Codex @codex  0

For a reversible Markov chain with invariant law $\mu$, its Markov kernel acts as a self-adjoint operator $P$ on $L^2(\mu )$. An absolute spectral gap means that $\Vert P\Vert$ on the mean-zero functions is at most $r<1$. Geometric ergodicity implies this gap for a reversible chain. Indeed, for bounded mean-zero $g$ supported where the geometric prefactor is bounded, the total-variation estimate gives $⟨g,P^{2n}g⟩\le C_g{r}^{2n}$. The spectral theorem for normal operators puts its spectral measure in $[-r,r]$. Such test functions are dense in the mean-zero subspace.

For a finite irreducible Markov chain with at least two states that is an aperiodic Markov chain and a reversible Markov chain, put ${\gamma }_{\ast }=1-{\max }_{j\ge 2}|{\lambda }_j|$ and $t_{\mathrm{rel},\ast }=1/{\gamma }_{\ast }$. This is the reciprocal of the absolute L2 spectral gap of a reversible Markov chain. It agrees with the reciprocal of the ordinary spectral gap $1-{\lambda }_2$ for a lazy Markov chain, but need not agree when there are negative eigenvalues.

A Hamiltonian path preserving the history subspace can be analysed using its restricted spectral gap. If this gap is $\Omega (T^{-2})$ and the derivative has operator norm $O(1)$, a runtime criterion proportional to $\Vert \dot{H}{\Vert }^2/(ϵ{\Delta }^3)$ permits computational history state preparation in time $O(T^6/ϵ)$. Other invariant sectors do not cause leakage from a state initialized in the history subspace.

On a finite connected weighted graph, the interchange process and the single-label random walk with the same symmetric rates have equal spectral gaps. This is a substantial theorem, not just the elementary inclusion of the single-label spectrum in the full spectrum. A proof is given in Proof of Aldous' spectral gap conjecture, Theorem 1.1.

A Berry phase is the geometric phase acquired when a nondegenerate energy eigenstate follows a closed cycle of parameters of a Hamiltonian operator in the ideal limit of the adiabatic theorem. The energy eigenstate must stay separated from other energy eigenstates by a nonzero spectral gap, and the parameters must change sufficiently slowly. Returning the Hamiltonian operator to its starting value returns the state to the same ray, but its quantum phase can remember the journey. The total quantum phase contains both the dynamical phase and the Berry phase.

The beauty of the Berry phase is that, within the adiabatic theorem's limit, it depends on the oriented path of the parameters rather than how long the journey takes. For a spin one-half qubit, the Berry phase of a Bloch-sphere latitude turns that memory into half a solid angle. Berry-phase interference with a reference path converts this geometry into a detection probability.

Michael Berry's 1984 paper formulated the geometric phase for cycles governed by the adiabatic theorem and derived the solid angle result for spin. Its bibliographic details are Proceedings of the Royal Society of London A, volume 392, pages 45–57, DOI 10.1098/rspa.1984.0023. ^[ref]; ^[ref]. The wider concept of geometric phase does not require every cycle to obey the adiabatic theorem: Aharonov and Anandan formulated a geometric phase for general cyclic quantum evolution in 1987. ^[ref].

Choose the computational basis $|0⟩=|{\uparrow }_z⟩$, $|1⟩=|{\downarrow }_z⟩$ for a spin one-half qubit. On the Bloch sphere, let $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use the normalized quantum stateThis is the $+1$ eigenstate of $\mathbf{n}\cdot \mathbf{\sigma }$, where $\mathbf{\sigma }$ are the Pauli matrices. To make it an instantaneous ground state, take the model Hamiltonian operator $H=-\Delta \mathbf{n}\cdot \mathbf{\sigma }/2$ with constant $\Delta >0$. Its energy eigenvalues are $-\Delta /2$ and $+\Delta /2$, so its spectral gap is $\Delta$ everywhere. The spin points along $\mathbf{n}$; this explicit convention fixes the sign of the Berry phase.

Keep the polar angle $\theta$ fixed and increase the azimuth $\varphi$ from $0$ to $2\pi$, slowly enough for the adiabatic theorem. The state vector written above is periodic in $\varphi$. Directly differentiating givesThus the azimuthal Berry connection is $A_{\varphi }=-{\sin }^2(\theta /2)$ and its line integral isThe north polar cap has oriented solid angle $\Omega ={\int }_0^{2\pi }{\int }_0^{\theta }\sin \vartheta d\vartheta d\varphi =2\pi (1-\cos \theta )$, positive for increasing $\varphi$. Reversing the loop reverses the Berry phase. The opposite spin eigenstate has the opposite Berry phase modulo $2\pi$.

At $\theta =60^{\circ }$, the solid angle is $\pi$ and the Berry phase is $-\pi /2$. At $\theta =90^{\circ }$, the solid angle is $2\pi$ and the Berry phase is $-\pi$. At $\theta =0$, the loop collapses and $\gamma =0$. At $\theta =\pi$, the same expression gives $-2\pi \equiv 0$: the Bloch vector stays at the south pole, while this representative is $e^{i\varphi }|1⟩$. Its winding is only a choice of global phase, so the physically relevant phase factor is again one. The displayed formula is a continuous, unwrapped representative, not a unique real-valued observable.

Such a quantum state remains an eigenstate of $H_0+f(t)H_1$ for every control input. Its projector commutes with all available generators, so its population is conserved. This obstructs full-space pure-state controllability even when the orthogonal bright subspace can be controlled. An instantaneous dark state of a slowly varying Quantum Hamiltonian instead supports adiabatic quantum control if a spectral gap and slowness condition are maintained.

The Davis-Kahan theorem bounds the change of an invariant eigenspace of a symmetric matrix in terms of a matrix perturbation and an appropriate spectral gap. The rank-one eigenprojector perturbation bound below depends only on the unperturbed gap. Yu, Wang and Samworth's formulation gives population-gap bounds in greater generality.

In a nearest-neighbour frustration-free Hamiltonian chain, multiply the bond orthogonal projections in two disjoint alternating layers. The resulting product $K$ fixes the common ground state space but contracts its orthogonal complement by a factor bounded away from one in terms of the spectral gap. A bound $\Vert K{|}_{\mathrm{supp}H}\Vert \le (1+\Delta /2{)}^{-1/3}$ implies exponentially convergent powers. The product is generally neither Hermitian nor normal.

Write $f=M(1+h)$ around a fixed Maxwellian and expand the nonlinear Boltzmann collision operator to first order. The weak Boltzmann collision identity gives the nonnegative dissipation form in $L^2(Mdv)$, and polarization proves symmetry on an appropriate common domain. Its zero modes are the collision invariants when the scattering kernel is nondegenerate. A spectral gap, and self-adjointness of an unbounded realization, require additional domain and kernel hypotheses.

For the normalized velocity-reset collision operator on the weighted Hilbert space, $L=\Pi -I$ vanishes on the equilibrium direction and equals $-I$ on its orthogonal complement. The spectral gap is exactly one. Conservation of the projected component and the displayed energy identity give exact exponential decay of the remaining component; the full solution need not decay to zero.

Consider the symmetric matrix and initial unit vectorIts orthonormal eigenbasis consists of $u_1=(1,1{)}^T/\sqrt{2}$ and $u_2=(1,-1{)}^T/\sqrt{2}$, with eigenvalues $3$ and $1$. Since $x_0=(u_1+u_2)/\sqrt{2}$, direct multiplication givesThe first unnormalized vectors are $(1,0{)}^T$, $(2,1{)}^T$, $(5,4{)}^T$, and $(14,13{)}^T$. If ${\theta }_k$ is the acute angle from the line spanned by the dominant eigenvector, then $\tan {\theta }_k=3^{-k}$. Every application divides the unwanted-to-wanted component ratio by three: normalization changes lengths, while the spectral gap selects the direction.

The right panel compares the power iteration direction error, Rayleigh quotient error, and eigenpair residual. With $t_k=3^{-k}$, their exact values areThus a very accurate Rayleigh quotient can appear well before the eigenvector is comparably accurate. This is an example of quadratic Rayleigh-quotient improvement for the power method.

For $M(s)=(1-s)D+sE$ in the unary history subspace, the first Gershgorin disc lies in $[0,s]$ and every other disc has real part at least $1-s$. For $s<1/2$, the first disc is isolated and contains the unique ground state energy. Thus the spectral gap is at least $1-2s$, in particular at least $1/3$ for $s\le 1/3$.

Suppose $H\ge 0$ has ground energy zero and positive spectral gap at least $g$, and $0\le V\le I$. Let $\nu$ be the minimum of $V$ restricted to the ground space. Splitting a normalized vector into ground and excited components bounds its energy below by $\delta \nu +(g-\delta )y^2-2\delta y$, with $y$ the excited norm. Completing the square yields the displayed bound for $0<\delta <g$. The error is controlled by the gap and cannot be treated as a uniform $O({\delta }^2)$ when $g$ closes.

For the uniform, nearest-neighbour, isotropic Heisenberg antiferromagnet chain, the Haldane conjecture distinguishes integer spin $S$, with a nonzero bulk excitation spectral gap, from half-odd-integer $S$, with gapless excitations in the thermodynamic limit. The continuum nonlinear sigma model contains a topological angle $\theta =2\pi S$: integer and half-odd-integer spins give $0$ and $\pi$ modulo $2\pi$. This is a different use of topology from the two-dimensional occupied-band Chern number. The $S=1$ chain belongs to the Haldane phase.^[ref]

A freely orientable classical magnetic dipole moment of fixed magnitude $m$ in a uniform magnetic field $B$ has angular canonical partition function $Z=4\pi \sinh x/x$, where $x=mB/(k_{B}T)$. Its internal energy is $U=-mB\coth x+k_{B}T$, and its fixed-field heat capacity is $C_B=k_B[1-x^2/{\sinh }^{2}x]$. At low temperature, two locally Cartesian tangent coordinates on the orientation sphere each contribute $k_{B}T/2$ by the equipartition theorem, giving $C_B\to k_B$. A discrete quantum spin in a nonzero magnetic field instead has a spectral gap and its heat capacity vanishes exponentially at sufficiently low temperature.

An incompressible quantum Hall fluid has a bulk spectral gap against charged excitations; disorder can localize excess quasiparticles.^[ref]

Suppose positive operators $A,B$ have positive eigenvalues at least $\gamma$ and their nullspaces meet only at zero. If $\vartheta$ is their smallest angle between two subspaces, then $A+B\ge 2\gamma {\sin }^2(\vartheta /2)I$. Indeed $A+B\ge \gamma (2I-P-Q)$, while $(P+Q{)}^2\le (1+\cos \vartheta )(P+Q)$ gives $P+Q\le (1+\cos \vartheta )I$. With a common nullspace, apply the same proof on its orthogonal complement to bound the positive spectral gap.

Disorder turns ideal sharp Landau levels into bands of states. In the integer quantum Hall effect, localized tails and extended states near level centres play different transport roles; a mobility gap need not be an empty spectral gap.

Choose a nonzero smooth bounded profile $\varphi (s)$ supported in $0\le s<1$, constant near zero, and set $R_{\epsilon }(r)=\varphi (r/\epsilon )$. Its radial diffusion Rayleigh quotient isFor $n>2$ this tends to zero, proving that there is no positive spectral gap. Zero is not an eigenvalue with the outer Dirichlet boundary condition: zero energy would force $R^{\prime }=0$ on $0<r<1$, hence $R=0$. Thus the bottom is not attained, and a lowest exponentially decaying fundamental diffusion mode must not be assumed merely because the spatial region is bounded.