Pure States

A pure state is a quantum state that cannot be written as a nontrivial probabilistic mixture of distinct quantum states. In the standard Hilbert-space formulation, the same state can be represented in three equivalent ways:

1. as a ray $[\psi]$ through a nonzero vector $|\psi\rangle$;

2. as a normalized ket, with global phase understood to be redundant;

3. as the rank-one density operator

   $$\rho_\psi=|\psi\rangle\langle\psi|.$$

Pure does not mean classically certain. A pure state may give nontrivial probabilities for most observables. Pure also does not mean stationary, unentangled, localized, semiclassical, or known exactly by an experimenter. It describes the position of a state in the convex set of quantum states.

For an ordinary density operator, the following criteria are equivalent:

$$\begin{gathered} \rho\text{ is pure}, \qquad \operatorname{rank}\rho=1, \qquad \rho^2=\rho, \\ \operatorname{Tr}(\rho^2)=1, \qquad S(\rho)=0, \qquad \rho\text{ is extremal}. \end{gathered}$$

Each criterion highlights a different aspect: geometry, linear algebra, entropy, or preparation statistics.

Maximal Quantum Specification

A pure state is sometimes called a maximal quantum specification. This means that within the quantum state model, the state contains no unresolved classical random choice among distinct states. It does not mean that every observable has a predetermined sharp value.

For example, the spin state $|+x\rangle$ is pure. A measurement of $S_x$ is certain, but a measurement of $S_z$ gives

$$P(+z)=P(-z)=\frac12.$$

The uncertainty is not evidence that the preparation was mixed. It is a prediction of one pure state for a measurement whose eigenbasis does not contain that state.

Operationally, a state is pure when its density operator cannot be expressed as

$$\rho = t\rho_1+(1-t)\rho_2, \qquad 0<t<1,$$

with $\rho_1\ne\rho_2$. This convex definition avoids language about an observer’s knowledge and makes purity a mathematical property of the state assignment.

Pure State as a Ray

Let $\mathcal H$ be the Hilbert space of the system. A normalized vector $|\psi\rangle$ represents a pure state, but so does every vector

$$e^{i\chi}|\psi\rangle, \qquad \chi\in\mathbb R.$$

More generally, the physical ray is

$$[\psi] = \left\lbrace \lambda|\psi\rangle: \lambda\in\mathbb C\setminus\{0\} \right\rbrace.$$

A non-unit scalar changes the norm of the representative but not its ray. A global phase preserves both the ray and normalization. The zero vector is excluded because it cannot be normalized and defines no probability rule.

The probability for an effect $E_a$ can be written without choosing a normalized representative:

$$p(a|[\psi]) = \frac{ \langle\psi|E_a|\psi\rangle }{ \langle\psi|\psi\rangle }.$$

Both numerator and denominator acquire $|\lambda|^2$ under $|\psi\rangle\mapsto\lambda|\psi\rangle$, so the result depends only on the ray.

The operational cancellation of global phase is developed in Rays and Global Phase. The quotient geometry of all rays belongs to Projective Hilbert Space.

Pure State as a Rank-One Density Operator

Choose a normalized representative. Its density operator is

$$\rho_\psi = |\psi\rangle\langle\psi|.$$

This operator is Hermitian because

$$\rho_\psi^\dagger = \left(|\psi\rangle\langle\psi|\right)^\dagger = |\psi\rangle\langle\psi|.$$

It is positive because for every $|v\rangle$,

$$\langle v|\rho_\psi|v\rangle = |\langle\psi|v\rangle|^2 \ge0.$$

It has unit trace:

$$\operatorname{Tr}\rho_\psi = \langle\psi|\psi\rangle = 1.$$

Its image is the one-dimensional span of $|\psi\rangle$ because

$$\rho_\psi|v\rangle = |\psi\rangle\langle\psi|v\rangle.$$

Thus $\rho_\psi$ has rank one. It also removes the arbitrary global phase exactly:

$$\begin{aligned} |\psi'\rangle\langle\psi'| &= e^{i\chi}|\psi\rangle e^{-i\chi}\langle\psi| \\ &= |\psi\rangle\langle\psi|. \end{aligned}$$

The mapping from rays to rank-one density operators is therefore one-to-one. It is often the most convenient bridge from introductory state vectors to the general state formalism.

The Projector Criterion

For a normalized ket,

$$\begin{aligned} \rho_\psi^2 &= |\psi\rangle\langle\psi| \psi\rangle\langle\psi| \\ &= |\psi\rangle \langle\psi|\psi\rangle \langle\psi| \\ &= |\psi\rangle\langle\psi| \\ &= \rho_\psi. \end{aligned}$$

Hence every pure-state density operator is idempotent:

$$\rho^2=\rho.$$

Conversely, suppose $\rho$ is a positive trace-one operator satisfying this condition. Its eigenvalues obey

$$\lambda_k^2=\lambda_k.$$

Therefore every eigenvalue is either zero or one. Since

$$\sum_k\lambda_k=1,$$

exactly one eigenvalue is one. The operator has rank one and is a pure-state projector.

Idempotence here is a criterion for a density operator. A general projector onto an $r$-dimensional subspace satisfies $P^2=P$, but its trace is $r$. Unless $r=1$, it is not itself a normalized density operator. The normalized operator $P/r$ is a mixed state on that subspace.

The Purity Criterion

The purity of a density operator is

$$\gamma(\rho) = \operatorname{Tr}(\rho^2).$$

If the spectral decomposition is

$$\rho = \sum_k\lambda_k|k\rangle\langle k|, \qquad \lambda_k\ge0, \qquad \sum_k\lambda_k=1,$$

then

$$\gamma(\rho) = \sum_k\lambda_k^2.$$

Because $0\le\lambda_k\le1$,

$$\lambda_k^2\le\lambda_k.$$

It follows that

$$\operatorname{Tr}(\rho^2)\le1.$$

Equality holds only when each nonzero eigenvalue satisfies $\lambda_k^2=\lambda_k$. Unit trace then forces one eigenvalue to equal one and all others to vanish. Thus

$$\rho\text{ is pure} \quad\Longleftrightarrow\quad \operatorname{Tr}(\rho^2)=1.$$

In dimension $d$, the lower bound is

$$\frac1d \le \operatorname{Tr}(\rho^2) \le1.$$

The minimum occurs for the maximally mixed state $I/d$. In an infinite-dimensional Hilbert space there is no normalized operator proportional to the identity, and no positive dimension-independent lower bound above zero. Purity can approach zero without being zero.

Purity is nonlinear in $\rho$. Estimating it from data is therefore not the same as measuring one copy with an ordinary observable; multi-copy protocols or reconstructed state models are typically used.

The Entropy Criterion

The von Neumann entropy is

$$S(\rho) = -\operatorname{Tr}(\rho\log\rho) = -\sum_k\lambda_k\log\lambda_k,$$

with $0\log0$ defined by continuity as zero. Every term is nonnegative. The entropy vanishes exactly when the spectrum is

$$(1,0,0,\ldots),$$

which is precisely the pure-state spectrum. Therefore

$$\rho\text{ is pure} \quad\Longleftrightarrow\quad S(\rho)=0.$$

Entropy and purity order states in related but not identical ways once the dimension exceeds two. Two mixed states can have the same purity and different entropies. The broader interpretation and conventions belong to Entropy Overview.

Pure States Are Extremal

The density operators form a convex set. If $\rho_1$ and $\rho_2$ are states, then

$$t\rho_1+(1-t)\rho_2, \qquad 0\le t\le1,$$

is also a state. Physically, this combination describes a classical random choice between two preparations when the choice label is ignored.

A point is extremal if it cannot be written as a nontrivial convex combination of two distinct points. Pure states are exactly the extremal states.

To see one direction, suppose

$$|\psi\rangle\langle\psi| = t\rho_1+(1-t)\rho_2, \qquad 0<t<1.$$

For every $|v\rangle$ orthogonal to $|\psi\rangle$,

$$0 = t\langle v|\rho_1|v\rangle + (1-t)\langle v|\rho_2|v\rangle.$$

Both terms are nonnegative, so each must vanish. Positivity then implies that $\rho_1$ and $\rho_2$ have support only in the span of $|\psi\rangle$. Unit trace forces

$$\rho_1=\rho_2=|\psi\rangle\langle\psi|.$$

Thus the decomposition was trivial.

For the converse, a mixed density operator has at least two nonzero eigenvalues and its spectral decomposition

$$\rho = \sum_k\lambda_k|k\rangle\langle k|$$

is a nontrivial convex combination of distinct pure states. Hence it is not extremal.

The extremal definition is the formulation that extends most naturally to general observable algebras.

A Pure State Has No Nontrivial Ensemble Decomposition

Suppose an ensemble of normalized kets realizes a pure density operator:

$$|\psi\rangle\langle\psi| = \sum_i p_i |\phi_i\rangle\langle\phi_i|, \qquad p_i>0.$$

Choose any $|v\rangle$ orthogonal to $|\psi\rangle$. Taking its expectation value gives

$$0 = \sum_i p_i |\langle v|\phi_i\rangle|^2.$$

Every summand is nonnegative, so each one vanishes. Therefore every $|\phi_i\rangle$ lies in the one-dimensional span of $|\psi\rangle$:

$$|\phi_i\rangle = e^{i\chi_i}|\psi\rangle.$$

An apparatus may use classical randomness while always preparing different representatives of the same ray, but it cannot reproduce a pure state by randomly choosing genuinely distinct rays.

Mixed density operators behave differently: they generally have infinitely many inequivalent ensemble decompositions. That nonuniqueness and its operational meaning are developed in Ensembles and Preparation Procedures.

Qubit Example

Every normalized pure qubit state can be written, up to global phase, as

$$|\psi(\theta,\phi)\rangle = \cos\frac\theta2,|0\rangle + e^{i\phi} \sin\frac\theta2,|1\rangle.$$

Its density matrix in the computational basis is

$$\rho_\psi = \begin{pmatrix} \cos^2(\theta/2) & e^{-i\phi} \cos(\theta/2)\sin(\theta/2) \\ e^{i\phi} \cos(\theta/2)\sin(\theta/2) & \sin^2(\theta/2) \end{pmatrix}.$$

Equivalently,

$$\rho_\psi = \frac12 \left( I+\boldsymbol r\cdot\boldsymbol\sigma \right),$$

where

$$\boldsymbol r = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta).$$

The Bloch vector has unit length:

$$|\boldsymbol r|=1.$$

For a general qubit density operator,

$$\operatorname{Tr}(\rho^2) = \frac12 \left(1+|\boldsymbol r|^2\right).$$

Thus the surface of the Bloch ball consists of pure states and the interior consists of mixed states. This geometry is developed in the canonical Bloch Sphere treatment.

Qutrit Example

Consider

$$|\psi\rangle = \frac{|0\rangle+i|1\rangle+|2\rangle}{\sqrt3}.$$

The corresponding density matrix is

$$\rho_\psi = \frac13 \begin{pmatrix} 1 & -i & 1\\ i & 1 & i\\ 1 & -i & 1 \end{pmatrix}.$$

It has one eigenvalue equal to one and two eigenvalues equal to zero. The three diagonal entries are all $1/3$, but that does not make the state mixed. They are probabilities for one basis measurement. The off-diagonal entries retain the relative-phase coherence needed to identify the pure ray.

By contrast,

$$\rho_{\mathrm{diag}} = \frac13 I_3$$

has the same diagonal probabilities in that basis but eigenvalues $(1/3,1/3,1/3)$ and purity $1/3$.

Wave-Mechanical Example

A normalized wavefunction in $L^2(\mathbb R)$ represents a pure state. For a Gaussian wave packet,

$$\psi(x) = \frac{1}{(2\pi\sigma^2)^{1/4}} \exp\left[ -\frac{(x-x_0)^2}{4\sigma^2} + \frac{i p_0x}{\hbar} \right],$$

one has

$$\int_{-\infty}^{\infty} |\psi(x)|^2\,dx=1.$$

The density operator has integral kernel

$$\rho(x,x') = \psi(x)\psi^*(x').$$

Its diagonal is the position probability density,

$$\rho(x,x)=|\psi(x)|^2,$$

while its off-diagonal values contain phase relations needed for predictions in other representations. The wave packet can spread and remain pure under closed unitary evolution. The canonical dynamics is developed in Gaussian Wave Packets.

Ideal position and momentum eigenkets are not normalizable Hilbert-space vectors and therefore are not density operators of the form $|x\rangle\langle x|$ or $|p\rangle\langle p|$ in the ordinary trace-class sense. They are generalized states used inside a rigged-Hilbert-space or distributional formalism.

Pure Is Not the Same as Stationary

An energy eigenstate of a time-independent Hamiltonian is pure when it is normalizable. Its vector representative evolves by a global phase, so its ray is stationary.

But a superposition of energy eigenstates can also be pure:

$$|\psi(0)\rangle = \frac{|E_0\rangle+|E_1\rangle}{\sqrt2}.$$

At time $t$,

$$|\psi(t)\rangle = \frac{ e^{-iE_0t/\hbar}|E_0\rangle + e^{-iE_1t/\hbar}|E_1\rangle }{\sqrt2}.$$

The relative phase changes, so suitable observables have time-dependent expectation values. Nevertheless,

$$\rho(t)^2=\rho(t)$$

at every time. Purity concerns whether the state is extremal, not whether its ray is constant under a chosen Hamiltonian.

Pure Is Not the Same as a Sharp Observable Value

A pure state $|\psi\rangle$ has zero variance for an observable $A$ exactly when it is an eigenstate of $A$ on the relevant domain:

$$(\Delta A)^2 = \langle\psi| (A-\langle A\rangle)^2 |\psi\rangle = 0.$$

Purity does not require this condition for every observable. In dimensions greater than one, no nonzero vector is generally an eigenvector of all noncommuting observables. A pure state supplies the most specific quantum probability rule, not a classical assignment of simultaneous definite values.

Pure-State Overlap

For two pure states

$$\rho_\psi=|\psi\rangle\langle\psi|, \qquad \rho_\phi=|\phi\rangle\langle\phi|,$$

their Hilbert–Schmidt overlap is

$$\begin{aligned} \operatorname{Tr}(\rho_\psi\rho_\phi) &= \operatorname{Tr} \left( |\psi\rangle\langle\psi|\phi\rangle\langle\phi| \right) \\ &= |\langle\psi|\phi\rangle|^2. \end{aligned}$$

It equals one exactly when the rays coincide and zero exactly when they are orthogonal. Intermediate values quantify nonorthogonality and also give the transition probability in the corresponding rank-one projective test.

Distinct nonorthogonal pure states cannot be perfectly distinguished from one copy. Purity does not make states classical labels with disjoint supports.

Closed Evolution Preserves Purity

For unitary evolution,

$$\rho(t) = U(t)\rho(0)U^\dagger(t).$$

If $\rho(0)^2=\rho(0)$, then

$$\begin{aligned} \rho(t)^2 &= U\rho(0)U^\dagger U\rho(0)U^\dagger \\ &= U\rho(0)^2U^\dagger \\ &= U\rho(0)U^\dagger \\ &= \rho(t). \end{aligned}$$

Thus closed-system unitary evolution maps pure states to pure states. It also preserves the complete spectrum of every density operator and hence preserves purity and von Neumann entropy for mixed states.

This conclusion concerns the complete closed system. A subsystem of that system can become mixed as it entangles with degrees of freedom that are later ignored.

General Operations Need Not Preserve Purity

A quantum operation with Kraus operators $K_\mu$ acts as

$$\Phi(\rho) = \sum_\mu K_\mu\rho K_\mu^\dagger.$$

For a pure input,

$$\Phi(\rho_\psi) = \sum_\mu |v_\mu\rangle\langle v_\mu|, \qquad |v_\mu\rangle=K_\mu|\psi\rangle.$$

The output is pure exactly when all nonzero vectors $|v_\mu\rangle$ are collinear after normalization. Otherwise the unresolved alternatives produce a mixed output.

For a selected measurement outcome described by a single Kraus operator $M_a$, a pure input has the conditional state

$$|\psi_a\rangle = \frac{M_a|\psi\rangle} {\sqrt{\langle\psi|M_a^\dagger M_a|\psi\rangle}},$$

provided the outcome probability is nonzero. This conditional state is pure. If the outcome is not recorded, one sums over outcomes and the resulting state can be mixed.

The ideal projective special case belongs to State Update Rule. The complete language of channels and selected branches belongs to Quantum Operations.

Pure Joint States and Mixed Subsystems

Purity is always relative to the system whose state is being described. The Bell state

$$|\Phi^+\rangle = \frac{|00\rangle+|11\rangle}{\sqrt2}$$

is a pure state of the joint system $AB$:

$$\rho_{AB} = |\Phi^+\rangle\langle\Phi^+|, \qquad \operatorname{Tr}(\rho_{AB}^2)=1.$$

The reduced state of subsystem $A$ is

$$\rho_A = \operatorname{Tr}_B\rho_{AB} = \frac12 I_A.$$

Therefore

$$\operatorname{Tr}(\rho_A^2)=\frac12.$$

The joint state is pure while each qubit state is maximally mixed. There is no contradiction: tracing out $B$ discards correlations needed to predict joint measurements.

This is one of the main reasons density operators are indispensable even when the larger system is assigned a pure ket. The canonical local-state construction is in Reduced States.

Schmidt Coefficients Diagnose Subsystem Purity

For a finite-dimensional bipartite pure state, the Schmidt decomposition is

$$|\Psi\rangle = \sum_{r=1}^{R} s_r|u_r\rangle_A|v_r\rangle_B, \qquad s_r>0, \qquad \sum_r s_r^2=1.$$

The reduced states are

$$\rho_A = \sum_r s_r^2|u_r\rangle\langle u_r|,$$

and

$$\rho_B = \sum_r s_r^2|v_r\rangle\langle v_r|.$$

Their purities agree:

$$\operatorname{Tr}(\rho_A^2) = \operatorname{Tr}(\rho_B^2) = \sum_r s_r^4.$$

This value equals one exactly when one Schmidt coefficient is nonzero. Hence, for a bipartite pure state, the following are equivalent:

$$\begin{aligned} |\Psi\rangle\text{ is a product state} &\Longleftrightarrow R=1 \\ &\Longleftrightarrow \rho_A\text{ is pure} \\ &\Longleftrightarrow \rho_B\text{ is pure}. \end{aligned}$$

The theorem and its deeper entanglement consequences are developed in Schmidt Decomposition Overview.

Pure Does Not Mean Unentangled

An entangled state can be pure as a state of the complete composite system. The adjectives answer different questions:

• pure or mixed asks whether the state is extremal in the convex state space;
• product or entangled asks whether the state factors across a specified tensor-product decomposition.

A pure product state, a pure entangled state, a mixed separable state, and a mixed entangled state are all possible. The categories are not substitutes for one another.

Purity and Preparation

A declared pure-state model is a claim about repeatable preparation statistics, not about one isolated specimen carrying a visible label. If the same pure density operator is prepared repeatedly, informationally complete measurements on many copies can in principle distinguish it from nearby mixed states.

In practice, finite data never establish exact mathematical purity. State estimation must account for sampling variation, imperfectly calibrated measurements, drift, leakage outside the modeled Hilbert space, and model mismatch. Constraining an estimator to pure states can reduce parameters, but it can also hide real noise by assumption.

For a $d$-level system, a normalized pure ray has

$$2d-2$$

real parameters, whereas a general density operator has

$$d^2-1.$$

This parameter reduction is useful only when the pure-state model is physically justified. The inference problem and its trust boundaries belong to State Tomography.

Infinite-Dimensional and Algebraic Caveats

For a separable Hilbert space, an ordinary normal state represented by a trace-class density operator is pure exactly when the density operator has rank one. The criteria

$$\rho^2=\rho, \qquad \operatorname{Tr}(\rho^2)=1$$

remain valid.

Several finite-dimensional shortcuts do not transfer unchanged:

• there is no maximally mixed density operator proportional to the identity on an infinite-dimensional Hilbert space;
• entropy may be infinite for valid mixed states;
• ideal continuum eigenkets are generalized vectors, not normalized pure density operators;
• domain questions matter for expectations of unbounded observables.

In the algebraic formulation, a state is a positive normalized linear functional on an observable algebra. A pure algebraic state is an extremal functional. Depending on the algebra and representation, it need not be described by a rank-one density operator on one preselected Hilbert space. That broader formulation is developed in States as Positive Linear Functionals.

What Purity Does Not Tell You

Knowing that a state is pure does not by itself tell you:

• which ray it is;
• which observables have sharp values;
• whether the state is stationary under a given Hamiltonian;
• whether it is entangled across a chosen subsystem split;
• whether it is easy to prepare or verify;
• whether it resembles a classical state;
• whether its wavefunction is localized;
• whether a particular measurement outcome will occur;
• whether a laboratory preparation is exactly described by the ideal model.

Purity is one structural property of the state. It does not replace the state itself or the specification of dynamics and measurements.

A Practical Purity Audit

When deciding whether a state is pure, proceed in this order:

1. Specify the system. A joint state and a subsystem state can have different purity.
2. Identify the representation. Is the object a normalized ket, a density operator, a reduced state, or an algebraic functional?
3. For a ket, check normalization. A nonzero normalized ket represents a pure ray by construction.
4. For a density operator, check positivity and trace first. Purity tests are meaningful only after physicality is established.
5. Use an invariant criterion. Check rank, spectrum, $\rho^2=\rho$, or $\operatorname{Tr}(\rho^2)=1$.
6. Distinguish preparation from decomposition. A pure density operator has no ensemble decomposition into distinct rays.
7. Check subsystem reduction. A pure joint ket may yield a mixed reduced density operator.
8. State experimental limitations. A numerical estimate near one is not a proof of exact purity.

Common Mistakes

Equating pure with certain

A pure state gives certainty only for projectors containing its ray. Other measurements can have broad outcome distributions.

Equating a ket sum with a mixture

A normalized superposition ket is pure. A mixture is a convex combination of density operators and generally lacks the same coherence.

Using one diagonal representation as the test

Every density operator is diagonal in an eigenbasis. Purity is determined by its eigenvalues, not by whether off-diagonal entries happen to appear in one basis.

Treating every projector as a state

A rank-$r$ projector has trace $r$. Only a rank-one projector is already a normalized density operator.

Inferring global purity from local purity data

A mixed reduced state can come from an entangled pure joint state. Conversely, local states do not uniquely identify the global state.

Equating pure with unentangled

The Bell states are pure and entangled. Purity and entanglement refer to different structures.

Claiming exact purity from finite tomography

Finite data support confidence regions or model comparisons, not exact equality $\operatorname{Tr}(\rho^2)=1$ without assumptions.

Connections

• Quantum States gives the general operational state concept.
• Density Operators develops the representation for all normal states.
• Pure vs Mixed States continues the comparison from the density-operator side.
• Ensembles and Preparation Procedures explains the nonuniqueness of mixed-state ensembles.
• Reduced States constructs subsystem density operators.
• State Update Rule gives conditional pure-state updates for ideal measurements.

References

1. P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, Chapters I and III.
2. J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955, Chapters I and IV.
3. A. S. Holevo, Probabilistic and Statistical Aspects of Quantum Theory, 2nd ed., Edizioni della Normale, 2011, Chapters 1 and 2.
4. L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014, Chapters 2 and 3.
5. A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995, Chapters 2, 3, and 5.
6. M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, Sections 2.4 and 8.2.
7. J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, Chapters 1 and 2.
8. I. Bengtsson and K. Życzkowski, Geometry of Quantum States, 2nd ed., Cambridge University Press, 2017, Chapters 4 and 8.
9. O. Bratteli and D. W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed., Springer, 1987, Section 2.3.

Exercises

1. Equivalence of spectral purity tests. Let $\rho$ be a finite-dimensional density operator with eigenvalues $\lambda_k$. Prove that the following are equivalent:

   $$\rho^2=\rho, \qquad \operatorname{Tr}(\rho^2)=1, \qquad \operatorname{rank}\rho=1.$$

Solution

Positivity and unit trace give

$$\lambda_k\ge0, \qquad \sum_k\lambda_k=1.$$

If $\rho^2=\rho$, then every eigenvalue satisfies $\lambda_k^2=\lambda_k$, so it is zero or one. Unit trace permits exactly one eigenvalue equal to one; hence the rank is one.

If the rank is one, the only nonzero eigenvalue must equal one, so $\operatorname{Tr}(\rho^2)=1$.

Finally,

$$1-\operatorname{Tr}(\rho^2) = \sum_k\lambda_k(1-\lambda_k).$$

Every summand is nonnegative. If the trace purity is one, each summand vanishes, so every eigenvalue is zero or one. Again, unit trace leaves exactly one eigenvalue equal to one, and therefore $\rho^2=\rho$.

2. Qubit purity. Consider

   $$\rho = \begin{pmatrix} p & c\\ c^* & 1-p \end{pmatrix}.$$

   Assuming $\rho$ is positive, derive the condition on $p$ and $c$ for the state to be pure. Relate it to $\det\rho$.

Solution

Direct calculation gives

$$\operatorname{Tr}(\rho^2) = p^2+(1-p)^2+2|c|^2.$$

Setting this equal to one yields

$$|c|^2=p(1-p).$$

The determinant is

$$\det\rho = p(1-p)-|c|^2.$$

Thus a physical qubit density operator is pure exactly when $\det\rho=0$. This determinant shortcut is special to trace-one two-level states; in higher dimensions, zero determinant alone does not imply rank one.

3. No distinct-state ensemble for a pure state. Suppose

   $$|\psi\rangle\langle\psi| = \sum_i p_i|\phi_i\rangle\langle\phi_i|, \qquad p_i>0.$$

   Prove that every $|\phi_i\rangle$ belongs to the same ray as $|\psi\rangle$.

Solution

For any $|v\rangle$ orthogonal to $|\psi\rangle$,

$$0 = \langle v|\psi\rangle\langle\psi|v\rangle = \sum_i p_i|\langle v|\phi_i\rangle|^2.$$

All terms are nonnegative and all $p_i$ are positive, so $\langle v|\phi_i\rangle=0$ for every $i$ and every vector orthogonal to $|\psi\rangle$. Therefore each $|\phi_i\rangle$ lies in $\operatorname{span}\{|\psi\rangle\}$. Normalization gives

$$|\phi_i\rangle=e^{i\chi_i}|\psi\rangle.$$

The ensemble may randomize representatives, but not physical rays.

4. Projector onto a subspace. Let $P$ project onto an $r$-dimensional subspace of a $d$-dimensional Hilbert space. Show that $P/r$ is a density operator, compute its purity and entropy, and determine when it is pure.

Solution

$P/r$ is positive and

$$\operatorname{Tr}\frac Pr = \frac{\operatorname{Tr}P}{r} = 1.$$

Since $P^2=P$,

$$\operatorname{Tr} \left[ \left(\frac Pr\right)^2 \right] = \frac{\operatorname{Tr}P}{r^2} = \frac1r.$$

The nonzero eigenvalues are all $1/r$, so

$$S(P/r)=\log r.$$

The state is pure exactly when $r=1$. Idempotence of $P$ does not make $P/r$ idempotent for $r>1$.

5. Channel purity criterion. Let a channel have Kraus operators $K_\mu$ and pure input $|\psi\rangle$. Show that the output is pure if and only if all nonzero vectors $K_\mu|\psi\rangle$ are proportional to one vector.

Solution

Set

$$|v_\mu\rangle=K_\mu|\psi\rangle.$$

Then

$$\Phi(|\psi\rangle\langle\psi|) = \sum_\mu|v_\mu\rangle\langle v_\mu|.$$

If every nonzero $|v_\mu\rangle$ is proportional to one vector $|w\rangle$, the sum is a positive scalar times $|w\rangle\langle w|$. Trace preservation normalizes that scalar, so the output has rank one.

Conversely, the support of the sum of positive rank-one operators is the span of the vectors $|v_\mu\rangle$. If the output has rank one, this span is one-dimensional, and all nonzero $|v_\mu\rangle$ must be proportional.

6. Partially entangled pair. For

   $$|\Psi\rangle = \sqrt p,|00\rangle + \sqrt{1-p},|11\rangle, \qquad 0\le p\le1,$$

   compute $\rho_A$, its purity, and its entropy. For which $p$ is the joint state a product state? When is the subsystem maximally mixed?

Solution

Tracing over $B$ gives

$$\rho_A = p|0\rangle\langle0| + (1-p)|1\rangle\langle1|.$$

Its purity is

$$\operatorname{Tr}(\rho_A^2) = p^2+(1-p)^2.$$

Its entropy is

$$S(\rho_A) = -p\log p-(1-p)\log(1-p).$$

The reduced state is pure, and the joint state is a product, for $p=0$ or $p=1$. It is maximally mixed at $p=1/2$, where the purity is $1/2$ and the entropy is $\log2$.

7. Unitary preservation and unread measurement. Start with $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$. First apply an arbitrary unitary $U$ and show that the state remains pure. Then perform a projective measurement in the computational basis and discard the outcome. Find the post-measurement state when $U=I$ and compute its purity.

Solution

The unitary output is

$$\rho_U = U|+\rangle\langle+|U^\dagger.$$

Because $U^\dagger U=I$,

$$\rho_U^2 = U|+\rangle\langle+|U^\dagger = \rho_U.$$

For $U=I$, an unread computational-basis measurement gives

$$\rho' = |0\rangle\langle0|\rho|0\rangle\langle0| + |1\rangle\langle1|\rho|1\rangle\langle1| = \frac12|0\rangle\langle0| + \frac12|1\rangle\langle1|.$$

Therefore

$$\operatorname{Tr}[(\rho')^2]=\frac12.$$

Each selected outcome would condition the system into a pure basis state, but discarding the outcome produces a mixed state.

8. Same local states, different pure joint states. Show that

   $$|\Phi^+\rangle = \frac{|00\rangle+|11\rangle}{\sqrt2}$$

   and

   $$|\Phi^-\rangle = \frac{|00\rangle-|11\rangle}{\sqrt2}$$

   have identical reduced states for both qubits. Give a joint measurement that distinguishes them with certainty.

Solution

For either sign, the joint density operator has diagonal terms

$$\frac12|00\rangle\langle00| + \frac12|11\rangle\langle11|.$$

The off-diagonal terms contain $|00\rangle\langle11|$ and its adjoint. Their partial trace over either subsystem vanishes because $\langle0|1\rangle=0$. Thus

$$\rho_A=\rho_B=\frac12 I$$

for both states.

The two joint kets are orthogonal:

$$\langle\Phi^+|\Phi^-\rangle=0.$$

A projective measurement in the Bell basis therefore distinguishes them with certainty. Identical local mixed states do not imply identical global pure states.