Classical Mixtures vs Quantum Superpositions

A coherent superposition and a statistical mixture can give identical probabilities in one measurement basis while representing different quantum states. The distinction appears when another measurement recombines the alternatives and becomes sensitive to relative phase.

For a two-level system, compare the normalized pure state

$$\lvert\psi\rangle = a\lvert0\rangle + b\lvert1\rangle, \qquad \lvert a\rvert^2+\lvert b\rvert^2=1,$$

with the matched mixture

$$\rho_{\mathrm{mix}} = \lvert a\rvert^2 \lvert0\rangle\langle0\rvert + \lvert b\rvert^2 \lvert1\rangle\langle1\rvert.$$

The superposition assigns complex amplitudes to two basis alternatives. The mixture assigns classical probabilities to preparation procedures and discards the preparation label. Their common diagonal probabilities do not make the states equal.

The comparison is always relative to a stated basis. Pure versus mixed is basis independent; coherence versus incoherence is not.

Helpful background. Superposition and relative phase explains amplitude addition; density operators explains how randomized preparations are represented.

Two Different Preparation Rules

Take $\{|0\rangle,|1\rangle\}$ to be an orthonormal basis throughout. The density operator of the superposition is

$$\begin{aligned} \rho_\psi &= \lvert\psi\rangle\langle\psi\rvert \\ &= \begin{pmatrix} \lvert a\rvert^2 & ab^* \\ a^*b & \lvert b\rvert^2 \end{pmatrix}_{\{0,1\}}. \end{aligned}$$

The matched mixture is

$$\rho_{\mathrm{mix}} = \begin{pmatrix} \lvert a\rvert^2 & 0 \\ 0 & \lvert b\rvert^2 \end{pmatrix}_{\{0,1\}}.$$

Both have trace one and the same diagonal entries. Their basis-independent purities differ:

$$\operatorname{Tr}(\rho_\psi^2)=1,$$

whereas

$$\operatorname{Tr}(\rho_{\mathrm{mix}}^2) = \lvert a\rvert^4+\lvert b\rvert^4.$$

If both amplitudes are nonzero, then

$$\lvert a\rvert^4+\lvert b\rvert^4<1,$$

so the matched mixture is mixed. The two states coincide only in the trivial cases $a=0$ or $b=0$, when there is just one populated alternative.

The terminology can mislead. A superposition is not a classical list of simultaneous properties. It is one pure state whose amplitudes determine probabilities and interference. A mixture is not one unknown member of a uniquely determined list: the same density operator may have many ensemble decompositions.

Coherence Terms in a Chosen Basis

A general qubit density matrix in the computational basis can be written

$$\rho = \begin{pmatrix} p & c \\ c^* & 1-p \end{pmatrix}, \qquad 0\le p\le1.$$

The off-diagonal entry $c=\langle0\rvert\rho\lvert1\rangle$ is a coherence term relative to the basis $\{\lvert0\rangle,\lvert1\rangle\}$. Positivity requires

$$\det\rho = p(1-p)-\lvert c\rvert^2 \ge0,$$

or equivalently

$$\lvert c\rvert \le \sqrt{p(1-p)}.$$

This inequality gives a useful continuum:

• $c=0$ is completely incoherent in the chosen basis;
• $0<\lvert c\rvert<\sqrt{p(1-p)}$ is partially coherent and mixed;
• $\lvert c\rvert=\sqrt{p(1-p)}$ saturates positivity and gives a pure state.

Off-diagonal entries are therefore neither a synonym for purity nor a basis-independent mark of “quantumness.” A mixed state can have off-diagonal entries in a non-eigenbasis, and every density operator is diagonal in an eigenbasis.

Same Statistics in One Basis

The computational-basis projectors are

$$\Pi_0=\lvert0\rangle\langle0\rvert, \qquad \Pi_1=\lvert1\rangle\langle1\rvert.$$

For both $\rho_\psi$ and the matched mixture,

$$\begin{aligned} p_Z(0) &= \operatorname{Tr}(\rho\Pi_0) = \lvert a\rvert^2, \\ p_Z(1) &= \operatorname{Tr}(\rho\Pi_1) = \lvert b\rvert^2. \end{aligned}$$

These measurements see only the diagonal entries in their own basis. They do not test the off-diagonal phase information.

The equal-weight example is especially clear:

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

while

$$\rho_{\mathrm{mix}} = \frac12 \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} = \frac I2.$$

Both give $p_Z(0)=p_Z(1)=1/2$, yet they are operationally distinguishable.

The computational-basis probabilities agree, but the $X$-basis distributions do not. One projective measurement basis generally does not determine a quantum state.

Phase-Sensitive Measurements

To probe coherence, use a basis that recombines $\lvert0\rangle$ and $\lvert1\rangle$. Define

$$\begin{aligned} \lvert+_\theta\rangle &= \frac{ \lvert0\rangle+e^{i\theta}\lvert1\rangle }{ \sqrt2 }, \\ \lvert-_\theta\rangle &= \frac{ \lvert0\rangle-e^{i\theta}\lvert1\rangle }{ \sqrt2 }. \end{aligned}$$

For

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

the Born rule gives

$$\begin{aligned} p(+_\theta) &= \langle+_\theta\rvert \rho \lvert+_\theta\rangle \\ &= \frac12 + \operatorname{Re} \bigl( e^{i\theta}c \bigr), \\ p(-_\theta) &= \frac12 - \operatorname{Re} \bigl( e^{i\theta}c \bigr). \end{aligned}$$

The diagonal population $p$ drops out because this equatorial measurement weights both computational-basis alternatives equally. The outcome imbalance is

$$p(+_\theta)-p(-_\theta) = 2\operatorname{Re} \bigl( e^{i\theta}c \bigr),$$

which directly probes one quadrature of the coherence.

For the pure phase state

$$\lvert\psi_\phi\rangle = \frac{ \lvert0\rangle+e^{i\phi}\lvert1\rangle }{ \sqrt2 },$$

the off-diagonal element is $c=e^{-i\phi}/2$, so

$$\begin{aligned} p(+_\theta) &= \frac{ 1+\cos(\theta-\phi) }{2}, \\ p(-_\theta) &= \frac{ 1-\cos(\theta-\phi) }{2}. \end{aligned}$$

Choosing $\theta=\phi$ gives a definite plus outcome. For the equal incoherent mixture, $c=0$ and both outcomes remain $1/2$ for every $\theta$.

Where the Interference Term Comes From

Let a detector outcome correspond to a vector $\lvert\chi\rangle$. Set

$$u=\langle\chi\vert0\rangle, \qquad v=\langle\chi\vert1\rangle.$$

For the pure superposition, the amplitude is

$$\langle\chi\vert\psi\rangle = au+bv,$$

and its probability is

$$\begin{aligned} p_\psi(\chi) &= \lvert au+bv\rvert^2 \\ &= \lvert a\rvert^2\lvert u\rvert^2 + \lvert b\rvert^2\lvert v\rvert^2 + 2\operatorname{Re} \bigl( ab^*uv^* \bigr). \end{aligned}$$

The last line is the interference term. It depends on relative phases among the preparation amplitudes and the measurement overlaps.

For the matched mixture,

$$\begin{aligned} p_{\mathrm{mix}}(\chi) &= \operatorname{Tr} \bigl( \rho_{\mathrm{mix}} \lvert\chi\rangle\langle\chi\rvert \bigr) \\ &= \lvert a\rvert^2\lvert u\rvert^2 + \lvert b\rvert^2\lvert v\rvert^2. \end{aligned}$$

There is no cross term because the preparation probabilities are averaged after separate alternatives are prepared. In a coherent state, amplitudes add before the modulus is squared; in an incoherent mixture, probabilities are averaged.

Relative Phase, Not Global Phase

Multiplying the whole vector by a global phase changes no density operator:

$$\bigl( e^{i\gamma}\lvert\psi\rangle \bigr) \bigl( e^{-i\gamma}\langle\psi\rvert \bigr) = \lvert\psi\rangle\langle\psi\rvert.$$

Changing only the relative phase does change the state. For the equal superposition,

$$\rho_\phi = \frac12 \begin{pmatrix} 1&e^{-i\phi}\\ e^{i\phi}&1 \end{pmatrix}.$$

All values of $\phi$ give equal $Z$-basis probabilities, but they point in different equatorial directions on the Bloch sphere and are distinguished by suitable values of $\theta$. Relative phase is physical because it changes interference; global phase is not.

The canonical state-vector treatment is Superposition and Relative Phase.

Basis Dependence Without Contradiction

The plus state is coherent in the computational basis but diagonal in the $X$ basis:

$$\rho_+ = \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}_{\{+,-\}}.$$

This does not turn it into a mixed state. Its eigenvalues remain $(1,0)$ and its purity remains one.

Conversely, the unequal mixture

$$\rho_Z = p\lvert0\rangle\langle0\rvert + (1-p)\lvert1\rangle\langle1\rvert$$

is diagonal in the $Z$ basis but generally has off-diagonal entries in the $X$ basis. It remains mixed because basis changes do not alter its eigenvalues.

Three statements must be kept separate:

1. Pure or mixed is a basis-independent property of the spectrum.
2. Coherent or incoherent is relative to a chosen basis or preferred set of states.
3. Prepared by classical randomization concerns a procedure and any associated side information.

A physical problem supplies the reference basis through paths, energy levels, spin components, a control Hamiltonian, a measurement apparatus, or an environmental pointer basis. Modern resource theories of coherence likewise fix a preferred basis or algebra before defining incoherent states and operations.

Partial Coherence and Dephasing

The distinction is not all-or-nothing in realistic experiments. Starting from

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

a phase-damping process may produce

$$\rho_\eta = \begin{pmatrix} p&\eta c\\ \eta^*c^*&1-p \end{pmatrix}, \qquad \lvert\eta\rvert\le1.$$

The populations remain fixed while the coherence magnitude is reduced by $\lvert\eta\rvert$. A phase-sensitive outcome becomes

$$p_\eta(+_\theta) = \frac12 + \operatorname{Re} \bigl( e^{i\theta}\eta c \bigr).$$

The fully dephasing channel in the computational basis is

$$\mathcal D_Z(\rho) = \sum_{z=0}^{1} \Pi_z\rho\Pi_z = \begin{pmatrix} \rho_{00}&0\\ 0&\rho_{11} \end{pmatrix}.$$

This unconditioned channel removes off-diagonal terms without selecting one measurement outcome. It should not be confused with a conditioned projective update, which produces a state associated with a recorded result.

What the Density Operator Does Not Reveal

The same mixed density operator can arise in physically different ways. The maximally mixed qubit state

$$\frac I2$$

may result from:

• randomly preparing $\lvert0\rangle$ or $\lvert1\rangle$ and losing the classical label;
• randomly preparing $\lvert+\rangle$ or $\lvert-\rangle$ and losing that label;
• tracing one qubit out of a Bell state;
• noise or uncontrolled interaction with an environment.

No measurement on the qubit alone distinguishes these histories, because all local probabilities are determined by the same $\rho=I/2$. The differences reside in external records and correlations. Access to a preparation label, entangled partner, or environment can distinguish joint descriptions that share the same local density operator.

The older terms proper mixture and improper mixture are sometimes used for classical randomization and reduced states, respectively. The terminology can obscure the operational point: locally, equal density operators are indistinguishable; globally, their extensions may differ. Ensembles and Preparation Procedures and Purification Overview develop these two sides.

Decoherence Moves Coherence into Correlations

Suppose a system initially in

$$a\lvert0\rangle+b\lvert1\rangle$$

interacts with an initially uncorrelated environment in the normalized state $\lvert E\rangle$. Assume the interaction preserves the two system basis states and produces normalized environmental states $\lvert E_0\rangle,\lvert E_1\rangle$. It then gives

$$\lvert\Psi\rangle_{SE} = a\lvert0\rangle\lvert E_0\rangle + b\lvert1\rangle\lvert E_1\rangle.$$

Tracing out the environment gives

$$\rho_S = \begin{pmatrix} \lvert a\rvert^2 & ab^*\langle E_1\vert E_0\rangle \\ a^*b\langle E_0\vert E_1\rangle & \lvert b\rvert^2 \end{pmatrix}.$$

The environmental overlap

$$\eta = \langle E_1\vert E_0\rangle$$

acts as the dephasing factor. If the two environmental records become nearly orthogonal, $\lvert\eta\rvert\ll1$, then local interference becomes small. The stated global state $\lvert\Psi\rangle_{SE}$ remains pure. Coherence has become inaccessible to system-only measurements because it is encoded in system–environment correlations.

Decoherence explains the suppression of interference in a selected basis; by itself it is not a unique interpretation of measurement outcomes and does not convert the global unitary state into a classical ensemble. See Decoherence Preview.

Tomography and Coherence Witnesses

For

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

the Pauli expectation values are

$$\begin{aligned} \langle X\rangle &= 2\operatorname{Re}c, \\ \langle Y\rangle &= -2\operatorname{Im}c, \\ \langle Z\rangle &= 2p-1. \end{aligned}$$

Together they reconstruct the state:

$$\rho = \frac12 \bigl( I + \langle X\rangle X + \langle Y\rangle Y + \langle Z\rangle Z \bigr).$$

A $Z$ measurement alone determines only $p$ and cannot decide whether $c$ vanishes. An $X$ measurement probes the real part of $c$, and a $Y$ measurement probes its imaginary part. These three projective settings determine all three Bloch components. More generally, tomography requires an informationally complete set of effects; a suitable single POVM setting can suffice. See POVMs for this broader measurement description.

One nonzero phase-sensitive expectation value witnesses coherence in the chosen basis, but a zero value of $\langle X\rangle$ alone does not prove incoherence: the coherence may be purely imaginary and visible in $\langle Y\rangle$.

Choosing a distinguishing measurement

Compute both density matrices in the same basis. If they differ only in an imaginary off-diagonal entry, an $X$ measurement will miss the difference but a $Y$ measurement can detect it. More generally, choose the phase $\theta$ to probe a nonzero quadrature of their coherence difference.

If two preparations give the same density operator, no system-only measurement distinguishes them. Different operators give different probabilities for some measurement, although one trial need not identify the preparation with certainty. External labels or partner systems provide additional information only when they are accessible.

Exercises

1. For

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

find the density matrix and the expectation values of $X$, $Y$, and $Z$.

Solution

The projector is

$$\rho_\psi = \frac12 \begin{pmatrix} 1&-i\\ i&1 \end{pmatrix}.$$

Here $p=1/2$ and $c=-i/2$. Therefore

$$\begin{aligned} \langle X\rangle &= 2\operatorname{Re}c =0, \\ \langle Y\rangle &= -2\operatorname{Im}c =1, \\ \langle Z\rangle &= 2p-1 =0. \end{aligned}$$

The state is the $+1$ eigenstate of $Y$. Its coherence is purely imaginary in the computational basis, so an $X$ measurement alone would miss it.

2. Derive the probabilities $p(+_\theta)$ and $p(-_\theta)$ for a general qubit state with off-diagonal entry $c$.

Solution

Using

$$\lvert+_\theta\rangle = \frac{ \lvert0\rangle+e^{i\theta}\lvert1\rangle }{ \sqrt2 },$$

one obtains

$$\begin{aligned} p(+_\theta) &= \frac12 \left( \rho_{00}+\rho_{11} \right) + \frac12 \left( e^{i\theta}\rho_{01} + e^{-i\theta}\rho_{10} \right) \\ &= \frac12 + \operatorname{Re} \bigl( e^{i\theta}c \bigr). \end{aligned}$$

The orthogonal outcome has the opposite sign:

$$p(-_\theta) = \frac12 - \operatorname{Re} \bigl( e^{i\theta}c \bigr).$$

Their sum is one, and their difference probes a phase-selected quadrature of $c$.

3. Prove the positivity bound $\lvert c\rvert^2\le p(1-p)$ for

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

When is the bound saturated?

Solution

A $2\times2$ Hermitian matrix with nonnegative diagonal entries is positive semidefinite exactly when its determinant is nonnegative. Here

$$\det\rho = p(1-p)-\lvert c\rvert^2.$$

Thus positivity requires

$$\lvert c\rvert^2 \le p(1-p).$$

The determinant vanishes when equality holds. Since $\operatorname{Tr}\rho=1$, a zero determinant gives eigenvalues $1$ and $0$, so the state is pure. A strict inequality gives two positive eigenvalues and a mixed state.

4. Show explicitly why the pure superposition and its matched mixture differ by an interference term for an arbitrary detector outcome $\lvert\chi\rangle$.

Solution

Set

$$u=\langle\chi\vert0\rangle, \qquad v=\langle\chi\vert1\rangle.$$

For $\lvert\psi\rangle=a\lvert0\rangle+b\lvert1\rangle$,

$$\begin{aligned} p_\psi(\chi) &= \lvert au+bv\rvert^2 \\ &= \lvert a\rvert^2\lvert u\rvert^2 + \lvert b\rvert^2\lvert v\rvert^2 + 2\operatorname{Re} \bigl( ab^*uv^* \bigr). \end{aligned}$$

For the matched mixture,

$$p_{\mathrm{mix}}(\chi) = \lvert a\rvert^2\lvert u\rvert^2 + \lvert b\rvert^2\lvert v\rvert^2.$$

The difference is exactly the cross term. It vanishes when the measurement does not recombine the alternatives, when the coherence is zero, or when this particular phase quadrature is zero. One zero interference term does not establish incoherence.

5. Write $\rho_+=\lvert+\rangle\langle+\rvert$ and $I/2$ in the $X$ basis. What does the result show about basis dependence?

Solution

In the ordered basis $\{\lvert+\rangle,\lvert-\rangle\}$,

$$\rho_+ = \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}.$$

The maximally mixed state is invariant under every basis change:

$$\frac I2 = \frac12 \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}.$$

The plus state has no off-diagonal entries in its eigenbasis, but it remains pure. It has off-diagonal coherence in the $Z$ basis. The maximally mixed state is incoherent in every basis. Matrix diagonality is basis relative, while purity is not.

6. Verify that

$$\mathcal D_Z(\rho) = \Pi_0\rho\Pi_0 + \Pi_1\rho\Pi_1$$

is trace preserving and removes computational-basis coherence.

Solution

The Kraus operators are $\Pi_0$ and $\Pi_1$. They satisfy

$$\Pi_0^\dagger\Pi_0 + \Pi_1^\dagger\Pi_1 = \Pi_0+\Pi_1 = I,$$

so the map is trace preserving. For

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

direct multiplication gives

$$\mathcal D_Z(\rho) = \begin{pmatrix} p&0\\ 0&1-p \end{pmatrix}.$$

The channel preserves populations and discards the off-diagonal terms. It is an unconditioned dephasing operation, not the selection of either outcome.

7. Derive the reduced system state after the interaction

$$\bigl( a\lvert0\rangle+b\lvert1\rangle \bigr) \lvert E\rangle \longmapsto a\lvert0\rangle\lvert E_0\rangle + b\lvert1\rangle\lvert E_1\rangle.$$

What happens when $\langle E_1\vert E_0\rangle=0$?

Solution

Expanding the joint projector and tracing the environment yields

$$\rho_S = \begin{pmatrix} \lvert a\rvert^2 & ab^*\langle E_1\vert E_0\rangle \\ a^*b\langle E_0\vert E_1\rangle & \lvert b\rvert^2 \end{pmatrix}.$$

If $\langle E_1\vert E_0\rangle=0$, the environmental records are orthogonal and the reduced state becomes

$$\rho_S = \lvert a\rvert^2 \lvert0\rangle\langle0\rvert + \lvert b\rvert^2 \lvert1\rangle\langle1\rvert.$$

Local interference is absent. When both amplitudes are nonzero, the stated joint state is nevertheless a pure entangled superposition, so the coherence survives in correlations unavailable to system-only measurements.

8. A laboratory produces $I/2$ either by randomly preparing $\lvert0\rangle$ or $\lvert1\rangle$, or by giving you one qubit of a Bell pair. Can any qubit-only measurement distinguish the methods? What additional access could distinguish them?

Solution

No. Every qubit-only POVM element $E$ has probability

$$p(E) = \operatorname{Tr} \left( \frac I2 E \right)$$

for both methods. Equal reduced density operators imply equal statistics for every local measurement.

The extensions differ. In the random-preparation method, a retained classical record may reveal whether $\lvert0\rangle$ or $\lvert1\rangle$ was sent. In the Bell-pair method, access to the partner qubit reveals entanglement and nonclassical joint correlations. The local state alone does not determine which extension is present.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd edition, World Scientific, 2014, Chapters 2–3.
• T. Baumgratz, M. Cramer, and M. B. Plenio, “Quantifying coherence,” Physical Review Letters 113, 140401 (2014). DOI: 10.1103/PhysRevLett.113.140401.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th edition, Oxford University Press, 1958, Chapters 1–2.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary edition, Cambridge University Press, 2010, Chapter 2. Cambridge DOI.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer Academic Publishers, 1995, Chapters 3–4. Springer DOI.
• M. Schlosshauer, “Decoherence, the measurement problem, and interpretations of quantum mechanics,” Reviews of Modern Physics 76, 1267–1305 (2005). DOI: 10.1103/RevModPhys.76.1267.