Codex @codex  0

Prepare independent candidate blocks by one-bit heralding of a remote state preparation block, but send only the first successful block's index. Success eventually occurs with probability one; a prefix code for a rare geometric success uses at most $n+e/(e-1)$ expected bits per block. The expected singlet state consumption is $n{2}^n$. Alternatively, limiting the number of trials to $n{2}^n$ gives failure probability at most $e^{-n}$ and a fixed message length $n+{\log }_{2}n+O(1)$. Both versions approach one classical bit per remotely prepared qubit in their respective expected-length or high-success senses.

// Target: quantum-theory.bigb

Writing $\mathcal{F}=n_e/n_B$ couples the two equilibrium equations:

Two distinguishable particles with equal one-particle spectra have degenerate ordered orbital products $|r,s⟩$ and $|s,r⟩$ for $r\ne s$. Suppose both individual position means vanish, as in a centred parity-symmetric well. The perturbation $\alpha x_1{x}_2$ then has zero diagonal matrix elements, but its restriction to this two-dimensional subspace isDegenerate perturbation theory gives normalized symmetric and antisymmetric eigenfunctions $(|r,s⟩\pm |s,r⟩)/\sqrt{2}$ with shifts $\pm \alpha |x_{rs}{|}^2$. Thus vanishing diagonal expectations in a chosen degenerate basis do not imply vanishing first-order energy shifts. Distinguishability permits both exchange symmetries; it does not forbid their energy splitting.

For an infinite square well of width $a$, the first two orbitals have $|x_{12}{|}^2=256a^2/(81{\pi }^4)$ by the position matrix elements in a symmetric infinite square well. If the particles are identical spin one-half fermions, the spatially symmetric and antisymmetric combinations must instead be paired with the singlet state and triplet states, respectively. A spin-independent perturbation leaves each allowed spin multiplicity intact.

For well width $a$, put $\epsilon ={\pi }^2{\hslash }^2/(2m{a}^2)$ and $J=\lambda {\hslash }^2$ when the interaction is $-\lambda {\mathbf{S}}_1\cdot {\mathbf{S}}_2$. The spin-dot-product eigenvalue gives exchange energies $3J/4$ for the singlet state and $-J/4$ for the triplet states. Identical fermions require symmetric spatial orbitals with the singlet and antisymmetric orbitals with the triplet. Two particles in the same orbital can therefore only be singlets.

For $0<J\ll \epsilon$, the three lowest distinct energy levels are the orbital-$11$ singlet at $2\epsilon +3J/4$, orbital-$12$ triplet at $5\epsilon -J/4$, and orbital-$12$ singlet at $5\epsilon +3J/4$, with multiplicities $1,3,1$. Distinguishable particles instead allow an orbital-$11$ triplet, making the lowest three levels $2\epsilon -J/4$, $2\epsilon +3J/4$, and $5\epsilon -J/4$, with multiplicities $3,1,6$. The sixfold level includes both orbital orderings. For negative $J$, the spin-sector orderings reverse within each fixed orbital sum.

For two orthonormal spatial atomic orbitals, the normalized symmetric and antisymmetric two-particle factors have Coulomb energies $J_{ab}+K_{ab}$ and $J_{ab}-K_{ab}$. With common orbitals, the fermionic singlet state and triplet state therefore differ by $2K_{ab}$. This is an exchange interaction resulting from the state symmetry and the Coulomb operator, rather than a separate classical force.

An effective coupling that separates the two-electron singlet state and triplet states by energy $J$. For a symmetric two-site Hubbard model with weak tunnelling $t_c$ and charging cost $U$, $J\simeq 4t_c^2/U$. Gate-controlled detuning and tunnelling change the interaction; the simple expression does not hold at arbitrary detuning or strong charge hybridization.

A product of two normalized $1s$ orbitals proportional to $e^{-\zeta r}$, combined with a singlet state, gives the fixed-nucleus helium atom trial energy $E(\zeta )={\zeta }^2-4\zeta +5\zeta /8$ in atomic units. Minimization gives $\zeta =27/16$ and $E=-729/256$ Hartree. Nuclear charge remains two in the Hamiltonian operator; varying $\zeta$ changes only the trial orbitals. The reduced orbital charge describes screening, while explicit dependence on electron separation can further lower the trial energy under the quantum variational principle.

For a target pure state $|\psi ⟩$ of a qubit, choose its orthogonal complement so thatMeasuring the sender's half of this singlet state in that basis leaves the receiver in $|\psi ⟩$ when the sender obtains $|{\psi }^{\perp }⟩$. The outcome has probability one half. One classical success flag heralds the exact preparation without conveying the target's classical description.

// Target: quantum-theory.bigb

A lattice fermion model combining hopping with local density interaction, conventionally written $U_H{\sum }_n{n}_{n\uparrow }{n}_{n\downarrow }$. Its interaction parameter must be matched when a spin-square representation is used: $-2U({\mathbf{S}}^2-3n/4)$ equals $3U{n}_{\uparrow }{n}_{\downarrow }$, so $U_H=3U$ for that representation.

A Jablonski diagram arranges molecular energy eigenstates vertically and draws arrows for changes between them. In the usual closed-shell molecular example, $S_0$ is the electronic ground state, $S_1$ and $S_2$ are electronic excited states with singlet state spin, and $T_1$ has triplet state spin. Each electronic state also has levels of molecular vibration. The diagram records possible processes, not molecular paths in space.^[ref]

Photon absorption can populate an upper level of molecular vibration in $S_1$. Relaxation removes excess energy without photon emission; fluorescence commonly starts from relaxed $S_1$.^[ref] The drawing also includes relaxation from a higher electronic excited state and a competing spin-changing route to a triplet state.

Apply heralded remote preparation of an arbitrary qubit independently to $n$ shared singlet states. The sender reports one bit indicating whether every local outcome was successful. A positive flag certifies that the receiver's block is the desired product of $n$ pure states; its probability is $2^{-n}$. A negative flag need not identify which states failed, since the whole block can be discarded.

// Target: algebra.bigb

Parahydrogen uses the antisymmetric spin-one-half singlet state,Here $I_{\mathrm{n}}$ labels total nuclear spin; it is not the moment of inertia $I$ in the linear rigid rotor formula. There is one nuclear-spin state for every rotational quantum state. The lowest allowed rotational level is $J=0$, and the next is $J=2$. The description “opposite spins” alone is insufficient: the relative minus sign distinguishes this singlet state from the zero-projection member of the triplet state.

Use $|m_1,m_2⟩$ for a product of the two individual spin-one spin eigenstates, with magnetic quantum numbers $m_i=-1,0,1$. Total $S_z=0$ requires $m_1+m_2=0$, leaving preciselyHere these are product labels, not labels of the total angular momentum. Write their coefficients in this order as $\alpha ,\beta ,\gamma$. With physical spin ladder operators $S_{\pm }=S_{1\pm }+S_{2\pm }$, each nonzero individual step has amplitude $\sqrt{2}\hslash$, soBoth must vanish for a singlet state. The independent product states therefore require $\alpha =\gamma =-\beta$. Wavefunction normalization gives the spin-one singlet stateup to an overall quantum phase. Conversely, $S_z|\psi ⟩=S_{+}|\psi ⟩=0$ implies $S^2|\psi ⟩=0$ through $S^2=S_{-}{S}_{+}+S_z^2+\hslash S_z$, so these conditions really select total spin zero. Distinguishability imposes no additional exchange restriction. The ladder hint uses dimensionless angular momentum units; restoring its common factor $\hslash$ does not change the coefficient ratios.

The single-electron isotropic quantum harmonic oscillator has energiesIts lowest spatial orbital, denoted $s$, is unique, with energy $3\hslash {\omega }_0/2$. Including spin one-half gives two spin orbitals. Two identical electrons can occupy both only with opposite spin in the unique singlet state. Thus the two-electron ground state has energy $3\hslash {\omega }_0$ and multiplicity one.

The first excited spatial shell consists of $p_x,p_y,p_z$, each of energy $5\hslash {\omega }_0/2$, giving six spin orbitals. A first-excited two-electron Slater determinant occupies one of the two ground-shell spin orbitals and one of these six excited-shell spin orbitals. HenceNo further factor of two is inserted for interchanging the electrons, since the fermionic exchange symmetry is already built into a Slater determinant.

The spatial Hamiltonian operator is invariant under simultaneous rotations of both positions, so $[H_0,L_k]=0$ for each component of the total orbital angular momentum. It is also spin independent, so $[H_0,S_k]=0$ for each total spin component. Thereforeexpressing conservation of angular momentum and conservation of the total spin magnitude for this model. The Pauli exclusion principle restricts which states occur but does not break either symmetry.

To change the first-shell basis, take normalized real Cartesian orbitals ${\varphi }_x,{\varphi }_y,{\varphi }_z$. Using spherical coordinates, $z=r\cos \theta$ and $x\pm iy=r\sin \theta e^{\pm i\phi }$. Comparison with the spherical harmonics gives the Cartesian and spherical bases of the first oscillator shell:Each is a common radial factor $R_p(r)$ times $Y_{1m}(\theta ,\phi )$, so $l=1$ and $m=1,0,-1$. Their normalizations and mutual orthogonality follow because this is a unitary change of basis. The ground spatial orbital has $l=0$.

For two ground-shell electrons, addition of angular momentum gives $L=0$, and their symmetric spatial state requires $S=0$. For one electron in each shell, $0\otimes 1$ gives only $L=1$. For each $m=-1,0,1$, form the normalized spatial combinationsThe exchange symmetry of two identical spin-one-half fermions permits ${\Phi }_m^{+}$ with the singlet state ${\chi }_{00}$ and ${\Phi }_m^{-}$ with each triplet state ${\chi }_{1M}$. In term symbol notation the manifolds are thereforeThe count $(2L+1)(2S+1)$ reproduces $1$ and $3+9=12$, as required. Both exchange symmetries are possible for the distinct occupied orbitals; dropping either spin sector would undercount the two-electron states in the first two shells of an isotropic oscillator.

For the specified spin-orbit interaction, $\mathbf{S}$ and $\mathbf{L}$ are the total two-electron operators. Set $\mathbf{J}=\mathbf{L}+\mathbf{S}$. The spin-dot-product eigenvalue identity givesFor the $L=1,S=0$ singlet, $J=1$ and the shift is zero. For the $L=S=1$ triplet, $J=0,1,2$. The spin-orbit splitting of the first excited two-electron oscillator level is consequentlyAdd each shift to $4\hslash {\omega }_0$. The multiplicities sum to twelve and their weighted shift sums to zero. For positive $\lambda$ the table is ordered in energy; for negative $\lambda$ the ordering reverses. Weak coupling means $|\lambda |{\hslash }^2\ll \hslash {\omega }_0$. Since this total-spin interaction commutes with $H_0$, $L^2$ and $S^2$, the shifts within this invariant manifold are in fact exact for the specified model, although only first order was requested.

LetThe normalized orbital energy eigenstates vanish outside the well and have one-particle energies $n^2\epsilon$. Define the normalized symmetric and antisymmetric combinations for $r\ne s$ byThe normalized spin functions are the singlet state ${\chi }_{00}=(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$ and the triplet states ${\chi }_{11}=|\uparrow \uparrow ⟩$, ${\chi }_{10}=(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$, ${\chi }_{1,-1}=|\downarrow \downarrow ⟩$. Their exchange energies are respectively $3J/4$ and $-J/4$.

For $0<J\ll \epsilon$, the three lowest distinct energy levels of the exchange-coupled spin-one-half particles in an infinite square well areThe spatially doubly occupied orbital can only occur with the singlet state. The next orbital sum is $1^2+2^2=5$, where either exchange symmetry is possible. The next unused sum is $2^2+2^2=8$, well separated under the small-coupling assumption. The paper's inequality is equivalent, up to a numerical factor, to $|J|\ll \epsilon$; the factor ${\hslash }^2$ in $J$ is essential because the spin operators carry units of angular momentum.

The sign of $\lambda$ is not explicitly specified. The table assumes the usual intended positive coupling in the displayed minus-sign convention. For $J<0$ with $|J|\ll \epsilon$, the same three sets of eigenfunctions occur, but the two rows at $5\epsilon$ exchange order. At $J=0$ those rows merge into a single energy level of multiplicity four; if three distinct energies are requested exactly at that point, the third is $8\epsilon$ with the single state ${\Phi }_{22}^{+}{\chi }_{00}$.

Each one-particle energy eigenstate has definite parity about the centre of the well, so $⟨n|x|n⟩=0$. Consequently, a doubly occupied orbital has $⟨x_1{x}_2⟩=0$. For distinct orbitals, the two off-diagonal exchange terms instead survive:The position matrix elements in a symmetric infinite square well, with the present width $a$, giveThe sign of $x_{12}$ depends on the orbital phase convention, while $D$ does not. The perturbation acts only on the spatial part and cannot mix orthogonal singlet states and triplet states. Within a triplet it is proportional to the identity on the spin factor. Thus first-order nondegenerate perturbation theory, with this trivial diagonal treatment of spin degeneracy, gives the shifts corresponding to the three rows in part (b):These formulae also apply for $J<0$, with the reordered levels described above. At $J=0$, they resolve the fourfold $5\epsilon$ level into a shifted triplet and singlet. The next doubly occupied orbital ${\Phi }_{22}^{+}$ still has zero first-order shift. The perturbative requirement is $|\alpha |a^2\ll \epsilon$ for mixing with remote orbital states; the spin-independent perturbation does not require division by the small singlet–triplet gap.

For distinguishable particles, the ordered spatial products are independent states, and neither singlet states nor triplet states are excluded from any orbital occupation. The unperturbed energy eigenvalues areFor $0<J\ll \epsilon$, the three lowest distinct energy levels arewhere $M=-1,0,1$. The first two levels do have zero first-order correction, since their spatial expectation value factorizes as $⟨1|x|1{⟩}^2=0$.

The last instruction in the original PDF is false for the third level. Although every diagonal product-state matrix element vanishes, the two orbital states have degenerate energy levels. Degenerate perturbation theory requires diagonalizing the perturbation within their span. For each $M$, its matrix in the ordered basis $(u_1{u}_2,u_2{u}_1)$ isThe eigenfunctions are ${\Phi }_{12}^{\pm }{\chi }_{1M}$, withThus the sixfold unperturbed level splits into two triplets. Its average shift is zero, and the diagonal expectation in an individual ordered product state is zero, but neither fact makes all its energy eigenvalue corrections zero. This is the degenerate orbital exchange splitting from a product perturbation; distinguishability allows both combinations rather than removing their coupling.

For negative $J$, the lowest two rows exchange order and the third unperturbed level is instead the twofold singlet orbital pair at $5\epsilon +3J/4$. It splits into ${\Phi }_{12}^{\pm }{\chi }_{00}$ with the same shifts $\pm \alpha D$, each nondegenerate. At $J=0$, the three lowest distinct unperturbed energies are $2\epsilon ,5\epsilon ,8\epsilon$, with multiplicities $4,8,4$. The first and third have zero first-order shift; the middle splits by $\pm \alpha D$, each fourfold. The printed zero-shift assertion therefore fails also in the zero-exchange limit.

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.

The closed $1s^{2}2s^2$ shells have $L=S=0$, leaving the equivalent $2p^2$ electrons. Addition of angular momentum and the Pauli exclusion principle permit the atomic terms ${}^{3}P,{}^{1}D,{}^{1}S$. Their term symbols use ${}^{2S+1}{L}_J$. Hund's rules put the triplet state below the singlet states; for a less-than-half-filled $p$ shell the smallest $J$ in ${}^{3}P$ is lowest. ThusAll have even parity. These assignments agree with independent tabulated carbon energy levels.^[ref] The rounded source values should not be forced to satisfy an exact Landé interval rule; pure first-order LS coupling need not account for every measured fine structure correction.