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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.

The filtered decomposition is not necessarily frustration-free. Commuting with $P_0$ makes $|{\varphi }_0⟩$ an eigenvector of each term; it does not make that eigenvector a lowest-energy state of each term. This is the distinction expressed by commuting with a ground projector does not imply frustration freeness.

For an explicit positive, two-local counterexample on three qubits, let $P_0^{(i)}=|0⟩⟨0{|}_i$ and $P_1^{(i)}=|1⟩⟨1{|}_i$, and use the two distinct interaction sets $\{1,2\}$ and $\{2,3\}$:Both are positive semidefinite and commute with their sum. The total energy on a computational basis vector isThus $H=h_{12}+h_{23}$ has unique ground state $|000⟩$, energy $1$, and spectral gap $1$. Since $[H,h_{12}]=[H,h_{23}]=0$, spectral filtering of Hamiltonian terms leaves both terms unchanged for every normalized filter:But $h_{12}|000⟩=|000⟩$, whereas $h_{12}|010⟩=0$. The global ground state therefore does not minimize $g^{(12)}$, proving that the resulting decomposition is not a frustration-free Hamiltonian.

There is a useful positive result if the original decomposition already is a frustration-free Hamiltonian. If $a_Z={\lambda }_{\min }(h_Z)$ and $h_Z|{\varphi }_0⟩=a_Z|{\varphi }_0⟩$, then $h_Z-a_{Z}I\ge 0$. Averaging its unitary conjugates with nonnegative $w$ gives $g^{(Z)}-a_{Z}I\ge 0$, while $g^{(Z)}|{\varphi }_0⟩=a_Z|{\varphi }_0⟩$. Thus filtering preserves existing frustration freeness; it does not create it for an arbitrary gapped Hamiltonian.

The Bose-Einstein distribution gives $n_i=[\exp ((E_i-\mu )/(k_{B}T))-1{]}^{-1}$, with chemical potential below the lowest energy when its discrete occupation is finite. In the continuum approximation,If the zero-energy ground state is macroscopically occupied, its discrete occupation $N_0$ must be added to the first integral; it contributes no energy to the second. This distinction is essential when applying the formulas to Bose-Einstein condensation.

For fixed particle number, Bose-Einstein condensation means that a macroscopic fraction occupies the single-particle ground state. As $\mu \uparrow 0$, the excited-state population reaches its maximumAt the transition this equals $N$, soBelow $T_c$, the chemical potential is approximately zero in the large-$N$ continuum description. Additional particles occupy the ground state, givingFor finite $N$, the actual chemical potential remains slightly negative; the saturation statement is the macroscopic approximation used here.

Ordinary equilibrium blackbody photons do not undergo this fixed-number condensation. Their number adjusts through absorption and emission, while their chemical potential stays zero; cooling reduces the equilibrium photon number rather than forcing an excess into the ground state. This answer concerns the thermal blackbody model, not specially constrained systems where an effective photon number is maintained.

The sudden approximation keeps the wavefunction unchanged at the instant of switching: a finite step of potential creates no delta-function impulse in the Schrodinger equation. Just before switching the old ground state differs from the printed real wavefunction only by an overall phase, which does not affect probabilities.

Let ${\tilde{\psi }}_0(x)=(\pi \hslash {)}^{-1/4}{e}^{-(x-d{)}^2/(2\hslash )}$, $d=qE$, be the new ground state. Its overlap with the initial state is, up to that phase,Since $x^2+(x-d{)}^2=2(x-d/2{)}^2+d^2/2$, the Gaussian integral gives $A_0=e^{-d^2/(4\hslash )}$. The Born rule and the ground-state overlap after sudden oscillator displacement thus giveThis is an overlap probability, not a claim that the original wavefunction instantaneously relaxes to the new ground state. It becomes one when $qE=0$.

Set $d=qE$. With the stated unit mass and oscillator frequency, complete the square:Translation by $d$ preserves the spectrum of the quantum harmonic oscillator, and the last term is a constant energy shift. The displaced quantum harmonic oscillator therefore hasIts normalized eigenfunctions are the original oscillator eigenfunctions evaluated at $x-d$, in particular the ground state is $(\pi \hslash {)}^{-1/4}{e}^{-(x-d{)}^2/(2\hslash )}$. The energy gaps remain $\hslash$.

The $-\mu \psi$ term means that the generator is the grand potential, rather than just the physical energy. With boundary variations vanishing, the Hamiltonian functional isIts functional derivative isso the Gross–Pitaevskii equation is $i\hslash {\psi }_t=\delta H/\delta {\psi }^{\ast }$. Treating $\psi$ and ${\psi }^{\ast }$ as independent variables in this variation gives the factor $g$, not $2g$.

For repulsive interactions $g>0$ and $\mu >0$, complete the square in the local energy:The gradient contribution is nonnegative, so the homogeneous ground state has arbitrary constant complex argument andThe infinite-volume constant may be subtracted to define a relative grand potential; it does not affect the functional derivative. At fixed particle number one instead minimizes the physical energy $H+\mu N$, with $\mu$ as its Lagrange multiplier. For $g>0$, $\mu \le 0$ the grand-potential minimum over $n\ge 0$ is the vacuum $n=0$. For $g<0$ the displayed unconstrained functional is unbounded below; the positive-density stable uniform state assumed in the following parts requires repulsion.

Invert the operator definitions to obtain $a=\sqrt{m\omega /(2\hslash )}\hat{x}+i\hat{p}/\sqrt{2\hslash m\omega }$ and the adjoint formula for $a^{†}$. The canonical commutation relation yields $[a,a^{†}]=1$, with $[a,a]=[a^{†},a^{†}]=0$. Substitution givesWrite $N=a^{†}a\ge 0$. Since $[N,a]=-a$ and $[N,a^{†}]=a^{†}$, these are lowering and raising operators. Applying $a$ to a ground state would lower its energy unless it vanishes, so $a|0⟩=0$. From the normalized unique ground state constructThe commutator gives the factorial normalization and $a|n⟩=\sqrt{n}|n-1⟩$. For any energy eigenstate, repeated lowering cannot pass below $N=0$; positivity of each lowered norm forces its $N$ eigenvalue to be a nonnegative integer and lowering eventually reaches the ground state. Its uniqueness then yields the entire eigenstate ladder. The usual oscillator on the real line has discrete complete spectrum because its potential is confining, so these states span its Hilbert space.

For the perturbation $\lambda W$ with $W=\hslash \omega (a^2+a^{†2})$, nondegenerate time-independent perturbation theory gives $E_n^{\prime }=E_n+\lambda W_{nn}+{\lambda }^2{\sum }_{m\ne n}|W_{mn}{|}^2/(E_n-E_m)+O({\lambda }^3)$. The first-order term vanishes. Only $m=n-2,n+2$ contribute, with squared matrix elements $(\hslash \omega {)}^{2}n(n-1)$ and $(\hslash \omega {)}^2(n+1)(n+2)$ respectively. ThusThe absent downward transition for $n=0,1$ is automatically represented by $n(n-1)=0$. Direct substitution also givesFor $|\lambda |<1/2$ this is another oscillator with mass $m^{\prime }=m/(1-2\lambda )$ and frequency ${\omega }^{\prime }=\omega \sqrt{1-4{\lambda }^2}$. ThereforeIts expansion is $E_n[1-2{\lambda }^2+O({\lambda }^4)]$, agreeing with the perturbative result and sharpening its remainder. At the excluded endpoints confinement is lost; beyond them the quadratic Hamiltonian is not bounded below.