6.1.1 Mixed States#

Prompts

What is the difference between a pure state and a mixed state? Why can’t a mixed state be described by a single ket \(\vert\psi\rangle\)?
Construct the density matrices for a qubit superposition \((\vert 0\rangle + \vert 1\rangle)/\sqrt{2}\) and a 50-50 classical mixture. How do they differ?
How does purity \(\operatorname{Tr}(\hat{\rho}^2)\) distinguish pure from mixed states? What is the geometric meaning on the Bloch ball?
Can two different ensembles produce the same density matrix? What does this imply about the physical status of the ensemble vs the density matrix?

Lecture Notes#

Overview#

When a quantum system is prepared in a single pure state \(\vert\psi\rangle\), the system is fully characterized. But what happens when we possess only incomplete information—perhaps due to classical ignorance, or because the system has interacted with an inaccessible environment? The density matrix captures this, unifying the description of pure and mixed states under a single formalism.

The Density Matrix#

Density Matrix

Pure state: For a known state \(\vert\psi\rangle\), the density operator is the rank-1 projector:

(100)#\[\hat{\rho} = \vert\psi\rangle\langle\psi\vert\]

Mixed state (ensemble): When the system is in state \(\vert\psi_i\rangle\) with probability \(p_i\) (\(p_i \geq 0\), \(\sum_i p_i = 1\)):

(101)#\[\hat{\rho} = \sum_i p_i \vert\psi_i\rangle\langle\psi_i\vert\]

Conditions for a valid density matrix:

Hermitian: \(\hat{\rho}^\dagger = \hat{\rho}\)
Trace one: \(\operatorname{Tr}(\hat{\rho}) = 1\)
Positive semidefinite: \(\hat{\rho} \geq 0\) (all eigenvalues \(\lambda_i \geq 0\))

Diagonal elements of \(\hat{\rho}\) in a basis are populations (outcome probabilities); off-diagonal elements are coherences (interference terms).

Example: Superposition vs Mixture

Problem. Compare the density matrices for (i) the superposition \(\vert +\rangle = (\vert 0\rangle + \vert 1\rangle)/\sqrt{2}\) and (ii) the classical mixture of \(\vert 0\rangle\) and \(\vert 1\rangle\) with equal probability.

Solution.

\[\begin{split}\hat{\rho}_{\text{super}} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \qquad \hat{\rho}_{\text{mix}} = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \frac{\hat{I}}{2}\end{split}\]

Both have identical diagonal elements (equal probability of measuring \(\vert 0\rangle\) or \(\vert 1\rangle\)), but \(\hat{\rho}_{\text{super}}\) has off-diagonal coherences while \(\hat{\rho}_{\text{mix}}\) does not. Measuring \(\hat{\sigma}^x\) distinguishes them: \(\langle\hat{\sigma}^x\rangle = 1\) for the superposition but \(0\) for the mixture.

Purity#

Purity

The purity of a density matrix is:

(102)#\[\mathcal{P} = \operatorname{Tr}(\hat{\rho}^2) = \sum_i \lambda_i^2\]

where \(\lambda_i\) are the eigenvalues of \(\hat{\rho}\).

Pure state: \(\hat{\rho}^2 = \hat{\rho}\) (idempotent), so \(\operatorname{Tr}(\hat{\rho}^2) = 1\).
Maximally mixed (\(d\)-dimensional): all \(\lambda_i = 1/d\), so \(\operatorname{Tr}(\hat{\rho}^2) = 1/d\).
General: \(1/d \leq \operatorname{Tr}(\hat{\rho}^2) \leq 1\).

Derivation: Purity Bounds

Pure state: \(\hat{\rho} = \vert\psi\rangle\langle\psi\vert\) gives \(\hat{\rho}^2 = \vert\psi\rangle\langle\psi\vert\psi\rangle\langle\psi\vert = \hat{\rho}\), so \(\operatorname{Tr}(\hat{\rho}^2) = 1\).

Lower bound: By Cauchy-Schwarz, \(\sum_i \lambda_i^2 \geq (\sum_i \lambda_i)^2/d = 1/d\), with equality when all \(\lambda_i = 1/d\).

Mixed state: If at least two \(\lambda_i > 0\), then \(\sum \lambda_i^2 < (\sum \lambda_i)^2 = 1\).

Ensemble Ambiguity#

Ensemble Ambiguity

Different statistical ensembles can produce the same density matrix. For example, the maximally mixed qubit \(\hat{\rho} = \hat{I}/2\) arises equally from mixing \(\{\vert 0\rangle, \vert 1\rangle\}\) or \(\{\vert +\rangle, \vert -\rangle\}\) with equal probabilities. No measurement can distinguish the two preparations.

The density matrix is the fundamental physical object—not the ensemble that generated it.

Any density matrix \(\hat{\rho} = \sum_i \lambda_i \vert u_i\rangle\langle u_i\vert\) has a unique spectral decomposition in terms of its eigenvectors \(\{\vert u_i\rangle\}\). All other decompositions (with non-orthogonal states) represent different ensembles yielding the same physics.

Discussion

A qubit in the state \(\hat{\rho} = \hat{I}/2\) can be prepared by mixing \(\vert 0\rangle\) and \(\vert 1\rangle\) with equal probability, or by mixing \(\vert +\rangle\) and \(\vert -\rangle\)—no measurement can distinguish these preparations. Yet in statistical mechanics, we routinely assign a specific thermal ensemble \(\{p_n, \vert n\rangle\}\) to a system at temperature \(T\).

If the density matrix is the fundamental object, why do physicists still talk about specific ensembles?
Can a system “know” how it was prepared, or does all preparation information get erased once the state is described by \(\hat{\rho}\)?
How does this connect to entropy: does \(S(\hat{\rho})\) measure our ignorance, or something objective about the system?

The Bloch Ball#

Bloch Ball Representation

A qubit density matrix can be written as:

(103)#\[\hat{\rho} = \frac{1}{2}(\hat{I} + \boldsymbol{r} \cdot \hat{\boldsymbol{\sigma}})\]

where \(\boldsymbol{r} = (\langle\hat{\sigma}^x\rangle, \langle\hat{\sigma}^y\rangle, \langle\hat{\sigma}^z\rangle)\) is the Bloch vector satisfying \(\vert\boldsymbol{r}\vert \leq 1\).

\(\vert\boldsymbol{r}\vert = 1\): pure state (Bloch sphere surface)
\(\vert\boldsymbol{r}\vert < 1\): mixed state (Bloch ball interior)
\(\boldsymbol{r} = \boldsymbol{0}\): maximally mixed (\(\hat{\rho} = \hat{I}/2\), ball center)

The purity is \(\operatorname{Tr}(\hat{\rho}^2) = \frac{1}{2}(1 + \vert\boldsymbol{r}\vert^2)\).

Derivation: Purity from Bloch Vector

Using \(\hat{\rho} = \frac{1}{2}(\hat{I} + \boldsymbol{r}\cdot\hat{\boldsymbol{\sigma}})\) and the Pauli identity \((\boldsymbol{r}\cdot\hat{\boldsymbol{\sigma}})^2 = \vert\boldsymbol{r}\vert^2 \hat{I}\):

\[\hat{\rho}^2 = \frac{1}{4}(\hat{I} + 2\boldsymbol{r}\cdot\hat{\boldsymbol{\sigma}} + \vert\boldsymbol{r}\vert^2 \hat{I})\]

Taking the trace (\(\operatorname{Tr}(\hat{I}) = 2\), \(\operatorname{Tr}(\hat{\sigma}^i) = 0\)):

\[\operatorname{Tr}(\hat{\rho}^2) = \frac{1}{4}(2 + 2\vert\boldsymbol{r}\vert^2) = \frac{1}{2}(1 + \vert\boldsymbol{r}\vert^2)\]

Expectation Values#

Expectation Value from Density Matrix

For any observable \(\hat{A}\):

(104)#\[\langle \hat{A} \rangle = \operatorname{Tr}(\hat{\rho} \hat{A})\]

This formula applies to both pure and mixed states. For pure states, it reduces to \(\langle\psi\vert\hat{A}\vert\psi\rangle\).

Time Evolution#

Von Neumann Equation

The density matrix evolves as:

(105)#\[\mathrm{i}\hbar\frac{\mathrm{d}\hat{\rho}}{\mathrm{d}t} = [\hat{H}, \hat{\rho}]\]

Equivalently, \(\hat{\rho}(t) = \hat{U}(t)\hat{\rho}(0)\hat{U}^\dagger(t)\) where \(\hat{U}(t) = \mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\).

Derivation: Von Neumann Equation

Differentiating \(\hat{\rho}(t) = \hat{U}(t)\hat{\rho}(0)\hat{U}^\dagger(t)\) and using \(\frac{\mathrm{d}\hat{U}}{\mathrm{d}t} = -\frac{\mathrm{i}}{\hbar}\hat{H}\hat{U}\):

\[\frac{\mathrm{d}\hat{\rho}}{\mathrm{d}t} = -\frac{\mathrm{i}}{\hbar}\hat{H}\hat{U}\hat{\rho}(0)\hat{U}^\dagger + \hat{U}\hat{\rho}(0)\hat{U}^\dagger\frac{\mathrm{i}}{\hbar}\hat{H} = -\frac{\mathrm{i}}{\hbar}[\hat{H}, \hat{\rho}(t)]\]

Purity is Conserved

Under unitary evolution, \(\operatorname{Tr}(\hat{\rho}^2)\) is constant:

\[\operatorname{Tr}((\hat{U}\hat{\rho}\hat{U}^\dagger)^2) = \operatorname{Tr}(\hat{\rho}^2)\]

Unitary evolution cannot convert a pure state into a mixed state. Apparent mixing requires interaction with an inaccessible environment (decoherence, §6.4.1).

Summary#

Density matrix \(\hat{\rho} = \sum_i p_i\vert\psi_i\rangle\langle\psi_i\vert\) describes both pure (\(\operatorname{Tr}(\hat{\rho}^2) = 1\)) and mixed (\(\operatorname{Tr}(\hat{\rho}^2) < 1\)) states.
Three conditions: Hermitian, trace one, positive semidefinite.
Ensemble ambiguity: Different ensembles can yield the same \(\hat{\rho}\); the density matrix is the fundamental object.
Bloch ball (qubit): \(\hat{\rho} = \frac{1}{2}(\hat{I} + \boldsymbol{r}\cdot\hat{\boldsymbol{\sigma}})\), with \(\vert\boldsymbol{r}\vert \leq 1\).
Von Neumann equation: \(\mathrm{i}\hbar\dot{\hat{\rho}} = [\hat{H}, \hat{\rho}]\) governs unitary evolution; purity is conserved.

Homework#

1. Density Matrix Properties. Show that if \(\hat{\rho} = \sum_i p_i |\psi_i\rangle\langle\psi_i|\) with \(p_i \geq 0\) and \(\sum_i p_i = 1\), then:

(a) \(\hat{\rho}\) is Hermitian: \(\hat{\rho}^\dagger = \hat{\rho}\).

(b) \(\hat{\rho}\) has non-negative eigenvalues.

2. Pure vs. Mixed States. (a) Show that a pure state \(\hat{\rho} = |\psi\rangle\langle\psi|\) satisfies \(\hat{\rho}^2 = \hat{\rho}\) and \(\mathrm{Tr}(\hat{\rho}^2) = 1\).

(b) Show that for a mixed state, \(\mathrm{Tr}(\hat{\rho}^2) < 1\). (Hint: write \(\hat{\rho}\) in its eigenbasis and use the constraint \(\sum p_i = 1\) with at least two \(p_i > 0\).)

3. Expectation Values from Density Matrix. For an observable \(\hat{A}\), show that \(\langle\hat{A}\rangle = \mathrm{Tr}(\hat{\rho}\hat{A})\) reproduces the quantum mechanical expectation value for a pure state \(\hat{\rho} = |\psi\rangle\langle\psi|\).

4. Bloch Vector. A qubit density matrix can be written as

\[ \hat{\rho} = \frac{1}{2}(\hat{I} + \boldsymbol{r}\cdot\hat{\boldsymbol{\sigma}})\]

where \(\boldsymbol{r} = (r_x, r_y, r_z)\) is the Bloch vector.

(a) Show that \(r^i = \langle\hat{\sigma}^i\rangle = \mathrm{Tr}(\hat{\rho}\hat{\sigma}^i)\).

(b) Show that \(|\boldsymbol{r}| \leq 1\) with equality iff the state is pure.

5. Von Neumann Equation. The time evolution of the density matrix is governed by

\[ \mathrm{i}\hbar\frac{\mathrm{d}\hat{\rho}}{\mathrm{d}t} = [\hat{H}, \hat{\rho}]\]

(a) Derive this equation from \(\hat{\rho}(t) = \hat{U}(t)\hat{\rho}(0)\hat{U}^\dagger(t)\) by differentiating.

(b) Show that \(\mathrm{Tr}(\hat{\rho}^2)\) is conserved under unitary evolution. What does this imply — can unitary evolution turn a pure state into a mixed state?

6. Coherences and Classical Mixing. Consider two states: (i) the superposition \(|+\rangle = (|0\rangle + |1\rangle)/\sqrt{2}\), and (ii) the 50-50 mixture \(\hat{\rho}_{\rm mix} = (|0\rangle\langle 0| + |1\rangle\langle 1|)/2\).

(a) Write the density matrices for both.

(b) Compute \(\langle\hat{\sigma}^x\rangle\) for both. Explain the physical difference.

(c) Why can’t any sequence of local measurements on state (ii) distinguish it from “I prepared either \(\vert 0\rangle\) or \(\vert 1\rangle\) with equal probability but forgot which”?

7. Energy Basis Representation. In the energy eigenbasis \(\{|n\rangle\}\), write out the matrix elements of \(\hat{\rho}(t)\) under evolution with time-independent \(\hat{H}\). Show that the diagonal elements (populations) are time-independent, while the off-diagonal elements (coherences) oscillate at frequency \((E_n - E_m)/\hbar\). What happens to coherences when averaged over a long time?

8. Preparation and Equivalence. Two experimenters prepare qubits differently: Alice prepares \(\vert 0\rangle\) or \(\vert 1\rangle\) with probability 1/2 each; Bob prepares \(|+\rangle = (|0\rangle+|1\rangle)/\sqrt{2}\) or \(|-\rangle = (|0\rangle-|1\rangle)/\sqrt{2}\) with probability 1/2 each. Show that both ensembles give the same density matrix \(\hat{\rho} = \hat{I}/2\), and therefore no experiment can distinguish the two preparation procedures.