Pure vs Mixed States

A quantum state is pure when its density operator is a rank-one projector,

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

It is mixed when no single ray represents the state. In finite dimensions, a density operator with spectral decomposition

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

is pure exactly when one eigenvalue is $1$ and all others vanish. It is mixed when at least two eigenvalues are nonzero.

This distinction is intrinsic to the density operator. It does not depend on the basis in which the matrix is written or on a particular preparation ensemble used to realize it.

Helpful background. Density operators defines positivity, trace, and spectral decomposition. We use finite dimensions except in the final qualification.

Pure states

For a normalized vector $\lvert\psi\rangle$,

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

has rank one and obeys

$$\rho_\psi^2=\rho_\psi.$$

It assigns probability one to the ray $\lvert\psi\rangle$:

$$\operatorname{Tr} \left( \rho_\psi^2 \right) =1.$$

It need not look diagonal in the basis being used. For example,

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

is pure, although its computational-basis density matrix has nonzero off-diagonal entries:

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

Every pure-state projector becomes

$$\operatorname{diag}(1,0,\ldots,0)$$

in a basis containing its state vector. Diagonal versus nondiagonal is therefore not a basis-independent test of mixedness.

Mixed states

A mixed state has more than one nonzero eigenvalue:

$$\rho = \sum_{r=1}^{R} \lambda_r \lvert r\rangle\langle r\rvert, \qquad R>1.$$

The rank $R$ counts the dimension of the support of $\rho$. A mixed state can arise from a randomized preparation,

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

or by reducing a larger entangled state. The same density operator can have many ensemble decompositions, so mixedness does not identify a unique list of underlying pure preparations.

The maximally mixed state on a $d$-dimensional Hilbert space is

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

Its rank is $d$, and its spectrum is uniform. Other mixed states interpolate between a rank-one boundary point and this maximally mixed center in ways made precise by purity and entropy.

Equivalent criteria

For a finite-dimensional density operator, the following conditions are equivalent:

1. $\rho$ is pure.
2. $\rho$ has rank one.
3. $\rho^2=\rho$.
4. $\operatorname{Tr}(\rho^2)=1$.
5. The spectrum is $\{1,0,\ldots,0\}$.
6. $\rho$ is an extreme point of the convex state space.

The state is mixed exactly when these pure-state conditions fail.

Spectral proof

Because $\rho$ is positive and trace one,

$$\lambda_r\ge0, \qquad \sum_r\lambda_r=1.$$

Its square has eigenvalues $\lambda_r^2$, so

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

The identity

$$1-\operatorname{Tr}(\rho^2) = 2\sum_{r<s} \lambda_r\lambda_s$$

shows that the trace of $\rho^2$ equals one exactly when every product $\lambda_r\lambda_s$ with $r\ne s$ vanishes. Since the eigenvalues sum to one, precisely one eigenvalue must then equal one.

Likewise,

$$\rho^2=\rho$$

forces every eigenvalue to satisfy

$$\lambda_r^2=\lambda_r.$$

Hence each $\lambda_r$ is $0$ or $1$, and trace one leaves exactly one nonzero eigenvalue.

Entropy criterion

The von Neumann entropy

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

vanishes exactly for a pure density operator. A finite-dimensional mixed state has $S(\rho)>0$. The conventions, base dependence, and examples are developed in Entropy Overview.

Purity

The purity of a state is

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

In $d$ dimensions,

$$\frac{1}{d} \le \gamma(\rho) \le1.$$

The upper bound is attained exactly by pure states. The lower bound is attained exactly by $I_d/d$.

To prove the lower bound, use the eigenvalues and Cauchy–Schwarz:

$$\left( \sum_{r=1}^{d}\lambda_r \right)^2 \le d\sum_{r=1}^{d}\lambda_r^2.$$

Since the sum on the left is one,

$$\gamma(\rho)\ge\frac1d.$$

If the state has rank $R$, the sharper support-dependent bound is

$$\frac1R \le \gamma(\rho) \le1.$$

For fixed $R>1$, the upper inequality is strict, though the purity can approach one when all but one eigenvalue approach zero. The lower value $1/R$ occurs for a state maximally mixed on its support.

Invariance and mixing

Purity is invariant under unitary evolution:

$$\begin{aligned} \gamma(U\rho U^\dagger) &= \operatorname{Tr} \left( U\rho^2U^\dagger \right) \\ &= \gamma(\rho). \end{aligned}$$

It is also a convex function. For

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

one finds

$$t\gamma(\rho_1) +(1-t)\gamma(\rho_2) -\gamma(\rho_t) = t(1-t) \operatorname{Tr} \left[ (\rho_1-\rho_2)^2 \right] \ge0.$$

Randomly forgetting which of two preparations occurred cannot increase purity beyond the probability-weighted average of the component purities. General channels can increase or decrease purity: a reset channel can turn $I/d$ into a pure state, whereas dephasing can turn a coherent pure state into a mixture. Conditioning on an outcome can also change purity, but a normalized conditional update is not generally a linear trace-preserving channel.

Purity is useful but not a complete description of a mixed state. For dimensions greater than two, different spectra can share the same value of $\operatorname{Tr}(\rho^2)$. For a qubit, purity does fix the two eigenvalues, but it does not fix their eigenvectors.

Why pure states are extreme

If a rank-one state is written as $\rho=t\rho_1+(1-t)\rho_2$ with $0<t<1$, its expectation on any vector orthogonal to its ray is zero. Positivity forces both component states to have zero support there, so each must be the same rank-one projector. Conversely, a mixed state’s spectral decomposition gives a nontrivial convex combination. This proves the extreme-point criterion.

Superposition Is Not Mixture

The state $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$ is pure, with purity one. An equal mixture of $|0\rangle$ and $|1\rangle$ is $I/2$, with purity $1/2$. Both give equal computational-basis probabilities, but an $X$ measurement gives $+$ with probabilities one and one half respectively. Classical mixtures versus quantum superpositions develops the interference comparison.

The number of preparation labels alone does not determine purity. For $\rho=p|a\rangle\langle a|+(1-p)|b\rangle\langle b|$ with normalized kets,

$$\operatorname{Tr}\rho^2=p^2+(1-p)^2 +2p(1-p)|\langle a|b\rangle|^2.$$

Orthogonal preparations give the smallest purity at fixed $p$; identical rays keep the state pure. Exercise 3 gives a nonorthogonal example.

A partially coherent qubit family

Consider

$$\rho(c) = \frac12 \begin{pmatrix} 1&c\\ c^*&1 \end{pmatrix}.$$

Positivity requires

$$\lvert c\rvert\le1.$$

The eigenvalues are

$$\lambda_\pm = \frac{1\pm\lvert c\rvert}{2},$$

and the purity is

$$\gamma(\rho(c)) = \frac{1+\lvert c\rvert^2}{2}.$$

Therefore:

• $\lvert c\rvert=1$ gives a pure equal-weight superposition;
• $0<\lvert c\rvert<1$ gives a partially coherent mixed state;
• $c=0$ gives the maximally mixed qubit.

The magnitude of $c$ controls mixedness in this chosen family, while its phase selects the direction of the coherence in the equatorial plane.

Bloch-ball criterion

Every qubit state has the form

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right), \qquad \lVert\mathbf r\rVert\le1.$$

Its eigenvalues are

$$\lambda_\pm = \frac{ 1\pm\lVert\mathbf r\rVert }{2},$$

and

$$\gamma(\rho) = \frac{ 1+\lVert\mathbf r\rVert^2 }{2}.$$

Thus

$$\begin{aligned} \lVert\mathbf r\rVert=1 &\quad\Longleftrightarrow\quad \rho\text{ is pure}, \\ \lVert\mathbf r\rVert<1 &\quad\Longleftrightarrow\quad \rho\text{ is mixed}. \end{aligned}$$

Pure qubit states lie on the surface of the Bloch ball; mixed states lie in its interior; $I_2/2$ lies at the center. The geometric and measurement interpretations are developed in Bloch Sphere.

Preparation uncertainty and entanglement

There are two physically different routes to the same local density operator.

Classical preparation record

A source may toss a fair classical coin and prepare $\lvert0\rangle$ or $\lvert1\rangle$. If the record is unavailable, the assigned state is

$$\rho_A = \frac{I_2}{2}.$$

Someone who retains the record can sort the ensemble into pure subensembles.

Reduction of an entangled state

The Bell state

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

is globally pure, but

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

Here the mixed local state arises because the information distinguishing correlated alternatives resides in the joint system. Access to subsystem $A$ alone cannot reveal whether $I_2/2$ arose from a classical source record or from this entangled preparation. Access to the record, the purifying subsystem, or suitable joint correlations can reveal the difference.

The terminology proper mixture and improper mixture is sometimes used for these two contexts. The labels concern the global preparation story, not two different kinds of local density operator. All local probabilities are determined by the same $\rho_A$.

Moreover, because ensemble decompositions are nonunique, one should not infer that a mixed density operator means the system secretly occupies one preferred pure state in one preferred ensemble. The operational and foundational interpretation must be stated separately from the density operator itself.

Reduced Density Matrices and Purification Overview develop the two constructions.

How purity is determined

Purity is nonlinear in $\rho$, so it is not the expectation value of one fixed observable on one copy of an arbitrary unknown state. Common routes include:

• reconstructing $\rho$ by state tomography and then evaluating $\operatorname{Tr}(\rho^2)$;
• using prior structure that reduces the number of unknown parameters;
• performing a collective measurement on two independently and identically prepared copies.

For two copies in the product state $\rho\otimes\rho$, let $S$ be the swap operator,

$$S \left( \lvert\alpha\rangle\otimes\lvert\beta\rangle \right) = \lvert\beta\rangle\otimes\lvert\alpha\rangle.$$

Then

$$\operatorname{Tr} \left[ S(\rho\otimes\rho) \right] = \operatorname{Tr}(\rho^2).$$

This identity gives purity a direct two-copy operational meaning. It does not imply that a single measurement outcome certifies purity; finite data always require statistical inference.

Derived Purity Identities

For a product state, purity factorizes:

$$\gamma(\rho_A\otimes\rho_B) =\gamma(\rho_A)\gamma(\rho_B).$$

For a globally pure bipartite state, a reduced state has purity one exactly when the global state is a product. For a globally mixed state, reduced purity by itself is not an entanglement measure.

Two common reparameterizations are the linear entropy and inverse purity,

$$S_L=1-\gamma, \qquad d_{\mathrm{eff}}=\frac{1}{\gamma}.$$

Some authors normalize $S_L$ by $d/(d-1)$; the convention must be stated. The effective dimension is generally not an integer and contains no more spectral information than purity.

For a Gibbs state $\rho_\beta=e^{-\beta H}/Z(\beta)$,

$$\gamma(\rho_\beta) =\frac{Z(2\beta)}{Z(\beta)^2}.$$

This shortcut requires the relevant partition functions to be finite. In finite dimensions, as temperature approaches zero from above, a nondegenerate ground state gives $\gamma\to1$; an equal mixture on an unresolved $g$-fold ground space gives $\gamma\to1/g$.

Infinite-dimensional caveat

For a positive trace-class operator with unit trace,

$$0<\operatorname{Tr}(\rho^2)\le1,$$

and equality to one still characterizes a pure state. There is no dimension-independent positive lower bound analogous to $1/d$, because the purity can be made arbitrarily small by spreading weight over more orthogonal states.

The formal operator $I/\operatorname{Tr}I$ is not a density operator on an infinite-dimensional Hilbert space: the identity is not trace class. A “maximally mixed state on the whole infinite-dimensional space” therefore does not exist in the finite-dimensional sense.

Common Mistakes

Reading mixedness from matrix appearance. Every density operator is diagonal in an eigenbasis. A pure state can have many nonzero off-diagonal entries in another basis; its rank and purity are unchanged.

Using a zero determinant as a purity test. In dimension greater than two, a mixed state can have a zero eigenvalue. Check rank one or, for an already valid density operator, unit purity.

Calling purity a complete state description. Many different states have the same purity. Even a qubit’s Bloch direction is not determined by its Bloch radius.

Assuming equal marginals make independent copies. The swap identity requires the product preparation $\rho\otimes\rho$. Correlations between the two preparations can change its expectation, as the last exercise shows.

Where deeper treatment lives

Density Operators is the canonical state-space definition. Ensembles and Preparation Procedures develops ensemble nonuniqueness. Purification Overview shows how every mixed state can be embedded in a larger pure state. Entropy Overview introduces a spectral mixedness measure beyond purity.

For the superposition comparison, see Superposition and Relative Phase and Classical Mixtures vs Quantum Superpositions.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific (2014), Ch. 3.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), Sec. 2.4.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer Academic (1995), Ch. 5.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), Ch. 2.

Exercises

1. A proposed qutrit state is $R=\operatorname{diag}(2/3,2/3,-1/3)$. A calculation finds $\operatorname{Tr}R=\operatorname{Tr}(R^2)=1$ and declares it pure. Diagnose the error and give a measurement outcome that exposes it. Which hypothesis of the purity criterion was omitted?

Solution

The two traces are indeed one, since $2/3+2/3-1/3=1$ and $4/9+4/9+1/9=1$. But $R$ is not positive: the projector onto its third basis ket would have probability $-1/3$. The criterion $\operatorname{Tr}(\rho^2)=1$ assumes that $\rho$ is already a valid density operator. In its spectral proof, positivity prevents cancellation among the products $\lambda_r\lambda_s$. Here $R^2\ne R$, so idempotency also fails. Trace and purity are useful diagnostics, not substitutes for positivity.

2. Derive the finite-dimensional lower bound

$$\operatorname{Tr}(\rho^2)\ge\frac1d$$

and identify the equality case.

Solution

Cauchy–Schwarz gives

$$\left( \sum_{r=1}^{d}\lambda_r \right)^2 \le d\sum_{r=1}^{d}\lambda_r^2.$$

The left side is one, so

$$\operatorname{Tr}(\rho^2) = \sum_r\lambda_r^2 \ge \frac1d.$$

Equality in Cauchy–Schwarz requires all eigenvalues to be equal. Therefore

$$\lambda_r=\frac1d$$

for every $r$, and $\rho=I_d/d$.

3. A source prepares $|0\rangle$ or $|+\rangle$ with equal probability. Find its density matrix and purity. Why is the purity greater than that of an equal mixture of $|0\rangle$ and $|1\rangle$?

Solution

The density matrix is

$$\rho=\frac12|0\rangle\langle0|+\frac12|+\rangle\langle+| =\begin{pmatrix}3/4&1/4\\1/4&1/4\end{pmatrix}.$$

Directly squaring gives $\operatorname{Tr}\rho^2=3/4$. Equivalently, the overlap is $|\langle0|+\rangle|^2=1/2$, so the ensemble formula gives $1/4+1/4+1/4=3/4$. The preparation alternatives are not fully distinguishable. An equal mixture of orthogonal kets instead has purity $1/2$.

4. For

$$\rho(c) = \frac12 \begin{pmatrix} 1&c\\ c^*&1 \end{pmatrix},$$

derive its eigenvalues and purity. For which $c$ is it pure?

Solution

The characteristic equation gives

$$\lambda_\pm = \frac{1\pm\lvert c\rvert}{2}.$$

Positivity requires $\lvert c\rvert\le1$. The purity is

$$\begin{aligned} \gamma &= \lambda_+^2+\lambda_-^2 \\ &= \frac{1+\lvert c\rvert^2}{2}. \end{aligned}$$

It equals one exactly when $\lvert c\rvert=1$. The state is mixed for $\lvert c\rvert<1$, including the maximally mixed case $c=0$.

5. Consider the qutrit state

$$\rho_p = \operatorname{diag}(p,1-p,0), \qquad 0\le p\le1.$$

Show why $\det\rho_p=0$ does not prove purity.

Solution

The determinant vanishes for every $p$ because one eigenvalue is zero. However,

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

For $0<p<1$, both $p$ and $1-p$ are nonzero and the purity is less than one. The state has rank two and is mixed. Only the endpoints $p=0$ and $p=1$ are pure.

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

Solution

Taking the partial trace gives

$$\rho_A=\rho_B=\frac{I_2}{2}.$$

Therefore

$$\operatorname{Tr}(\rho_A^2) = \operatorname{Tr} \left( \frac{I_2}{4} \right) = \frac12.$$

Each subsystem is mixed even though the joint Bell state is pure.

7. Starting from

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right),$$

derive

$$\operatorname{Tr}(\rho^2) = \frac{ 1+\lVert\mathbf r\rVert^2 }{2}.$$

Solution

Use

$$\left( \mathbf r\cdot\boldsymbol\sigma \right)^2 = \lVert\mathbf r\rVert^2 I.$$

Then

$$\rho^2 = \frac14 \left[ \left( 1+\lVert\mathbf r\rVert^2 \right)I + 2\mathbf r\cdot\boldsymbol\sigma \right].$$

Since $\operatorname{Tr}I=2$ and every Pauli matrix is traceless,

$$\operatorname{Tr}(\rho^2) = \frac{ 1+\lVert\mathbf r\rVert^2 }{2}.$$

8. Prove the two-copy identity

$$\operatorname{Tr} \left[ S(\rho\otimes\rho) \right] = \operatorname{Tr}(\rho^2)$$

using an orthonormal basis.

Solution

Let $S\lvert i\rangle\lvert j\rangle =\lvert j\rangle\lvert i\rangle$. Then

$$\begin{aligned} \operatorname{Tr} \left[ S(\rho\otimes\rho) \right] &= \sum_{i,j} \langle ij\rvert S(\rho\otimes\rho) \lvert ij\rangle \\ &= \sum_{i,j} \langle j\rvert\rho\lvert i\rangle \langle i\rvert\rho\lvert j\rangle \\ &= \sum_i \langle i\rvert\rho^2\lvert i\rangle \\ &= \operatorname{Tr}(\rho^2). \end{aligned}$$

The swap expectation on independently prepared copies is therefore the purity. This independence matters: the correlated state $\tfrac12|00\rangle\langle00|+\tfrac12|11\rangle\langle11|$ has swap expectation one, although each marginal is $I/2$ with purity $1/2$.