advanced34 min studyLesson 412 in Physics

Kraus operators and complete positivity

Build channels from unitary dilations and test trace preservation separately from complete positivity.

On this page

What you’ll understand

  • Derive Kraus operators for a fixed environment preparation.
  • Check the operator trace-preservation identity.
  • Distinguish positivity on a system from complete positivity with an ancilla.

A quantum channel specifies a deterministic transformation of density operators, including inputs entangled with another system. Checking only that isolated input states remain positive misses this last requirement. Kraus operators make the stronger condition transparent, but their normalization serves a separate purpose: it fixes total probability. Keeping these two tests distinct prevents a formally positive expression from being mistaken for a valid deterministic experiment.

From an environment to an operator sum

Assume a fixed environment state initially independent of the system, and a joint unitary. For a pure environment preparation, insert an orthonormal environment basis into the final partial trace. Each matrix element of the joint unitary between the initial environment vector and an output basis vector is an operator acting on the system alone. These are Kraus operators. A mixed environment can be purified, or its initial classical weights can be absorbed into an enlarged list of operators.

The construction relies on one environment preparation for every allowed input. If preparation procedures correlate system and environment differently for different inputs, a single channel on the entire system state space need not describe them. Conversely, every finite-dimensional completely positive, trace-preserving map has a dilation with a sufficiently large initially uncorrelated environment. The dilation is a mathematical representation, not a unique microscopic reconstruction: distinct environments, bases, and interactions can produce identical system input–output statistics.

E(ρ)=∑αKαρKα†,Kα=E⟨α∣U∣e⟩E,∑αKα†Kα=I\mathcal E(\rho)=\sum_\alpha K_\alpha\rho K_\alpha^\dagger,\qquad K_\alpha={}_E\langle\alpha|U|e\rangle_E,\qquad \sum_\alpha K_\alpha^\dagger K_\alpha=I
Kraus operators, U, and density matrices are dimensionless. Unitarity and the complete environment basis imply the final identity, which guarantees trace preservation for every input. The dilation assumes a fixed product preparation.↔ Scroll wide equations horizontally.

Positivity must survive an untouched ancilla

Each term in a Kraus sum is positive when the input is positive: its expectation in any vector is the input expectation in a transformed vector. The same argument works after tensoring every Kraus operator with the identity of an arbitrary reference system. This is complete positivity. Ordinary positivity tests only the system, whereas complete positivity also protects probabilities when that system is one member of an entangled pair. Trace preservation alone says nothing about either positivity test.

Transposition illustrates the distinction. It preserves trace and the eigenvalues of each individual density matrix, so it is positive. Yet transposing only one half of a Bell-state projector produces a matrix with a negative eigenvalue. Therefore transposition cannot be a channel acting independently on an unknown member of an arbitrary entangled pair. This obstruction is stronger than saying the map is unfamiliar experimentally; it is a direct inconsistency with nonnegative joint measurement probabilities.

(T⊗I)(∣Φ+⟩⟨Φ+∣)=12(1000001001000001),spec⁡={12,12,12,−12}(T\otimes I)(|\Phi^+\rangle\langle\Phi^+|)=\frac12\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix},\quad \operatorname{spec}=\left\{\frac12,\frac12,\frac12,-\frac12\right\}
Here T transposes the first qubit in the computational basis, Φ⁺ = (|00⟩ + |11⟩)/√2, and I denotes the identity map on the reference. All entries are dimensionless. The negative eigenvalue proves that T is not completely positive.↔ Scroll wide equations horizontally.

Normalization and records are different questions

For a deterministic operation, cyclicity of trace turns output normalization into the operator identity in the first display. Reversing the order of the operators would test unitality instead, the preservation of the identity operator. Amplitude damping is trace preserving but not unital, so these conditions cannot be exchanged. A selected measurement outcome may instead have a trace-decreasing completely positive map; its output trace is the probability of that outcome before the conditional state is normalized.

Kraus decompositions are not unique. Unitary mixing of a fixed list leaves its summed channel unchanged, although a chosen interpretation of individual environment records can change. A random phase-flip channel is a simple example: apply the identity with one probability and a Pauli phase flip with the complementary probability. Each branch is unitary, but forgetting which branch occurred generally produces a mixed state. No single pure-state Bloch trajectory represents that averaged output for arbitrary input states.

Check a phase-flip channel by hand

Let K0=3/4IK_0=\sqrt{3/4}I and K1=1/4ZK_1=\sqrt{1/4}Z, with Z=diag⁡(1,−1)Z=\operatorname{diag}(1,-1). For input ∣+⟩|+\rangle, verify normalization and calculate the output coherence and purity.

  1. Since Z†Z=IZ^\dagger Z=I, the normalization sum is (3/4)I+(1/4)I=I(3/4)I+(1/4)I=I. The Kraus form supplies complete positivity independently of this normalization check.

  2. The identity branch leaves the input projector unchanged, while the Z branch reverses its off-diagonal signs. Their weighted sum has diagonal entries 1/21/2 and off-diagonal entries (3/4−1/4)/2=1/4(3/4-1/4)/2=1/4.

  3. The eigenvalues are 3/43/4 and 1/41/4, so purity is 9/16+1/16=5/89/16+1/16=5/8. The plus probability is 3/43/4, matching the probability of the branch that preserves the plus state.

The result

The map is completely positive and trace preserving; the output coherence is 1/4 and its purity is 5/8. Randomly mixing unitary branches can reduce purity.

Explore the model

Interactive lab

Interactive lab · quantum

Inspect the two pure branches, not their mixture

Reset and set both controls to θ = 90° and φ = 0°; predict Bloch x = 1. Reset and set both to θ = 90° and φ = 180°; predict Bloch x = −1, while both basis probabilities stay at 1/2. Keep playback paused. These are the identity and Z outputs for a plus input. The instrument cannot average branches, accept Kraus matrices, or display their mixed output.

Inspect the two pure branches, not their mixtureA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forIndividual phase-flip branches are pure; the discarded branch record is essential to the channel mixture.

Check your understanding

Which identity guarantees trace preservation of a Kraus sum for every input?

Try it yourself

Consider B(ρ)=0.8ρ+0.2XρX\mathcal B(\rho)=0.8\rho+0.2X\rho X, with X∣0⟩=∣1⟩X|0\rangle=|1\rangle and X∣1⟩=∣0⟩X|1\rangle=|0\rangle. For input diag⁡(3/4,1/4)\operatorname{diag}(3/4,1/4), find the output and its purity.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The X branch swaps the populations; it does not create off-diagonal terms for this input.

Compare with the solution

The output zero population is 0.8(3/4)+0.2(1/4)=0.65=13/200.8(3/4)+0.2(1/4)=0.65=13/20, and the one population is 7/207/20. Coherences remain zero. The purity is (169+49)/400=109/200=0.545(169+49)/400=109/200=0.545. The operators 0.8I\sqrt{0.8}I and 0.2X\sqrt{0.2}X obey the normalization sum, and their Kraus form protects entangled inputs as well. The output trace is one, which alone would not establish complete positivity.

Take it with you

  • Complete positivity includes any untouched entangled reference.
  • Trace preservation uses the ordered sum K†K.
  • A channel and its Kraus decomposition do not uniquely specify an environment record.

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