research36 min studyLesson 799 in Mathematics

Tensor products and completely positive maps

Require positivity under matrix amplification, construct a unital Stinespring dilation, and diagnose the transpose map with an entangled positive input.

On this page

What you’ll understand

  • Distinguish positivity from complete positivity under auxiliary matrix systems.
  • Track unital observable maps versus trace-preserving state maps.
  • Verify a finite-dimensional Stinespring dilation and a partial-transpose counterexample.

A positive map respects nonnegative observables on one system. To remain compatible with an untouched auxiliary system, it must preserve positivity after every matrix amplification. This stronger requirement is complete positivity. We work over complex unital C*-algebras, and use observable maps in the Heisenberg direction unless explicitly taking their dual action on density matrices. Separating these conventions prevents the unital and trace-preserving conditions from being exchanged accidentally.

Tensor completion and matrix amplification are separate choices

The algebraic tensor product combines finite sums of elementary products, but completing arbitrary C*-algebras requires a C*-norm. The minimal tensor norm comes from faithful Hilbert-space representations and their spatial tensor product. The maximal norm allows commuting representations on a common Hilbert space. These completions need not agree for arbitrary algebras. Matrix algebras and commutative C*-algebras are nuclear, so their minimal and maximal tensor norms agree with any other C*-algebra; no such claim is made for every algebra.

A linear map Phi from A to B is completely positive when the entrywise map on matrices over A preserves positivity for every positive integer matrix size. Size one is ordinary positivity, so complete positivity implies positivity. The converse fails because matrix-level positivity detects correlations with an auxiliary system. The transpose on M₂ is positive by ordinary spectral reasoning but fails at amplification size two. Testing only separate positive elementary tensors would miss the entangled input causing the failure.

Φ completely positive  ⟺  (id⁡Mk⊗Φ)(X)≥0∀k≥1, X∈Mk(A)+\Phi\text{ completely positive}\iff(\operatorname{id}_{M_k}\otimes\Phi)(X)\ge0\quad\forall k\ge1,\ X\in M_k(A)_+
Matrix amplification acts entrywise. Positivity at k=1 alone does not establish complete positivity.↔ Scroll wide equations horizontally.

Dilations realize positive maps as compressed representations

Stinespring's theorem says a completely positive map from a unital C*-algebra A to B(H) can be written as V-star pi(a)V for a *-representation pi on a larger Hilbert space and a bounded V from H into that space. When the map is unital, V is an isometry. This theorem is stated rather than fully proved; its positivity-based quotient construction resembles GNS with matrix-valued rather than scalar-valued averages. A compressed representation need not remain multiplicative after compression.

In finite dimensions, an observable map of the form sum of K_j-star a K_j is completely positive. It is unital exactly when the sum of K_j-star K_j equals the identity. The dual state map sends rho to the sum of K_j rho K_j-star, and that same identity makes it trace-preserving. Distinguish the two map directions before checking a formula. A map may erase coherence while preserving probabilities and positivity, which is useful precisely because it need not be an algebra homomorphism.

Φ(a)=V∗π(a)V,Φ(1)=I⇒V∗V=I,Φ(a)=∑jKj∗aKj  in finite dimensions\Phi(a)=V^*\pi(a)V,\qquad \Phi(1)=I\Rightarrow V^*V=I,\qquad \Phi(a)=\sum_jK_j^*aK_j\ \text{ in finite dimensions}
The Stinespring representation is taken unital on its dilation space. For the finite sum, unitality is equivalent to ∑Kⱼ*Kⱼ=I.↔ Scroll wide equations horizontally.

Entanglement exposes positivity that scalar tests miss

Let Omega be the normalized two-qubit vector with equal coefficients on 00 and 11. Its rank-one density operator is positive. Transposing only the second subsystem turns it into one half of the flip operator, which exchanges the two tensor factors. The antisymmetric unit vector is an eigenvector with eigenvalue minus one half. Thus the amplified transpose sends a positive input to a nonpositive output, proving that ordinary transpose is not completely positive.

This counterexample does not imply transposing a single positive matrix creates negative eigenvalues; it does not. The obstruction appears only with the untouched auxiliary factor and its correlated input. It also does not say every positive map has a Kraus form: complete positivity is what supports that representation in the matrix setting. Tensor products, channels, and entangled tests therefore need to be developed together rather than assuming scalar positivity automatically survives adjoining another system.

Dilate a dephasing map

On M₂(C), let p=e₁₁, q=e₂₂ and Φ(a)=pap+qaq. Construct an isometric V into C²⊗C² and evaluate Φ on [[2,1+i],[1−i,3]].

  1. Define Vξ=pξ⊗e₁+qξ⊗e₂. Orthogonality of the two auxiliary basis vectors gives V*V=p+q=I, so V is an isometry.

  2. Take π(a)=a⊗I. Expanding V*π(a)V removes cross terms between the auxiliary labels and leaves pap+qaq, establishing the displayed dilation.

  3. The map keeps only diagonal entries, so the chosen matrix is sent to diag(2,3). It is unital and completely positive. It is not multiplicative: Φ(e₁₂)Φ(e₂₁)=0 while Φ(e₁₂e₂₁)=p.

The result

Dephasing is a compressed representation with an isometric dilation, not an algebra homomorphism. Its dual density-matrix map is trace-preserving.

Explore the model

Interactive lab

Interactive lab · quantum

Coherence can disappear while basis probabilities stay

Pause, reset, then set θ=90 degrees and φ=0. Predict basis probabilities 1/2 and Bloch vector (1,0,0). For a fresh comparison, pause and reset, set θ=90 and φ=90, and predict the same probabilities with Bloch vector (0,1,0). Manually dephasing either state gives I/2. The lab only displays pure states; it cannot display that mixed output or apply the channel.

Coherence can disappear while basis probabilities stayA 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 forEqual diagonal probabilities need not mean equal input states, and a completely positive map can discard their coherence.

Check your understanding

Which calculation proves that transpose T on M₂ is positive but not completely positive?

Try it yourself

Write the partial transpose of the Bell projection as a four-by-four matrix in basis 00,01,10,11 and test it on (|01⟩−|10⟩)/√2.

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

Give me a hint

Its diagonal nonzero entries stay at 00 and 11; the two off-diagonal Bell terms move into the 01–10 block.

Compare with the solution

The matrix is (1/2)[[1,0,0,0],[0,0,1,0],[0,1,0,0],[0,0,0,1]]. Multiplying the antisymmetric vector (0,1,−1,0)/√2 gives minus one half times the same vector. Its quadratic form is therefore −1/2, so the matrix is not positive. Before partial transpose the Bell density operator was a rank-one positive projection, giving the required complete-positivity counterexample.

Take it with you

  • Complete positivity tests all auxiliary matrix sizes, not only one-system positive inputs.
  • Unital observable maps are dual to trace-preserving state maps under the stated finite-dimensional convention.
  • Stinespring compression and entangled counterexamples explain why positivity alone is insufficient.

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