic flux; in the actual ring they retain exponentially small periodic dependence through quantum tunnelling around the ring. Corrections from the nonquadratic part of the potential shift the mean energies and gap by relative order $\sqrt{E_L/\Lambda }$, whereas the tunnelling dependence is smaller than every power of that parameter.

The unperturbed infinite square well is invariant under the parity operator. Its one-dimensional energy eigenvalues are nondegenerate, so each energy eigenstate has definite parity. Its position expectation value therefore vanishes: $|{\varphi }_n(x){|}^2$ is even, while $x$ is odd. The first-order nondegenerate perturbation theory correction is consequentlyThe electric field does mix opposite-parity quantum states, and second-order nondegenerate perturbation theory contains two factors of $eEx$. Its leading contribution is thus proportional to $E^2$: this is a quadratic Stark effect. More generally, applying the parity operator changes $H(E)$ into $H(-E)$, so their corresponding energy eigenvalues agree. An analytic expansion about zero electric field contains only even powers of $E$. We assume $e|E|a$ is small compared with the unperturbed excitation scale.

Use zero-based labels to agree with ${\varphi }_0$ in the later parts. The normalized infinite square well energy eigenstates and energy eigenvalues areThe position matrix elements in a symmetric infinite square well, obtained by integrating the product of the two wavefunctions against $x$, areThe integral in the question uses positive integer labels; replacing them by $k+1,n+1$ is essential here. Only opposite-parity states, $k+n$ odd, contribute, as also follows from the parity operator selection rule.

Second-order nondegenerate perturbation theory now gives the requested infinite sum:In fully explicit form this isThe symbol $m$ here denotes the electron mass, not a summation index. For the ground state $n=0$, every denominator in the first form is negative, so the ground energy level moves downward. For excited states, contributions from lower and higher energy levels have opposite signs.

The thermal oscillator occupation is the Bose-Einstein distribution. With the rate notation ${\Gamma }_{\eta \sigma }$ for a transition from $\sigma$ to $\eta$, the master equation for the up population isAt thermal equilibrium, assuming ${\Gamma }_0>0$, vanishing time derivative is detailed balance between the two spin states. HenceorTogether with normalization, this gives $p_{\uparrow }=1/(1+e^{\beta \hslash \Omega })$ and $p_{\downarrow }=e^{\beta \hslash \Omega }/(1+e^{\beta \hslash \Omega })$, the Boltzmann distribution for the energy eigenvalues $\pm \hslash \Omega /2$. At zero temperature, absorption vanishes and emission relaxes the spin into its down ground state.

The nonzero-rate condition is necessary: if $\lambda =0$ or $g(\Omega )=0$, both rates vanish, so every initial pair of populations is stationary within this approximation. In that case stationarity alone cannot imply a thermal distribution. The equilibrium conclusion therefore presumes coupling to resonant oscillator modes.

Choose the local positive $\tilde{z}$ axis along the classical spin. For nonzero field, a proper rotation isIt preserves the Spin commutation relations and turns the field term into $-H_{f}S{\sum }_j{\tilde{S}}_j^z$. At zero field choose the classical direction $+x$, so $s=1$. Expanding about the original $z$ axis without this rotation would put the reference state away from the classical minimum and would not give a controlled small-deviation expansion.

To the required order,The longitudinal exchange gives $-J{S}^2+JS(n_j+n_{j+1})+O(S^0)$ on each bond. The transverse exchange givesFinally the field contributes its classical constant and $H_{f}S{n}_j$. Each site belongs to two bonds, so after removing the order-$S^2$ constant,These are the order-$S$ terms; the discarded products of number operators and the next square-root terms are order $S^0$ for occupations held fixed as $S\to \infty$. A different transverse-axis phase can reverse the sign of $B$ without changing the spectrum, but the sign above follows from the stated rotation and oscillator convention.

Let $N_c=c^{†}c$, $N_d=d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Substitution of the Bogoliubov transformation, with normal ordering using $d{d}^{†}=1-d^{†}d$ and $c{c}^{†}=1-c^{†}c$, givesFor example, the $v^{2}d{d}^{†}$ term in $a^{†}a$ contributes $v^2-v^2{N}_d$; these constants must be retained to obtain the ground-state energy. The two-orbital fermionic pairing Hamiltonian is consequentlyTo eliminate the pairing term, choose a canonical rotation of two fermionic modes satisfyingWith $u=\cos \theta$, $v=\sin \theta$, take $0<\theta <\pi /4$, so $\tan 2\theta =\Delta /\epsilon$. Setting $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$ givesThis positive choice fixes a convenient convention for the vacuum state in part (d). It yieldsBoth number operators independently have eigenvalues $0,1$. Hence the energy eigenvalues and state degeneracies areThere is one fermionic Bogoliubov quasiparticle of energy $E$ per occupied transformed orbital. The two Nambu spinor matrix eigenvalues $\pm E$ should not be mistaken for two independently occupiable levels of opposite energy; the normal ordering constant and the two positive quasiparticle excitation costs give the four physical energy eigenvalues above.

An independent diagonalization of a matrix checks both the pairing sign and the additive constant. In the ordered orthonormal basis $|0⟩,|11⟩$, the even-fermion parity block isbecause $ab|11⟩=-|0⟩$ and $b^{†}{a}^{†}|0⟩=-|11⟩$. Its characteristic polynomial is $x^2-2\epsilon x-{\Delta }^2$, giving $x=\epsilon \pm E$. The odd-fermion parity Fock states $a^{†}|0⟩,b^{†}|0⟩$ are unaffected by pairing and both have energy $\epsilon$.

Equivalently, with the Nambu spinor $\Psi =(a,b^{†}{)}^{\mathsf{T}}$,Replacing $\Psi$ by $R(c,d^{†}{)}^{\mathsf{T}}$ gives $E{c}^{†}c-Ed{d}^{†}+\epsilon =E(N_c+N_d)+\epsilon -E$, reproducing the result by normal ordering.

Neglect hyperfine structure, spin-orbit coupling and the diamagnetic term, and set $b={\mu }_{B}B$, $\epsilon =\alpha a_0^2$. The Zeeman energy is $b(L_{\hat{B}}+2S_{\hat{B}})$. The electron spin is free, so every spatial result below is duplicated with an added $\pm b$.

For a magnetic field parallel to $z$, the old magnetic quantum number $m$ remains good and the first-order energy levels areThe two $s$ orbitals have zero orbital angular momentum, so only their electron spin contributes at this order.

For a magnetic field parallel to $x$, the expectation of $L_x$ in each individual $p_m$ orbital vanishes. One must still check the degenerate $m=\pm 1$ subspace: its projected $L_x$ is zero as well. Thus there is no first-order orbital splitting in a sufficiently weak transverse magnetic field. Spin still contributes $\pm b$.

The transverse orbital Zeeman mixing can be seen without assuming its size. Put $E_0=-\mathrm{Ry}/4+3\epsilon$, $E_t=-\mathrm{Ry}/4+21\epsilon$, and $\Delta =E_t-E_0=18\epsilon$. In the real $p_x,p_y,p_z$ orbital basis, $p_x$ is unaffected while $p_y,p_z$ have off-diagonal matrix elements of modulus $b$. Diagonalizing this two-dimensional Hamiltonian gives the spatial energy eigenvaluesWhen $b\ll \Delta$, their shifts relative to the earlier levels are $0,+b^2/\Delta ,-b^2/\Delta$, with $\pm b$ added for electron spin. The displayed square-root expression also handles $b$ comparable to $\Delta$, provided coupling to other principal shells stays negligible. The $2s$ orbital retains zero orbital correction.

These quadratic orbital estimates keep the enhanced contribution from mixing the split $2p$ levels. Minimal magnetic minimal coupling also produces $e^2{B}^2{r}_{\perp }^2/(8m_e)$; its expectation value is positive and must be added for a full order-$B^2$ calculation. For example it gives $e^2{B}^2{a}_0^2/(4m_e)$ for $1s$. It is parametrically smaller than the $2p$ mixing term when $\epsilon \ll \mathrm{Ry}$.

Differentiating the specified rotating-frame amplitudes gives the time-independent Schrödinger equationThe characteristic polynomial is ${\lambda }^2-({\delta }^2+{\Omega }^2)/4=0$, henceFor $\Omega \ne 0$, corresponding normalized eigenvectors are proportional to $(\Omega ,2{\lambda }_{\pm }-\delta {)}^T$. Their arbitrary complex coefficients evolve as $e^{-i{\lambda }_{\pm }t}$; this solves the amplitude equations for any initial quantum state.

Restoring the common energy origin $\overline{E}=(E_1+\hslash \omega +E_2)/2$, the two dressed state energy eigenvalues areThus their positive energy separation increases byEquivalently, the branches connected to $E_1^{\prime }=E_1+\hslash \omega$ and $E_2$ shift by $+\hslash {\Omega }^2/(4\delta )$ and $-\hslash {\Omega }^2/(4\delta )$ respectively. Their signed difference $E_1^{\prime }-E_2$ changes by $\hslash {\Omega }^2/(2\delta )$, which is the printed expression with a signed convention. For negative Rabi detuning, an actual positive separation still increases; for example at $\delta =-10\Omega$ it cannot have the negative change implied by reading the printed expression as an unsigned gap. These opposite AC Stark shifts describe level repulsion.

Let the dimensionful angular momentum operators obey $\mathbf{F}=\mathbf{I}+\mathbf{J}$ and take the interaction Hamiltonian as $H^{\prime }=A\mathbf{I}\cdot \mathbf{J}$. Sincethe energy eigenvalues at fixed $I,J$ arewhere $F=|I-J|,\dots ,I+J$. Subtraction gives the Landé interval ruleMany atomic conventions use dimensionless angular momentum operators and call the energy-valued constant itself $A$; then the right-hand side is $AF$. Each energy level has $2F+1$ magnetic sublevels with the same energy in zero field.

For fine structure, the effective interaction is $a_{LS}\mathbf{L}\cdot \mathbf{S}$ with dimensionless angular momentum operators. It originates in the spin-orbit interaction, including the magnetic interaction of electron spin with orbital motion and its relativistic correction. Within an isolated LS coupling atomic term, the labels are the electron configuration, $L,S,J,M_J$, and the interval between adjacent $J$ energy levels is $a_{LS}J$. This is smaller than the electrostatic separation of atomic terms.

For magnetic-dipole hyperfine structure, the nuclear magnetic dipole moment interacts with the magnetic field generated by the electrons. Contributions include the Fermi contact interaction of penetrating $s$ electrons, orbital magnetic fields, and the dipolar field of electron spin. Projected onto one electronic $J$ energy level, their rotationally invariant magnetic-dipole part has the form $a_{IJ}\mathbf{I}\cdot \mathbf{J}$. The labels are $I,J,F,M_F$, together with the approximate electronic labels such as $L,S$. This splitting is generally smaller than fine structure, reflecting the much smaller nuclear magnetic dipole moment. An electric nuclear quadrupole moment can add a different interaction and violate the simple Landé interval rule.

At the larger electrostatic energy scale, an effective two-spin exchange interaction can also be written $a_s{\mathbf{s}}_1\cdot {\mathbf{s}}_2$ within fixed orbitals. Its origin is the Coulomb interaction combined with the Pauli exclusion principle, rather than a large direct magnetic interaction. It separates singlet state and triplet state atomic terms; before spin-orbit coupling, useful labels include the electron configuration, $L,M_L,S,M_S$. General electrostatic atomic term separations need not obey a single Landé interval rule as $L$ varies.

For a $2p$ atomic orbital, $l=1$. The radial factor obeysSubstitution into the radial Schrödinger equation for a central potential givesThus the energy eigenvalue isThe normalization constant cancels; $E_{\mathrm{h}}$ is the Hartree energy.

Expand the trial in a complete orthonormal eigenbasis of the Quantum Hamiltonian, $\tilde{\Psi }={\sum }_n{c}_n{\varphi }_n$, with $\hat{H}{\varphi }_n=E_n{\varphi }_n$. Orthonormality givesEvery summand in the numerator is nonnegative because $E_n\ge E_0$. This proves the quantum variational principle for the stated discrete expansion; the trial energy is a weighted mean of exact energy eigenvalues.

Insert the expansions into the energy eigenvalue equation $\hat{H}{\Psi }_i=E_i{\Psi }_i$. Through first order,The zeroth-order coefficients reproduce part (i). Equating the coefficients of $\lambda$ and moving them to one side givesTerms involving ${\hat{H}}^{(2)}$ first enter at order ${\lambda }^2$ and do not affect this equation of first-order nondegenerate perturbation theory.

In atomic units, $\hslash =m_e=e=4\pi {\epsilon }_0=1$, and the hydrogen atom hasThe allowed quantum numbers areHere the fixed electron spin quantum number is $s=1/2$. The energy eigenvalues are in hartrees; restoring units gives ${\hslash }^{2}l(l+1)$ and $\hslash m$ for the orbital operators and the corresponding factors for spin. The spin-independent hydrogen Hamiltonian does not split the two spin projections.

Within the well, the potential is zero and the Hamiltonian is purely kinetic:The infinite walls are encoded in the operator's boundary conditions $\Psi (0)=\Psi (a)=0$, rather than differentiated as part of the interior expression. Sincewe obtainThese positive energy eigenvalues grow quadratically with the quantum number; the ground state has a nonzero confinement energy.

Set $x=r\varphi$ and $a=2\pi r$. The wrapped spatial form isHence the differential expression for the quantum rotor acts asThe box expression becomes $E_n=n^2{\hslash }^2/(8m{r}^2)$, so consistency givesthe moment of inertia of a point mass a distance $r$ from the axis. Equivalently, ${\partial }_x=r^{-1}{\partial }_{\varphi }$ transforms the kinetic operator directly. Being an eigenfunction of this differential expression is not yet sufficient: the ring boundary conditions must also hold.

For a wavefunction normalized with the angular measure $\mathrm{d}\varphi$, use ${\Phi }_n(\varphi )=\sqrt{r}{\Psi }_n(r\varphi )={\pi }^{-1/2}\sin (n\varphi /2)$. This change of normalization convention does not change the energy eigenvalue.

Differentiate the Gaussian before substituting into the Hamiltonian:ThusAn energy eigenvalue must be independent of $x$, so the coefficient of $x^2$ vanishes. Normalizability requires $\alpha >0$, givingThe constant term is the quantum harmonic oscillator's zero-point energy; a Gaussian width ${\alpha }^{-1/2}$ balances its kinetic and potential contributions.