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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.

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.

A phase-selected Hermitian field observable. With annihilation operator $a$ and $[a,a^{†}]=1$, define $X_1=(a+a^{†})/2$ and $X_2=(a-a^{†})/(2i)$. Then $[X_1,X_2]=i/2$ and the Fock vacuum has variance $1/4$ in either quadrature. This dimensionless convention gives $X_{\vartheta }=X_1\cos \vartheta +X_2\sin \vartheta$. Balanced homodyne detection selects $\vartheta$ through the reference wave phase.

Start from electrons moving in the periodic equilibrium ionic potential. Displace ion $l$ by ${\mathbf{u}}_l$ and expand its contribution as $V(\mathbf{r}-{\mathbf{R}}_l-{\mathbf{u}}_l)\simeq V(\mathbf{r}-{\mathbf{R}}_l)-{\mathbf{u}}_l\cdot \nabla V(\mathbf{r}-{\mathbf{R}}_l)$. Quantizing the harmonic normal modes expresses ${\mathbf{u}}_l$ as a sum of creation and annihilation operators with displacement scale $\sqrt{\hslash /(2{\rho }_{\mathrm{ion}}V{\omega }_{\mathbf{q}})}$ for a single simple branch. Expanding the electronic field in bloch states then gives a band-projected Fröhlich HamiltonianHere $c,c^{†}$ obey canonical anticommutation relations and remove/create an electron with crystal momentum $\mathbf{k}$ and spin $\sigma$; $b,b^{†}$ are boson operators for lattice modes. The matrix element $M_{\mathbf{q}}$ contains the derivative of the ionic potential, phonon polarization, displacement normalization and electronic wavefunction overlaps. With phases chosen appropriately, $M_{-\mathbf{q}}=M_{\mathbf{q}}^{\ast }$ ensures a Hermitian Hamiltonian. The displayed one-band, one-branch notation suppresses additional branch and band labels. “Fröhlich Hamiltonian” here means the linear electron-phonon model used in the question; it is not a requirement to use specifically the polar-optical-phonon coupling of a polar semiconductor. A harmonic lattice, small displacements and a weak enough coupling to organize perturbation theory are the underlying approximations.

The Fröhlich canonical transformation removes real linear phonon operators from the electron-only effective problem to first order. Choose an anti-Hermitian $S$ with $[S,H_0]=-V_{e\text{-}ph}$. For $A_{\mathbf{kq}\sigma }=c_{\mathbf{k}+\mathbf{q},\sigma }^{†}{c}_{\mathbf{k}\sigma }$ and ${\Delta }_{\mathbf{kq}}={ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$, one possible generator isIts Hermitian conjugate is its negative after relabelling momenta. The Baker--Campbell--Hausdorff formula givesThe commutator describes virtual emission and absorption of a phonon. normal ordering yields an electron-electron term as well as one-electron energy renormalizations and phonon terms. Retain the former, absorb single-particle shifts into ${ϵ}_{\mathbf{k}}$ and eliminate the explicit phonon sector. For elastic pair scattering the two virtual time orderings have the combined amplitudeUsing the usual factor one-half to count unordered pairs gives the coefficient in the printed Hamiltonian, with $\Delta ={ϵ}_{\mathbf{k}}-{ϵ}_{\mathbf{k}-\mathbf{q}}$. The two outgoing momenta are $\mathbf{k}-\mathbf{q}$ and ${\mathbf{k}}^{\prime }+\mathbf{q}$, so total momentum and the spins are conserved. A fully off-shell effective operator can instead use symmetrized denominators for both external legs; the displayed expression is the usual elastic-scattering reduction. Eliminating nearly resonant real phonons requires more care than this second-order construction.

For $|\Delta |<{\Omega }_{\mathbf{q}}$ the denominator is negative: virtual phonon exchange is attractive in this low-energy channel. Physically, an electron changes the ionic density and a second electron can gain energy by responding to the delayed distortion. This is a retarded interaction, not an unscreened instantaneous attraction at every energy transfer. At large transfer the sign changes, and screened Coulomb repulsion must also be included when deciding the net low-energy pairing strength. The connection between lattice dynamics and superconductivity is supported by the isotope effect, but the mathematical attraction alone does not determine a measured transition temperature.^[ref]

For the pair calculation, make the model restrictions explicit. The filled Fermi sea is inert, and $|0⟩$ denotes that sea with no added particles, not an empty electronic vacuum. Project the interaction onto the previously empty shell $0\le {\xi }_{\mathbf{k}}\le D$, where ${\xi }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)-E_F$ and $D=\hslash {\omega }_D$. Measure the added-pair energy relative to $2E_F$; this is the negative $E$ in the final formula. If absolute one-electron energies are retained in the preceding eigenvalue equation, its eigenvalue must instead be $E_{\mathrm{abs}}=2E_F+E$. A shell-restricted state would not in general be an exact eigenstate of the completely unprojected, all-momentum contact Hamiltonian.

Write $g={\gamma }^2>0$ and $|\mathbf{k}⟩=c_{-\mathbf{k}\downarrow }^{†}{c}_{\mathbf{k}\uparrow }^{†}|0⟩$. Within the zero-total-momentum pair subspace the two opposite-spin terms cancel the interaction's factor one-half, while same-spin s-wave terms cancel by fermionic antisymmetry. Thus the projected Hamiltonian acts asFor $|\Psi ⟩={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}|\mathbf{k}⟩$ and $C={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}$, matching each basis coefficient givesThis is the stated coefficient formula in the consistent Fermi-level energy convention. Its even dependence on $\mathbf{k}$ produces an even orbital state with a spin singlet. The constant $C$ is fixed up to an overall phase by normalization; for a bound state, $|C{|}^{-2}=g^2{\sum }_{⟨\mathbf{k}⟩}(E-2{\xi }_{\mathbf{k}}{)}^{-2}$. It cannot be set to zero for the nontrivial bound solution.

Put $E=-B$ with $B>0$ and impose the sum defining $C$:Here $\rho (E_F)$ counts orbital states for one spin species, as appropriate to the unrestricted $\mathbf{k}$ sum in this pair basis, and is approximately constant over the thin shell. If a two-spin total density of states is used, it must be divided by two in this integral. State-counting and the volume convention in $g$ and $\rho$ must be consistent. With $\xi =g\rho (E_F)$ as the paper's dimensionless coupling, inversion yieldsThe first approximation is the constant-density-of-states, restricted-shell model; the final exponential additionally assumes weak attraction. The binding energy is nonanalytic at zero coupling, so a finite-order power expansion in $g$ cannot produce it.

Cooper's problem asks whether two electrons added above an inert Fermi sea bind despite Pauli exclusion of lower states. An arbitrarily weak net attraction binds a zero-momentum singlet below the free-pair threshold, because the near-Fermi-surface integral diverges logarithmically as $B\downarrow 0$. Its wavefunction draws on many time-reversed momentum states and becomes dominated by a narrow region near the Fermi surface at weak coupling; the corresponding pair is spatially extended. This establishes a Cooper instability of the unpaired reference state.^[ref] The full superconducting ground state needs a coherent many-electron treatment in which overlapping pairs and the reorganized sea are included. BCS theory supplies that step; the two-electron binding scale alone is neither a calculation of the many-body gap nor a derivation of zero resistance.^[ref]

A single-mode coherent state is an annihilation operator eigenstate:Its photon statistics are Poisson distribution: $⟨n⟩=\mathrm{Var}n=|\alpha {|}^2$. For optical quadratures $X=(a+a^{†})/2$, $P=(a-a^{†})/(2i)$, one has $[X,P]=i/2$ and $\mathrm{Var}X=\mathrm{Var}P=1/4$. Its field mean oscillates classically while retaining vacuum fluctuations.^[ref][ref]

A classical mixture of coherent states has a nonnegative Glauber-Sudarshan representation. A nonclassical state of light cannot be represented that way. Examples are a squeezed vacuum state, with quadrature variances $e^{-2r}/4$ and $e^{2r}/4$, and a nonvacuum number state, with zero photon number variance. Noise below the vacuum level in one quadrature, sub-Poisson number noise, or photon antibunching can witness nonclassicality. For $|n⟩$, $g^{(2)}(0)=1-1/n<1$; coherent light has $g^{(2)}(0)=1$. Thermal light has excess noise but is a classical mixture: noncoherent does not by itself mean nonclassical.

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 the quantum harmonic oscillator, the canonical commutation relation $[x,p]=i\hslash$ gives $[a,a^{†}]=1$ and $H_0=\hslash {\omega }_0(a^{†}a+1/2)$. Inverting the definition of the annihilation operator givesWith normalized energy eigenstates and consistent phases, $a|0⟩=0$ and $a^{†}|0⟩=|1⟩$. Their orthogonality therefore givesMore generally, $⟨n|x|0⟩=x_{01}{\delta }_{n1}$, the oscillator electric-dipole selection rule out of the ground state.

To derive the transition amplitude, insert the expansion in stationary energy eigenstates into the Schrödinger equation with $H=H_0+V_1(t)$. The derivatives of the explicit phases cancel the $H_0$ terms. Projection onto $⟨n|$ leaves the exact coefficient equationsThe initial conditions are $c_0(-\infty )=1$ and $c_{n\ne 0}(-\infty )=0$. To first order in the perturbation, replace the amplitudes on the right by these initial values. Since $E_n-E_0=n\hslash {\omega }_0$, integration gives time-dependent perturbation theory in the formThis neglects depletion and higher-order transitions; it is justified when the total transition probability is small. Here $⟨0|x|0⟩=0$, so there is also no first-order diagonal amplitude change from this particular perturbation.

For the exponential electric field, the electric-dipole interaction is $V_1=-qEx{e}^{t/\tau }$, with $\tau >0$. Only $n=1$ couples directly to the initial state, andThe Born rule givesThe real part $1/\tau >0$ makes the lower-limit contribution vanish. ThusSet $u={\omega }_0\tau$ and $\eta =qE{x}_{01}/(\hslash {\omega }_0)$. The dimensionless form is $p_1/{\eta }^2=u^2/(1+u^2)$, increasing monotonically from zero to one, with half the asymptotic value at $u=1$. A sufficient weak-field condition for the entire curve is $|\eta |\ll 1$.

For ${\omega }_0\tau \ll 1$, the force has acted for too little time to appreciably change the state at the instant of measurement. The short-time amplitude is approximately $iqE{x}_{01}\tau /\hslash$, soFor ${\omega }_0\tau \gg 1$, instead,The nonzero slow-switch limit is a probability in the original $H_0$ basis. It does not contradict the adiabatic theorem. At $t=0$ the instantaneous oscillator is displaced by $d=qE/(m{\omega }_0^2)$:Its ground state is $|0_E⟩=|0⟩+\eta |1⟩+O({\eta }^2)$ in the original basis. Adiabatic following therefore predicts exactly the finite original-basis limit found above.

The exponential switch-on of a driven quantum oscillator also permits an exact check. A linear force preserves a coherent state, with original-basis displacement $\alpha (0)=i\eta u/(1+iu)$. Thus the exact original-basis probability is $|\alpha {|}^2{e}^{-|\alpha {|}^2}$, whose weak-field limit is the expression above. In the instantaneous displaced basis the mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/(1+u^2)$ and does tend to zero for slow switching. The two bases measure different notions of excitation.

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 real angles write $u_p=\cos {\theta }_p$ and $v_p=\sin {\theta }_p$. For one independent pair $p>0$, the transformation and its conjugate readUsing the canonical anticommutation relations for the $\alpha$ modes givesThe remaining anticommutators vanish for independent momentum pairs. The displayed two-by-two matrix is unitary and its inverse is obtained by changing $i$ to $-i$.

When the same formula is indexed over both signs of momentum, its first row at $-p$ must agree with the conjugate of its second row at $p$. Thus a consistent global convention requiresThis is the odd-momentum fermionic Bogoliubov angle convention. If one checks only the first-row formulas, their cross anticommutator is $i\sin ({\theta }_p+{\theta }_{-p})$; its vanishing gives the condition modulo $\pi$, with possible signs absorbable in the $\alpha$ modes. Requiring both rows to describe the same operators fixes the convention above. Merely checking a separate unitary matrix at each momentum misses this redundancy: choosing ${\theta }_p={\theta }_{-p}=\pi /4$ in both first rows would give $\{c_p,c_{-p}\}=i$.

At $p=0$, the annihilation operator and its conjugate belong to one mode. The full displayed formula is consistent only with $v_0=0$; choose ${\theta }_0=0$ and ${\alpha }_0=c_0$. This mode is not a two-mode pair.

Since every quasiparticle energy is positive, the ground state is the vacuum of all the ${\alpha }_p$ annihilation operators, including ${\alpha }_0=c_0$. Its zero-momentum mode is therefore empty. For each positive momentum, the inverse Bogoliubov transformation givesBoth annihilate the normalized pair state $(u_p+i{v}_p{c}_p^{†}{c}_{-p}^{†})|0⟩$. For example, applying ${\alpha }_p$ produces the two cancelling terms $i{u}_p{v}_p{c}_{-p}^{†}|0⟩$. ThusThe factors have even fermion parity and commute for disjoint momentum pairs. Each normalized factor is a superposition of zero and two fermions, whose pair-occupation probability is $v_p^2$. ThereforeThere is no zero-mode contribution. For fixed $L$, the sum converges since $v_p^2=O(p^{-4})$.

A normalized simple excited state is $|k⟩={\alpha }_k^{†}|\Omega ⟩$ for $k>0$. Its excitation energy is $E_k$, and its total momentum is $k$: both terms in ${\alpha }_k^{†}=u_k{c}_k^{†}+i{v}_k{c}_{-k}$ add momentum $k$. Acting on the $k,-k$ pair givesThe affected pair now contains exactly one fermion instead of mean number $2v_k^2$; all other pairs are unchanged. ConsequentlyThis is the fermion number of a Bogoliubov excitation. Its change in mean fermion number lies between zero and one for $\mu >0$, with the value one at $\Delta =0$. The Bogoliubov quasiparticle coherently mixes adding a fermion and removing the opposite-momentum fermion; it is not an eigen-excitation of the original fermion number operator. Indeed,so fermion number is not conserved when $\Delta \ne 0$, while fermion parity is conserved. The generally noninteger change is an expectation value in a coherent paired background, rather than a fractional eigenvalue of the fermion number operator. In the weak-pairing limit at fixed $k$, it tends to one, as expected for an ordinary added fermion.

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