Proper and Improper Mixtures

A mixed density operator can describe uncertainty about a randomly prepared state, or the local state of an entangled system. These are traditionally called a proper mixture and an improper mixture, respectively. The adjective describes the physical preparation or larger state, not a different kind of density matrix. Both give valid local predictions.

Helpful background. Density operators describe ensembles; the partial trace extracts predictions for one subsystem.

One local density matrix, two joint states

A fair coin can select the qubit preparations $|0\rangle$ and $|1\rangle$, producing $\rho_A=I/2$ when its result is ignored. The same $\rho_A$ occurs when $A$ is half of the Bell state $(|00\rangle+|11\rangle)/\sqrt2$. No measurement on $A$ alone can distinguish these preparations. Access to the preparation record or to joint correlations can distinguish the larger descriptions.

Proper Mixture: Ignorance About Preparation

A proper mixture represents an ensemble generated by a classical random preparation procedure. For example, suppose a source prepares

$$\lvert0\rangle \quad \text{with probability }p,$$

and

$$\lvert1\rangle \quad \text{with probability }1-p.$$

If the preparation label is ignored, the state assigned to the system is

$$\rho_{\mathrm{proper}} = p\lvert0\rangle\langle0\rvert + (1-p)\lvert1\rangle\langle1\rvert.$$

Here there is a classical record, at least in principle, of which preparation occurred. An observer who obtains that record can refine the state assignment to $\lvert0\rangle\langle0\rvert$ or $\lvert1\rangle\langle1\rvert$.

The mixture is “proper” relative to that preparation description because the probabilities express ignorance over alternatives in the ensemble.

Improper Mixture: Reduced State of an Entangled System

Now consider the Bell state

$$\lvert\Phi^+\rangle_{AB} = \frac{1}{\sqrt2} \left( \lvert00\rangle+\lvert11\rangle \right).$$

The joint state is pure:

$$\rho_{AB} = \lvert\Phi^+\rangle\langle\Phi^+\rvert.$$

The reduced state of subsystem $A$ is

$$\rho_A = \operatorname{Tr}_B\rho_{AB} = \frac12 \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right) = \frac12 I.$$

This reduced state is mixed, but not because qubit $A$ was prepared by a local coin flip. In the pure Bell-state description, $A$ has no standalone pure state. The mixedness comes from entanglement with $B$.

Such a reduced state is called an improper mixture. It gives correct local probabilities, but it should not be read as classical ignorance about a locally prepared pure state unless additional assumptions are introduced.

For the general subsystem construction, see Reduced Density Operators and Partial Trace.

Same Local State, Different Correlations

Compare the Bell state with the classically correlated mixed state

$$\rho_{\mathrm{cc}} = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert.$$

Both have the same reduced state on $A$:

$$\rho_A=\frac12I.$$

They also have the same reduced state on $B$. But they are not the same joint state.

For the Bell state,

$$\langle X\otimes X\rangle_{\Phi^+}=1, \qquad \langle Z\otimes Z\rangle_{\Phi^+}=1.$$

For the classically correlated state,

$$\langle X\otimes X\rangle_{\mathrm{cc}}=0, \qquad \langle Z\otimes Z\rangle_{\mathrm{cc}}=1.$$

The local density operators cannot detect the difference. Measuring $X$ on both qubits can: the product of the two outcomes is always $+1$ for the Bell state, but is $+1$ or $-1$ with equal probabilities for $\rho_{\mathrm{cc}}$. With equal prior probabilities for the two candidate joint states, guessing “Bell” after $+1$ and “classical mixture” after $-1$ succeeds with probability $3/4$. The measurement distinguishes the statistics; it does not identify the state with certainty in one trial.

This example is the cleanest diagnostic: a reduced state is complete for local predictions but incomplete for questions about global purity, entanglement, and correlations.

Ensemble Decompositions Are Not Unique

The same density operator can be decomposed into many ensembles. For example,

$$\frac12I = \frac12\lvert0\rangle\langle0\rvert + \frac12\lvert1\rangle\langle1\rvert,$$

but also

$$\frac12I = \frac12\lvert+\rangle\langle+\rvert + \frac12\lvert-\rangle\langle-\rvert.$$

Both decompositions are mathematically valid. Neither decomposition alone tells you which preparation procedure actually occurred.

A proper mixture requires a physical ensemble description or preparation record. It is not created merely by choosing a decomposition of $\rho$ on paper.

This is one reason density matrices are powerful and subtle: $\rho$ determines all expectation values for the system, but it does not uniquely encode its preparation history.

Entanglement and the decohered local state

For a normalized pure initial system state and a pure, initially uncorrelated environment, a controlled interaction can give

$$\left( \sum_a c_a\lvert a\rangle \right) \lvert E_0\rangle \longmapsto \sum_a c_a \lvert a\rangle \lvert E_a\rangle.$$

The reduced state is

$$\rho_S = \sum_{a,b} c_ac_b^* \langle E_b|E_a\rangle \lvert a\rangle\langle b\rvert.$$

When the environmental records are approximately orthogonal,

$$\langle E_b|E_a\rangle\approx0 \qquad (a\ne b),$$

one obtains

$$\rho_S \approx \sum_a |c_a|^2 \lvert a\rangle\langle a\rvert.$$

This looks like a classical ignorance mixture over the pointer alternatives. But in the unitary model, the global state is still

$$\sum_a c_a \lvert a\rangle \lvert E_a\rangle.$$

If at least two nonzero amplitudes multiply distinguishable environment states, the joint pure state is entangled and the mixed reduced state is improper in this description. Merely copying a record to another quantum system does not change that conclusion about the global state. Conditioning on a measured record gives a conditional state; it is a different question from what the unconditioned global state is.

This example does not imply that every loss of coherence requires entanglement with the physical environment. An initially mixed environment can also produce random phase shifts; see the explicit mixed-environment example.

This is the core reason decoherence explains local interference suppression without, by itself, solving every measurement-outcome question. See What Decoherence Does Not Solve for the dedicated boundary page.

Measurement Records and Conditioning

Measurement theory often shifts between proper and improper descriptions, so it is useful to separate three levels.

First, before a record is read, a system-apparatus model may produce an entangled state:

$$\sum_a c_a \lvert a\rangle_S \lvert A_a\rangle.$$

The reduced state of $S$ alone is an improper mixture if $A$ is ignored.

Second, if the apparatus record is actually available but not consulted by a particular observer, that observer may use a nonselective state:

$$\rho_S = \sum_a p_a\rho_{S|a}.$$

Relative to an external description that includes the classical record, this can be a proper mixture over recorded outcomes.

Third, if the observer conditions on a specific outcome $a$, the state assignment becomes the selective conditional state $\rho_{S|a}$.

The same matrix can appear in more than one role. The difference is not the matrix alone; it is what physical record exists and which observer or model has access to it. See Selective and Nonselective Measurements and State-Update Rules for the operational update language.

Purification Perspective

Every finite-dimensional mixed state can be represented as the reduced state of a larger pure state. If

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

is a spectral decomposition, then

$$\lvert\Psi\rangle_{SR} = \sum_k \sqrt{p_k} \lvert k\rangle_S \lvert k\rangle_R$$

is a purification, and

$$\operatorname{Tr}_R \lvert\Psi\rangle\langle\Psi\rvert = \rho.$$

This constructs a possible larger pure-state description. It does not prove that a particular preparation actually contains that purifying system or those correlations. A device that samples pure states with classical probabilities supplies another physical realization of the same local density operator.

The mathematical density operator is the same in both uses. The interpretation depends on the physical embedding.

Operational Equivalence and Global Inequivalence

For observables $M_A$ acting only on subsystem $A$,

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

This identity is why a proper mixture and an improper mixture with the same $\rho_A$ cannot be distinguished by measurements on $A$ alone.

To distinguish their origins, one needs additional access:

• the preparation record;
• the purifying system;
• the environment that carries decoherence records;
• joint correlations;
• interference experiments that recombine the larger system coherently.

If those degrees of freedom are inaccessible, the reduced density operator is the operational state for local predictions.

Exercises

Bell-state reduced mixture

Compute the reduced state of qubit $A$ for

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \left( \lvert00\rangle+\lvert11\rangle \right).$$

Is the resulting mixture proper or improper in the pure Bell-state description?

Solution

The joint density operator is

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

Tracing over $B$ removes the cross terms because $\langle0|1\rangle=0$:

$$\rho_A = \frac12 \lvert0\rangle\langle0\rvert + \frac12 \lvert1\rangle\langle1\rvert = \frac12I.$$

In the pure Bell-state description, this is an improper mixture: the local mixedness comes from entanglement with $B$, not from a local classical coin flip.

Correlation test

For

$$\rho_{\mathrm{cc}} = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert,$$

show that $\langle X\otimes X\rangle_{\mathrm{cc}}=0$ while $\langle X\otimes X\rangle_{\Phi^+}=1$.

Solution

For the classically correlated state,

$$X\otimes X\lvert00\rangle=\lvert11\rangle, \qquad X\otimes X\lvert11\rangle=\lvert00\rangle.$$

Therefore

$$\langle00|X\otimes X|00\rangle=0, \qquad \langle11|X\otimes X|11\rangle=0,$$

so

$$\langle X\otimes X\rangle_{\mathrm{cc}}=0.$$

For the Bell state,

$$X\otimes X\lvert\Phi^+\rangle = \lvert\Phi^+\rangle,$$

so

$$\langle X\otimes X\rangle_{\Phi^+}=1.$$

The same local reduced states can therefore hide different joint correlations.

Two decompositions of the maximally mixed state

Show that

$$\frac12I = \frac12\lvert0\rangle\langle0\rvert + \frac12\lvert1\rangle\langle1\rvert = \frac12\lvert+\rangle\langle+\rvert + \frac12\lvert-\rangle\langle-\rvert.$$

Why does this not identify the actual preparation?

Solution

Using

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

we have

$$\lvert+\rangle\langle+\rvert = \frac12 \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix}, \qquad \lvert-\rangle\langle-\rvert = \frac12 \begin{pmatrix} 1&-1\\ -1&1 \end{pmatrix}.$$

Averaging gives

$$\frac12\lvert+\rangle\langle+\rvert + \frac12\lvert-\rangle\langle-\rvert = \frac12I.$$

This equality is an algebraic decomposition of the same density operator. The actual preparation depends on the source and its record, not on the decomposition one chooses afterward.

Decoherence and improper mixtures

In the state

$$\lvert\Psi\rangle = c_0\lvert0\rangle\lvert E_0\rangle + c_1\lvert1\rangle\lvert E_1\rangle,$$

assume normalized environment states, $\langle E_1|E_0\rangle=0$, $|c_0|^2+|c_1|^2=1$, and $c_0c_1\ne0$. Compute $\rho_S$ and explain why it is not automatically a proper mixture.

Solution

Tracing over the environment gives

$$\rho_S = |c_0|^2\lvert0\rangle\langle0\rvert + |c_1|^2\lvert1\rangle\langle1\rvert.$$

It has the same local form as a classical mixture over $\lvert0\rangle$ and $\lvert1\rangle$. But the global state is still the entangled superposition $\lvert\Psi\rangle$. Its reduced state is an improper mixture in this specified pure, entangled description. A conditional state given a measured record answers a different question; it does not retrospectively turn the unconditioned Bell-like state into a separable ensemble.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific (2014).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010).
• M. Schlosshauer, Decoherence and the Quantum-to-Classical Transition, Springer (2007). Publisher.
• W. H. Zurek, “Decoherence, einselection, and the quantum origins of the classical,” Reviews of Modern Physics 75, 715–775 (2003). DOI.