Density Operators

A density operator $\rho$ is a positive, trace-one operator representing a quantum state. It includes state vectors as a special case but also describes statistical preparation uncertainty, subsystems entangled with other degrees of freedom, thermal equilibrium states, and states evolving under noise.

Required background. State Vectors supplies the pure-state language, and Expectation Values supplies the trace-rule target. Familiarity with traces, eigenvalues, and positive semidefinite matrices is assumed.

In a finite-dimensional Hilbert space, the defining conditions are

$$\rho\ge0, \qquad \operatorname{Tr}\rho=1.$$

Positivity entails Hermiticity, although in calculations it is often useful to check all three matrix conditions explicitly:

$$\rho^\dagger=\rho, \qquad \rho\ge0, \qquad \operatorname{Tr}\rho=1.$$

The operator $\rho$ is the state. A density matrix is its matrix in a chosen basis.

Density operators for different preparations

A normalized vector $\lvert\psi\rangle$ describes a pure state. Being closed does not itself imply purity; a closed system can also have a mixed preparation. Three common situations require a more general state language.

Randomized preparations

Suppose a source prepares $\lvert\psi_k\rangle$ with probability $p_k$. If the classical label $k$ is unavailable, predictions are governed by

$$\rho = \sum_k p_k\, \lvert\psi_k\rangle\langle\psi_k\rvert, \qquad \sum_k p_k=1.$$

Here the probabilities are nonnegative. To see why the projectors are averaged, consider a normalized detector state $|\phi\rangle$. The law of total probability gives

$$p(\phi)=\sum_k p_k|\langle\phi|\psi_k\rangle|^2 =\left\langle\phi\left|\sum_k p_k|\psi_k\rangle\langle\psi_k|\right|\phi\right\rangle =\langle\phi|\rho|\phi\rangle.$$

One operator therefore supplies the probabilities for every detector direction, without retaining the preparation label. Averaging the kets themselves would introduce interference between separate runs and would even depend on their arbitrary global phases. Preparation ensembles develops this distinction and the role of an accessible record. Unless all preparations with nonzero weight represent the same ray, no single state vector reproduces every measurement probability.

Subsystems of entangled states

A composite system may be in a pure state while one subsystem is not. For the Bell state

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

the state of the first qubit is

$$\rho_A = \operatorname{Tr}_B \left( \lvert\Phi^+\rangle\langle\Phi^+\rvert \right) = \frac{I_2}{2}.$$

There is no vector in $\mathcal H_A$ whose projector equals $I_2/2$.

Thermal and open systems

Equilibrium statistical mechanics uses density operators such as

$$\rho_\beta = \frac{e^{-\beta H}}{Z}, \qquad Z=\operatorname{Tr}(e^{-\beta H}).$$

Here $\beta=1/(k_BT)$ and the partition function must be finite and nonzero. This is automatic in finite dimensions at finite $\beta$, but not for every infinite-dimensional model. Density operators also describe open systems. A mixed state can always be represented mathematically as part of a larger pure state, but that purification need not identify the actual physical preparation.

Two qubit-state comparisons

Consider the coherent pure state

$$\lvert+\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}.$$

Its density matrix in the computational basis is

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

By contrast, an equal incoherent mixture of $\lvert0\rangle$ and $\lvert1\rangle$ is

$$\rho_{\mathrm{mix}} = \frac12 \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}.$$

Both states give $0$ and $1$ with equal probability in a computational basis measurement. They differ in the $x$ basis:

$$\operatorname{Tr}(\rho_+\lvert+\rangle\langle+\rvert) =1,$$

whereas

$$\operatorname{Tr} \left( \rho_{\mathrm{mix}} \lvert+\rangle\langle+\rvert \right) = \frac12.$$

The off-diagonal entries in $\rho_+$ encode the relative-phase coherence that the second measurement reveals.

Definition and equivalent tests

For any vector $\lvert\varphi\rangle$, positivity means

$$\langle\varphi\rvert \rho \lvert\varphi\rangle \ge0.$$

In finite dimensions, the spectral theorem gives

$$\rho = \sum_{r=1}^{d} \lambda_r \lvert r\rangle\langle r\rvert,$$

where the eigenvectors may be chosen orthonormal. The density-operator conditions are equivalent to

$$\lambda_r\ge0, \qquad \sum_{r=1}^{d}\lambda_r=1.$$

Consequently,

$$0\le\lambda_r\le1, \qquad 0\le\rho\le I.$$

The trace-one condition fixes total probability. Positivity ensures that every measurement effect has a nonnegative probability. Hermiticity ensures real expectation values for observables.

Infinite-dimensional qualification

On an infinite-dimensional Hilbert space, a density operator is a positive trace-class operator with unit trace. Trace class is essential: a positive bounded operator need not have a finite trace. The spectral decomposition may contain countably many nonzero eigenvalues accumulating at zero. Because trace-class operators are compact, every nonzero spectral value is an eigenvalue; zero can be a spectral accumulation point. This differs from an observable such as position, whose spectrum is continuous. There is no normalized state proportional to the identity on an infinite-dimensional space.

Matrix representation

In an orthonormal basis $\{\lvert i\rangle\}$,

$$\rho_{ij} = \langle i\rvert\rho\lvert j\rangle.$$

The diagonal entry

$$\rho_{ii} = \langle i\rvert\rho\lvert i\rangle$$

is the probability of obtaining basis outcome $i$. Off-diagonal entries encode coherence relative to that basis. They are not probabilities and they change under a basis transformation.

Positivity constrains the coherences. Every $2\times2$ principal minor obeys

$$\lvert\rho_{ij}\rvert^2 \le \rho_{ii}\rho_{jj}.$$

Thus a state cannot have arbitrarily large off-diagonal entries while its populations remain fixed.

If the columns of the unitary matrix $V$ are the new orthonormal basis kets expressed in the old basis, then the matrix representing the same operator changes passively as

$$[\rho]' = V^\dagger[\rho]V.$$

This is different from physically applying a unitary to the state, which produces a new operator $U\rho U^\dagger$.

A qubit validity test

Every trace-one Hermitian qubit matrix can be written as

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

It is positive if and only if

$$0\le a\le1, \qquad \lvert c\rvert^2 \le a(1-a).$$

Checking only the diagonal entries is insufficient. For example,

$$R = \begin{pmatrix} \tfrac12&\tfrac34\\ \tfrac34&\tfrac12 \end{pmatrix}$$

is Hermitian and has trace one, but its eigenvalues are

$$\frac54, \qquad -\frac14.$$

The negative eigenvalue makes $R$ unphysical.

Pure states as density operators

A normalized vector defines the rank-one projector

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

It satisfies

$$\rho_\psi^2=\rho_\psi, \qquad \operatorname{Tr}\rho_\psi=1.$$

The global phase cancels:

$$(e^{i\theta}\lvert\psi\rangle) (e^{i\theta}\lvert\psi\rangle)^\dagger = \lvert\psi\rangle\langle\psi\rvert.$$

Thus the density-operator representation automatically respects the fact that pure states are rays.

In a basis where

$$\lvert\psi\rangle = \sum_i \psi_i\lvert i\rangle,$$

the matrix entries are

$$(\rho_\psi)_{ij} = \psi_i\psi_j^*.$$

The projector has one eigenvalue equal to one and all remaining eigenvalues equal to zero.

Mixed states

A density operator is mixed when it is not rank one. Equivalently, its spectral decomposition contains at least two nonzero eigenvalues:

$$\rho = \sum_r \lambda_r \lvert r\rangle\langle r\rvert.$$

This spectral decomposition is an orthogonal ensemble representation, but it is not generally the only ensemble representation. A given $\rho$ may also admit

$$\rho = \sum_k p_k \lvert\psi_k\rangle\langle\psi_k\rvert$$

with nonorthogonal $\lvert\psi_k\rangle$. No measurement performed on the system alone can distinguish two preparation procedures that yield the same density operator.

This nonuniqueness is why an ensemble is extra preparation data rather than an intrinsic decomposition of the state. Pure versus mixed is instead decided by basis-independent properties of $\rho$: rank one, idempotency, and unit purity are equivalent tests for a pure state.

Maximally mixed state

On a $d$-dimensional Hilbert space,

$$\rho_*=\frac{I_d}{d}$$

is the maximally mixed state. It assigns equal probability $1/d$ to every vector in any orthonormal measurement basis. It is basis-independent:

$$U\rho_*U^\dagger=\rho_*$$

for every unitary $U$.

“Maximally mixed” is always relative to a specified Hilbert space or support. The operator $P/r$ is maximally mixed on an $r$-dimensional subspace with projector $P$, but it is not maximally mixed on a larger ambient space.

Convex structure of state space

If $\rho_1$ and $\rho_2$ are density operators and $0\le t\le1$, then

$$\rho = t\rho_1+(1-t)\rho_2$$

is also a density operator. Positivity and trace one are preserved. The set of states is therefore convex.

Its extreme points are exactly the rank-one projectors. A mixed state can be expressed as a nontrivial convex combination of other states, while a pure state cannot. This geometric statement is independent of which particular ensemble decomposition is used.

For a $d$-dimensional system, a Hermitian matrix has $d^2$ real parameters. The trace constraint removes one, so density operators form a convex body of real dimension

$$d^2-1$$

inside the affine space of trace-one Hermitian operators. Positivity carves out the physical region. For $d=2$, that region is the Bloch ball.

Expectation values

For a bounded observable $A$ (in particular, any finite-dimensional observable), the expectation value in state $\rho$ is

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

For a pure-state projector,

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

For an ensemble,

$$\rho = \sum_k p_k \lvert\psi_k\rangle\langle\psi_k\rvert,$$

linearity gives

$$\langle A\rangle_\rho = \sum_k p_k \langle\psi_k\rvert A\lvert\psi_k\rangle.$$

For an unbounded self-adjoint observable, a finite expectation requires a finite absolute first moment of its spectral probability measure. The notation $\operatorname{Tr}(\rho A)$ then needs the corresponding moment interpretation; an ordinary operator-product trace can require stronger assumptions. See trace-rule expectations.

The trace rule is basis-independent. In a basis that diagonalizes $\rho$, it is the probability-weighted average of the diagonal matrix elements of $A$; in any other basis, cyclicity of the trace gives the same answer.

Measurement probabilities

For a projective measurement with projectors $\{P_a\}$,

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

For a general POVM with effects $\{E_i\}$,

$$E_i\ge0, \qquad \sum_i E_i=I,$$

the probability rule is

$$p(i) = \operatorname{Tr}(\rho E_i).$$

These numbers are valid probabilities:

$$p(i)\ge0, \qquad \sum_i p(i)=1.$$

Positivity of $\rho$ and $E_i$ ensures nonnegativity, while completeness of the POVM and $\operatorname{Tr}\rho=1$ ensure normalization.

For a rank-one projector

$$P_\phi = \lvert\phi\rangle\langle\phi\rvert,$$

the rule reduces to

$$\begin{aligned} \operatorname{Tr}(\rho_\psi P_\phi) &= \lvert \langle\phi\vert\psi\rangle \rvert^2. \end{aligned}$$

Thus density operators extend the Born rule rather than replace it.

Reduced states

For a joint state $\rho_{AB}$, the state of subsystem $A$ is

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

It is itself positive and trace one. Its defining operational property is

$$\operatorname{Tr}_{AB} \left[ \rho_{AB}(M_A\otimes I_B) \right] = \operatorname{Tr}_A(\rho_A M_A)$$

for every bounded observable $M_A$. A pure joint state may have a mixed reduced state, as the Bell example showed. Such mixedness reflects entanglement, not merely an unknown classical preparation label.

The Partial Trace article develops the basis-independent definition, product-basis contraction, and computational checks for this construction.

Unitary evolution

If a closed system evolves by a unitary operator $U$, then

$$\rho \longmapsto \rho' = U\rho U^\dagger.$$

This map preserves positivity, trace, rank, and eigenvalues. In particular, unitary evolution cannot turn a pure state into a mixed state or vice versa.

For a finite-dimensional differentiable unitary evolution generated by a Hermitian Hamiltonian $H(t)$, the density operator obeys the von Neumann equation

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

This is the density-operator counterpart of the Schrödinger equation. An unbounded Hamiltonian requires additional regularity for this differential equation; the unitary formula remains the more general starting point. Nonunitary reduced dynamics can change the spectrum because correlations with an environment are being discarded. At the operator-theoretic level, Strongly Continuous Unitary Groups and Stone’s Theorem supply the precise generator statement.

Common Mistakes

Checking only the diagonal. Coherence can be too large even when every population lies in $[0,1]$. Check Hermiticity, trace, and positive semidefiniteness; the invalid qubit matrix above passes the first two tests.

Using an unsuitable positivity test. Eigenvalues or a factorization designed for semidefinite matrices work. An ordinary Cholesky routine requiring strict positivity can reject a valid rank-deficient pure state. Diagnose numerical errors at the matrix’s scale before clipping eigenvalues.

Inferring a unique preparation from a matrix. Different ensembles can give the same density operator. Access to a classical preparation record or to another correlated system is additional information.

Confusing a basis change with loss of coherence. A pure state can be diagonal in its own eigenbasis without becoming mixed. Purity and the density-operator eigenvalues do not depend on the basis.

Where deeper treatment lives

The chapter overview turns the definitions and checks on this page into a reading route. Continue to Entangled States for separability, Partial Trace for subsystem states, and Quantum Operations for physical transformations. The No-Broadcasting Theorem shows how noncommutativity becomes an operational obstruction.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific (2014), Ch. 3.
• A. S. Holevo, Probabilistic and Statistical Aspects of Quantum Theory, 2nd ed., Edizioni della Normale (2011), Ch. 1.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), Secs. 2.4 and 8.2.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), Ch. 2.

Exercises

1. Let $\lvert\psi\rangle$ be normalized. Verify that

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

is positive, has trace one, and is idempotent.

Solution

For any $\lvert\varphi\rangle$,

$$\begin{aligned} \langle\varphi\rvert \rho_\psi \lvert\varphi\rangle &= \langle\varphi\vert\psi\rangle \langle\psi\vert\varphi\rangle \\ &= \lvert\langle\psi\vert\varphi\rangle\rvert^2 \ge0. \end{aligned}$$

In a basis containing $\lvert\psi\rangle$, the projector has diagonal entries $1,0,\ldots,0$, so its trace is one. Finally,

$$\rho_\psi^2 = \lvert\psi\rangle \langle\psi\vert\psi\rangle \langle\psi\rvert = \rho_\psi.$$

2. Determine when

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

is a valid density matrix.

Solution

For real $a$ and complex $c$, Hermiticity and trace one are already built in. Positivity requires the diagonal entries and determinant to be nonnegative:

$$\begin{aligned} 0 &\le a\le1, \\ \det\rho &= a(1-a)-\lvert c\rvert^2 \ge0. \end{aligned}$$

Therefore the necessary and sufficient condition is

$$0\le a\le1, \qquad \lvert c\rvert^2\le a(1-a).$$

3. Compare the equal mixture of $\lvert0\rangle,\lvert1\rangle$ with the pure state $\lvert+\rangle$. Compute the probability of the $+$ outcome in an $x$-basis measurement.

Solution

The two density matrices are

$$\rho_{\mathrm{mix}} = \frac{I_2}{2}, \qquad \rho_+ = \lvert+\rangle\langle+\rvert.$$

For $P_+=\lvert+\rangle\langle+\rvert$,

$$\operatorname{Tr}(\rho_{\mathrm{mix}}P_+) = \frac12,$$

whereas

$$\operatorname{Tr}(\rho_+P_+) =1.$$

Equal computational-basis populations do not make a mixture equivalent to a coherent superposition.

4. Show that the maximally mixed qubit has both ensemble decompositions

$$\frac{I_2}{2} = \frac12\lvert0\rangle\langle0\rvert + \frac12\lvert1\rangle\langle1\rvert$$

and

$$\frac{I_2}{2} = \frac12\lvert+\rangle\langle+\rvert + \frac12\lvert-\rangle\langle-\rvert.$$

Solution

Completeness of either orthonormal basis gives

$$\lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert = I_2,$$

and

$$\lvert+\rangle\langle+\rvert + \lvert-\rangle\langle-\rvert = I_2.$$

Dividing both identities by two gives the claimed decompositions. They are different preparation ensembles but the same density operator.

5. For

$$\rho = \begin{pmatrix} \tfrac34&0\\ 0&\tfrac14 \end{pmatrix},$$

compute $\langle X\rangle$, $\langle Y\rangle$, and $\langle Z\rangle$.

Solution

Because $\rho$ is diagonal, its trace with either off-diagonal Pauli matrix vanishes:

$$\langle X\rangle=0, \qquad \langle Y\rangle=0.$$

For $Z=\operatorname{diag}(1,-1)$,

$$\langle Z\rangle = \operatorname{Tr}(\rho Z) = \frac34-\frac14 = \frac12.$$

6. Prove that unitary evolution preserves the density-operator conditions for $\rho'=U\rho U^\dagger$.

Solution

For any $\lvert\varphi\rangle$,

$$\begin{aligned} \langle\varphi\rvert\rho'\lvert\varphi\rangle &= \langle\varphi\rvert U\rho U^\dagger \lvert\varphi\rangle \\ &= \langle\chi\rvert\rho\lvert\chi\rangle \ge0, \end{aligned}$$

where $\lvert\chi\rangle=U^\dagger\lvert\varphi\rangle$. Cyclicity of the trace gives

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

Hermiticity is also preserved:

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

7. Compute the reduced density operator of either qubit in $\lvert\Phi^+\rangle$.

Solution

The joint projector is

$$\lvert\Phi^+\rangle\langle\Phi^+\rvert = \frac12 \left( \lvert00\rangle\langle00\rvert + \lvert00\rangle\langle11\rvert + \lvert11\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \right).$$

Tracing over the second qubit removes the cross terms and gives

$$\rho_A = \frac12 \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right) = \frac{I_2}{2}.$$

The same calculation holds for $\rho_B$.

8. Explain why

$$R = \begin{pmatrix} \tfrac12&\tfrac34\\ \tfrac34&\tfrac12 \end{pmatrix}$$

cannot represent a state by finding a vector assigned a negative expectation value.

Solution

The vector

$$\lvert-\rangle = \frac{ \lvert0\rangle-\lvert1\rangle }{\sqrt2}$$

is an eigenvector of $R$ with eigenvalue $-1/4$. Therefore

$$\langle-\rvert R\lvert-\rangle = -\frac14<0.$$

This violates positivity. If $P_-=\lvert-\rangle\langle-\rvert$ were used as a measurement outcome, the trace rule would assign the impossible probability

$$\operatorname{Tr}(R P_-) = -\frac14.$$