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Stinespring Dilation: Construction and Normality

Complete positivity permits a map on observables to be realized by a representation on a larger Hilbert space followed by compression. This is Stinespring’s theorem. The canonical Stinespring Dilation article explains its finite-dimensional environment interpretation; here the construction starts with a unital C*-algebra and isolates the extra hypothesis behind the familiar tensor-product formula.

Required background. Completely Positive Maps supplies matrix positivity and normality. Helpful background. Algebraic Tensor Products and Completion explains quotient and completion constructions.

Complete positivity defines a dilation inner product

Section titled “Complete positivity defines a dilation inner product”

Let A\mathcal A be a unital C*-algebra and let T:A→B(H)T:\mathcal A\to B(\mathcal H) be a bounded completely positive map. The theorem provides a Hilbert space K\mathcal K, a unital *-representation π:A→B(K)\pi:\mathcal A\to B(\mathcal K), and a bounded V:H→KV:\mathcal H\to\mathcal K such that

T(a)=V†π(a)V,∥V∥2=∥T(1)∥.T(a)=V^\dagger\pi(a)V,\qquad \|V\|^2=\|T(1)\|.

If T(1)=IT(1)=I, then VV is an isometry. Neither finite dimension nor preservation of products by TT is assumed.

Use inner products linear in the second argument. On A⊙H\mathcal A\odot\mathcal H, define

⟨∑iai⊗ui,∑jbj⊗vj⟩T=∑i,j⟨ui,T(ai†bj)vj⟩.\left\langle\sum_i a_i\otimes u_i,\sum_j b_j\otimes v_j\right\rangle_T =\sum_{i,j}\langle u_i,T(a_i^\dagger b_j)v_j\rangle.

The matrix [ai†aj][a_i^\dagger a_j] is positive in Mn(A)M_n(\mathcal A). Complete positivity therefore makes this form nonnegative. Quotient by its zero-norm subspace and complete to obtain K\mathcal K. Ordinary positivity of TT would not justify this matrix step; this is exactly where the stronger assumption enters.

Set π(a)[b⊗u]=[ab⊗u]\pi(a)[b\otimes u]=[ab\otimes u]. To prove this is well defined and bounded, observe that

c=∥a∥21−a†a≥0.c=\|a\|^2 1-a^\dagger a\ge0.

For z=∑ibi⊗uiz=\sum_i b_i\otimes u_i, the difference ∥a∥2⟨z,z⟩T−⟨az,az⟩T\|a\|^2\langle z,z\rangle_T-\langle az,az\rangle_T is the quadratic form of the positive matrix [T(bi†cbj)][T(b_i^\dagger c b_j)]. Consequently

∥π(a)z∥T≤∥a∥∥z∥T.\|\pi(a)z\|_T\le\|a\|\|z\|_T.

Null vectors stay null, and left multiplication extends continuously. Products and adjoints follow from the defining form, so π\pi is a *-representation.

Define Vu=[1⊗u]Vu=[1\otimes u]. Then

⟨u,V†π(a)Vv⟩=⟨1⊗u,a⊗v⟩T=⟨u,T(a)v⟩.\langle u,V^\dagger\pi(a)Vv\rangle =\langle 1\otimes u,a\otimes v\rangle_T =\langle u,T(a)v\rangle.

This proves the compression formula and V†V=T(1)V^\dagger V=T(1). Conversely, any such compression is CP: matrix amplification gives a compression of the positive operator matrix [π(aij)][\pi(a_{ij})]. The argument proves both the representation and its physical positivity property without choosing Kraus operators.

Minimality fixes the unused environment freedom

Section titled “Minimality fixes the unused environment freedom”

A dilation is minimal if

K=span⁡‾{π(a)Vu:a∈A, u∈H}.\mathcal K=\overline{\operatorname{span}}\{\pi(a)Vu:a\in\mathcal A,\ u\in\mathcal H\}.

The construction above is minimal by definition. For two minimal dilations, send π1(a)V1u\pi_1(a)V_1u to π2(a)V2u\pi_2(a)V_2u and extend linearly. The compression identity shows that all inner products of such finite sums agree. Thus the map is well defined and isometric; minimality makes its extension a surjective unitary intertwining both representations and the two embeddings.

Minimal uniqueness is uniqueness of the dilation up to this intertwining unitary. An arbitrarily enlarged environment contains unused subspaces and is not fixed by the theorem.

For a normal CP map T:B(Hout)→B(Hin)T:B(\mathcal H_{\mathrm{out}})\to B(\mathcal H_{\mathrm{in}}), the minimal representation can be chosen normal. A normal representation of this type-I factor is an amplification:

K=Hout⊗E,T(A)=V†(A⊗IE)V.\mathcal K=\mathcal H_{\mathrm{out}}\otimes\mathcal E,\qquad T(A)=V^\dagger(A\otimes I_{\mathcal E})V.

For separable input and output Hilbert spaces, a minimal E\mathcal E can be chosen separable. Expanding in its orthonormal basis produces the countable Kraus theorem proved in Kraus Representations. Normality is what connects an abstract C*-algebra representation to an ordinary normal-state channel and this spatial form; an arbitrary CP map on B(H)B(\mathcal H) need not have that property.

For the trace-preserving predual channel, VV is isometric and Φ(ρ)=Tr⁡E(VρV†)\Phi(\rho)=\operatorname{Tr}_{\mathcal E}(V\rho V^\dagger). An isometry is not automatically a unitary on the same space.

Worked example: a one-operator channel that is not unitary

Section titled “Worked example: a one-operator channel that is not unitary”

On ℓ2(N0)\ell^2(\mathbb N_0) let S∣n⟩=∣n+1⟩S|n\rangle=|n+1\rangle. Since S†S=IS^\dagger S=I,

Φ(ρ)=SρS†,T(A)=S†AS\Phi(\rho)=S\rho S^\dagger,\qquad T(A)=S^\dagger AS

is a channel with a one-dimensional Stinespring environment. Yet SS†=I−∣0⟩⟨0∣SS^\dagger=I-|0\rangle\langle0|, so SS is not surjective. The channel’s output never occupies ∣0⟩|0\rangle. A unitary conjugation on this same Hilbert space cannot have that restriction for every input.

Unitary environment implementations remain possible with suitable extra space, but extending an isometry to a unitary requires matching the orthogonal complements of its initial and final subspaces. The finite-dimensional instruction “complete the columns” must include that condition when dimensions are infinite.

Nonunital dilation. If T(1)≤IT(1)\le I, prove that the Stinespring embedding is a contraction and identify the corresponding trace inequality.

Solution

∥Vu∥2=⟨u,T(1)u⟩≤∥u∥2\|Vu\|^2=\langle u,T(1)u\rangle\le\|u\|^2. For the normal predual and ρ≥0\rho\ge0, Tr⁡Φ(ρ)=Tr⁡[ρT(1)]≤Tr⁡ρ\operatorname{Tr}\Phi(\rho)=\operatorname{Tr}[\rho T(1)]\le\operatorname{Tr}\rho. Missing trace represents a discarded outcome, not a failure of complete positivity.

Compression and multiplicativity. For unital TT, write P=VV†P=VV^\dagger. Show the identity

T(a†a)−T(a)†T(a)=V†π(a)†(I−P)π(a)V.T(a^\dagger a)-T(a)^\dagger T(a) =V^\dagger\pi(a)^\dagger(I-P)\pi(a)V.
Solution

Insert T(a)=V†π(a)VT(a)=V^\dagger\pi(a)V and V†V=IV^\dagger V=I, then subtract. The right-hand side is positive. It vanishes for every aa precisely when the embedded subspace is invariant under every π(a)\pi(a); since adjoints are included, that subspace reduces the representation and the compression is multiplicative.

  • Holevo, Alexander S. Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd ed. De Gruyter, 2019. doi:10.1515/9783110642490. Normal channels and their spatial representations.
  • Stinespring, W. Forrest. “Positive functions on C*-algebras.” Proceedings of the American Mathematical Society 6, 211–216 (1955). doi:10.1090/S0002-9939-1955-0069403-4. Original representation theorem and positivity construction.