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At fixed particle number $N$, write $n_1=n$ and $n_2=N-n$. The contact interaction between two ring modes has energy eigenvaluesFor $g>0$, the nonnegative product $n(N-n)$ is minimized at the endpoints. ConsequentlyFor $N\ge 1$ this is a two-dimensional degenerate ground state subspace. $N=0$ instead has the unique Fock vacuum. In this projected model, putting all particles into one traveling mode avoids the extra exchange contribution paid when both distinct modes are occupied.

For the alternative coefficients, put $\alpha =\beta =A<0$ and $\gamma =0$. Then $E(n)=A[N(N-1)-2n(N-n)]$ is again minimized at $n=0,N$, with $E_{\min }=AN(N-1)$. Evenness of $N$ is unnecessary for this result; the balanced Fock state maximizes this attractive-interaction energy instead of minimizing it.

A normalized oscillator coherent state is an eigenstate of the annihilation operator, $a|\alpha ⟩=\alpha |\alpha ⟩$. Expanding it in Fock states givesSince $x=\sqrt{\hslash /(2M\omega )}(a+a^{†})$, $⟨a^2⟩={\alpha }^2$, and $⟨a{a}^{†}⟩=|\alpha {|}^2+1$, the variance is $⟨x^2⟩-⟨x{⟩}^2=\hslash /(2M\omega )$. Henceindependently of the complex amplitude $\alpha$.

For $A=\alpha a^{†}$ and $B=-{\alpha }^{\ast }a$, the commutator is the scalar $[A,B]=|\alpha {|}^{2}I$. The Baker--Campbell--Hausdorff formula therefore terminates and gives the displacement operatorBecause $a|0⟩=0$, applying it to the Fock vacuum yields exactly the Fock state expansion above:For a general complex $\alpha$, its complex conjugate is required in the annihilation term. The expression with $-\alpha a$ printed in the paper is valid when $\alpha$ is real.

The quantum state is continuous through a sudden finite-force quantum quench, so it is still $|0⟩$ at $t=0^{+}$. DefineThe new Hamiltonian operator is $H=\hslash \omega (b^{†}b+1/2)-\hslash \omega {\lambda }^2$. The old Fock vacuum obeys $b|0⟩=-\lambda |0⟩$, so the coherent state after shifting an oscillator equilibrium rotates with amplitude $-\lambda e^{-i\theta }$ relative to the new equilibrium. In the original oscillator basis, the full state isOne way to obtain the phase is to writeand apply the displacement operator composition rule. The coherent-state oscillator expectation values giveThe position uncertainty stays $\sqrt{\hslash /(2M\omega )}$. These mean values obey the forced classical equation, while the nonzero variance retains the quantum width.

Using the consistently normalized Jordan–Wigner transformation from part (c), write ${\sigma }_1^x{\sigma }_2^x+{\sigma }_1^y{\sigma }_2^y=2(s_1^{+}{s}_2^{-}+s_1^{-}{s}_2^{+})$. Since $f_1^{†}{P}_1=f_1^{†}$ and $f_1{P}_1=-f_1$, the parity string givesThereforeThe orthogonal transformation $b=(f_1+f_2)/\sqrt{2}$, $c=(f_1-f_2)/\sqrt{2}$ preserves the canonical anticommutation relations and diagonalizes this bilinear Hamiltonian:The Fock vacuum is $|\downarrow \downarrow ⟩$, with energy $2h$. Occupying $b$ alone gives the symmetric opposite-spin state with energy $-2J$; occupying $c$ alone gives the antisymmetric state with energy $2J$; occupying both gives $|\uparrow \uparrow ⟩$, up to a global phase, with energy $-2h$. This reproduces the entire spectrum in part (b).

The Holstein–Primakoff transformation replaces each finite-dimensional spin algebra by a bosonic creation operator and annihilation operator, while restricting the local occupation number to $0\le n_j\le 2S$. The number $n_j=a_j^{†}{a}_j$ counts deviations from the fully polarized reference quantum state. Its zero-boson Fock vacuum has ${\tilde{S}}^z=S$.

The representation is exact on this physical subspace: the ordered square root gives ${\tilde{S}}^{+}|n⟩=\sqrt{n(2S-n+1)}|n-1⟩$ and ${\tilde{S}}^{-}|n⟩=\sqrt{(n+1)(2S-n)}|n+1⟩$, reproducing the spin commutation relations. For $n_j\ll S$,so the leading Hamiltonian is quadratic in bosons and describes spin waves.

The square root generates boson interactions beyond leading order and its operator ordering matters. Truncation also enlarges the Hilbert space to unphysical occupations unless the low-density approximation remains self-consistent. Large fluctuations, particularly near a gap closing or in low dimensions, can invalidate the linear spin-wave approximation even when its algebraic manipulation is straightforward.

Use the unitary discrete Fourier transformThe transformed operators obey $[a_k,a_q^{†}]={\delta }_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. ThuswhereThe factor one half in the full momentum sum avoids counting an anomalous pair twice. For $H_f>0$, ${\alpha }_k>|{\Delta }_k|$ for every allowed momentum. SetReal even angles ensure consistency between opposite momenta, and ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$ preserves the canonical commutation relations. The anomalous coefficient becomes ${\Delta }_k\cosh 2{\theta }_k-{\alpha }_k\sinh 2{\theta }_k=0$. Normal ordering then giveswithThe field-polarized planar ferromagnetic spin wave has energy gap $ϵ(0)=S\sqrt{H_f(H_f+2J)}$. The momenta run over the whole Brillouin zone once, including the self-inverse momenta $0$ and, for even $L$, $\pi$; these use single-mode squeezing rather than introducing an extra independent partner.

The zero-field case has a zero-mode obstruction to bosonic Bogoliubov diagonalization. Its nonzero-momentum modes have $ϵ(k)=2\sqrt{2}JS|\sin (k/2)|$, but ${\alpha }_0={\Delta }_0=JS$, so no finite squeezing angle exists for the uniform mode. With $Q_0=(a_0+a_0^{†})/\sqrt{2}$, its Hamiltonian is $JS{Q}_0^2-JS/2$, not an identically zero oscillator. For example, its matrix element between the vacuum and the two-boson number state is $JS/\sqrt{2}$, whereas $ϵ(0)b_0^{†}{b}_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero field; at zero field treat the collective orientation separately or introduce a field regulator before taking a limit.

For nonzero field, the spin-wave ground state is the vacuum of all $b_k$. The inverse Bogoliubov transformation gives $⟨a_k^{†}{a}_q⟩={\delta }_{kq}{\sinh }^2{\theta }_k$ and vanishing one-point oscillator averages. Translational invariance therefore gives the depletionRotating back to the physical axes yields, to the retained spin-wave order,The zero transverse components also follow from rotation by $\pi$ about the physical $x$ axis in a symmetry-preserving ground state. The nonzero depletion describes zero-point quantum fluctuations: the original boson vacuum is not the ground state because the anomalous terms create pairs. It is nonnegative and must satisfy $\delta S\ll S$ for the expansion to be reliable. At strong field it tends to zero, with $\delta S\sim J^2/(8H_f^2)$ in the infinite-chain limit.

For fixed $H_f>0$, the momentum sum has a finite thermodynamic limit obtained by $L^{-1}{\sum }_k\to (2\pi {)}^{-1}{\int }_{-\pi }^{\pi }dk$. At zero field the uniform mode already has no oscillator vacuum, as explained in part d. Even after removing that mode, the nonzero modes behave asThus assigning a finite ordered moment by simply setting $h=0$ in the infinite momentum sum is invalid. The divergence indicates failure of this expansion about a fixed planar orientation; it is not a negative physical magnetization. At finite size an exact zero-field state chosen to have definite total $S^z$ has zero planar one-point magnetization by the continuous rotation symmetry. Establishing its full correlations requires a treatment beyond the divergent ordered-state approximation.

Promote the mode expansion of a free field coefficients to annihilation operators and creation operators, replacing $a^{\ast }$ by $a^{†}$. The canonical commutation relations areTogether with the normalization in part a, they give $[\varphi (t,\mathbf{x}),\pi (t,\mathbf{y})]=i{\delta }^3(\mathbf{x}-\mathbf{y})$ and vanishing equal-time commutators between two fields or two conjugate momenta.

Insert the mode expansion of a free field in the Hamiltonian. The spatial integral produces $(2\pi {)}^3{\delta }^3(\mathbf{p}\pm \mathbf{q})$. The $aa$ and $a^{†}{a}^{†}$ coefficients cancel because ${\omega }_{\mathbf{p}}^2={\mathbf{p}}^2+m^2$. The remaining terms giveThis is a continuum of quantum harmonic oscillators. In a finite periodic volume $V$, $(2\pi {)}^3{\delta }^3(0)=V$. The Fock vacuum satisfies $a_{\mathbf{p}}|0⟩=0$, so its vacuum energy density isNormal ordering would subtract this constant; here it must be retained because it is the quantity being calculated.^[ref]

With no pairing, the Hamiltonian is diagonal in the fermionic occupation numbers. The Pauli exclusion principle allows each momentum mode to be either empty or occupied once. Without a fixed fermion number constraint, occupying a mode lowers the energy precisely when its single-particle energy is negative. Therefore, with $|0⟩$ the Fock vacuum, a ground state isChoose a fixed ordering of the fermionic creation operators; changing it alters only the overall phase. Any zero-energy modes may be occupied or empty without changing the ground-state energy.

To identify the occupied modes, complete the square:Here $\beta$ is the dispersion parameter, not inverse temperature. If ${\beta }^2<4\mu$, every mode has positive energy and the unique ground state is the Fock vacuum, with $E_{\mathrm{GS}}=0$. If ${\beta }^2=4\mu$, there are still no negative-energy modes, but $p=\pm \sqrt{\beta /2}$ have zero energy if they lie on the allowed momentum grid. Each such mode doubles the ground-state degeneracy.

For ${\beta }^2>4\mu$, defineBoth are positive, and the occupied Fermi sea consists of the two intervals $p_{-}<|p|<p_{+}$. With $p=2\pi n/L$, the requested finite-size ground-state energy is explicitlyThe factor two counts opposite momenta. Endpoints are excluded from this sum because they have zero energy; when an endpoint is allowed, its optional occupation adds ground-state degeneracy. A narrow negative-energy interval can contain no allowed finite-size momenta, in which case the sum is empty. The zero-momentum mode is always empty since $ϵ(0)=\alpha \mu >0$.

For a stable free Klein-Gordon field with a complete timelike Killing flow and a positive self-adjoint one-particle energy generator, its positive spectral subspace selects positive-frequency solutions of positive Klein-Gordon inner product. Their annihilation operators define the ground-state Fock vacuum. Degenerate modes can be mixed by a unitary matrix without changing this splitting or vacuum. Merely being stationary near infinity is insufficient if the Killing field becomes spacelike in an ergoregion of a stationary spacetime; zero modes, instability or additional boundary conditions also require separate treatment. The choice of a ground state does not prevent other states, such as thermal states, from using the same particle operators.

// Target: statistical-physics.bigb

Fix a detected position and polarization of waves. Write the electric field operator as $\hat{E}(t)={\hat{E}}^{(+)}(t)+{\hat{E}}^{(-)}(t)$, where the positive-frequency part contains annihilation operators and ${\hat{E}}^{(-)}=({\hat{E}}^{(+)}{)}^{†}$. An ideal absorbing photodetector responds towith density matrix $\rho$. The diagonal gives the single-count rate up to the detector coupling. Multiple absorptions require higher, normal-ordered field products. Roy J. Glauber used these correlations to develop the quantum account of optical coherence.^[ref]

The ordering follows from a detection amplitude. If absorption takes an initial pure state $|i⟩$ to an unobserved final state $|f⟩$, its field factor is $⟨f|{\hat{E}}^{(+)}|i⟩$. Squaring its complex modulus and summing over a complete orthonormal basis of final states gives $⟨i|{\hat{E}}^{(-)}{\hat{E}}^{(+)}|i⟩$. Averaging over initial preparations replaces this expression by a trace. Thus a detector that absorbs cannot count the Fock vacuum, although the vacuum has quantum fluctuations.

For one selected normal mode, ${\hat{E}}^{(+)}(t)=\mathcal{E}u(t)a$. All Fock states with nonzero occupation, coherent states, and thermal light in this same mode havewhere both denominators are nonzero. Its magnitude is one even when the expected value of the field vanishes. Perfect first-order optical coherence therefore does not identify a coherent state. Pair detection supplies additional information.

For a stable relativistic quantum field theory in Minkowski space, a vacuum is a lowest-energy state for the chosen time evolution; a physical vacuum is normally also invariant under spacetime translations. It need not have zero fluctuations or vanish under every field operator. A free field's Fock vacuum is annihilated by its annihilation operators, while an interacting vacuum need not equal the free-theory Fock state. Multiple degenerate vacua permit spontaneous symmetry breaking. In curved spacetime the choice of time evolution and positive-frequency modes introduces additional vacuum ambiguity in a nonstationary spacetime.