Pure and Mixed States

For normal states on the full bounded-operator algebra, purity means that the density operator is a rank-one projection. Equivalently the trace of its square is one, or it cannot be expressed as a nontrivial convex mixture of distinct states. These equivalences survive infinite dimension, while the existence of a maximally mixed state and several geometric intuitions do not.

Helpful background. Pure versus mixed states supplies the preparation interpretation. Density operators supplies trace-class spectral expansions.

Equivalent purity criteria

Let $\rho\geq0$ be trace class with $\operatorname{Tr}\rho=1$. Write its positive eigenvalues as $p_n$, so $\sum_np_n=1$. Then

$$\operatorname{Tr}\rho^2=\sum_np_n^2\leq\sum_np_n=1.$$

Equality occurs exactly when $p_n(1-p_n)=0$ for every $n$. Normalization then forces exactly one eigenvalue equal to one and all others zero. Consequently the following are equivalent:

• $\rho=|\psi\rangle\langle\psi|$ for a normalized vector $\psi$.
• $\rho^2=\rho$.
• $\operatorname{Tr}\rho^2=1$.
• $\rho$ has rank one.

The convex characterization is proved in Extreme Points. A normal state is mixed when these conditions fail. Its purity $\operatorname{Tr}\rho^2$ is always strictly positive, even in infinite dimensions, since at least one eigenvalue is positive. It may, however, be arbitrarily close to zero.

For $d$ finite dimensions, Cauchy–Schwarz gives $\operatorname{Tr}\rho^2\geq1/d$, attained by $I/d$. In infinite dimension the infimum is zero and is not attained. There is no density operator proportional to the identity: its trace would diverge. “Maximally mixed” therefore requires a finite subspace, a constraint, or a separately specified algebraic meaning.

Mixtures have a convergence requirement

A countable preparation ensemble $(q_k,\psi_k)$ defines

$$\rho=\sum_kq_k|\psi_k\rangle\langle\psi_k|, \qquad q_k\geq0,\quad \sum_kq_k=1.$$

This series converges in trace norm, whether or not the vectors are orthogonal. Its purity can be evaluated by nonnegative summation:

$$\operatorname{Tr}\rho^2 =\sum_{j,k}q_jq_k|\langle\psi_j,\psi_k\rangle|^2.$$

If the ensemble includes nonparallel rays with positive probabilities, at least one overlap factor is strictly below one and the purity is less than one. Conversely, a pure-state density operator can have an ensemble decomposition only into its own ray, apart from zero-weight terms. Global phases do not create distinct rays.

This criterion concerns the operator after any preparation label is discarded. Adding a classical label as an explicit register changes the system being described. Likewise, a mixed subsystem state can arise from a pure entangled state of a larger system; purification gives the precise construction.

A mixture of two nonorthogonal rays

Take $\rho=pP_\psi+(1-p)P_\phi$, with normalized vectors, $0<p<1$, and $c=|\langle\psi,\phi\rangle|$. Direct expansion gives

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

For $c=1$ both vectors define the same ray, and the mixture is pure. For $c<1$ it is mixed. Orthogonal components maximize the impurity at fixed $p$.

The two nonzero eigenvalues, when the support is two-dimensional, are

$$\lambda_\pm =\frac{1\pm\sqrt{1-4p(1-p)(1-c^2)}}2.$$

Their sum is one and their product is $p(1-p)(1-c^2)$. This example shows why the weights of a nonorthogonal preparation ensemble are generally not the eigenvalues of its density operator. The spectral ensemble is distinguished by orthogonality, not by physical uniqueness.

Purity depends on which observables are retained

Consider $\psi=(|0\rangle+|1\rangle)/\sqrt2$ and restrict observables to matrices diagonal in the computational basis. The resulting functional agrees with the equal mixture of $|0\rangle$ and $|1\rangle$. On that diagonal algebra it is a nontrivial mixture of the two evaluation states, even though $P_\psi$ is rank one on the full matrix algebra.

Thus “rank one means pure” presumes the full algebra $\mathcal B(\mathcal H)$ and a normal state. Superselection or restriction to a subsystem changes the relevant algebra. General algebraic purity means extremality in the state space of that algebra; it is not a basis-independent statement about any arbitrarily chosen representing matrix.

Holevo treats the Hilbert-space state criteria; Bratteli and Robinson supply the algebraic distinction. No purity criterion alone decides an interpretation of mixed states or a claim about hidden preparation information.

Exercises

1. For a thermal geometric distribution $p_n=(1-q)q^n$, $n\geq0$, $0<q<1$, find the purity and its limiting values.

Solution

Summing the geometric series gives $\operatorname{Tr}\rho^2=(1-q)^2/(1-q^2)=(1-q)/(1+q)$. It approaches one as $q\to0$ and zero as $q\to1$. The latter limit is not an infinite-dimensional maximally mixed density operator; normalization escapes to arbitrarily high levels.

2. An equally weighted ensemble contains normalized vectors $\psi,\phi$ with $|\langle\psi,\phi\rangle|=1/2$. Is its purity $1/2$?

Solution

No. The formula gives $1-\tfrac12(1-\tfrac14)=5/8$. Only orthogonal equal-weight components yield purity $1/2$. This is the distinction between ensemble weights and spectral probabilities emphasized in Core Formalism.

References

• Bratteli, Ola, and Derek W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed., Springer, 1987 — algebraic states and purity. Publisher.
• Holevo, Alexander S., Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd revised and expanded ed., De Gruyter, 2019 — density operators, mixtures, and purity. Publisher.