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The perturbed Hamiltonian is $H^{\prime }=(Q^2+P^2)/2-g(t)Q$. Hamilton's equations give $\dot{Q}=P$, $\dot{P}=-Q+g(t)$, henceIn the paper's time units the natural frequency is one. The retarded Green function is $G_R(t)=\Theta (t)\sin t$: it vanishes before the source, is continuous at zero and has a unit derivative jump there. With Fourier convention $e^{-i\omega t}$,Both real-frequency poles are displaced below the real axis. Causality therefore selects the convolutionFor $t<0$ the exponential rises as $e^{{\omega }_0{t}^{\prime }}$ throughout the integral. For $t\ge 0$, split the integral at $t^{\prime }=0$. Evaluation yieldsThe response is nonzero before $t=0$ because the prescribed force is already nonzero then; this does not violate causality. Both $Q$ and $\dot{Q}$ are continuous at zero. The force eventually decays, leaving the free oscillation ${\omega }_0^2\sin t/(1+{\omega }_0^2)$. A general solution adds $A\cos t+B\sin t$, but the retarded condition of no incoming free oscillation as $t\to -\infty$ fixes $A=B=0$.

The oscillator equation is well-defined on the full $(Q,P)$ plane. Interpreted through the positive-$p$ chart of part (c), it describes the original real variables only while $P>|Q|$, with $P=\dot{Q}$. The requested retarded oscillator solution is a continuation in the new variables; it is not a claim that this canonical chart remains valid at all times for every ${\omega }_0$.

Let the coordination number be $z_c=4$, and choose the spin-space $z$ axis along $\mathbf{S}=S\hat{\mathbf{z}}$. The mean-field decoupling of a bond isThe original directed neighbor sum counts each bond twice, so its factor $1/2$ cancels that double counting. The resulting energy isThe constant prevents double counting of the mean interaction energy and is necessary if free energy is minimized with respect to $S$. With the source's unnormalized solid-angle measure, the single-site integral isThereforeAt $S=0$ take $\sinh x/x\to 1$. A normalized solid-angle measure would remove the factor $(4\pi {)}^N$ without changing the self-consistency equation. These are three-component classical spins on a two-dimensional positional lattice, not spins constrained to a plane.

The PDF's hint is a correct angular-integration identity for a term with coefficient $\beta J/2$. That coefficient is not the full single-site molecular field after summing all four neighbors and decoupling both entries of each bond. The net coefficient is $\beta z_{c}J$, as follows from the explicitly stated Hamiltonian; inserting the hint's coefficient as the total field would lose the coordination and double-counting factors.

The azimuth is a cyclic coordinate, so its canonical momentum is conserved:Initially the velocity is parallel to the axis in a field-free region, so $p_{\theta }=0$. Conservation of canonical angular momentum across a solenoid fringe therefore gives, inside the uniform field,This holds away from the coordinate singularity $\rho =0$; the Cartesian orbit will be smooth there.

Time independence also conserves the energy Hamiltonian,The velocity-linear magnetic term cancels in this Legendre transform. Thus $\gamma ={\gamma }_v=(1-v^2/c^2{)}^{-1/2}$ is constant, as expected because a static magnetic Lorentz force does no work. Mechanical angular momentum can change even though $p_{\theta }$ and the total speed are conserved.

The square lattice has coordination number of a lattice $z=4$. The mean-field decoupling isAccounting for the factor $1/2$ in the directed-neighbour sum gives the Hamiltonian in the mean-field approximation:The constant term is necessary to correct the double counting of the interaction. Let $h=4JS+B$. The one-site partition function isso $Z_{\mathrm{MF}}=e^{-2\beta JN{S}^2}{Z}_1^N$. The Helmholtz free energy is thereforeThe equilibrium mean-field approximation is obtained by minimizing this expression over the allowed $S$. The $s=0$ state supplies the extra $1$ in $Z_1$; omitting it would instead describe the two-state Ising model. This is the Blume-Capel mean-field transition at zero single-ion anisotropy.

The canonical commutation relation $[x^{\prime },p^{\prime }]=i\hslash I$ and hamiltonian $H^{\prime }=p^{\prime 2}/(2m)+m{\omega }^2{x}^{\prime 2}/2$ giveSubstitution in the Heisenberg equation of motion yields ${\dot{x}}^{\prime }=p^{\prime }/m$ and ${\dot{p}}^{\prime }=-m{\omega }^2{x}^{\prime }$. Differentiating once more gives the harmonic-oscillator equation for each operator. Its initial-value solution isDifferentiating these expressions at zero verifies both prescribed initial derivatives. They also preserve $[x^{\prime }(t),p^{\prime }(t)]=i\hslash I$.

The spin operators here are dimensionless: $S^j={\sigma }_j/2$, so $\Omega$ has units of inverse time and $\Phi$ is a dimensionless pulse area. Integrating the Schrödinger evolution across an infinitesimally short pulse gives $U_{\Phi }=e^{-i\Phi S^x}$. The finite free Hamiltonian contributes a vanishing integrated phase within that pulse. Between pulses the free evolution is $U_{\Omega }=e^{-i\Omega T{S}^z}$.

In the $S^z$ eigenbasis, $S^z=\mathrm{diag}(1,-1)/2$ and $({\sigma }_x{)}^2=I$. ConsequentlyStarting just before a pulse, the kick acts first and the free interval second. Operators act on states right to left, henceThere are exactly $N$ kicks, at times $0,T,\dots ,(N-1)T$; the pulse at $NT$ is excluded. Sampling just after kicks would change the displayed product order. This endpoint convention specifies the delta-kicked spin Floquet operator.

Put $g=k\hslash /(2m\omega )$, which has units of energy. In the original oscillator basis,Write $A=\tilde{a}$, $B=\tilde{b}$, $C_2=\cosh 2\theta$ and $S_2=\sinh 2\theta$. The transformed number and pair terms obeyHence the coefficient of $AB+A^{†}{B}^{†}$ is $g{C}_2-\hslash \omega S_2$, which vanishes for

This cancels the requested pair terms but does not diagonalize the printed Hamiltonian. Its number-conserving coupling also transforms:After the pair cancellation, the last creation terms have coefficient $-g{S}_2/2$, nonzero for $k\ne 0$. They create two quanta from the $A,B$ vacuum, so that vacuum is not an eigenstate of the literal Hamiltonian. This is an inconsistency in the ground-state assertion used by the next part, present in the printed PDF.

For the actual coordinate-coupled system, the normal coordinates $q_{\pm }=(x\pm y)/\sqrt{2}$ and their conjugate momenta diagonalize $H$ into independent oscillators withA normalizable ground state requires $|k|<m{\omega }^2$. Eliminating just the two indicated cross-pair terms is not sufficient to establish that ground state.

The stated annihilation conditions can be solved algebraically, even though (b) shows they are not ground-state conditions for the full printed Hamiltonian. Invert the canonical transformation:For $|{\psi }_{\lambda }⟩=\mathcal{N}{e}^{\lambda a^{†}{b}^{†}}|0,0⟩$, the commutators giveThus both $A$ and $B$ annihilate it precisely whenThe normalized two-mode squeezed vacuum isThe factorial from the exponential cancels those generated by the two creation operators. Normalization follows from a geometric series; the problem does not require its value, but it makes the subsequent partial trace explicit.

For comparison, the actual ground wave function of the printed stable Hamiltonian isIt follows directly from the two normal-mode Gaussian vacua. The proposed $|{\psi }_{\lambda }⟩$ is instead the exact ground state of the distinct pair-coupling Hamiltonian $\hslash \omega (a^{†}a+b^{†}b+1)+g(ab+a^{†}{b}^{†})$ under its stability condition. Omitting the remaining terms of $kxy$ silently would change the problem.

For a unitary transformation $e^S$, take $S^{†}=-S$ and $S=O(t)$. Operator conjugation by an exponential givesSeparate the unperturbed energy $\mathcal{E}(\Delta )$ from $H_t=O(t)$. To first order,Therefore eliminating the first-order hopping requiresThis is a Schrieffer-Wolff transformation: virtual transitions are removed from the leading Hamiltonian and reappear as higher-order effects within a retained occupation sector. The same canonical-elimination strategy relates localized-moment and effective-exchange descriptions in the original construction.^[ref]

Such a choice requires the hopping to connect separated unperturbed energies. If it connects exactly degenerate states, those states must remain in the retained degenerate block; a singular perturbative generator is not an acceptable replacement.

A linear number-difference energy gives only a one-body level renormalization. Let $\mathcal{E}(\Delta )=E_c+\gamma \Delta$, with $\gamma \ne 0$. Then $f=t/(2\gamma )$ is constant, so the exchange coefficient cancels. Since $[X,Y]=\Delta$,The same-direction commutators vanish as well, so this is the whole second-order term. This agrees with expanding the two-site single-particle splitting $\sqrt{{\gamma }^2+t^2}$ with the appropriate assignment of levels. A collection of noninteracting bosons remains noninteracting after a change of single-particle basis. The case $\gamma =0$ is resonant and cannot be treated by this elimination.

A nonlinear energy generally makes the virtual denominators occupancy-dependent. Then $f(\Delta )$ and $f(\Delta +2)$ need not agree, and the occupancy-preserving effective Hamiltonian contains density-dependent shifts and a spin-exchange coefficientmultiplying ${\mathbf{s}}_a\cdot {\mathbf{s}}_b$. Its sign depends on the function and sector; nonlinearity alone does not fix a universal exchange sign. The full second-order result can additionally contain pair hopping between different occupation sectors.

For a concrete repulsive example, take $\mathcal{E}=U{\Delta }^2/4$, $U>0$, and restrict to two bosons with one on each site. Here $f(0)=-t/U$, $f(2)=t/U$, $N_a=N_b=1$, givingThe triplet energy shift is $-4t^2/U$ and the singlet shift is zero. This is ferromagnetic bosonic superexchange in this example: the symmetric spin sector can lower its energy through virtual double occupancy. The sign must not be imported from the fermionic Hubbard problem. The example specifies an additional energy function solely to illustrate the general expression; it is not an asserted numerical input.

The occupation formula from part (b) givesConsequentlyThe independent occupation numbers $n_1,n_2$ are each zero or one. The Hamiltonian is independent of $n_1$, so its four states haveEach energy admits $n_1=0$ or $n_1=1$; these states differ by occupancy of the zero-energy mode. The ground energy is $-|t|$ for either sign of $t$. These degeneracies refer to the full two-mode fermionic Fock space. If one separately restricts to a fixed global fermion parity, only one state at each energy remains in that parity sector.

The annihilation operator associated with the new coordinate $X=x-x_0$ isAs the initial state is the old vacuum,Equivalently, applying $a^{\prime }$ to the number-state expansion in part (c) and shifting its summation index returns $-\lambda$ times the same state. It is therefore a coherent state of the displaced oscillator.

The new Hamiltonian is $H=U_{\min }+\hslash \omega (a^{\prime †}{a}^{\prime }+1/2)$. In the Schrödinger picture its evolution givesConsequentlyThe displayed prefactor is only a global phase. Equivalently, in the Heisenberg picture $a^{\prime }(t)=a^{\prime }(0)e^{-i\omega t}$. The mean occupation $|\alpha (t){|}^2={\lambda }^2$ is constant, and $\hslash \omega ({\lambda }^2+1/2)$ reproduces the energy above the new minimum found in part (b). Its mean position and momentum likewise reproduce that part's oscillating wavepacket centre.

Use the ordered orthonormal basis $(|1s⟩,|2p_x⟩)$. The projected Hamiltonian between pulses is $H_0=-\Delta {\sigma }_z$, so its free time-evolution operator through a period isAcross one instantaneous pulse, integrating the Schrödinger equation givesusing the exponential of a Pauli vector. The finite free Hamiltonian contributes vanishing evolution during the zero-duration kick. A period measured immediately before successive pulses therefore consists of the kick followed by free evolution. Composition givesThis is the Floquet operator at the specified phase of the pulse train. Actual hydrogenic energy levels are not opposite numbers: the displayed convention subtracts their midpoint, with $\Delta =(E_{2p}-E_{1s})/2$. Restoring that midpoint multiplies the result by an unobservable global phase. Likewise $E$ in the given Hamiltonian denotes the electric coupling per unit length; if it is instead a physical electric-field strength, its charge factor must be included in $E{x}_0$.

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).

Take the two magnetic moments along $z$ to be $\pm \mu$, so $H=-\mu B_z{\sigma }_z$ and the two energies are $\mp \mu B_z$. The quantum Gibbs state at inverse temperature $\beta =1/(k_{B}T)$ isIts Bloch vector is $(0,0,\tanh \beta \mu B_z)$. At $B_z=0$, or in the high-temperature limit, it approaches $I/2$; at $T\to 0^{+}$ with nonzero $\mu B_z$, it approaches the lower-energy pure state. Here $\mu$ is the magnitude or signed coefficient multiplying ${\sigma }_z$ in the magnetic Hamiltonian, rather than a moment with an additional unstated factor of two.

The Hamiltonian has the discrete symmetry described in solution, whereas either selected easy-axis ground state has a nonzero order parameter $⟨S^x⟩$ and is not invariant under it. This is spontaneous symmetry breaking: the equations preserve the symmetry while the selected macroscopic quantum state does not.

An infinitesimal longitudinal magnetic field selects one orientation. The defining order of limits isin the ordered zero-temperature phase, with the opposite sign for $h\to 0^{-}$. A symmetric finite-volume quantum state can instead have zero magnetization. Quantum ordered states are generally dressed by fluctuations rather than exact fully polarized product states, because unequal transverse exchange can create pairs of spin waves. Discrete easy-axis symmetry does not imply a gapless Goldstone mode; indeed solution finds a nonzero spin wave gap. Nor does this argument assert positive-temperature order in a one-dimensional short-range chain.^[ref]

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.

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 $ϵ(0)=\alpha \mu >0$.