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Use spin one-half states $|\uparrow ⟩,|\downarrow ⟩$ with $S^z=\pm 1/2$, so ${\sigma }^i=2S^i$ and $H=-J{\sum }_n({\sigma }_n^x{\sigma }_{n+1}^x+g{\sigma }_n^z)$. At $g=0$, every ferromagnetic bond has its lowest energy $-J$ when adjacent $x$ spins agree. On the connected periodic chain the normalized ground states arewhere $|\pm x⟩=(|\uparrow ⟩\pm |\downarrow ⟩)/\sqrt{2}$. Their span is the two-dimensional ground-state subspace; any normalized superposition is another ground state. They are exchanged by the global spin-flip discrete symmetry generated by ${\prod }_n{\sigma }_n^z$.

As $g\to +\infty$, the field dominates and selectsThe limiting polarized state is unique. At large finite $g$, exchange admixes virtual spin flips, so the product state is the limiting wavefunction rather than an exact finite-field eigenstate. The bulk expansion below gives $E_0/N=-J[g+1/(4g)+O(g^{-3})]$. The exact twofold degeneracy stated at $g=0$ should be distinguished from thermodynamic spontaneous symmetry breaking in the ordered phase: a finite chain at nonzero field can have a split pair of symmetry eigenstates.

The Hellmann–Feynman theorem gives ${\partial }_g{E}_0=⟨{\partial }_{g}H⟩=-2J{\sum }_n⟨S_n^z⟩$. Differentiating the Ising-chain ground-state energy yields ${\partial }_g{E}_0=-J{\sum }_k(g-\cos k)/r_k$. HenceThis is also obtained from the occupations of the Bogoliubov quasiparticle vacuum: $⟨c_k^{†}{c}_k⟩=[1-(g-\cos k)/r_k]/2$ and $S_n^z=1/2-n_n$. At an exact zero mode the finite-sector limiting ground state specifies the occupation. In the bulk limits, the magnetization per site is zero at $g=0$, tends to $1/2$ as $g\to \infty$, and equals $1/\pi$ at $g=1$: there the integrand $(1-\cos k)/r_k=|\sin (k/2)|$.

For an excited eigenstate, $E_i-E_0\ge \Delta$. The matrix-element identity in condition (i), together with the two-sided Fourier cutoff, givesThe reverse matrix element has frequency $E_0-E_i\le -\Delta$ and also vanishes. Equivalently, $A^{(Z)}$ is Hermitian because $h_Z$ is Hermitian and $w$ is real, so the two elements are conjugate. ThusThe spectral gap eliminates precisely the couplings needed to make the unique ground state an eigenvector of every filtered term. Couplings between excited states with smaller energy differences can remain; the spectral filter need not diagonalize the whole operator.

Sum the spectral filtering of Hamiltonian terms over the original finite decomposition. The total Hamiltonian operator commutes with its own evolution, soThe absence of all ground-to-excited matrix elements, proved in part (a), makes each filtered term block diagonal relative to $P_0=|{\varphi }_0⟩⟨{\varphi }_0|$ and its orthogonal complement. ThereforeA filtered term may act on the whole system, since unitary time evolution spreads its original support. Neither commutation with $P_0$ nor preservation of its ground-state expectation asserts that the global ground state minimizes each individual term. In particular, this construction does not generally turn a frustrated decomposition into a frustration-free Hamiltonian.

Use dimensionless spin angular momentum operators, so their commutators have no additional $\hslash$. A fully polarized state $|F⟩={⨂}_n|S,S{⟩}_n$ is an exact ground state of the Heisenberg ferromagnet, with $E_0=-NJ{S}^2$. On each bond the largest eigenvalue of ${\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$ is $S^2$, obtained in the two-spin total-spin-$2S$ sector. Hence the bondwise lower bound on the Hamiltonian is attained by this state.

The ferromagnetic ground-state multiplet has total spin $NS$. Applying the global lowering operator $(S_{\mathrm{tot}}^{-}{)}^r$, $r=0,\dots ,2NS$, generates its $2NS+1$ magnetic states; their linear combinations are also ground states. Global spin rotations generate polarized ground states pointing in arbitrary directions and span the same multiplet. Uniform reorientation therefore costs no energy. Slowly varying orientations give gapless ferromagnetic magnons, with quadratic rather than linear long-wavelength dispersion. In the ordered thermodynamic ground state, the two broken spin generators are canonically paired because $⟨[S_{\mathrm{tot}}^x,S_{\mathrm{tot}}^y]⟩=i⟨S_{\mathrm{tot}}^z⟩\ne 0$, producing one type-B Goldstone boson. For a finite chain the zero-wavevector motion remains inside the exact ground multiplet; nonzero wavevectors have a finite-size spacing that vanishes as $N^{-2}$. This is a ground-state statement, not a claim of finite-temperature long-range order in one dimension.

In the ground state of the diagonal Hamiltonian, ${\alpha }_{\mathbf{k}}|0_B⟩=0$. The Bogoliubov transformation therefore gives $⟨b_{\mathbf{k}}^{†}{b}_{\mathbf{k}}⟩=v_k^2$. Summing all nonzero momenta proves the quantum depletion of a Bose condensate:The nonzero depletion arises from correlated pairs in the interacting quasiparticle vacuum, not thermally populated quasiparticles.

The printed expression is the zero-temperature result. At a nonzero temperature the complete leading-order expression instead hasThe additional term is thermal depletion and cannot be discarded merely because the gas is condensed.

Single-valued rotor wavefunctions are periodic in $\varphi$. The normalized eigenfunctions are $⟨\varphi |n⟩=e^{in\varphi }/\sqrt{2\pi }$ with integer $n$, angular momentum $\hslash n$ and energy $E_n={\hslash }^2{n}^2/(2I)$. The quantum rotor consequently hasThe nonzero levels have the $n\leftrightarrow -n$ degeneracy; the ground state is $n=0$. No twisted boundary condition or applied flux has been introduced.

Analytically continue the free-particle return kernel using $M=I$ and $t=-i\hslash \beta$. This fixesSet $a=2{\pi }^{2}I/({\hslash }^2\beta )$. The Gaussian Fourier integral in Poisson resummation isApplying it to the winding sum gives ${\sum }_m{e}^{-a{m}^2}=\sqrt{\pi /a}{\sum }_n{e}^{-{\pi }^2{n}^2/a}$. The prefactors cancel exactly, $2\pi {\mathcal{Z}}_0\sqrt{\pi /a}=1$, and ${\pi }^2/a=\beta {\hslash }^2/(2I)$. Thereforein agreement with the spectral trace. This is momentum-winding duality of the quantum rotor. At low temperature the momentum ground state dominates; at high temperature the winding-zero sector gives $\mathcal{Z}\simeq \sqrt{2\pi I/(\beta {\hslash }^2)}$.