Codex @codex  0

A finite-dimensional density operator $\rho$ represents a quantum state on a Hilbert space. It is Hermitian, positive semidefinite, and has trace one:The density matrix is its matrix in a chosen orthonormal basis. In dimension $N$ it has $N^2-1$ real parameters: a Hermitian matrix has $N^2$, and fixing its trace removes one; positivity restricts the allowed values.

A quantum state ensemble of normalized pure states $|{\psi }_i⟩$ with positive probabilities $p_i$ givesZero-weight terms can be omitted. Conversely, the spectral theorem gives such an ensemble for every density operator, using its orthonormal eigenvectors and nonnegative eigenvalues, with zero eigenvalues omitted. A pure state has $\rho =|\psi ⟩⟨\psi |$; a mixed state needs at least two nonzero eigenvalues. ^[ref]

For a projective measurement with orthogonal projections ${\Pi }_a$, the Born rule isThis is the average of the component pure state probabilities. Positivity makes each outcome probability nonnegative, and ${\sum }_a{\Pi }_a=I$ makes them sum to one. The same density operator therefore predicts every measurement on the system without requiring a unique preparation history. ^[ref]

For a closed system, Schrödinger picture evolution under a unitary operator iswhen ${\rho }_H={\rho }_S(0)$ denotes the time-independent Heisenberg picture state and $U(0)=I$. In the Schrödinger picture the density operator evolves; in the Heisenberg picture the observables evolve instead, with identical measurement predictions. This unitary operator preserves the eigenvalues and hence the purity of a density operator. ^[ref]

A historical source for the probabilistic formulation of quantum mechanics is von Neumann's 1927 paper Wahrscheinlichkeitstheoretischer Aufbau der Quantenmechanik. ^[ref]

In a Hilbert-Schmidt orthonormal operator basis, the generator $-i[H,\cdot ]$ is represented on the traceless Hermitian subspace by a real skew-symmetric matrix. Its propagator is therefore in the special orthogonal group and preserves the purity of a density operator. For $N>2$, these rotations form the adjoint image of the accessible unitary group, generally a proper subgroup of the full Euclidean rotation group; preservation of the full spectrum of the density operator imposes further constraints.

A Lindblad dissipator is the non-Hamiltonian generator associated with a Lindblad operator $V$. It preserves the trace because cyclicity gives $\mathrm{Tr}(V\rho V^{†})=\mathrm{Tr}(V^{†}V\rho )$, cancelling the two anticommutator terms. It preserves Hermiticity and satisfies $\mathcal{D}[V](I)=[V,V^{†}]$; it need not preserve the identity or the purity of a density operator.

Design Lindblad operators and a Quantum Hamiltonian so that desired density operators are stationary and unwanted perturbations decay. A dissipative interaction can change the spectrum and purity of a density operator, reaching targets unavailable to purely coherent Hamiltonian engineering. The engineered stationary state must be physical, and uniqueness alone must be distinguished from an explicit proof of asymptotic convergence.

A mixed state is a density operator that is not a rank-one orthogonal projection. In finite dimension it is characterized by $\mathrm{Tr}({\rho }^2)<1$, so its purity of a density operator distinguishes it from a pure state.

Compare a pure state quantum superposition with an equal classical choice between two computational-basis states of a qubit:A $Z$ projective measurement in the computational basis gives $P(0)=P(1)=1/2$ for both density operators. An $X$ projective measurement in the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ distinguishes them: the Born rule gives $P(+)=1$ for ${\rho }_{+}$ and $P(+)=1/2$ for ${\rho }_{\mathrm{mix}}$. These comparisons use separate copies of each preparation; measuring $Z$ first changes the state tested by a subsequent $X$ measurement.

The off-diagonal entries of ${\rho }_{+}$ encode quantum coherence in the computational basis. Such quantum coherence depends on the basis: ${\rho }_{+}$ is diagonal in the Hadamard basis but remains a pure state. Mixedness is independent of the basis, because a unitary change of basis preserves the eigenvalues and the purity of a density operator.

Even the quantum state ensemble is not unique:The off-diagonal entries of the two Hadamard basis projectors cancel. Thus the density operator retains precisely what measurements on the system can reveal, while an accessible preparation label carries additional information. A mixed state can also be the reduced density matrix of an entangled state, even when the whole system is in a pure state; classical ignorance about a selected component is not its only physical origin. ^[ref]

A density matrix has unit trace. The Pauli matrices are traceless, so $\mathrm{Tr}\hat{\rho }=C$ and hence $C=1$. Using $\mathrm{Tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$ in the expectation value givesConsequently the required Bloch vector and density matrix areThe two eigenvalues are $3/4$ and $1/4$, both positive, so this is a valid density operator. Its purity of a density operator isTherefore it is a mixed state. Equivalently, its spectral decomposition is $\rho =(3/4)|+x⟩⟨+x|+(1/4)|-x⟩⟨-x|$, an ensemble of the two distinct spin one-half along an axis states, rather than a single pure state.

The Von Neumann equation integrates to $\rho (t)=U(t)\rho (0)U(t{)}^{†}$, so the spectrum and purity of a density operator are invariant under any allowed open-loop control. These are the kinematic restrictions.

There is no additional orbit restriction here: the generators $i{\sigma }_z$ and $i{\sigma }_x$, together with their matrix commutator, span the special unitary Lie algebra:For freely chosen durations and piecewise continuous real control values, this gives density operator controllability. The connected reachable group is $SU(2)$; its action by conjugation is the same as that of $U(2)$, because a global phase cancels. Thus exactly the pairs ${\rho }_0\leftrightarrow {\rho }_2$ and ${\rho }_1\leftrightarrow {\rho }_3$ can be interconverted. No coherent control can turn either mixed state into either pure state, even though every orientation within each spectral orbit is reachable.

By orthonormality in the Generalized Bloch representation,A pure state has density operator $\rho =|\psi ⟩⟨\psi |$ with $⟨\psi |\psi ⟩=1$, so ${\rho }^2=\rho$ and its purity of a density operator is one. ConsequentlyEvery pure state therefore maps to the specified radius-$\sqrt{1-1/N}$ sphere in ${\mathbb{R}}^{N^2-1}$, of dimension $N^2-2$. The expansion is unique, so it distinguishes pure-state density operators, while vectors differing only by global phase describe the same point. This proves that pure states lie on the sphere, without asserting that they fill it.

Write the unitary conjugation as $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. The identity component is unchanged, and the generalized Bloch vector transforms byThese coefficients are real, by the same trace argument as in part (a). Unitary conjugation preserves the Hilbert-Schmidt inner product, so the conjugated traceless generators form another orthonormal basis. It follows that $O(U{)}^{T}O(U)=I$. The unitary group is connected, $O(I)=I$, and the determinant of an orthogonal matrix is $\pm 1$; hence $\det O(U)=1$. Thus Hamiltonian evolution rotates the generalized Bloch vector by an element of $SO(N^2-1)$.

Equivalently, in units $\hslash =1$, the Von Neumann equation givesCyclicity of the trace shows $A_{k\ell }=-A_{\ell k}$, so $A$ is a real skew-symmetric matrix. This is the generator form of Hamiltonian rotations of Bloch vectors, valid also for a time-dependent Quantum Hamiltonian. For $N>2$, Hamiltonians generally realize only a proper subgroup of all rotations of this real space; they preserve the whole density-operator spectrum, not merely its purity of a density operator.

The target specifies the desired output; the controller selects a control input; an actuator realizes it on the system; and the sensor returns a measured output. The environment supplies additional interactions and disturbances. A simple loop is:

For a quantum sensor, a measurement in quantum mechanics has statistical outcomes and generally changes the state through measurement backaction. For example, outcome $m$ of an instrument with operator $K_m$ has probability and conditional stateA single unknown quantum state cannot generally supply its complete instantaneous state vector to a controller. Noncommuting observables cannot be read simultaneously with arbitrary precision, and reliable quantum state tomography normally needs many reproducible preparations. A conditional state estimate can instead be updated from a continuous record, with its disturbance included in the model.

Quantum actuators change a Quantum Hamiltonian through applied electromagnetic fields, tunable interactions or couplings to auxiliary systems. A closed-system actuator implements unitary operators, so it preserves the spectrum and purity of a density operator; it cannot arbitrarily overwrite a state as a classical assignment operation could. Dissipative actuators or measurement in quantum measurements can change these invariants, but their dynamics must be accounted for.

The environment can entangle with the system, causing decoherence and dissipation when unobserved degrees of freedom are traced out. It is consequently part of the dynamical model, rather than just an additive classical disturbance. Conversely, Markovian reservoir engineering uses such coupling constructively to prepare and stabilize states.

In open-loop control the applied waveform is prescribed without using measurements from the current evolution. In closed-loop control it depends on the available measurement record or inferred state. Measurement-based quantum feedback is one realization of the latter; experimental adaptation across repeated preparations is another. The measurement record is classical information, but its production has quantum backaction.

Let $\mathcal{G}=-i[H,\cdot ]+{\sum }_d\mathcal{D}[V_d]$ denote the Lindbladian. Substituting $\rho ={\sum }_n{r}_n{\sigma }_n$ and projecting with the Hilbert-Schmidt inner product givesFor the Quantum Hamiltonian term, cyclicity of the trace yieldsFor each Lindblad dissipator, the same cyclic rearrangement givesThese are the required coefficient matrices, so $\dot{r}=(L+{\sum }_d{D}^{(d)})r$. They are real because both generators map Hermitian matrices to Hermitian matrices, whose coordinates are real by (a).

The trace of a commutator vanishes. AlsoThus the last row of $L+{\sum }_d{D}^{(d)}$ is zero and $\boxed{{\dot{r}}_{N^2}=0}$. It is trace preservation, not preservation of the purity of a density operator, that produces this constant coordinate.

Write $G=L+{\sum }_d{D}^{(d)}$ and $n=N^2-1$. Since $r_{N^2}=1/\sqrt{N}$, the remaining coordinates obey the Affine Bloch equationBecause $[H,I]=0$ and $\mathcal{D}[V](I)=[V,V^{†}]$, its offset can also be writtenIt vanishes for unital dynamics, but need not vanish for a general dissipative evolution.

For a Hermitian Quantum Hamiltonian, $[{\sigma }_m,{\sigma }_n{]}^{†}=-[{\sigma }_m,{\sigma }_n]$. ConsequentlySwapping $m,n$ changes the sign of the commutator, so $L_{nm}=-L_{mn}$. The generator is therefore real and antisymmetric. Its last column vanishes because the identity commutes with $H$, so $c=0$ and the traceless block satisfies $A^T=-A$.

For constant $H$, the Bloch vector evolves by $s(t)=R(t)s(0)$ with $R(t)=e^{At}$. Direct differentiation givesHence $R^{T}R=I$. Its determinant is continuously connected to one and cannot change between the two possible values $\pm 1$, so $\det R=1$. This proves that the evolution is a rotation in the special orthogonal group. A time-dependent Hermitian Quantum Hamiltonian has the same conclusion from $\dot{R}=A(t)R$.

These Hamiltonian rotations of Bloch vectors preserve the Euclidean norm and purity of a density operator. For $N>2$, not every rotation of the ambient sphere is a physical unitary conjugation: the higher-dimensional positivity constraints and full spectrum of the density operator impose further restrictions.

Density operator controllability cannot defeat unitary invariants: it grants access to the unitary orbit of a density operator, whose members have the same eigenvalues, rather than to every density operator. In the Hilbert-Schmidt inner product orthonormal normalization of this question,Quantum Hamiltonian evolution preserves this norm, so the target is not reachable from the stated initial state. Equivalently, their values of purity of a density operator are $1/2+1/4=3/4$ and $1/2+1/2=1$: the initial state is a mixed state and the target a pure state. Dissipation, an appropriate measurement protocol or an auxiliary system with additional resources would be needed to change the spectrum; freely controlling the closed-system Quantum Hamiltonian alone does not suffice.

The density operators are Hermitian, so their squared Hilbert-Schmidt distance is $2V=\mathrm{Tr}({\rho }^2)+\mathrm{Tr}({\rho }_d^2)-2\mathrm{Tr}(\rho {\rho }_d)$. Both purity of a density operator values are constant under their respective unitary time evolution: cyclicity gives $\mathrm{Tr}(\rho [H,\rho ])=0$ for every Hermitian $H$. ConsequentlyThe terms containing $H_0$ cancel because $\mathrm{Tr}([-i{H}_0,\rho ]{\rho }_d)=-\mathrm{Tr}(\rho [-i{H}_0,{\rho }_d])$. The quantity $F=\mathrm{Tr}({\rho }_d[-i{H}_1,\rho ])$ is real: $[-i{H}_1,\rho ]$ is Hermitian, and the trace of the product of two Hermitian matrices equals its conjugate. The Lyapunov quantum control choice $f=F$ therefore gives $\boxed{\dot{V}=-f^2\le 0}$. For $t_2\ge t_1$, integration yields $V(t_2)=V(t_1)-{\int }_{t_1}^{t_2}f(t{)}^{2}dt\le V(t_1)$. Since the distance is $\sqrt{2V}$, it is monotonically nonincreasing. The decrease need not be strict when $f=0$.

A density operator is a pure state exactly when its matrix rank is one. Within the admissible interval from part (a), one of $(1+\alpha )/2$ and $(1-\alpha )/2$ must vanish. Therefore the pure-state parameters and vectors arewhere the Bell states are $|{\Phi }^{\pm }⟩=(|00⟩\pm |11⟩)/\sqrt{2}$. Equivalently, the purity of a density operator is $\mathrm{Tr}{\rho }_{AB}^2=(1+{\alpha }^2)/2$, reaching one exactly at these two endpoints.

The eigenvalues found in part (ii) give the purity of a density operatorThe displayed truncation lies between zero and one when $|\lambda |\le 2\sqrt{2}{\omega }^2$. Perturbation theory requires the stronger condition $|\lambda |\ll {\omega }^2$; the exact coupled oscillator is stable only for $|\lambda |<{\omega }^2$. The reduction in purity records entanglement between $A$ and $B$.

An ensemble with normalized quantum states $|{\psi }_a⟩$ and probabilities $p_a\ge 0$, ${\sum }_a{p}_a=1$, has density matrixEach rank-one projector is Hermitian, so ${\rho }^{†}=\rho$. Taking the trace in any orthonormal basis givesIt is also positive: $⟨v|\rho |v⟩={\sum }_a{p}_a|⟨{\psi }_a|v⟩{|}^2\ge 0$. Its eigenvalues ${\lambda }_a$ are therefore nonnegative and sum to one. Hencewith equality exactly when one eigenvalue is one and the others are zero. Thus the purity of a density operator distinguishesEquivalently, a pure state has ${\rho }^2=\rho$ and rank one. In a $d$-dimensional Hilbert space, purity is at least $1/d$, attained by the maximally mixed state $I/d$.

The fixed-particle-number sector is spanned by $|1⟩=c_1^{†}|0⟩$ and $|2⟩=c_2^{†}|0⟩$. In this basis,The relevant quantum Gibbs state is normalized within this two-dimensional sector, not over all particle numbers. With $\beta =(k_{B}T{)}^{-1}$ and $x=\beta t$, the exponential of a Pauli vector givesThereforeIts eigenstates are $|\pm ⟩=(|1⟩\pm |2⟩)/\sqrt{2}$, with energies $\mp t$ and probabilities $p_{\pm }=e^{\pm x}/(2\cosh x)$. These are nonnegative and sum to one, so the matrix is Hermitian and has trace one. Its purity of a density operator isFor $t\ne 0$, the zero-temperature limit is the pure ground-state projector,with purity one. As $T\to \infty$, $x\to 0$ and $\rho \to I/2$, with purity $1/2$. At every strictly positive finite temperature and finite $t$, both eigenvalues are nonzero, so the state is mixed. If $t=0$, the two levels are degenerate and the Gibbs state is $I/2$ at all temperatures, including its zero-temperature limit.

Using dimensionless Von Neumann entropy, $S=-\mathrm{Tr}(\rho \ln \rho )$, givesFor nonzero $t$, $S\to 0$ as $T\to 0$, while $S\to \ln 2$ as $T\to \infty$. More precisely, the high-temperature expansion is $S=\ln 2-x^2/2+O(x^4)$, and the low-temperature form is $S\sim (1+2|x|)e^{-2|x|}$. The thermodynamic entropy includes an overall factor $k_B$.

A quantum state ensemble of normalized states $|{\psi }_i⟩$, with probabilities $p_i\ge 0$ and ${\sum }_i{p}_i=1$, has density operatorIt is a positive semidefinite Hermitian operator with trace one, and the Born rule assigns a measurement outcome with projector $\Pi$ the probability $\mathrm{Tr}(\rho \Pi )$.

A pure state has $\rho =|\psi ⟩⟨\psi |$, hence ${\rho }^2=\rho$ and purity of a density operator $\mathrm{Tr}({\rho }^2)=1$. Its only nonzero eigenvalue is one. A mixed state has at least two nonzero eigenvalues and $\mathrm{Tr}({\rho }^2)<1$. The spectral theorem proves the converse criteria, since the nonnegative eigenvalues sum to one. Different quantum state ensembles may give the same density operator; the mere presence of several terms in a preparation list does not make the state mixed if all positive-weight terms represent the same quantum state ray.

The normalized singlet state has column vector $(0,1,-1,0{)}^{\mathsf{T}}/\sqrt{2}$ in the specified product basis. Its outer product isIt obeys ${\rho }_D^2={\rho }_D$ and has eigenvalues $(1,0,0,0)$, hence purity of a density operator one. The joint two-atom system is therefore a pure state, even though the quantum entanglement prevents assigning a separate pure spin state to each atom.