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The canonical commutation relations imply $[N,{\alpha }_{-k}^i]=k{\alpha }_{-k}^i$ for $N={\sum }_{k>0}{\alpha }_{-k}\cdot {\alpha }_k$. With zero level assigned to the oscillator vacuum, a Fock space basis state containing $r_{ki}$ creation operators of each mode has level ${\sum }_{k,i}k{r}_{ki}$. The level is therefore a nonnegative integer; it counts oscillator excitation weighted by frequency, not simply the number of creation operators.

The Lamb-Dicke regime means the wave-induced phase change across the occupied motional wavepacket is small. For one normal mode, let $x_0=\sqrt{\hslash /(2m\nu )}$ be the ground state's position standard deviation, and write $x=x_0(a+a^{†})$ using creation and annihilation operators. Here $m$ is mass and $\hslash$ the Planck constant divided by $2\pi$. The effective wave vector component $k_{\mathrm{eff}}$ is the single beam's component for a direct transition, or the difference of two beam components for a Raman transition. With dimensionless $\eta =|k_{\mathrm{eff}}|x_0$, the useful condition is $\eta \sqrt{2n+1}\ll 1$ over the populated number states, rather than merely $\eta \ll 1$.

Absorb the sign of $k_{\mathrm{eff}}$ into the pulse phase. Expanding the coupling factor givesThe constant term preserves $n$; the annihilation operator gives $a|n⟩=\sqrt{n}|n-1⟩$; the creation operator gives $a^{†}|n⟩=\sqrt{n+1}|n+1⟩$. The different square roots explain why removing a motional quantum becomes impossible at $n=0$, while adding one remains possible. Smallness of $\eta$ controls this spatial expansion; resolving the resulting transitions is a separate frequency requirement.

With 24 physical half-integer open-string oscillators, the first three string level operator eigenvalues are $0,1/2,1$. Their states are the vacuum, $a_{1/2}^{i†}|0⟩$, and $a_{1/2}^{i†}{a}_{1/2}^{j†}|0⟩$. The creation operators commute, so the third level is the symmetric square of the 24-dimensional transverse vector, of dimension $24\cdot 25/2=300$. Its trace is a scalar and its trace-free part has dimension 299. Including the vacuum shift gives ${\alpha }^{\prime }{M}^2=1/2,1,3/2$ when the two longitudinal directions are common directions with Neumann boundary conditions. The representation labels refer to the surviving ND rotation group, not full unbroken target Lorentz symmetry.

A Nambu spinor groups an annihilation operator with an opposite-momentum creation operator to express pairing in a quadratic Hamiltonian as a matrix problem. Its components are not independent: the spinor at $-k$ is related to conjugated components at $k$. This redundancy must be accounted for when summing energies and normal ordering constants. Component phases are a convention and change the displayed off-diagonal matrix entries.

Spatial integration of the free Hamiltonian density cancels particle-antiparticle pair terms through $E_p^2-{\mathbf{p}}^2-m^2=0$. The raw Hamiltonian is $\int d{\Pi }_p{E}_p(a^{†}a+b{b}^{†})$. Normal ordering removes the field-independent vacuum energy and gives $H=\int d{\Pi }_p{E}_p(a^{†}a+b^{†}b)$. Both creation operators increase the energy by $E_p$, so a negative-frequency field mode represents a positive-energy antiparticle, not a negative-energy state.

A normal-ordered product of free-field oscillator operators places all creation operators to the left of all annihilation operators, including the fermionic sign required to reorder fermions. Its vacuum expectation vanishes if any nontrivial oscillator factors remain. The Wick theorem expresses a free-field time-ordered product through these products and all Wick contractions.

The initial quantum state is a coherent state. Since $a|\alpha ⟩=\alpha |\alpha ⟩$ and the state is normalized, its mean occupation number is $N=⟨a^{†}a⟩=|\alpha {|}^2$. Each quantum harmonic oscillator energy eigenstate acquires the phase $e^{-i\omega (n+1/2)t}$, soThe prefactor is a global phase and does not change an expectation value. Inverting the definitions of the annihilation operator and creation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. If $\alpha =\sqrt{N}{e}^{i\phi }$, thenFor a classical harmonic oscillator with amplitude $A$, the conserved energy is $E=m{\omega }^2{A}^2/2$. Setting $E=\hslash \omega N$ gives exactly the same amplitude, $A=\sqrt{2\hslash N/(m\omega )}$. Thus these coherent-state oscillator expectation values follow the classical orbit.

The full quantum mean energy is nevertheless $⟨H⟩=\hslash \omega (N+1/2)$: the extra zero-point energy belongs to the nonzero position and momentum uncertainties required by the Heisenberg uncertainty relation, rather than to the motion of the packet centre. In particular, $N=0$ gives zero mean displacement but a finite ground-state width. A coherent state follows a classical mean trajectory while retaining finite position and momentum uncertainties.

For the requested oscillator response, treat $y$ as the fixed transverse position of a one-dimensional quantum harmonic oscillator along $z$, and put $\theta =ky$. The supplied radiation is transverse and already satisfies the Coulomb gauge. Its perturbation isThe annihilation operator and creation operator giveThus the only first-order transition out of the ground state is to $|1⟩$. More generally, the oscillator electric-dipole selection rule is $\Delta n=\pm 1$. This is a rule for matrix elements; exact energy conservation is replaced by a finite-time resonance profile during a switched finite-duration interaction.

To display both the resonant and nonresonant terms without spurious singularities, defineThe interaction picture coefficient in the given expansion is thenThe full Schrödinger picture coefficient of $|1⟩$ is $e^{-3i{\omega }_{0}t/2}{b}_1(t)$, while that of $|0⟩$ is $e^{-i{\omega }_{0}t/2}$ at this order. The normalization changes only at order $A_0^2$. In particular, at $\omega ={\omega }_0$ the first term is $C{e}^{i\theta }t$, finite for every finite time, and the perturbative result requires $|b_1|\ll 1$.

The fixed-$y$ interpretation is needed for the displayed target formulae. If $p^2$ were instead a three-dimensional kinetic operator with no transverse confinement, that model would have no normalizable transverse ground state, and $e^{\pm iky}$ would transfer transverse momentum $\pm \hslash k$. A treatment of that different model needs specified transverse states and recoil energies; they are absent from the question. The following parts use the intended localized-oscillator response.

The Pauli matrix ${\sigma }_x$ flips the spin with unit matrix element. A nonzero oscillator matrix element changes exactly one occupation number, through a creation operator or an annihilation operator:For a spin flip, the spin energy change divided by $\hslash$ is $-s_{\sigma }\Omega$. Thus creation and annihilation in mode $j$ have total frequency changes ${\omega }_j-s_{\sigma }\Omega$ and $-{\omega }_j-s_{\sigma }\Omega$, respectively. Distinct modes or different occupation changes lead to orthogonal final quantum states, so their probabilities add. Fermi's golden rule consequently givesThe Dirac delta function is even, accounting for the sign in the annihilation term. Since all ${\omega }_j,\Omega >0$, an initially up spin can only emit an oscillator quantum, while an initially down spin can only absorb one. Writing the oscillator density of states per unit angular frequency as $g(\omega )={\sum }_j\delta (\omega -{\omega }_j)$, the continuum limit givesThe $+1$ permits emission even from the oscillator vacuum; absorption requires a quantum already present. The notation $n_{\Omega }$ assumes a common or smoothly specified occupation at the resonant frequency. If several resonant modes have different occupations, it denotes their density of states weighted mean, so that $g(\Omega )n_{\Omega }={\sum }_j{n}_j\delta (\Omega -{\omega }_j)$ after continuum smoothing. This is the golden-rule relaxation of a spin coupled to an oscillator bath.

For the specified positive-exponent discrete Fourier transform, the inverse discrete Fourier transforms have negative exponents:Every sum runs over $q=2\pi j/L$, $j=0,\dots ,L-1$. Indeed, orthogonality of roots of unity givesso substituting the forward discrete Fourier transform into either proposed inverse recovers the original operator. Taking Hermitian conjugates supplies the inverse formulas for the creation operators.

Multiplying the inverse discrete Fourier transforms gives the local number operator combinationThis local expression includes off-diagonal momentum terms. Only after summing over sites does orthogonality of roots of unity giveIn a physical fixed-$s$ state the first boxed operator acts as $s$ at every site, by the local spin constraint in the Schwinger boson representation.

Use $⟨O⟩=⟨G|O|G⟩$. The given state is a two-species uniform boson number state: only the $q=0$ mode of each species is occupied. Applying an annihilation operator to an unoccupied mode gives zero, while $A_0^{†}{A}_0$ has eigenvalue $n_A$. ThusThe two species have independent fixed occupation numbers, so mixed expectation values factorize. The inverse discrete Fourier transforms yieldIn the momentum-space Hamiltonian operator, only $p=q=r=t=0$ contributes: each annihilation operator must remove an occupied boson, and each creation operator must restore it to the same mode. HenceThere is no replacement of $n_B$ by $n_B+1$ in this energy: in the real-space calculation the relevant neighboring-site commutator is zero. Both expressions assume distinct neighbouring sites, as in ii.

To match the prescribed spin length in the mean, impose $n_A+n_B=2sL=:N$. At fixed integer $N$, the energy is minimized by maximizing $n_A{n}_B=[N^2-(n_A-n_B{)}^2]/4$. Consequently the best choices within this relaxed family areWhen $N$ is even, $n_A=n_B=sL$ and $⟨H⟩=-J{s}^{2}L$. When $N$ is odd, either assignment of $(N-1)/2,(N+1)/2$ gives $⟨H⟩=-J(N^2-1)/(4L)$. Without fixing $N$, this auxiliary boson Hamiltonian operator has no lower bound: taking $n_A=n_B\to \infty$ sends its expectation value to $-\infty$.

Next, $S_i^x=(a_i^{†}{b}_i+b_i^{†}{a}_i)/2$. Each term changes the two species' occupation numbers, producing a state orthogonal to $|G⟩$. ThereforeFor distinct sites $i,j$ modulo $L$, expand the product of the two spin operators. The terms $S_i^{+}{S}_j^{+}$ and $S_i^{-}{S}_j^{-}$ again have zero expectation value. The remaining terms giveHere $⟨b_i{b}_j^{†}⟩=n_B/L+{\delta }_{ij}$ and the Kronecker delta vanishes at distinct sites. ConsequentlyThe same holds for the $y$ components. At the balanced choice and as $L\to \infty$, each transverse two-point correlation function tends to $s^2/2$ independently of separation, whereas the one-point transverse magnetization is zero. This is transverse spin order in a number eigenstate: the state has a definite total $S^z=(n_A-n_B)/2$ and retains the global U(1) symmetry of rotations about $z$, yet has the transverse correlations of an ordered state with no selected common azimuth. The sum of the $x$ and $y$ correlations tends to $s^2$, describing ferromagnetic long-range order within this approximation. The thermodynamic limit is understood with both the separation and $L$ growing, without identifying the two sites periodically.

Fixed-spin qualification. The proposed state is not actually in the physical fixed-$s$ Hilbert space for $s>0,L>1$. It satisfies the local spin constraint in the Schwinger boson representation only in expectation value. For example, with $L=2,s=1/2,n_A=n_B=1$,The nonzero component $a_0^{†}{b}_0^{†}|\mathrm{vac}⟩/2$ has local total occupation numbers $(2,0)$ and hence local spins $(1,0)$, instead of $(1/2,1/2)$. More generally the local total number operator $N_n$ has varianceEach species contributes a binomial distribution at a given site, with success probability $1/L$, and the two contributions are independent. In the physical subspace that variance must be zero. The proposed state even hasby the Casimir operator identity from a. Thus the calculated energy is a relaxed mean-field theory estimate, not a Rayleigh-Ritz variational principle bound for the stated fixed-$s$ quantum XY chain. Projecting onto $N_n=2s$ at every site would give a physical trial state and generally change its expectation values; in the two-site example the projected state has energy $-J$, whereas the unprojected value is $-J/2$. The calculated long-range order characterizes the supplied approximation and does not establish long-range order in the exact one-dimensional ground state.

Let $V=L^3$ be the box volume. The number operator is the sum of the occupation numbers of all momentum eigenstates, including the zero mode:In the Fourier series for the contact interaction, the two creation operators supply phase $e^{-i({\mathbf{k}}_1+{\mathbf{k}}_2)\cdot \mathbf{r}}$ and the two annihilation operators supply $e^{i({\mathbf{k}}_3+{\mathbf{k}}_4)\cdot \mathbf{r}}$. Fourier orthogonality with periodic boundary conditions givesThe four normalization factors together contribute $V^{-2}$, leaving the momentum-space contact-interaction HamiltonianThe Kronecker delta enforces momentum conservation. An equivalent expression that resolves the momentum transfer isBoth are already in normal ordering; no extra commutator constant is introduced.

Taking the Hermitian adjoint gives ${\phi }_{\mathbf{k}}=c_k{\psi }_{\mathbf{k}}+s_k{\psi }_{-\mathbf{k}}^{†}$. For nonzero wavevectors, the canonical commutation relations implyWhen the Kronecker delta is nonzero, $k=q$, so the hyperbolic Pythagorean identity $c_k^2-s_k^2=1$ givesFor the two annihilation operators, the only potentially nonzero terms areHere $\mathbf{q}=-\mathbf{k}$ implies equal magnitudes and hence equal $c,s$, because ${\theta }_k$ depends only on $|\mathbf{k}|$. ThusThe corresponding commutators of two creation operators vanish by the Hermitian adjoint. This verifies that the Bogoliubov transformation preserves the full bosonic canonical commutation relations.

Normalize the specified state by $⟨0|0⟩=1$ and write $⟨A⟩=⟨0|A|0⟩$. Its zero mode is a coherent state, so any normally ordered product of zero-mode creation operators and annihilation operators can be evaluated using ${\psi }_0\mapsto \sqrt{N_0}$ and ${\psi }_0^{†}\mapsto \sqrt{N_0}$. In particular $⟨({\psi }_0^{†}{)}^2{\psi }_0^2⟩=N_0^2$, rather than $N_0(N_0-1)$, which would apply to a zero-mode number state.

Sort the momentum-space contact-interaction Hamiltonian by its number of zero-mode factors. Four such factors give $u{N}_0^2/(2V)$. Three zero-mode factors and one nonzero momentum violate momentum conservation, so no such term survives. With two zero-mode factors there are three possibilities:

The sums over $\mathbf{k}\ne 0$ include both members of each opposite-momentum pair, consistently with the overall factor $1/2$. Terms with one zero-mode factor contain three nonzero-mode factors; their expected values vanish in the centered multimode squeezed vacuum. Indeed each $\psi$ is linear in $\phi ,{\phi }^{†}$, and an odd product has zero vacuum expectation value. Terms with no zero-mode factors are independent of $N_0$ at fixed ${\theta }_k$. ThereforeTo evaluate these vacuum expectation values, use the inverse Bogoliubov transformation and its Hermitian adjoint:Since ${\phi }_{\mathbf{k}}|0⟩=0$, the only nonzero two-factor vacuum expectation value is $⟨{\phi }_{\mathbf{k}}{\phi }_{\mathbf{q}}^{†}⟩={\delta }_{\mathbf{k},\mathbf{q}}$. The bosonic squeezed-vacuum contractions are consequentlyIn particular the three requested quantities are $-\sinh {\theta }_k\cosh {\theta }_k$, $-\sinh {\theta }_k\cosh {\theta }_k$ and ${\sinh }^2{\theta }_k$, respectively. The minus sign in the pair vacuum expectation values comes from the inverse Bogoliubov transformation and is essential to lowering the interaction energy. Substitution yields

Write $\hat{N}={\hat{a}}^{†}\hat{a}$. The creation and annihilation operators obeyConsequently,The resulting function is not zero: its squared norm isThus ${\hat{a}}^{†}{\Psi }_0$ is the normalized first excited state, up to phase. More generally the creation operator raises the energy by one quantum $\hslash \omega$: ${\hat{a}}^{†}|n⟩=\sqrt{n+1}|n+1⟩$. The factor is required because applying it to a higher state does not preserve normalization.

A relativistic quantum field describes both creation and annihilation of excitations; a one-particle wavefunction alone is insufficient when interactions can change particle number. For a free complex scalar field, one conventional mode expansion of a free field isThe apparently negative-frequency term creates a positive-energy antiparticle; it is not a physical particle of unbounded negative energy. The normal-ordered Hamiltonian and conserved global phase charge have the formsParticles and antiparticles have equal mass and opposite additive internal charges. Including both frequency sectors is also essential to the local field commutator: the positive- and negative-frequency terms combine so that spacelike-separated scalar observables commute, respecting microcausality.

A Dirac field similarly contains particle annihilation operators multiplying $u$ spinors and antiparticle creation operators multiplying $v$ spinors. Its canonical anticommutation relations implement Fermi-Dirac statistics, whereas scalar modes use commutators and Bose-Einstein statistics. These assignments are tied to the Spin-statistics theorem under the usual relativistic locality and positivity assumptions. Conserved charge constrains particle creation and annihilation, but particle and antiparticle numbers need not be separately conserved in an interacting theory.

An antiparticle need not be a distinct species. A neutral real scalar field is self-conjugate; the photon is also its own antiparticle. A Majorana fermion obeys a self-conjugacy condition and cannot carry an unbroken additive charge that distinguishes it from its antiparticle. Electric neutrality alone does not force self-conjugacy when other conserved quantum numbers distinguish the states.

Use the two-pair component of polarization-entangled down-conversion to account for identical bosons in the same collected modes. Write $a_H^{†}$ for a creation operator in mode $a$ with horizontal polarization, and similarly for the other modes. For the plus Bell state, the pair-creation operator is $K^{†}=a_H^{†}{b}_V^{†}+a_V^{†}{b}_H^{†}$. The normalized four-photon state isThese are normalized Fock states. In particular, $(a_H^{†}{)}^2|0⟩=\sqrt{2}|2H_a⟩$; omitting this factor would misnormalize the four-photon input.

The first polarizing beam splitter sends $a_H^{†}$ to $T_H^{†}$ and $a_V^{†}$ to $c_V^{†}$. Only the middle term can give exactly one trigger photon and leave another photon in $c$. In the first term both $a$ photons reach $T$ and none reaches $c$; in the last there is no trigger. Thus the relevant, still unnormalized component is

For the selected paths of the half-wave plate and beam splitter, takeHere $e$ labels the output towards $D_3$. An immaterial common reflection phase has been absorbed in the mode definitions. Keeping one photon at $D_3$ leavesAt the final polarizing beam splitter, the $H_{c^{\prime }}{H}_d$ alternative yields one $H$ at each of $D_1,D_2$, while $V_{c^{\prime }}{V}_d$ yields one $V$ at each. The two opposite-polarization alternatives send both photons to the same detector and fail fourfold coincidence counting. The accepted component is thereforeAfter conditioning and normalization, the three detected output modes haveThe optical mode matching must erase any timing, frequency or other record that distinguishes the two alternatives. Otherwise the same polarization populations could describe an incoherent mixed quantum state, rather than this quantum superposition. The Bouwmeester three-photon GHZ experiment uses this principle. The photonic GHZ preparation by fourfold coincidence is a measurement in the coincidence basis: detecting all four photons projects and consumes them. A click at $T$ by itself does not herald a guaranteed freely available three-photon state.

Put $b=\mathcal{E}{e}^{i(kr-\omega t)}$, allowing $\mathcal{E}$ to be complex. The Hermitian electric field is $\hat{E}=b\hat{a}+b^{\ast }{\hat{a}}^{†}$. The annihilation operator and creation operator change photon number, so their diagonal matrix elements in a Fock state vanish. Since $\hat{a}{\hat{a}}^{†}=\hat{n}+1$ and ${\hat{a}}^{†}\hat{a}=\hat{n}$,The terms containing ${\hat{a}}^2$ or ${\hat{a}}^{†2}$ connect orthogonal number states and contribute zero. Even as $n$ grows, there is no definite oscillating mean electric field; its quadrature fluctuations grow instead. A Fock state can represent fixed field energy, but not a classical sinusoidal wave of prescribed wave phase. This is distinct from representing a classical field with a randomly averaged phase.

Orthogonality of the number states makes the coherent state norm an exponential series:Using $\hat{a}|n⟩=\sqrt{n}|n-1⟩$ and reindexing $m=n-1$,It is not an eigenstate of the creation operator. Indeed its supposed eigenvalue would have to be ${\alpha }^{\ast }$, from its expectation, whereasThis nonzero residual holds even at $\alpha =0$.

Choose the normalization constant real and positive; its otherwise arbitrary phase is physically immaterial. Orthogonality of the Fock states givesActing with the annihilation operator and shifting the summation index givesThus a coherent state is an annihilation-operator eigenket. It is not also an eigenket of the creation operator: a common eigenket would make $[\hat{a},{\hat{a}}^{†}]$ annihilate it, contradicting $[\hat{a},{\hat{a}}^{†}]=I$. Taking the adjoint of the annihilation-operator equation instead gives the distinct, valid bra relationWith $\hat{n}={\hat{a}}^{†}\hat{a}$, the photon number expected value is $\mu =|\alpha {|}^2$. Normal ordering ${\hat{n}}^2=({\hat{a}}^{†}{)}^2{\hat{a}}^2+{\hat{a}}^{†}\hat{a}$ gives $⟨n^2⟩={\mu }^2+\mu$. ThereforeAt large mean photon number the relative fluctuations vanish, giving a classical-field limit even though quantum fluctuations remain. For the vacuum $\mu =0$, the relative ratio itself is undefined.

The Born rule applied to the number-state coefficient gives the Poisson distributionIt depends on the single parameter $\mu =⟨n⟩$, and has equal expected value and variance, consistent with the preceding moments.

For a Fock state $|n⟩$, raising or lowering the number makes an orthogonal state, so the optical quadrature has $⟨\hat{X}⟩=0$. Expanding its square givesThe first two terms have zero diagonal matrix elements, while the last two contribute $n+1$ and $n$. In particular, the vacuum retains quadrature variance $1/2$.