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The purity of a density operator is $\gamma =\mathrm{Tr}({\rho }^2)$. It equals one exactly for a pure state; in finite dimension $d$ it lies between $1/d$ and $1$. The minimum occurs exactly at the maximally mixed state $I/d$.

A worked qubit family interpolates between the mixed state $I/2$ and the pure state ${\rho }_{+}=|+⟩⟨+|$, where $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$:The eigenvectors are $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, with eigenvalues ${\lambda }_{\pm }=(1\pm p)/2$. Squaring and summing these eigenvalues givesHere $P_X(+)$ is the Born rule probability of the positive outcome in an $X$ projective measurement. The Bloch vector is $(p,0,0)$, since ${\rho }_p=(I+p{\sigma }_x)/2$ with ${\sigma }_x$ one of the Pauli matrices. At $p=0$ the purity is $1/2$; at $p=1/2$ it is $5/8$, with $P_X(+)=3/4$; at $p=1$ it is one. Within this family, $p=1$ is a pure state and every $p<1$ is a mixed state. Extending the formula to $p>1$ produces a negative eigenvalue $(1-p)/2$, so it ceases to be a density operator despite remaining Hermitian with trace one.

The Bloch vector length measures polarization: for any qubit density operator, $\gamma =(1+|\mathbf{r}{|}^2)/2$. This connects the interior of the Bloch ball with mixed states and its boundary, the Bloch sphere, with pure states. ^[ref]

For a short proof of the general bound, use the spectral theorem: the nonnegative eigenvalues ${\lambda }_j$ sum to one, so $\gamma ={\sum }_j{\lambda }_j^2$. ThenThe lower bound is attained only when all eigenvalues equal $1/d$; the upper bound is attained only when exactly one eigenvalue is nonzero, and hence equals one. Purity is therefore a simple quantitative test of mixedness, invariant under a unitary operator. It does not specify the entire density operator: even distinct pure states all have purity one. ^[ref]

Independent verifier circuits on $r$ witness registers induce a threshold acceptance effect built from tensor products of $M$ and $I-M$, where $M$ is the quantum verifier acceptance operator. In a product eigenbasis of $M$, the effect's eigenvalues are threshold probabilities for independent Bernoulli random variables with parameters given by the selected eigenvalues of $M$. Bounding every parameter on a NO instance bounds the effect's operator norm, even if the actual submitted witness is entangled. Independence is used only for the spectral calculation, not assumed for the witness.

The QR algorithm computes eigenvalues through repeated QR decompositions and similarity updates. Shifted and unshifted forms have different convergence rates; the unshifted QR algorithm uses a zero shift at every step.

At a double-zero eigenvalue with centre equations $\dot{u}=p$, $\dot{p}=0$ to linear order, stable filters $\dot{v}=-v+u^2$ and $\dot{w}=-\tau w+u^2/\tau$ have the displayed quadratic centre manifold graphs. Solving $p{\partial }_{u}h=-\alpha h+u^2/\alpha$ gives $h=u^2/{\alpha }^2-2up/{\alpha }^3+2p^2/{\alpha }^4$. Under the weighted scaling $u=O(\epsilon )$, $p=O({\epsilon }^2)$, the $p^2$ graph term affects an $uh$ coupling only at order ${\epsilon }^5$.

// Target: dynamical-systems.bigb

For $\dot{u}=u(\mu +u-v)$, $\dot{v}=\lambda u^2$ in $u>0$, the time change $d\tau /dt=u$ gives the affine linear system $u_{\tau }=\mu +u-v$, $v_{\tau }=\lambda u$. Shifting $w=v-\mu$ removes the forcing. Its eigenvalues are $(1\pm \sqrt{1-4\lambda })/2$. For $\lambda >1/4$, putting $\omega =\sqrt{4\lambda -1}/2$ givesFor fixed $\mu >0$ and initial $h,v_0$ tending to zero, the first return to small positive $u$ occurs after approximately $\pi /\omega =2\pi /\sqrt{4\lambda -1}$. This is a half-turn, not a full turn. At $\lambda =1/4$, $u=e^{\tau /2}[h+(\mu +h/2-v_0)\tau ]$ does not return and grows without bound. The loss of return diagnoses escape to infinity rather than a finite-amplitude homoclinic orbit. Physical time obeys $dt=d\tau /u$ and need not share the divergent desingularized travel time.

Expanding the Hamiltonian about a static classical solution removes the linear term by its Euler-Lagrange field equation. For a canonical real scalar field with potential $U$, the quadratic fluctuation operator is $\mathcal{H}=-{\partial }_X^2+U^{\prime \prime }({\phi }_{\mathrm{cl}})$. Positive-frequency normal modes become quantum harmonic oscillators under canonical quantization, yielding the Gaussian contribution to a one-loop soliton mass correction. A negative eigenvalue signals an instability, while a physical zero mode in field theory must be replaced by a collective coordinate.

For $\lambda >0$, the equilibrium points are the origin and $(s,s^2)$ with $s=(1\pm \sqrt{1+4\mu })/2$. The nonzero branches have Jacobian trace $s-\lambda$ and determinant $\lambda s(2s-1)$. They have a saddle-node bifurcation at $\mu =-1/4$, $s=1/2$; the origin exchanges stability with the lower branch at the transcritical bifurcation $\mu =0$. A Hopf bifurcation occurs on the upper branch at $\mu =\lambda (\lambda -1)$ for $\lambda >1/2$. The point $(\mu ,\lambda )=(-1/4,1/2)$ has a double zero eigenvalue and a nontrivial Jordan block. Its generic unfolding includes a saddle-loop curve and the intervening stable periodic orbit region.

A quadratic scan statistic maximizes a directional empirical second moment over candidate unit vectors. For a Gaussian random vector sample with a rank-one covariance spike in one of those directions, the matching statistic has its scale multiplied by the spike's eigenvalue. A chi-squared concentration inequality controls each direction.

Quantum mechanics describes a quantum system using a quantum state, with observables represented by self-adjoint operators. For a single particle in one dimension, a pure state can be described by a wavefunction $\psi (x,t)$: its squared modulus gives the probability density for position. The Schrödinger equation predicts how this probability amplitude evolves between interventions; measurement in quantum mechanics connects it to experimental outcomes through the Born rule. A wavefunction therefore predicts a distribution of possible positions rather than a single trajectory as in classical mechanics.

The infinite square well makes three central ideas visible with elementary sine functions. Confinement selects discrete energy eigenvalues; wavefunction normalization converts a shape into quantitative probabilities; and a quantum superposition can have a changing probability density even when each constituent energy eigenstate has a stationary one. This combination explains why quantum mechanics uses both linear algebra and differential equations: eigenfunctions describe the building blocks, while quantum phases determine how they interfere. The same tools help explain the discrete energies of chemical atoms and molecules, although their Quantum Hamiltonians require different potentials and often many particles.

In his first 1926 paper, Erwin Schrödinger formulated quantization as an eigenvalue problem for a hydrogen atom in nonrelativistic quantum mechanics. On its opening page he compared the emergence of integers with the number of nodes of a standing wave on a vibrating string. The infinite square well below realizes that appealing idea especially transparently: the allowed integers follow from the boundary conditions. This historical analogy concerns the origin of discrete energies; the Born rule supplies the statistical interpretation of the modern wavefunction. ^[ref].

Let a Hermitian operator $A$ have dyadic eigenvalues ${\lambda }_j=c_j/2^t$ in $[0,1)$ for known $t$, with controlled access to the unitary operator $U=e^{2\pi iA}$. This interval ensures that different eigenvalues have different eigenphases; exact representability alone would not exclude phase aliasing, as $0$ and $1$ both give phase zero. Let a real function $h$ obey $|h({\lambda }_j)|\le 1$, and assume the required quantum variable rotations are available. Coherent exact quantum phase estimation, a quantum variable rotation and uncomputation implementPostselection on flag one gives $h(A)|b⟩$ normalized, with probability $\Vert h(A)|b⟩{\Vert }^2$, provided this vector is nonzero. Erasing the eigenvalue label by uncomputation is essential to preserve coherence between different eigenvectors. On this dyadic spectrum, $U^{2^t}=I$ gives $U^{-1}=U^{2^t-1}$, so reversing the phase-estimation gates is possible using forward controlled-$U$ calls. The choice $h(\lambda )=\lambda$ multiplies by $A$; a scaled reciprocal gives the different filter used in the HHL algorithm.

A quantum state describes a preparation through its predicted measurement in quantum mechanics probabilities. In the Hilbert space formulation, a pure state is a quantum state ray: a normalized vector representative $|\psi ⟩$ satisfies $⟨\psi |\psi ⟩=1$, and representatives differing by a global phase describe the same quantum state. ^[ref]

Both pure states and mixed states are represented by a density operator on the chosen complex Hilbert space:Thus $\rho$ is Hermitian, positive semidefinite and of unit trace; in infinite dimension it must also be a trace-class operator. For a normalized pure state, $\rho =|\psi ⟩⟨\psi |$ is a rank-one orthogonal projection. A mixed state has no single such vector representative. Its density matrix is the matrix of $\rho$ in a chosen orthonormal basis. ^[ref] For a discrete POVM with effects $E_a\ge 0$ and ${\sum }_a{E}_a=I$, the Born rule readsThese constraints and Born rule probabilities describe arbitrary finite-dimensional preparations. ^[ref]

Randomly choosing normalized pure states with probabilities $w_j$ and forgetting the label givesThis quantum state ensemble averages probabilities; a quantum superposition adds probability amplitudes. Several ensembles can give the same density operator. Quantum entanglement can make a reduced density matrix mixed while the joint quantum state is pure. ^[ref]

The preparation, the choice of basis, and the measurement are separate ingredients. A wavefunction $\psi (x)=⟨x|\psi ⟩$ is the position representation of a pure state; a general mixed state requires a density operator. For a closed quantum system, a unitary operator evolves it as $\rho \mapsto U\rho U^{†}$, preserving its eigenvalues and purity of a density operator. ^[ref] The Bloch sphere and Bloch ball give the qubit geometry; quantum state tomography connects repeated preparations to an estimated density matrix; projective measurement treats the conditional state after an outcome.

The quantum variational principle estimates the lowest energy by choosing a trial wavefunction and computing its expectation value, without solving the Schrödinger equation exactly. Minimizing over a manageable family of trial wavefunctions gives an upper bound; enlarging that family can improve the estimate.

Let $H$ be a self-adjoint operator bounded below, and put $E_{\ast }=\inf \sigma (H)$. For $\Vert \psi \Vert =1$ in its operator domain $D(H)$,Trial states may also lie in the form domain $Q(H)=D((H-bI{)}^{1/2})$, where $b\le E_{\ast }$. Then use the quadratic form of a positive quantum Hamiltonian shifted by $b$:This extends the expectation value beyond $D(H)$ without requiring $H\psi$ to exist. Equality for a normalized state holds exactly in the energy eigenspace at $E_{\ast }$, if that eigenvalue exists. ^[ref].

A short proof makes the direction of the bound memorable. If $H$ has a complete orthonormal basis of energy eigenstates ${\varphi }_n$, writeThe expectation value is a weighted average of allowed energies:Equality requires every occupied energy eigenstate to have energy $E_{\ast }$. ^[ref]. With continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure; the same nonnegative-difference argument works.

For an unnormalized trial wavefunction, divide by its squared norm: $R_H[\psi ]=q_H[\psi ]/\Vert \psi {\Vert }^2$ is the form version of the Rayleigh quotient. The quantum variational principle supplies an upper bound on $E_{\ast }$; it supplies neither a lower bound on $E_{\ast }$ nor, by itself, a guarantee that a bound state exists.

To see the method at work, choose the real Gaussian wave packetThe Gaussian integral gives wavefunction normalization and the two useful position averages:Thus its position probability density is a normal distribution with standard deviation $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hslash =m=1$, integration by parts and ${\psi }_a^{\prime }=-ax{\psi }_a$ give the kinetic energyThese smooth, rapidly decaying trial wavefunctions belong to the operator domains of both polynomial Quantum Hamiltonians below. Narrowing the Gaussian wave packet increases its kinetic energy; widening it increases the potential energy in either confining potential.

For the quantum harmonic oscillator, take $\hslash =m=\omega =1$:The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The derivative $E_{\mathrm{ho}}^{\prime }(a)=(1-a^{-2})/4$ changes sign at $a=1$, where kinetic energy and potential energy are both $1/4$. Here the trial family contains an exact ground state:The zero-point energy remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to 0$ as $a\to 0$: the free Quantum Hamiltonian on the line has spectral bottom zero but no normalizable zero-energy energy eigenstate. A sequence of ever broader trial wavefunctions need not approach an actual bound state.

For a pure quartic oscillator, keep $\hslash =m=1$ and set the quartic coefficient to one:The trial values at $a=1,2,3$ are respectively $1,0.6875,0.833333\dots$. Setting the derivative to zero givesThe sign of this derivative changes from negative to positive at the optimum, giving the smallest energy within this Gaussian wave packet family:At this width, $⟨T⟩\approx 0.454280$ and $⟨V⟩\approx 0.227140$, so $2⟨T⟩=4⟨V⟩$, as expected from the power-law quantum virial theorem for a scale-optimized trial wavefunction. The number $0.681420$ is an upper estimate of the ground-state energy, not an asserted exact energy eigenvalue. Indeed,depends on $x$ for every $a>0$, so no member of this trial family is an energy eigenstate. Nevertheless, a single width parameter already turns a difficult Schrödinger equation into a useful, explicit energy estimate.

An observed dissipative channel combined with direct feedback can create a modified Lindblad operator and a compensating Quantum Hamiltonian. Choosing those parameters can make a prescribed pure state stationary and globally attractive. At exceptional parameter values, the dissipator can instead leave an entire family of states stationary, so algebraic target stationarity does not establish stabilization. Checking the Affine Bloch equation and its transverse eigenvalues distinguishes these possibilities.

A qubit state specifies measurement probabilities, rather than a pre-existing answer to every possible measurement. For a pure state, normalized vectors differing by a global phase represent the same physical state:Equivalently, the pure state is a ray: the one-dimensional complex linear subspace spanned by any nonzero representative. Dividing a nonzero vector by its norm fixes its length, but leaves the global phase free. The density operator is unchanged because $(e^{i\gamma }|\psi ⟩)(e^{i\gamma }|\psi ⟩{)}^{†}=|\psi ⟩⟨\psi |$. In contrast, multiplying only one nonzero component by $e^{i\varphi }$ changes a relative quantum phase and can change measurement probabilities. ^[ref].

After choosing a global phase, every pure state of a qubit can be writtenAt $\theta =0$ or $\pi$, the parameter $\varphi$ does not distinguish physical states: only one component survives. Elsewhere it is the relative quantum phase. The corresponding unit Bloch vector is $\mathbf{r}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ on the existing Bloch sphere. Its angles describe a space of states; for a photon polarization qubit they are not the photon's position or direction of travel. ^[ref].

The Born rule in the computational basis gives $P(0)={\cos }^2(\theta /2)$ and $P(1)={\sin }^2(\theta /2)$. These probabilities discard $\varphi$. In the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, adding or subtracting the probability amplitudes instead givesThis last equality follows by substituting the angular representative. In the basis $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, the same calculation givesThus both the real and imaginary parts of quantum coherence can be observed, using different measurement settings. For example, $(|0⟩+|1⟩)/\sqrt{2}$ and $(|0⟩-|1⟩)/\sqrt{2}$ both give $P(0)=P(1)=1/2$, but the first gives $P_X(+)=1$ and the second $P_X(+)=0$. A global phase could never cause this difference.

For a general qubit state, write its density matrix in the computational basis asThese are exactly the Hermitian, unit-trace, positive semidefinite constraints. Positivity means every eigenvalue is nonnegative, not that every matrix entry is nonnegative. The general definition is given in density matrix. ^[ref].

For this two-by-two matrix, the explicit constraint follows from its determinant, $\det \rho =p(1-p)-|c{|}^2$: with trace one, a nonnegative determinant and real eigenvalues rule out a negative eigenvalue. Its eigenvalues areThe pure states have $|c{|}^2=p(1-p)$ and eigenvalues $1,0$; the mixed states have strict inequality and two positive eigenvalues. For a normalized representative, $p=|\alpha {|}^2$ and $c=\alpha {\beta }^{\ast }$. A statistical preparation $\rho ={\sum }_j{w}_j|{\psi }_j⟩⟨{\psi }_j|$, with $w_j\ge 0$ and ${\sum }_j{w}_j=1$, averages probabilities, rather than adding probability amplitudes. Such a preparation is mixed if the positive-weight vectors do not all represent the same ray. A reduced density matrix of an entangled pair can also be mixed even when the pair has a pure state. ^[ref].

For a concrete check, the pure state $(\sqrt{3}|0⟩+|1⟩)/2$ and the mixed state obtained by preparing $|0⟩$ with probability $3/4$ and $|1⟩$ with probability $1/4$ haveBoth have trace one and the same computational-basis measurement probabilities. Their eigenvalues are respectively $(1,0)$ and $(3/4,1/4)$; the respective values of their purity of a density operator $\mathrm{Tr}({\rho }^2)$ are $1$ and $5/8$. Their Hadamard basis plus probabilities are respectively $1/2+\sqrt{3}/4\approx 0.9330$ and $1/2$. By contrast, setting $p=3/4,c=1/2$ would give determinant $-1/16$ and eigenvalue $(2-\sqrt{5})/4<0$: normalized diagonal entries alone do not make a valid qubit state.

The same constraints become especially beautiful in the Bloch ball representation. With the Pauli matrices $\mathbf{\sigma }=(X,Y,Z)$ and the identity operator $I$,The eigenvalues are $(1\pm |\mathbf{r}|)/2$, so the Bloch sphere is precisely the pure state boundary, and $I/2$ is the maximally mixed centre. For each axis $j$, the Born rule gives $P_j(\pm )=(1\pm r_j)/2$. This unifies a geometric picture, matrix positivity, and experimental frequencies in one description. For the equatorial family in the phase diagram, $p=1/2,c=e^{-i\varphi }/2$ and $\mathbf{r}=(\cos \varphi ,\sin \varphi ,0)$; replacing it by $I/2$ erases the transverse components and every fringe. See quantum state tomography for how independent ensembles measured along the three axes estimate these components. ^[ref].