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The oscillator operators satisfy $[a,a^{†}]=1$ and $[a,a]=[a^{†},a^{†}]=0$. Substitution into the kinetic and potential energies givesThe number operator $N=a^{†}a$ is nonnegative because $⟨\psi ,N\psi ⟩=\Vert a\psi {\Vert }^2$. The commutators $[N,a]=-a$ and $[N,a^{†}]=a^{†}$ lower and raise an eigenvalue by one. If an eigenvalue were a noninteger $\nu \ge 0$, repeated lowering would produce a nonzero eigenvector with eigenvalue between zero and one, and one further lowering would have negative eigenvalue; its norm is nonzero because $\Vert a\psi {\Vert }^2=\nu \Vert \psi {\Vert }^2$. This contradicts nonnegativity. Thus only nonnegative integers are allowed. The normalized Gaussian solves $a|0⟩=0$, and raising it supplies every level, withFor the three-dimensional unperturbed oscillator the product states are $|n_1,n_2,n_3⟩=|n_1⟩\otimes |n_2⟩\otimes |n_3⟩$, with energy $\hslash \omega (n_1+n_2+n_3+3/2)$.

Write the perturbation as $\lambda W$, where $W=(\hslash \omega /2){\sum }_{i<j}(a_i+a_i^{†})(a_j+a_j^{†})$. Its ground-state expectation is zero. Acting on the ground state, it produces $|110⟩,|101⟩,|011⟩$, each with coefficient $\hslash \omega /2$. They lie $2\hslash \omega$ above the ground state, so second-order perturbation theory givesIn the first excited subspace with basis $|100⟩,|010⟩,|001⟩$, the first-order perturbation matrix is $(\lambda \hslash \omega /2)\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)$. Its symmetric eigenvector has eigenvalue $\lambda \hslash \omega$, while the perpendicular two-dimensional space has eigenvalue $-\lambda \hslash \omega /2$. Thus the lowest excited triplet splits intoAs an independent check, diagonalizing the quadratic potential gives normal frequencies $\omega \sqrt{1+2\lambda }$ and $\omega \sqrt{1-\lambda }$ twice; their small-$\lambda$ expansions reproduce both results.

Periodic boundary conditions give wavevectors $\mathbf{k}=(2\pi /L)(n_1,n_2)$. Each state occupies area $(2\pi /L{)}^2$ in wavevector space. A ring of radius $k$ contains $L^2(2\pi kdk)/(2\pi {)}^2$ states. With $ϵ={\hslash }^2{k}^2/(2m)$, $kdk=mdϵ/{\hslash }^2$, givingThere is no spin degeneracy. At temperature $T>0$, set $z=e^{\mu /(k_{B}T)}<1$. The Bose occupation integral givesThe integral follows by expanding the denominator as ${\sum }_{\ell \ge 1}{z}^{\ell }{e}^{-\ell ϵ/(k_{B}T)}$ and integrating each positive term.

As $\mu \uparrow 0$, this density diverges. Thus the excited states never have a finite maximum capacity at a positive temperature: every finite prescribed density can be accommodated with a strictly negative chemical potential. The ground-state occupation then stays finite as area tends to infinity and its fraction vanishes. Hence there is no Bose-Einstein condensation at any positive temperature in this homogeneous ideal two-dimensional gas. At exactly zero temperature the ground state is occupied; the printed phrase about any temperature is understood as absence of a nonzero critical temperature, not exclusion of that zero-temperature limit.

For a nonzero normalized or unnormalized trial function in the Hamiltonian's quadratic-form domain, the Rayleigh-Ritz variational principle gives $E_0\le ⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Expanding in energy eigenstates, or using the spectral measure, makes the quotient a weighted average of spectral energies, all at least $E_0$. For an even one-dimensional potential, the ground state is even and the first excited bound state is odd by the nodal ordering theorem. Restricting to odd trial functions therefore gives an upper bound on the lowest odd energy and on $E_1$ when that bound state exists; a negative odd quotient also certifies its existence.

Here ${\hslash }^2/(2m)=1$. For the Gaussian trial function, integration by parts gives kinetic quotient $\int |{\psi }^{\prime }{|}^2/\int |\psi {|}^2=a/2$, and the potential quotient is $-V_0\sqrt{a/(1+a)}$. ThereforeFor $a>0$, its zero obeys $a(1+a)=4V_0^2$, giving the unique positive zeroAlso $E^{\prime }(a)=1/2-V_0/[2\sqrt{a}(1+a{)}^{3/2}]$, so its stationary points satisfy $a(1+a{)}^3=V_0^2$. The left side increases strictly from zero to infinity, proving that there is exactly one stationary point, a minimum.

As $a\downarrow 0$, $E(a)=-V_0\sqrt{a}+O(a)$ approaches zero from below; as $a\to \infty$, it tends to infinity. The sketch and variational inequality therefore give a negative energy for every $V_0>0$. Since this decaying potential has essential spectrum beginning at zero, at least one bound state exists for every positive well depth.

For small $V_0$, the minimizer is $a_{\ast }=V_0^2-3V_0^4+O(V_0^6)$. Substitution gives $E(a_{\ast })=-V_0^2/2+V_0^4/2+O(V_0^6)$. Meanwhile the potential is strictly greater than $-V_0$ except at a single point and the kinetic energy is nonnegative, so a normalized bound-state expectation is strictly greater than $-V_0$. Thus

The canonical commutation relations are $[\hat{x},\hat{p}]=i\hslash I$ (in several dimensions $[x_i,p_j]=i\hslash {\delta }_{ij}I$, with the position-position and momentum-momentum commutators zero). For Hermitian $x,p$, the commutator is anti-Hermitian, just as $i\hslash I$ is; there is no conflict with their self-adjointness. The annihilation operator and its adjoint obeyThus the quantum harmonic oscillator Hamiltonian is $\boxed{H=\hslash (N+1/2)}$. The commutators $[N,a]=-a$, $[N,a^{†}]=a^{†}$ and the normalized ground state constructNormalization follows inductively from $a{a}^{†}=N+1$. Positivity of $N$ and repeated lowering exclude a noninteger eigenvalue: lowering it until it becomes negative would produce a nonzero vector of negative $N$ eigenvalue. Every eigenstate therefore lowers to the unique ground state, so these normalized ladder states exhaust the oscillator eigenstates.

For two independent oscillators, the tensor products $|n_1,n_2⟩$ have $E=\hslash (n_1+n_2+1)$. The level $n=n_1+n_2$ has degeneracy $n+1$. In its degenerate subspace the perturbation $\lambda x_1{x}_2$ reduces tosince terms that raise or lower both numbers leave the subspace. For $n=0$ its matrix is zero. For $n=1$, in $|1,0⟩,|0,1⟩$, it is $(\lambda \hslash /2)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. For $n=2$, in $|2,0⟩,|1,1⟩,|0,2⟩$, it is $(\lambda \hslash /2)\left(\begin{array}{ccc}0& \sqrt{2}& 0\\ \sqrt{2}& 0& \sqrt{2}\\ 0& \sqrt{2}& 0\end{array}\right)$. Their eigenvalues giveThe $n=1$ states are the symmetric and antisymmetric combinations. At $n=2$, the shifted states are $(|2,0⟩\pm \sqrt{2}|1,1⟩+|0,2⟩)/2$, and the zero-shift state is $(|2,0⟩-|0,2⟩)/\sqrt{2}$.

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 $\mathcal{Z}\simeq \sqrt{2\pi I/(\beta {\hslash }^2)}$.

Because each term in the history-subspace propagation Hamiltonian is positive semidefinite, the uniform history vector has zero energy and is a ground state:Indeed every edge term annihilates the uniform coefficients. Conversely, a zero-energy vector must have equal coefficients across every edge of the connected path, so this computational history state is the unique ground state in $\mathcal{L}$. It coherently records every intermediate state of the quantum circuit.

The matrix $D=I-|e_0⟩⟨e_0|$ is an orthogonal projection. Its eigenvalues areIts unique ground state in $\mathcal{L}$ is $|e_0⟩=|{\psi }_0⟩|0{⟩}^{\otimes T}$, and its spectral gap is one.

In the history basis, $M(s)=(1-s)D+sE$ is a Hermitian tridiagonal matrix. Its first diagonal entry is $s/2$, its interior entries are $1$, its last entry is $1-s/2$, and each off-diagonal entry is $-s/2$.

The Gershgorin disc theorem confines the first row's eigenvalue to $[0,s]$, the interior rows to $[1-s,1+s]$, and the last row to $[1-s,1]$. For $0\le s\le 1/3$, the first disc is disjoint from all the others and therefore contains exactly one eigenvalue. Since $M(s)$ is positive semidefinite, that eigenvalue is the ground state energy. All other eigenvalues are at least $1-s$, while the lowest is at most $s$. Hence the Gershgorin gap bound for an adiabatic history path givesThis argument also covers $s=0$, where the first disc is the isolated point zero.

Part (a) shows that $\mathcal{L}$ is invariant throughout the interpolation. As the Hamiltonians are Hermitian, ${\mathcal{L}}^{\perp }$ is invariant as well: evolution starting in $\mathcal{L}$ has no leakage into other sectors. The quantum adiabatic theorem may therefore be applied using the restricted spectral gap, even if other sectors of the full Hamiltonian have different low-energy levels.

Combining (b)(iv) with the supplied bound for $s>1/3$ gives ${\min }_s\Delta (M(s))\ge c/(T+1{)}^2$ for a constant $c>0$. The matrix $E-D$ has diagonal entries $1/2,0,\dots ,0,-1/2$ and off-diagonal entries $-1/2$. Its Gershgorin discs all lie in $[-1,1]$, so $\Vert E-D\Vert \le 1$. Thus the runtime criterion for adiabatic preparation of a computational history state is satisfied bywith a sufficiently large constant.

Starting from the supplied initial ground state $|e_0⟩$, the resulting state ${\rho }_{\mathrm{out}}$ obeysHere the trace norm follows the question's convention; the conventional trace distance is half this norm. The Hamiltonian has $O(T)$ explicitly specified terms, each acting on at most five qubits, so its description and the runtime are polynomial in circuit size. If ${ϵ}^{-1}$ is polynomially bounded, the whole preparation has polynomial cost. The initial computational input is assumed to have been prepared, as in the premise of the quantum adiabatic theorem.

Use an $n$-qubit state as the witness. Uniformly choose a term $h_j$ and perform the two-outcome positive operator-valued measure $\{I-h_j,h_j\}$. Regard the second outcome as an energy flag. For witness $\rho$, its probability isThis is an efficient local-energy measurement verification: a unitary operator on the $k$ relevant qubits and one ancilla qubit can implement the measurement using the isometryThe local dimension is constant; the term matrices and the measurement can be approximated to inverse-polynomial accuracy. Uniform choice of $j$ and the classical threshold comparison also have polynomial overhead.

SetIn the nontrivial case $0\le a<b\le M$, $\delta$ is inverse-polynomial. Cases outside this energy range are decided directly from $0\le H\le MI$.

Ask for $r$ witness registers, measure an independently sampled local term on each, and accept if the mean number of energy flags is at most $\theta$. An honest product ground state on a YES instance has flag probability at most $p_{\mathrm{y}}$. The Hoeffding inequality gives failure at most $e^{-r{\delta }^2/2}$. For completeness, the required Bernoulli concentration estimate follows by bounding the centered log moment-generating function $f(\lambda )=\ln \mathbb{E}{e}^{\lambda (Y-p)}$: $f(0)=f^{\prime }(0)=0$ and $f^{\prime \prime }(\lambda )$ is a tilted Bernoulli variance, at most $1/4$. Hence $f(\lambda )\le {\lambda }^2/8$; exponential Markov's inequality optimized at $\lambda =4u$ yields $\mathrm{Pr}(\overline{Y}-p\ge u)\le e^{-2r{u}^2}$, and the lower-tail version follows in the same way.

Soundness must also cover entangled witness registers. The averaged flag effect is $R=H/M$. All its eigenvalues are at least $p_{\mathrm{n}}$ on a NO instance. The repeated verifier's acceptance effect is a polynomial in the commuting tensor-factor operators $R_j$ and $I-R_j$. In their product eigenbasis its eigenvalues are lower-tail probabilities of independent Bernoulli variables whose parameters are each at least $p_{\mathrm{n}}$. Their lower tail is bounded by the identical-parameter tail at $p_{\mathrm{n}}$, using the same uniform-variable coupling as in QMA parallel repetition with entangled witnesses. Therefore its operator norm is at most $e^{-r{\delta }^2/2}$, which bounds acceptance for any entangled witness.

Choosinggives completeness at least $2/3$ and soundness at most $1/3$. A slightly larger constant absorbs implementation errors; for example, approximate each flag effect in norm to at most $\delta /8$, retain a gap at least $3\delta /4$, and increase $r$ by a constant factor. The total witness length $rn$ and circuit size are polynomial. Thus

Here is a finite-dimensional form of the Kitaev geometrical lemma. Let $A,B$ be positive operators, with nullspaces $S_A,S_B$ and all positive eigenvalues at least $\gamma >0$. First assume $S_A\cap S_B=\{0\}$. Let $P,Q$ be the orthogonal projections onto those nullspaces, and define their smallest angle between two subspaces byIf either nullspace is zero, set $\cos \vartheta =0$. Then

To prove it, spectral decomposition gives $A\ge \gamma (I-P)$ and $B\ge \gamma (I-Q)$. Write $C=P+Q$ and $c=\cos \vartheta$. For every vector $z$,Thus $C^2\le (1+c)C$. Since $C$ is positive semidefinite, each of its eigenvalues lies in $[0,1+c]$, so $2I-C\ge (1-c)I$. Consequentlyas claimed.

If there is a common nullspace $S=S_A\cap S_B$, restrict to $S^{\perp }$ and define the angle between $S_A\cap S^{\perp }$ and $S_B\cap S^{\perp }$ there. The same proof bounds the smallest positive eigenvalue of $A+B$ by $2\gamma {\sin }^2(\vartheta /2)$, while $S$ is its zero-energy subspace. This version gives a spectral gap above a possibly degenerate ground state space.

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$ and define the spectral filtering of Hamiltonian terms byThe normalization $\hat{w}(0)=1$ means $\int w=1$. Since $w$ is real and nonnegative, $\hat{w}(-E)=\overline{\hat{w}(E)}$. Thus the stated positive-frequency cutoff also implies $\hat{w}(E)=0$ for $E\le -\Delta$. The integral exists in operator norm, because its integrand has norm at most $w(t)\Vert h_Z\Vert$.

Let $H|{\varphi }_i⟩=E_i|{\varphi }_i⟩$, with $E_i-E_0\ge \Delta$ for $i>0$. In this eigenbasis,Every matrix element coupling the unique ground state to an excited state vanishes, in either direction. HenceMoreover, linearity and conservation of $H$ under its own Heisenberg picture evolution giveEach filtered term is Hermitian, but it need not remain supported on its original set $Z$. The tail assumption supplies stronger long-time control than the integrability used here; no additional tail estimate is needed for the requested identities.

For each interaction neighbourhood, setThen the shell term is $g^{(Z,s)}=G_s^{(Z)}-G_{s-1}^{(Z)}$. Since $H_0=h_Z$, its conjugation leaves $h_Z$ fixed, so $G_0^{(Z)}=h_Z\int w=h_Z$. Since $H_D=H$, $G_D^{(Z)}$ is precisely the choice of $g^{(Z)}$ from (b). Therefore the telescoping local-shell decomposition givesNo limiting interchange is needed: the sum has finitely many shells. Each $G_s^{(Z)}$ is supported within the union of supports appearing in $H_s$ together with $Z$, because its unitary time evolution acts trivially outside that neighbourhood. The individual shell terms need not commute with the full ground state projector; the full filtered sum does.