6.3.3 Quantum Channels#

Prompts

What is a quantum channel? Why must physical evolution of density matrices be completely positive and trace-preserving (CPTP)?
State the Kraus representation theorem. How does the completeness relation \(\sum_k \hat{K}_k^\dagger \hat{K}_k = \hat{I}\) ensure trace preservation?
Compare the depolarizing, amplitude damping, and dephasing channels: what physical noise does each model, and what are their Kraus operators?
How does unitary evolution appear as a special case of a quantum channel, and why is such a channel reversible? Contrast it with an irreversible channel like amplitude damping — what happens to the system’s entropy, and why can the input state no longer be recovered?

Lecture Notes#

Overview#

A quantum channel is a linear map describing the most general physical evolution of a quantum system. Unlike unitary evolution (which is reversible), quantum channels include irreversible processes: measurement, decoherence, noise. The key insight is that any physical process must satisfy two fundamental requirements: trace preservation (probability is conserved) and complete positivity (the map makes physical sense even when the system is entangled with other systems we don’t measure).

Consider a concrete example: a qubit coupled to a thermal bath. As time evolves, superpositions decay and populations relax toward thermal equilibrium. The mathematical framework describing such evolution is a quantum channel. In this section, we define channels rigorously, derive their universal form (the Kraus representation), examine canonical examples (depolarizing, dephasing, amplitude damping), distinguish reversible unitary channels from irreversible ones that lose information to the environment, and show why these properties are both necessary and sufficient.

Kraus Representation: The Universal Form of Quantum Channels#

Theorem (Kraus): Any linear map \(\mathcal{E}\) on density matrices is completely positive and trace-preserving if and only if it can be written as:

(244)#\[ \mathcal{E}(\hat{\rho}) = \sum_{k=1}^r \hat{K}_k \hat{\rho} \hat{K}_k^\dagger \]

where the Kraus operators \(\{\hat{K}_k\}\) satisfy the completeness relation:

(245)#\[ \sum_k \hat{K}_k^\dagger \hat{K}_k = \hat{I} \]

This is the universal form: every physical channel on density matrices has this representation.

Why this form ensures the right properties:

Proof Sketch

Trace-preserving: by linearity of the trace (to move the sum outside and back in), cyclic invariance of the trace (to reorder each term), and the Kraus completeness relation \(\sum_k \hat{K}_k^\dagger \hat{K}_k = \hat{I}\),

\[\begin{split} \begin{split} \operatorname{Tr}\Bigl(\sum_k \hat{K}_k \hat{\rho} \hat{K}_k^\dagger\Bigr) &= \sum_k \operatorname{Tr}\bigl(\hat{K}_k \hat{\rho} \hat{K}_k^\dagger\bigr) \\ &= \sum_k \operatorname{Tr}\bigl(\hat{\rho} \hat{K}_k^\dagger \hat{K}_k\bigr) \\ &= \operatorname{Tr}\Bigl(\hat{\rho} \sum_k \hat{K}_k^\dagger \hat{K}_k\Bigr) \\ &= \operatorname{Tr}(\hat{\rho}\hat{I}) \\ &= \operatorname{Tr}(\hat{\rho}) \end{split} \end{split}\]

Completely positive: First, \(\mathcal{E}\) is positive. If \(\hat{\rho} \geq 0\), each term \(\hat{K}_k \hat{\rho} \hat{K}_k^\dagger\) is itself positive: for any vector \(\vert\psi\rangle\), write \(\vert\phi\rangle = \hat{K}_k^\dagger\vert\psi\rangle\), so that \(\langle\psi\vert\hat{K}_k \hat{\rho} \hat{K}_k^\dagger\vert\psi\rangle = \langle\phi\vert\hat{\rho}\vert\phi\rangle \geq 0\) because \(\hat{\rho} \geq 0\). Summing over \(k\), \(\mathcal{E}(\hat{\rho}) = \sum_k \hat{K}_k \hat{\rho} \hat{K}_k^\dagger \geq 0\) as a sum of positive operators.

Complete positivity requires more: \(\mathcal{E} \otimes \mathcal{I}_B\) must be positive for any auxiliary system \(B\). This follows from the same Kraus structure — the extended map has Kraus operators \(\hat{K}_k \otimes \hat{I}_B\), so that \((\mathcal{E} \otimes \mathcal{I}_B)(\hat{\rho}_{AB}) = \sum_k (\hat{K}_k \otimes \hat{I}_B)\,\hat{\rho}_{AB}\,(\hat{K}_k \otimes \hat{I}_B)^\dagger\). Applying the one-line argument above to these operators shows \((\mathcal{E} \otimes \mathcal{I}_B)(\hat{\rho}_{AB}) \geq 0\) for every joint state \(\hat{\rho}_{AB} \geq 0\).

Non-Uniqueness of Kraus Operators:

Given one set \(\{\hat{K}_k\}\), other valid sets exist related by unitary rotation:

\[ \hat{K}_k' = \sum_j U_{kj} \hat{K}_j \]

where \(\{U_{kj}\}\) is unitary. Different Kraus representations encode the same physical channel.

Discussion: Non-Uniqueness of Kraus Operators

The Kraus representation is non-unique, but the channel map \(\mathcal{E}(\hat{\rho})\) is always the same. What does this tell you about the relationship between Kraus operators and physical observables? Could two very different sets of Kraus operators describe the same noise process?

CPTP Maps: A Formal Perspective#

A quantum channel is formally characterized as a completely positive, trace-preserving (CPTP) map on density matrices. The Kraus representation theorem tells us that a map is CPTP if and only if it has the form above. To understand why both conditions are needed:

1. Trace preservation (already built into Kraus):

(246)#\[ \operatorname{Tr}(\mathcal{E}(\hat{\rho})) = \operatorname{Tr}(\hat{\rho}) = 1 \]

Probabilities are conserved; output is a valid density matrix.

2. Complete positivity (physical necessity):

A map must be positive not only on the system of interest but also when tensored with identity on any ancillary system:

(247)#\[ \mathcal{E} \otimes \mathcal{I}_n \text{ is positive for all } n \]

Why Complete Positivity Matters

Consider two qubits A and B, entangled. You apply a channel \(\mathcal{E}\) to A only. If \(\mathcal{E}\) were merely positive but not completely positive, the map \(\mathcal{E} \otimes \mathcal{I}_B\) could produce negative eigenvalues in the joint system (negative probability!). This is unphysical. Complete positivity ensures the channel makes sense even when the system is entangled with other systems we don’t measure.

Poll: Complete positivity and physical channels

Why must a quantum channel be completely positive? Consider two qubits A and B, prepared in an entangled state. You apply a channel \(\mathcal{E}\) to A alone, leaving B untouched. Which statement is correct?

(A) Complete positivity prevents the joint state from having negative eigenvalues (negative probability).

(B) Complete positivity ensures the channel is invertible on all Hilbert spaces.

(D) Complete positivity guarantees that the channel preserves entanglement.

Common Quantum Channels#

1. Unitary Channel.

(248)#\[ \mathcal{E}_U(\hat{\rho}) = \hat{U} \hat{\rho} \hat{U}^\dagger \]

Kraus form: Single operator \(\hat{K} = \hat{U}\) with \(\hat{U}^\dagger \hat{U} = \hat{I}\).

Properties: Reversible (inverse is \(\mathcal{E}_{U^\dagger}\)); entropy conserved; no information loss.

2. Depolarizing Channel.

With probability \(p\), the state becomes completely random; with probability \(1-p\), unchanged:

(249)#\[ \mathcal{E}_\text{depol}(\hat{\rho}) = (1-p)\hat{\rho} + p\frac{\hat{I}}{d} \]

where \(d = \dim(\mathcal{H})\).

Kraus operators (qubit, \(d=2\)): consistent with the mixing form (249),

(250)#\[\begin{split} \begin{split} \hat{K}_0 &= \sqrt{1 - \tfrac{3p}{4}}\,\hat{I}, \\ \hat{K}_1 &= \sqrt{\tfrac{p}{4}}\,\hat{X}, \\ \hat{K}_2 &= \sqrt{\tfrac{p}{4}}\,\hat{Y}, \\ \hat{K}_3 &= \sqrt{\tfrac{p}{4}}\,\hat{Z}. \end{split} \end{split}\]

Check: \(\sum_k \hat{K}_k^\dagger \hat{K}_k = (1-\tfrac{3p}{4})\hat{I} + 3\,\tfrac{p}{4}\hat{I} = \hat{I}\), and using the Pauli identity \(\hat{X}\hat{\rho}\hat{X} + \hat{Y}\hat{\rho}\hat{Y} + \hat{Z}\hat{\rho}\hat{Z} = 2\hat{I} - \hat{\rho}\) for any qubit state, the Kraus form reproduces \(\mathcal{E}(\hat{\rho}) = (1-p)\hat{\rho} + \tfrac{p}{2}\hat{I}\).

Derivation: Depolarizing Kraus Form Reproduces the Mixing Form

Building block — the Pauli identity. Write any qubit state in Bloch form \(\hat{\rho} = \tfrac{1}{2}\bigl(\hat{I} + \boldsymbol{r}\cdot\hat{\boldsymbol{\sigma}}\bigr)\), with \(\boldsymbol{r}\cdot\hat{\boldsymbol{\sigma}} = r_x\hat{X} + r_y\hat{Y} + r_z\hat{Z}\). Distinct Pauli operators anticommute and each squares to \(\hat{I}\), so conjugation by \(\hat{X}\) leaves \(\hat{I}\) and \(\hat{X}\) fixed but flips the sign of \(\hat{Y}\) and \(\hat{Z}\) (for example \(\hat{X}\hat{Y}\hat{X} = -\hat{Y}\hat{X}\hat{X} = -\hat{Y}\)), and likewise for conjugation by \(\hat{Y}\) and \(\hat{Z}\). Conjugating the Bloch form term by term,

\[\begin{split} \begin{split} \hat{X}\hat{\rho}\hat{X} &= \tfrac{1}{2}\bigl(\hat{I} + r_x\hat{X} - r_y\hat{Y} - r_z\hat{Z}\bigr), \\ \hat{Y}\hat{\rho}\hat{Y} &= \tfrac{1}{2}\bigl(\hat{I} - r_x\hat{X} + r_y\hat{Y} - r_z\hat{Z}\bigr), \\ \hat{Z}\hat{\rho}\hat{Z} &= \tfrac{1}{2}\bigl(\hat{I} - r_x\hat{X} - r_y\hat{Y} + r_z\hat{Z}\bigr). \end{split} \end{split}\]

Summing the three lines, the identity terms add to \(3\hat{I}\) while each Pauli component appears once with a \(+\) sign and twice with a \(-\) sign, leaving \(-\boldsymbol{r}\cdot\hat{\boldsymbol{\sigma}}\):

\[\begin{split} \begin{split} \hat{X}\hat{\rho}\hat{X} + \hat{Y}\hat{\rho}\hat{Y} + \hat{Z}\hat{\rho}\hat{Z} &= \tfrac{1}{2}\bigl(3\hat{I} - \boldsymbol{r}\cdot\hat{\boldsymbol{\sigma}}\bigr) \\ &= \tfrac{1}{2}\bigl(3\hat{I} - (2\hat{\rho} - \hat{I})\bigr) = 2\hat{I} - \hat{\rho}, \end{split} \end{split}\]

where the second line substitutes \(\boldsymbol{r}\cdot\hat{\boldsymbol{\sigma}} = 2\hat{\rho} - \hat{I}\), the Bloch form solved for the Pauli part.

Assembly — the Kraus sum. Each operator in (250) is a real multiple of \(\hat{I}\) or of a Pauli operator, hence Hermitian, so \(\hat{K}_k^\dagger = \hat{K}_k\). The \(k=0\) term gives \(\hat{K}_0\hat{\rho}\hat{K}_0^\dagger = (1-\tfrac{3p}{4})\,\hat{I}\hat{\rho}\hat{I} = (1-\tfrac{3p}{4})\hat{\rho}\), and each \(k\ge1\) term gives \(\tfrac{p}{4}\) times a Pauli conjugate of \(\hat{\rho}\):

\[\begin{split} \begin{split} \mathcal{E}_\text{depol}(\hat{\rho}) &= \sum_{k=0}^{3}\hat{K}_k\hat{\rho}\hat{K}_k^\dagger \\ &= \bigl(1 - \tfrac{3p}{4}\bigr)\hat{\rho} + \tfrac{p}{4}\bigl(\hat{X}\hat{\rho}\hat{X} + \hat{Y}\hat{\rho}\hat{Y} + \hat{Z}\hat{\rho}\hat{Z}\bigr) \\ &= \bigl(1 - \tfrac{3p}{4}\bigr)\hat{\rho} + \tfrac{p}{4}\bigl(2\hat{I} - \hat{\rho}\bigr) \\ &= (1 - p)\,\hat{\rho} + \tfrac{p}{2}\hat{I}, \end{split} \end{split}\]

which is the mixing form (249) at \(d = 2\).

Bloch-vector action: \(\boldsymbol{r}\to (1-p)\boldsymbol{r}\) — every component of the Bloch vector shrinks uniformly toward the origin. This is isotropic depolarization: the Bloch ball contracts by factor \((1-p)\) with no preferred axis.

Physical interpretation: Generic noise that does not single out any direction; models the worst-case scenario where errors on all three Pauli axes contribute equally.

3. Amplitude Damping (\(T_1\) Process).

Excited state decays to ground state at rate \(\Gamma = 1/T_1\):

(251)#\[ \mathcal{E}_\text{amp}(\hat{\rho}) = \hat{K}_0 \hat{\rho} \hat{K}_0^\dagger + \hat{K}_1 \hat{\rho} \hat{K}_1^\dagger \]

with:

(252)#\[\begin{split} \begin{split} \hat{K}_0 &= \begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1-\gamma} \end{pmatrix}, \\ \hat{K}_1 &= \begin{pmatrix} 0 & \sqrt{\gamma} \\ 0 & 0 \end{pmatrix} \end{split} \end{split}\]

where \(\gamma = 1 - \mathrm{e}^{-t/T_1}\) is the decay probability.

Physical interpretation: Energy dissipation; ground state is absorbing (fixed point).

4. Phase Damping (\(T_2\) Process, Dephasing).

Coherences decay without energy loss at rate \(\Gamma_\phi = 1/T_2\):

(253)#\[ \mathcal{E}_\text{phase}(\hat{\rho}) = \hat{K}_0 \hat{\rho} \hat{K}_0^\dagger + \hat{K}_1 \hat{\rho} \hat{K}_1^\dagger \]

with:

(254)#\[\begin{split} \begin{split} \hat{K}_0 &= \begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1-\gamma} \end{pmatrix}, \\ \hat{K}_1 &= \begin{pmatrix} 0 & 0 \\ 0 & \sqrt{\gamma} \end{pmatrix} \end{split} \end{split}\]

where \(\gamma = 1 - \mathrm{e}^{-t/T_2}\).

Physical interpretation: Random phase fluctuations; off-diagonal elements (coherences) decay; diagonal (populations) preserved.

Reversibility and Irreversibility#

Reversible (Unitary) Channel:

Single Kraus operator \(\hat{K} = \hat{U}\) (unitary).
Entropy conserved: \(S(\mathcal{E}(\hat{\rho})) = S(\hat{\rho})\).
Inverse channel exists: \(\mathcal{E}^{-1}(\hat{\rho}') = \hat{U}^\dagger \hat{\rho}' \hat{U}\).
State can be recovered perfectly.

Irreversible (Non-Unitary) Channel:

Multiple Kraus operators (\(r > 1\)) or degenerate structure.
Entropy is not monotonic in general: a unital channel — one with \(\mathcal{E}(\hat{I}) = \hat{I}\), such as depolarizing or phase damping — cannot decrease entropy, \(S(\mathcal{E}(\hat{\rho})) \geq S(\hat{\rho})\), whereas a non-unital channel such as amplitude damping can raise or lower it (damping toward the pure ground state can even purify a mixed input).
No inverse: original state cannot be recovered (information lost).
Examples: Depolarizing, amplitude damping.

Summary#

Quantum channels: The most general physical evolution of a density matrix is a completely positive, trace-preserving (CPTP) map, covering unitary dynamics, measurement, decoherence, and noise.
Kraus representation: Every CPTP map has the universal form \(\mathcal{E}(\hat{\rho}) = \sum_k \hat{K}_k \hat{\rho} \hat{K}_k^\dagger\) with the completeness relation \(\sum_k \hat{K}_k^\dagger \hat{K}_k = \hat{I}\), which enforces trace preservation. The Kraus operators are not unique: sets related by a unitary mixing \(\hat{K}_k' = \sum_j U_{kj} \hat{K}_j\) describe the same channel.
Complete positivity: Trace preservation conserves probability; complete positivity (\(\mathcal{E} \otimes \mathcal{I}_n\) positive for every ancilla) is required because a merely positive map could produce negative eigenvalues when the system is entangled with an untouched ancilla.
Canonical channels: The unitary channel (reversible), the depolarizing channel (isotropic Bloch-ball contraction \(\boldsymbol{r} \to (1-p)\boldsymbol{r}\)), amplitude damping (\(T_1\) relaxation toward the ground state), and phase damping (\(T_2\) loss of coherence with populations preserved) model the standard forms of qubit noise.
Reversibility: A unitary channel — built from a single Kraus operator — conserves entropy and is reversible, so the input state can be recovered exactly. A channel with several Kraus operators is generally irreversible: information leaks to the environment and the original state can no longer be reconstructed.

Homework#

1. Kraus Completeness Relation. Consider a quantum channel with Kraus operators \(\{\hat{K}_1, \hat{K}_2, \hat{K}_3\}\). Show that the completeness relation \(\sum_k \hat{K}_k^\dagger \hat{K}_k = \hat{I}\) is necessary and sufficient for the channel to be trace-preserving. Start with the trace of the output \(\operatorname{Tr}(\mathcal{E}(\hat{\rho})) = \sum_k \operatorname{Tr}(\hat{K}_k \hat{\rho} \hat{K}_k^\dagger)\) and use the cyclic property of trace.

2. Depolarizing Channel on Bloch Sphere. A qubit depolarizing channel with parameter \(p\) is given by:

\[ \mathcal{E}_\text{depol}(\hat{\rho}) = (1-p)\hat{\rho} + \frac{p}{2}\hat{I} \]

Express this in Kraus form using Pauli operators \(\{\hat{I}, \hat{X}, \hat{Y}, \hat{Z}\}\).

(a) Write down all four Kraus operators explicitly.

(b) Verify the completeness relation \(\sum_k \hat{K}_k^\dagger \hat{K}_k = \hat{I}\).

(c) Interpret the action on the Bloch vector: if \(\hat{\rho} = \frac{1}{2}(\hat{I} + \boldsymbol{r} \cdot \hat{\boldsymbol{\sigma}})\) with Bloch vector \(\boldsymbol{r}\), show that the output has Bloch vector \(\boldsymbol{r}' = (1-p)\boldsymbol{r}\).

3. Amplitude Damping and Spontaneous Emission. A two-level atom decays from excited state \(\vert 1\rangle\) to ground state \(\vert 0\rangle\) with lifetime \(T_1\). After time \(t\), the amplitude damping channel is given by Kraus operators:

\[\begin{split} \begin{split} \hat{K}_0 &= \begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1-\gamma} \end{pmatrix}, \\ \hat{K}_1 &= \begin{pmatrix} 0 & \sqrt{\gamma} \\ 0 & 0 \end{pmatrix} \end{split} \end{split}\]

where \(\gamma = 1 - \mathrm{e}^{-t/T_1}\).

(a) Verify the completeness relation \(\hat{K}_0^\dagger \hat{K}_0 + \hat{K}_1^\dagger \hat{K}_1 = \hat{I}\).

(b) Apply the channel to a pure state \(\vert\psi\rangle = \alpha\vert 0\rangle + \beta\vert 1\rangle\). Write the output density matrix \(\mathcal{E}_\text{amp}(\vert\psi\rangle\langle\psi\vert)\) and interpret physically.

(c) Show that the ground state \(\vert 0\rangle\) is a fixed point of the channel: \(\mathcal{E}_\text{amp}(\vert 0\rangle\langle 0\vert) = \vert 0\rangle\langle 0\vert\).

4. Completely Positive Trace-Preserving (CPTP) Maps. Consider the map \(\mathcal{E}\) defined by:

\[ \mathcal{E}(\hat{\rho}) = \frac{1}{2}\hat{\rho} + \frac{1}{2}\hat{Z} \hat{\rho} \hat{Z} \]

where \(\hat{Z} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\).

(a) Show that this map is linear and trace-preserving.

(b) Express this map in Kraus form with operators \(\hat{K}_0\) and \(\hat{K}_1\).

(c) Verify complete positivity by checking that the map remains positive when tensored with identity: test on the maximally entangled state \(\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\) and confirm the resulting density matrix is positive semidefinite.

5. Amplitude Damping as Kraus Map. For the amplitude damping channel (Problem 3), consider a pure state \(\hat{\rho}_\text{in} = \vert+\rangle\langle+\vert\) where \(\vert+\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\).

(a) Compute \(\mathcal{E}_\text{amp}(\hat{\rho}_\text{in})\) for \(\gamma = 0.5\).

(b) What is the purity of the output state? Is information lost?

6. Composing Channels. Consider two channels:

Phase flip: \(\mathcal{E}_Z(\hat{\rho}) = (1-q)\hat{\rho} + q\hat{Z}\hat{\rho}\hat{Z}\)
Amplitude damping: \(\mathcal{E}_\text{amp}\) from Problem 3 with \(\gamma = 0.1\)

(a) Express the phase-flip channel in Kraus form.

(b) The composition \(\mathcal{E}_\text{comp} = \mathcal{E}_\text{amp} \circ \mathcal{E}_Z\) applies phase flip first, then amplitude damping. Write the Kraus operators of the composed channel.

(c) Show that the composed channel is still CPTP. Is the composition commutative? Does \(\mathcal{E}_\text{amp} \circ \mathcal{E}_Z = \mathcal{E}_Z \circ \mathcal{E}_\text{amp}\)? Explain physically why or why not.

7. Distinguishing Quantum Noise Channels. Three noise channels (all with parameter \(p = 0.5\)) act on a qubit:

Bit-flip (X noise): \(\mathcal{E}_X(\hat{\rho}) = (1-p)\hat{\rho} + p\hat{X}\hat{\rho}\hat{X}\)
Phase-flip (Z noise): \(\mathcal{E}_Z(\hat{\rho}) = (1-p)\hat{\rho} + p\hat{Z}\hat{\rho}\hat{Z}\)
Depolarizing: \(\mathcal{E}_\text{depol}(\hat{\rho}) = (1-p)\hat{\rho} + \frac{p}{2}\hat{I}\) (random Pauli, normalized)

(a) Apply each channel to the input state \(\hat{\rho}_\text{in} = \frac{1}{2}(\hat{I} + \hat{X})\) (eigenstate of \(\hat{X}\)). Compare the outputs: which channel preserves the most information?

(b) Conceptual: The depolarizing channel is called “universal noise” because it cannot distinguish the computational basis. Explain why bit-flip and phase-flip channels can be distinguished by an appropriate measurement on the output, but depolarizing cannot.