Density-Matrix Formulation

The density-matrix formulation takes density operators as the primary quantum states. Pure-state vectors remain available through $\rho_\psi=\lvert\psi\rangle\langle\psi\rvert$, but mixed preparations, reduced subsystem states, unread measurement outcomes, and open-system processes can all be stated without leaving the same operator language.

The aim is to combine state preparation, evolution, measurement, and conditioning in one calculation. Density operators explains the state space in detail.

Helpful background. Minimal postulates gives the common framework; the trace rule explains how states and observables produce averages.

States, effects, and outcome branches

The formulas apply directly in finite dimension and, with trace-class and continuity qualifications, to normal states on separable Hilbert spaces. Finite or countable outcome sets are used for compact notation. Continuous outcomes require operator-valued measures and instrument-valued measures.

The formulation distinguishes three kinds of positive operator:

• a normalized state $\rho$ has trace one;
• a subnormalized state or branch $\widetilde\rho$ has trace between zero and one;
• an effect $E$ satisfies $0\leq E\leq I$ and pairs with states to produce probabilities.

These objects can look similar as matrices but play different operational roles. The notation and trace condition identify which role is intended.

Postulate 1: State as Density Operator

A normalized state is a positive trace-class operator $\rho$ satisfying

$$\rho\geq0, \qquad \operatorname{Tr}\rho=1.$$

Positivity means

$$\langle\phi\rvert \rho \lvert\phi\rangle \geq0$$

for every $\lvert\phi\rangle\in\mathcal H$. In finite dimension, the spectral theorem gives

$$\rho = \sum_j \lambda_j \lvert j\rangle\langle j\rvert,$$

where

$$\lambda_j\geq0, \qquad \sum_j\lambda_j=1.$$

Thus the eigenvalues of a density matrix form a probability distribution.

Pure states as a special case

A normalized vector defines

$$\rho_\psi = \lvert\psi\rangle \langle\psi\rvert.$$

The projector is unchanged by global phase:

$$\lvert\psi\rangle \longmapsto e^{i\alpha}\lvert\psi\rangle \quad\Longrightarrow\quad \rho_\psi \longmapsto \rho_\psi.$$

In finite dimension, equivalent purity conditions are

$$\begin{aligned} \rho\text{ pure} &\Longleftrightarrow \operatorname{rank}\rho=1\\ &\Longleftrightarrow \rho^2=\rho\\ &\Longleftrightarrow \operatorname{Tr}\rho^2=1. \end{aligned}$$

These tests do not require choosing a state vector.

Convexity and preparations

If preparation $j$ is selected with classical probability $q_j$, the state is

$$\rho = \sum_jq_j\rho_j, \qquad q_j\geq0, \qquad \sum_jq_j=1.$$

The same $\rho$ can generally have inequivalent ensemble decompositions. All predictions internal to the system are determined by $\rho$, not by a preferred decomposition. If a classical record of $j$ is retained, the larger classical–quantum state contains more information than the averaged operator alone.

See Ensembles and Preparation Procedures for the detailed treatment.

Subnormalized branches

A positive operator $\widetilde\rho$ with

$$0\leq \operatorname{Tr}\widetilde\rho \leq1$$

can encode an event together with its probability. If its trace is nonzero, then

$$\rho_{\mathrm{cond}} = \frac{\widetilde\rho} {\operatorname{Tr}\widetilde\rho}$$

is normalized. Keeping branches subnormalized until the final conditioning step makes sequential calculations cleaner and avoids losing probability factors.

Postulates 2 and 3: Effects and the Trace Rule

A finite or countable measurement is represented by a POVM $\{E_a\}$:

$$E_a\geq0, \qquad \sum_aE_a=I.$$

The probability of outcome $a$ in state $\rho$ is

$$p(a) = \operatorname{Tr}(\rho E_a).$$

The state–effect pairing is linear in each argument. If a preparation is mixed,

$$\operatorname{Tr} \left[ \left( \sum_jq_j\rho_j \right) E_a \right] = \sum_jq_j \operatorname{Tr}(\rho_jE_a).$$

If two outcomes are classically merged, their effect is the sum of their effects, and their probabilities add. These linearity properties make the trace rule compatible with ordinary probabilistic randomization and coarse graining.

Why the result is a probability

Because $\rho$ and $E_a$ are positive,

$$\operatorname{Tr}(\rho E_a) = \operatorname{Tr} \left( \rho^{1/2}E_a\rho^{1/2} \right) \geq0.$$

Completeness gives

$$\begin{aligned} \sum_a p(a) &= \operatorname{Tr} \left( \rho\sum_aE_a \right)\\ &= \operatorname{Tr}\rho =1. \end{aligned}$$

For a pure state and rank-one projector,

$$\begin{aligned} \operatorname{Tr} \left( \rho_\psi \lvert a\rangle\langle a\rvert \right) &= \lvert\langle a\mid\psi\rangle\rvert^2. \end{aligned}$$

The squared-amplitude Born rule is therefore a special case of the trace rule.

Sharp observables

For

$$A = \sum_a aP_a,$$

the projectors $\{P_a\}$ form a PVM and

$$p(a) = \operatorname{Tr}(\rho P_a).$$

For bounded $A$, the expectation value is

$$\langle A\rangle_\rho = \operatorname{Tr}(\rho A).$$

Unbounded $A$ requires a finite absolute first spectral moment; the operator-product trace needs care. The derivation and precise conditions are in Trace Rule for Expectation Values.

Postulate 4: Unitary Evolution

A closed system evolves by unitary conjugation:

$$\rho(t_2) = U(t_2,t_1) \rho(t_1) U(t_2,t_1)^\dagger.$$

This map preserves all state conditions. Hermiticity follows from taking the adjoint. Positivity follows because

$$\langle\phi\rvert U\rho U^\dagger \lvert\phi\rangle = \langle\chi\rvert \rho \lvert\chi\rangle \geq0,$$

where $\lvert\chi\rangle=U^\dagger\lvert\phi\rangle$. Trace preservation follows from cyclicity:

$$\operatorname{Tr} \left( U\rho U^\dagger \right) = \operatorname{Tr} \left( \rho U^\dagger U \right) =1.$$

Unitary conjugation also preserves the eigenvalues of $\rho$. Consequently, it preserves rank, purity, and von Neumann entropy. Closed evolution can rotate the eigenvectors but cannot purify a mixed state without conditioning or changing the system boundary.

For a time-independent self-adjoint Hamiltonian,

$$U(t) = \exp\!\left( -\frac{i}{\hbar}Ht \right),$$

and, in finite dimension or under suitable trace-norm differentiability and domain conditions, the differential form is

$$i\hbar \frac{d\rho}{dt} = [H,\rho].$$

In finite dimension, $[H,\rho]=0$ makes the density operator stationary. In general, stationarity means that $\rho$ commutes with every $U(t)$ (equivalently, the spectral projections of $H$). Eigenvectors in a spectral ensemble may acquire phases while their projectors remain fixed.

Schrödinger and Heisenberg descriptions

One may evolve the state,

$$\rho' = U\rho U^\dagger,$$

or equivalently evolve an effect or observable backward,

$$E_a' = U^\dagger E_aU.$$

The trace pairing agrees:

$$\operatorname{Tr}(\rho'E_a) = \operatorname{Tr}(\rho E_a').$$

This duality is one reason density-operator notation makes picture changes especially transparent.

Postulate 5: Composition and Reduction

For distinguishable systems,

$$\mathcal H_{AB} = \mathcal H_A \otimes \mathcal H_B.$$

Independent preparations have product states

$$\rho_{AB} = \rho_A\otimes\rho_B,$$

but general joint density operators need not be convex mixtures of product states. Entangled states lie outside that separable subset. In infinite dimensions, include trace-norm limits of convex product mixtures in the definition of separability.

The state available for predictions on $A$ alone is

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

It is characterized by the identity

$$\begin{aligned} \operatorname{Tr}_{AB} \left[ \rho_{AB}(X_A\otimes I_B) \right] &= \operatorname{Tr}_A(\rho_AX_A) \end{aligned}$$

for every bounded operator $X_A$.

This identity guarantees that local Born probabilities can be calculated from the reduced state:

$$\begin{aligned} \operatorname{Tr}_{AB} \left[ \rho_{AB}(E_a\otimes I_B) \right] &= \operatorname{Tr}_A(\rho_AE_a). \end{aligned}$$

A pure joint state can have a mixed reduced state. Mixedness therefore does not by itself imply classical ignorance about a pure state of the subsystem. See Reduced Density Matrices.

Projective Measurement Update

Let $\{P_a\}$ be a PVM. The ideal Lüders branch for outcome $a$ is

$$\widetilde\rho_a = P_a\rho P_a.$$

Its trace is the Born probability:

$$\begin{aligned} \operatorname{Tr}\widetilde\rho_a &= \operatorname{Tr}(P_a\rho P_a)\\ &= \operatorname{Tr}(\rho P_a) =p(a). \end{aligned}$$

If $p(a)>0$, the selective state is

$$\rho_a = \frac{P_a\rho P_a} {\operatorname{Tr}(\rho P_a)}.$$

The division is not defined for a zero-probability outcome. The formalism supplies no conditional state for an event that cannot occur under the stated model.

If the measurement occurs but the outcome is ignored, the nonselective state is

$$\rho' = \sum_aP_a\rho P_a.$$

Selective and nonselective updates answer different information questions. Averaging the normalized conditional states with their outcome probabilities recovers the nonselective state:

$$\rho' = \sum_{a:p(a)>0} p(a)\rho_a.$$

For a degenerate outcome, $P_a$ projects onto an eigenspace rather than one ray. The Lüders update preserves coherence within that eigenspace. A more disturbing apparatus can have the same PVM effects and different postmeasurement states.

The update rule is developed canonically in State Update Rule.

Generalized Operations

Density operators make open dynamics and generalized measurement compact.

Quantum channels

A quantum channel $\mathcal E$ maps normalized input states to normalized output states. It is linear, completely positive, and trace preserving. In finite dimension it has a Kraus representation

$$\mathcal E(\rho) = \sum_\mu K_\mu\rho K_\mu^\dagger,$$

with

$$\sum_\mu K_\mu^\dagger K_\mu = I.$$

Trace preservation follows directly:

$$\begin{aligned} \operatorname{Tr}\mathcal E(\rho) &= \sum_\mu \operatorname{Tr} \left( \rho K_\mu^\dagger K_\mu \right)\\ &= \operatorname{Tr}\rho =1. \end{aligned}$$

Complete positivity ensures that $\mathcal E\otimes\mathcal I_R$ remains positive when the system is entangled with an arbitrary reference $R$. Ordinary positivity on isolated input states is not enough for a universally valid local physical process.

For reduced dynamics this channel description holds, for example, when the environment begins in a fixed state independent of and uncorrelated with the system input. Pre-existing correlations can obstruct a channel defined on every system state. Unitary evolution is the one-Kraus special case

$$\mathcal E_U(\rho) = U\rho U^\dagger.$$

Quantum instruments

An instrument is a family of linear, completely positive, trace-nonincreasing maps $\{\mathcal I_a\}$ whose sum is a channel:

$$\sum_a\mathcal I_a = \mathcal E.$$

Each map produces an unnormalized branch,

$$\widetilde\rho_a = \mathcal I_a(\rho),$$

and

$$p(a) = \operatorname{Tr}\mathcal I_a(\rho).$$

The corresponding effect satisfies

$$\operatorname{Tr}\mathcal I_a(\rho) = \operatorname{Tr}(\rho E_a)$$

for every state $\rho$. The effects determine the classical outcome distribution; the maps determine both outcomes and disturbance.

A finite-dimensional instrument can be written

$$\mathcal I_a(\rho) = \sum_\mu M_{a\mu} \rho M_{a\mu}^\dagger,$$

with

$$E_a = \sum_\mu M_{a\mu}^\dagger M_{a\mu}.$$

The overall completeness relation is

$$\sum_{a,\mu} M_{a\mu}^\dagger M_{a\mu} = I.$$

See Quantum Instruments, Kraus Representation, and Completely Positive Maps for the mature open-systems treatment.

Worked Example: Selective and Unread Qubit Measurement

Write a qubit state in the computational basis as

$$\rho = \begin{pmatrix} a&c\\ c^*&1-a \end{pmatrix},$$

with positivity requiring

$$0\leq a\leq1, \qquad \lvert c\rvert^2 \leq a(1-a).$$

Measure the PVM

$$P_0 = \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad P_1 = \begin{pmatrix} 0&0\\ 0&1 \end{pmatrix}.$$

The outcome probabilities are

$$p(0)=a, \qquad p(1)=1-a.$$

The unnormalized Lüders branches are

$$\begin{aligned} \widetilde\rho_0 &= \begin{pmatrix} a&0\\ 0&0 \end{pmatrix},\\ \widetilde\rho_1 &= \begin{pmatrix} 0&0\\ 0&1-a \end{pmatrix}. \end{aligned}$$

When $0<a<1$, the conditional states are

$$\rho_0=P_0, \qquad \rho_1=P_1.$$

If the outcome is unread, the state is

$$\rho' = \widetilde\rho_0 + \widetilde\rho_1 = \begin{pmatrix} a&0\\ 0&1-a \end{pmatrix}.$$

The measurement removes the off-diagonal coherence $c$ in this basis but preserves the outcome populations. Before the unread measurement,

$$\operatorname{Tr}\rho^2 = a^2+(1-a)^2+2\lvert c\rvert^2.$$

Afterward,

$$\operatorname{Tr}(\rho')^2 = a^2+(1-a)^2.$$

The purity decreases by $2\lvert c\rvert^2$ unless the state was already diagonal. For the input $\lvert+\rangle\langle+\rvert$, one has $a=1/2$ and $c=1/2$, so

$$\lvert+\rangle\langle+\rvert \longmapsto \frac{I}{2}$$

under the unread computational-basis measurement.

This example distinguishes:

• the POVM, which supplies $p(0)$ and $p(1)$;
• the selective branches, which encode outcome and probability together;
• the normalized conditional states;
• the nonselective channel, which removes coherence when the record is ignored.

Equivalence with Pure-State Formulas

Whenever $\rho=\lvert\psi\rangle\langle\psi\rvert$:

• the trace Born rule reduces to

  $$\operatorname{Tr}(\rho E) = \langle\psi\rvert E\lvert\psi\rangle;$$

• unitary conjugation reduces to

  $$U\rho U^\dagger = \lvert U\psi\rangle \langle U\psi\rvert;$$

• the Lüders branch reduces to

  $$P_a\rho P_a = \lvert\widetilde\psi_a\rangle \langle\widetilde\psi_a\rvert,$$

  where

  $$\lvert\widetilde\psi_a\rangle = P_a\lvert\psi\rangle;$$

• tensor-product pure states reduce to projectors onto product vectors.

The density-matrix formulation therefore contains the pure-state formulation rather than competing with it.

Exercises

1. Validate a qubit state. Determine the real values of $r$ for which

$$\rho_r = \frac12 \begin{pmatrix} 1&r\\ r&1 \end{pmatrix}$$

is a density matrix, and identify the pure cases.

Solution

The matrix is Hermitian and has trace one. Its eigenvalues are

$$\lambda_\pm = \frac{1\pm r}{2}.$$

Positivity requires

$$-1\leq r\leq1.$$

The purity is

$$\operatorname{Tr}\rho_r^2 = \frac{1+r^2}{2}.$$

It equals one exactly when $\lvert r\rvert=1$. Thus $r=1$ and $r=-1$ are the pure states $\lvert+\rangle\langle+\rvert$ and $\lvert-\rangle\langle-\rvert$.

2. Recover the vector Born rule. Show directly that

$$\operatorname{Tr} \left[ \lvert\psi\rangle\langle\psi\rvert \lvert a\rangle\langle a\rvert \right] = \lvert\langle a\mid\psi\rangle\rvert^2.$$

Solution

Multiply the rank-one operators:

$$\lvert\psi\rangle\langle\psi\mid a\rangle \langle a\rvert = \langle\psi\mid a\rangle \lvert\psi\rangle\langle a\rvert.$$

The trace of $\lvert u\rangle\langle v\rvert$ is $\langle v\mid u\rangle$, so

$$\begin{aligned} \operatorname{Tr} \left[ \lvert\psi\rangle\langle\psi\rvert \lvert a\rangle\langle a\rvert \right] &= \langle\psi\mid a\rangle \langle a\mid\psi\rangle\\ &= \lvert\langle a\mid\psi\rangle\rvert^2. \end{aligned}$$

3. Unitary preservation. Prove that unitary conjugation preserves $\operatorname{Tr}\rho^2$.

Solution

Let $\rho'=U\rho U^\dagger$. Then

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

Cyclicity and unitarity give

$$\operatorname{Tr}(\rho')^2 = \operatorname{Tr} \left( \rho^2U^\dagger U \right) = \operatorname{Tr}\rho^2.$$

Thus closed-system evolution preserves purity.

4. Selective versus nonselective update. A qubit in $\lvert+\rangle$ is measured in the computational basis with the Lüders instrument. Find the two branches, conditional states, and unread state.

Solution

The input is

$$\rho_+ = \frac12 \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix}.$$

The branches are

$$\widetilde\rho_0 = \frac12P_0, \qquad \widetilde\rho_1 = \frac12P_1.$$

Each has trace $1/2$. The conditional states are $P_0$ and $P_1$. Ignoring the outcome gives

$$\rho' = \frac12P_0+\frac12P_1 = \frac{I}{2}.$$

The unread measurement removes the computational-basis coherence.

5. Same effects, different instruments. For $P_a=\lvert a\rangle\langle a\rvert$, compare

$$\mathcal I_a(\rho) = P_a\rho P_a$$

with

$$\mathcal J_a(\rho) = \operatorname{Tr}(\rho P_a) \lvert+\rangle\langle+\rvert.$$

Show that the effects agree and the conditional states differ.

Solution

For either map,

$$\operatorname{Tr}\mathcal I_a(\rho) = \operatorname{Tr}\mathcal J_a(\rho) = \operatorname{Tr}(\rho P_a).$$

Thus both instruments have effect $P_a$ and identical outcome probabilities. For a nonzero branch, $\mathcal I_a$ gives conditional state $P_a$, while $\mathcal J_a$ gives $\lvert+\rangle\langle+\rvert$. The POVM does not determine disturbance.

6. Reduce a Bell state. For

$$\lvert\Phi^+\rangle = \frac{ \lvert00\rangle+\lvert11\rangle }{\sqrt2},$$

compute the reduced density operator of subsystem $A$.

Solution

Expand the joint projector:

$$\rho_{AB} = \frac12 \Big( \lvert00\rangle\langle00\rvert + \lvert00\rangle\langle11\rvert + \lvert11\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \Big).$$

The partial trace over $B$ removes the cross terms because $\langle1\mid0\rangle=0$:

$$\rho_A = \frac12 \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right) = \frac{I}{2}.$$

The joint state is pure while the subsystem state is maximally mixed.

7. A depolarizing channel. For a $d$-dimensional system, define

$$\mathcal D_q(\rho) = q\rho + (1-q)\frac{I}{d},$$

with $0\leq q\leq1$. Show that it preserves trace and find its fixed states for $q\neq1$.

Solution

Trace preservation follows from

$$\begin{aligned} \operatorname{Tr}\mathcal D_q(\rho) &= q\operatorname{Tr}\rho + (1-q) \operatorname{Tr}\frac{I}{d}\\ &= q+(1-q) =1. \end{aligned}$$

A fixed state satisfies

$$\rho = q\rho + (1-q)\frac{I}{d}.$$

For $q\neq1$, division by $1-q$ gives

$$\rho=\frac{I}{d}.$$

The maximally mixed state is the unique fixed density operator in that range. The linear extension to arbitrary operators is $\mathcal D_q(X)=qX+(1-q)\operatorname{Tr}(X)I/d$. Complete positivity follows because this is a convex mixture of the identity channel and the replacement channel. For the latter, Kraus operators $K_{jk}=|j\rangle\langle k|/\sqrt d$ satisfy $\sum_{jk}K_{jk}XK_{jk}^\dagger=\operatorname{Tr}(X)I/d$ and $\sum_{jk}K_{jk}^\dagger K_{jk}=I$.

8. Coarse-grain an instrument. Outcomes $a$ in a subset $C$ are reported only as “$C$.” Show that the coarse-grained branch and probability are

$$\widetilde\rho_C = \sum_{a\in C} \mathcal I_a(\rho)$$

and

$$p(C) = \sum_{a\in C}p(a).$$

Solution

The coarse-grained operation is the sum of the mutually exclusive outcome maps. Its trace is

$$\begin{aligned} \operatorname{Tr}\widetilde\rho_C &= \sum_{a\in C} \operatorname{Tr}\mathcal I_a(\rho)\\ &= \sum_{a\in C}p(a). \end{aligned}$$

Thus $p(C)=\operatorname{Tr}\widetilde\rho_C$. If this probability is nonzero, the state conditioned only on the coarse record is

$$\rho_C = \frac{ \sum_{a\in C}\mathcal I_a(\rho) }{ \sum_{a\in C}p(a) }.$$

Coarse graining retains less classical information than learning the fine outcome. It does not undo the fine measurement: in general $\sum_{a\in C}P_a\rho P_a$ differs from $P_C\rho P_C$, where $P_C=\sum_{a\in C}P_a$.

References

• P. Busch, P. J. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer, 2016 — modern effect and instrument formalism.
• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Springer, 1983 — states, effects, operations, and instruments.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010 — density operators, channels, measurements, and composite systems.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995 — operationally careful treatment of mixed states and measurement.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955 — foundational operator and density-matrix formulation.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018 — rigorous finite-dimensional states, maps, and complete positivity.