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The quantum variational principle states that the energy expectation of an admissible trial wavefunction is at least the ground state energy:For the discrete, complete orthonormal basis assumed here, expand in Hamiltonian operator eigenfunctions, $f={\sum }_n{c}_n{\varphi }_n$, with $\hat{H}{\varphi }_n=E_n{\varphi }_n$ and $E_n\ge E_0$. ThenEvery summand is nonnegative. Equality holds precisely when all occupied components lie in the ground eigenspace. Thus minimizing trial energies supplies upper bounds on $E_0$, with exact equality if the trial family contains a ground state.

The $3d$ and $4s$ atomic orbitals of chromium are close enough in energy that electron-electron terms can overturn the simple filling order. The configuration $[\mathrm{Ar}]3d^{5}4s^1$ permits five singly occupied $3d$ orbitals and a parallel-spin $4s$ electron. Relative to $3d^{4}4s^2$, it removes the $4s$ pair and permits more parallel-spin exchange interactions, lowering the Coulomb-and-exchange contribution enough to offset the orbital promotion cost.

As qualitative bookkeeping, six parallel electron spins give $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ same-spin pairs, compared with $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ in high-spin $3d^{4}4s^2$. The individual exchange integrals are unequal, so this is an explanation of the tendency, rather than a quantitative atomic energy calculation. The observed ground state configuration is $\boxed{[\mathrm{Ar}]3d^{5}4s^1}$.^[ref]

Relative to the ground state, the quantum harmonic oscillator levels are $vhc\tilde{\omega }$. Put ${\theta }_v=hc\tilde{\omega }/k$. Summing the geometric series gives the vibrational partition function $q=(1-e^{-{\theta }_v/T}{)}^{-1}$. Therefore the vibrational internal energy isThe omitted zero-point energy would add $Nhc\tilde{\omega }/2$ if absolute oscillator energies were used.

Under the stated vibrational-ground-state approximation and a nondegenerate electronic ground state, only translation and rotation contribute. ThereforeA temperature-independent ground-state energy, including zero-point energy, contributes no entropy. This estimate deliberately omits thermal vibrational excitation; it need not equal a tabulated value that includes all modes.

The octahedral splitting ${\Delta }_o$ separates the metal $t_{2g}$ and $e_g$ orbital sets. The Racah parameter $B$ characterizes interelectronic repulsion and hence the spacing of many-electron d-shell terms; it is not another name for ${\Delta }_o$. A Tanabe-Sugano diagram plots term excitation energies $E/B$ against ${\Delta }_o/B$, for specified values of the other repulsion-parameter ratios. Each ordinate is measured relative to whichever term is the ground state at that field strength. The original many-electron treatment diagonalizes the relevant d-shell energy matrices rather than identifying every absorption with a one-electron $t_{2g}\to e_g$ gap.^[ref]

The ground state changes from high-spin $t_{2g}^3{e}_g^2$, ${}^6{A}_{1g}$, to low-spin $t_{2g}^5$, ${}^2{T}_{2g}$. The larger splitting eventually compensates the additional pairing cost. Because all plotted excitation energies are re-referenced to the new ground state, the branches change slope at the crossover. The break is not an infinite physical excitation energy or a discontinuity of the underlying Hamiltonian.

The fixed localized determinant is unsuitable near equilibrium. It does not develop the correct shared bonding orbital and is not a pure singlet: for one alpha and one beta orbital, $⟨S^2⟩/{\hslash }^2=1-|⟨A|B⟩{|}^2$, which is nonzero unless the occupied spatial orbitals coincide. At dissociation it is an equal singlet/triplet mixture, though their energies become degenerate there. Near equilibrium the physical ground state is a bonding singlet. An optimized UHF calculation can recover the restricted, spin-pure solution by making its two occupied spatial orbitals equal; that does not rescue the particular fixed atomic-orbital determinant written here. A spin-adapted multideterminant description is needed to combine good bonding near equilibrium with correct singlet dissociation.

The midpoint-odd perturbation couples the ground state only to even-index states. Direct integration givesInsertion in the second-order expression yieldsIt is negative because all coupled states lie above the ground state. This is the coefficient of ${\lambda }^2$; the physical energy shift includes the additional factor $q^2{\epsilon }^2$.

An example is the doubly occupied antibonding configuration $({\sigma }_u{)}^2$, rather than the bonding $({\sigma }_g{)}^2$ configuration dominant near equilibrium. It can be a self-consistent stationary determinant because symmetry makes the corresponding occupied-virtual Fock mixing vanish, while its bonding energy and nodal structure are inappropriate for the ground state. Solving the Hartree-Fock stationarity equations does not impose ground-state occupancy automatically.

The stationary determinant should be stable against orbital rotations: its orbital-energy Hessian must have no negative direction. A suitable local minimum, preferably the lowest accessible determinant of the appropriate symmetry and spin, can approximate the ground state; a stationary excited determinant or saddle generally cannot. Stability is still not a guarantee of small electron-correlation error.

For fixed electron-electron interaction and electron number, a nondegenerate ground-state electron density determines the external potential up to an additive constant, and hence determines the Hamiltonian and ground-state observables.

Suppose two potentials $v,v^{\prime }$ differing by more than a constant have different ground states $\Psi ,{\Psi }^{\prime }$ but the same density $\rho$. Strict Rayleigh-Ritz variational principle inequalities giveAdding produces the contradiction $E+E^{\prime }<E+E^{\prime }$. This proves the first Hohenberg-Kohn theorem under the stated nondegeneracy assumptions. A constant potential shift changes energy by $N$ times that constant without changing the density.^[ref]

A frustrated triangle has two ferromagnetic metal–radical bonds and an antiferromagnetic metal–metal bond. With the Hamiltonian convention above, take $J/k_B=+800$ K and $J^{\prime }/k_B=-8$ K. In the maximally aligned ground state the two metal–radical products are $1\times 1/2$ and the metal–metal product is $1\times 1$. ThereforeHere $F_C=-1$ denotes complete satisfaction of all exchange interactions; the small deviation $1+F_C=0.0198$ quantifies the weak frustrated contribution in this normalization. The numerical answer assumes antiferromagnetic metal–metal exchange. The printed magnitudes alone do not establish that sign. If the metal–metal exchange were instead ferromagnetic, all three bonds would be satisfied and $F_C=-1$. The ground-state scalar products also follow directly from $S_{12}=2$, $S=5/2$, so a separate classical approximation is unnecessary for this fully polarized multiplet.

Use $C=0.125g^{2}S(S+1)$ for one ion per formula unit, with high-spin $S=5/2,2,3/2,1$ for Mn, Fe, Co, Ni. ThenHigh-spin Mn(II), with an approximately orbitally nondegenerate $d^5$ ground state, stays closest to the free-electron value. Fe, Co and Ni have stronger orbital admixture, particularly Co in this effective Curie-law comparison. Spin-orbit coupling creates magnetic anisotropy, which can make reversal of the weak-ferromagnetic domains slower and enhance field-cooled/zero-field-cooled separation. A field-cooled sample selects domains as it orders; a zero-field-cooled sample initially has competing domains and can remain pinned when a small measuring field is applied.

This is a qualitative rationalization, not a strict proportionality between $g-2$ and the magnetization separation. The anisotropy tensor, domain-wall pinning and experimental timescale also matter; scalar $g$ values alone do not quantitatively fix the divergence.

For competing exchanges on a body-centred cubic antiferromagnet, use the Heisenberg model convention $H={\sum }_{⟨ij⟩}{J}_{ij}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$, with each bond counted once, dimensionless spins, and positive antiferromagnetic $J$. There are eight nearest and six next-nearest neighbours. Bond counting in the two classical collinear structures givesFor type II the eight $J_1$ contributions cancel, while all six $J_2$ bonds are antiparallel. Their crossing yields $c=2/3$. One can also exclude a lower spiral within the classical bilinear model: writing $u_i=\cos (q_{i}a/2)$ givesFor a favourable sign of the product, $u_x^2+u_y^2+u_z^2\ge 3|u_x{u}_y{u}_z{|}^{2/3}\ge 3|u_x{u}_y{u}_z|$. Hence when $J_2\ge 2J_1/3$ its minimum is at $u_x=u_y=u_z=0$, the type-II ordering vectors. When $J_2\le 2J_1/3$, compare with the type-I value and use $3-\sum u_i^2\le 3(1-|u_x{u}_y{u}_z|)$ to obtain the complementary bound. ThusThis is a classical/mean-field energy comparison; an exact quantum ground state is not generally a collinear product state.

For type I a magnetic cell is the simple cube of side $a$, with opposite moments at $0$ and $a(1,1,1)/2$. Its magnetic structure factor is proportional to $1-e^{i\pi (h+k+l)}$. Allowed magnetic reflections have integer $h,k,l$ with $h+k+l$ odd. The first two families are $(100)$ and $(111)$, with symmetry-equivalent members understood.

For type II choose magnetic primitive translations $a(1,1,0)$, $a(1,0,1)$ and $a(0,1,1)$, enclosing four spins in volume $2a^3$. Equivalently, a side-$2a$ cube is a convenient nonprimitive magnetic cell. On each original simple-cubic sublattice the moment at $a\mathbf{n}$ is proportional to $(-1{)}^{n_x+n_y+n_z}$. Accordingly,The translated body-centre basis gives $1+e^{i\pi (h+k+l)}=1\pm i$, so there is no further extinction. The first two families are $(1/2,1/2,1/2)$ and $(3/2,1/2,1/2)$. The neutron polarization factor may additionally suppress a reflection for a particular moment direction; the stated rules are the positional selection rules.

Each sublattice has number density $n_A=1/a^3$ and saturation magnetization $M_s=g{\mu }_{B}S/a^3$. Replacing neighbouring spins by their averages makes their exchange energy equivalent to a molecular magnetic field. Its minus sign expresses antiferromagnetic exchange, not an externally applied field. Since $⟨S_B⟩=M_B/(n_{A}g{\mu }_B)$,For $M_B=-M_A$, $B_{\mathrm{mf}}=({\lambda }_{AB}-{\lambda }_{AA})M_A$. Linearizing the Brillouin magnetization formula givesThe g-factor cancels because exchange is specified in energy units. A convention with $2J{\mathbf{S}}_i\cdot {\mathbf{S}}_j$ doubles these exchange energies and $T_N$.

Let $m=M_A/M_s$, $A=(S+1)/(3S)$ and let $b_B>0$ denote the cubic coefficient of the Brillouin magnetization formula. The argument is $y=(T_N/T)m/A$. Thus $m=Ay-b_B{y}^3$ impliesConsequently $\boxed{\beta =1/2}$ in mean-field theory. The positive branch falls from saturation at zero temperature to zero at $T_N$ and remains zero above it. An elastic magnetic neutron-reflection intensity scales as $I\propto M_A^2$; a log–log fit against $T_N-T$ has slope $2\beta$. Background subtraction and a suitable critical interval are needed. Fluctuations can change the measured exponent from its mean-field value; uniform bulk magnetization vanishes for a compensated antiferromagnet and is not itself this order parameter.

For $J>0$ each bond minimizes its energy when its two Ising spins are opposite, giving bond energy $-J$. An even ring allows every bond to do this simultaneously, including the closing bond required by the periodic boundary conditions. Its two ground states areThe two configurations are related by spin inversion symmetry. There is no geometrical frustration.

Put $r=\tanh K$, so $r\to 1^{-}$ as the temperature tends to zero. For even $N$, the finite antiferromagnetic Ising-ring correlation becomesThe two perfectly alternating ground states give exactly the same product $S_1{S}_{n+1}=(-1{)}^n$; averaging them preserves its magnitude even though the one-point correlation function remains zero.

For odd $N$ the signs instead giveBoth numerator and denominator vanish. Set $r=1-\delta$ and expand at fixed $N,n$:HenceThere is a direct ground state interpretation. The mandatory parallel bond of an odd antiferromagnetic Ising ring is uniformly distributed among its $N$ bonds in the limiting canonical ensemble. Along the directed $n$-bond arc from site 1 to site $n+1$, the product is $(-1{)}^n$ if the defect lies outside that arc, and $-(-1{)}^n$ if it lies inside. The respective probabilities are $1-n/N$ and $n/N$, giving the displayed two-point correlation function. Global spin inversion symmetry doubles the number of ground states but leaves each product unchanged.

The reduction on an odd ring survives at separations comparable with $N$, despite the vanishing defect density $1/N$. For fixed $n$, however, taking $N\to \infty$ after the zero-temperature limit recovers $(-1{)}^n$ for either parity, consistently with the diverging correlation length. The linear envelope at finite odd $N$ comes from one freely located frustration defect; it is distinct from the exponential decay caused by thermal defects in the thermodynamic limit.

Write $Q=c^{†}c+d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Expanding the number operators and restoring the displayed order with the canonical anticommutation relations givesThe minus sign in the first line follows from $c^{†}{d}^{†}+dc=-(cd+d^{†}{c}^{†})$. Substitution into the Hamiltonian operator yieldsLet $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$. A convenient real Bogoliubov transformation hasThese obey $u^2+v^2=1$, $u^2-v^2=\epsilon /E$ and $2uv=\Delta /E$. Equivalently $u=\cos \theta$, $v=\sin \theta$ with $0<\theta <\pi /4$ and $\tan (2\theta )=\Delta /\epsilon$. The coefficient of $P$ vanishes, while the coefficient of $Q$ is $E$. The constant isThus the diagonalized two-mode fermion pairing Hamiltonian isThe two transformed fermionic occupation numbers independently take values zero or one. The energy eigenvalues and their eigenspace dimensions are thereforeThe positive quasiparticle energy $E$ makes the transformed vacuum state the unique ground state. The constant $\mu$ is essential for the correct ground-state energy.

As a separate sign and spectrum check, use the original fermionic Fock space basis. Both singly occupied states are energy eigenstates with energy $\epsilon$, since the pairing terms annihilate them. In the ordered even-fermion parity basis $(|0⟩,a^{†}{b}^{†}|0⟩)$,The off-diagonal signs follow from $b^{†}{a}^{†}=-a^{†}{b}^{†}$ and $ab{a}^{†}{b}^{†}|0⟩=-|0⟩$. Its characteristic polynomial is $z^2-2\epsilon z-{\Delta }^2$, giving the remaining two eigenvalues $\epsilon \pm E$, in agreement with the Bogoliubov transformation.

A tall, broad central potential barrier has $\varphi \gg 1$. Let $\sigma =+1$ denote even parity and $\sigma =-1$ odd parity, and write $\theta =(n+\frac{1}{2})\pi +{\delta }_{\sigma }$, where $n=0,1,\dots$. Since $\tan [(n+\frac{1}{2})\pi +\delta ]=-\cot \delta$, the condition in part (d) givesThus the leading correction to the isolated-well Bohr-Sommerfeld quantization condition isThis is an implicit condition, with all action integrals evaluated at the corrected energy. To convert it into an explicit leading energy correction, let $E_n^{(0)}$ solve $\theta (E_n^{(0)})=(n+\frac{1}{2})\pi$ and define the full classical oscillation period in one well by $\mathcal{T}(E)=2m{\int }_{x_1}^{x_2}dx/p(x)$. Differentiation of the action integral has no endpoint terms because $p(x_1)=p(x_2)=0$, and givesConsequently, to leading exponentially small order within the WKB approximation,The even member lies below its odd partner. For an approximately harmonic oscillator well, $\mathcal{T}\simeq 2\pi /\omega$, giving the familiar double-well tunneling splitting $\Delta E\simeq (\hslash \omega /\pi )e^{-\varphi }$. The explicit energy formula retains only its leading exponential correction; varying $\varphi$ and ${\theta }^{\prime }$ with the shifted energy also contributes at higher orders.

The ground state has an even, everywhere same-sign wavefunction with a peak in each well and a small nonzero value beneath the central potential barrier. Its derivative vanishes at zero. The first excited state is odd, has opposite-sign lobes in the two wells, and has exactly one interior node of a wavefunction, at zero. Both wavefunctions decay outside the wells. In terms of localized normalized wavefunctions ${\phi }_L,{\phi }_R$, with overlap $s=⟨{\phi }_L|{\phi }_R⟩$, their qualitative shapes are the lowest parity states of a symmetric double well:The overall sign of the odd wavefunction is arbitrary. The sketch uses localized Gaussian profiles to show the parity and the central node of a wavefunction, rather than claiming exact energy eigenfunctions for an unspecified potential energy.

Treat $|a⟩=|0⟩$ and $|b⟩=|1⟩$ as an isolated ground-state hyperfine qubit. Their transition angular frequency is ${\omega }_0=2\pi \times 320.7\mathrm{MHz}$. The radiofrequency magnetic field couples them through an allowed magnetic dipole transition. After choosing its phase reference, take the two-level laboratory Hamiltonian operator to bewhere $X,Z$ are Pauli operators. Here $\Omega$ is defined by the on-resonance rotating-frame off-diagonal matrix element $\hslash \Omega /2$.

Use a quantum rotating frame and the rotating-wave approximation, valid for $\Omega ,|\delta |\ll {\omega }_0$, with Rabi detuning $\delta ={\omega }_0-\omega$. Ignoring off-resonant coupling to other levels givesAt resonance $\delta =0$, the unitary time evolution isThus the Born rule gives coherent Rabi oscillations,The stated amplitudes are in the rotating frame; the laboratory carrier phases are tracked by the drive reference.

These logical states are orthogonal, individually resolvable hyperfine levels in the same electronic ground state. They avoid fast optical decay, can be coherently coupled by radiofrequency control, and have a state-selective optical readout through $|c⟩$. The excited level $|c⟩$ is useful for fluorescence detection rather than for the storage superposition in this model. Long lifetimes alone do not guarantee a useful qubit: control isolation and adequate quantum coherence are also required.

A good atomic qubit needs two reproducible, orthogonal internal states whose populations and relative quantum phase survive for many gate times. Both should be long lived, with weak spontaneous emission and little decoherence from fluctuating fields or atomic motion. Their transition must nevertheless permit fast, selective, coherent control with little leakage to other states. Reliable initialization and state-selective quantum measurement are also required; optical pumping and fluorescence readout can supply these functions.

For rubidium-87, use the two ground state hyperfine levels in the $5{}^2{S}_{1/2}$ electronic configuration. The electronic angular momentum is $J=1/2$, and addition of angular momentum to the given nuclear spin $I=3/2$ gives $F=|I-J|,\dots ,I+J=1,2$. A suitable Rubidium-87 clock qubit isThese are coupled electron-nucleus angular momentum eigenstates, rather than two bare nuclear-spin projections. Their hyperfine structure supplies a resolvable splitting ${\nu }_{01}\simeq 6.835\mathrm{GHz}$. Both belong to the electronic ground state, avoiding a short optical excited-state lifetime. Since the weak-field Zeeman shift is proportional to $g_F{m}_{F}B$, both have zero linear shift at $B=0$: their rubidium-87 clock transition suppresses first-order magnetic-field dephasing near zero field. A quadratic Zeeman shift of a clock transition and differential optical shifts remain. A small stable bias field can define the quantization axis and help resolve unwanted transitions, at the cost of some residual magnetic sensitivity.^[ref]

Drive the magnetic dipole transition with a resonant microwave field of suitable polarization, here $\Delta m_F=0$, or use two phase-coherent optical fields for a far-detuned stimulated Raman transition with their frequency difference resonant with ${\nu }_{01}$. The optical excited state is then only virtually populated, reducing spontaneous emission. In the rotating frame and rotating-wave approximation, define the drive phase $\varphi$ by the effective HamiltonianHere $X,Y,Z$ are the Pauli matrices on the chosen qubit, and $\Omega$ is the Rabi angular frequency. Pulse duration controls the rotation gate angle, while phase selects the axis in the equatorial plane of the Bloch sphere. A $\pi$ pulse exchanges the populations; a $\pi /2$ pulse makes an equal quantum superposition from either basis state. Controlled Rabi detuning, a differential energy shift, or compositions of these pulses give $z$ rotations. For example, $R_z(\alpha )R_y(\beta )R_z(\gamma )$ realizes any one-qubit unitary operator up to a global phase. Thus the atomic encoding admits arbitrary single-qubit quantum gates.