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Consider an arbitary positive semidefinite operator ρ, acting on ℂA ⊗ ℂB ⊗ ℂC, for A,B,C finite. Also, let P be an orthogonal projector on ℂB ⊗ ℂC . For the sake of concision, I will write R = 1A ⊗ P ; this of course is also an orthogonal projector. Consider the completely positive transformation

M(ρ) = (1 − R) ρ (1 − R) + R ρ R .

As R is an orthogonal projector, it is easy to show that || M(ρ) ||2 ≤ || ρ ||2 . This is because we may represent ρ as matrix in a basis consisting of the eigenvectors of R; if we divide ρ into block according to rows/columns representing vectors in the image or the kernel of R, the effect of the map M is to set the non-diagonal blocks to zero.

I am interested in how the map M may similarly affect the operator 2 norm of reduced operators on ℂA ⊗ ℂB. So I would like to know:

Is it also true that || trC( M(ρ) ) ||2 ≤ || trC(ρ) ||2  —  where trC is the trace operator acting on ℂC, taken in tensor product with 1A ⊗ 1B ?

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To clarify, what precisely is the trace operator on $\mathbb{C}^{C}$? Do you mean the trace on $M_{|C|}(\mathbb{C})$? If so, is this trace normalized so that the trace of the identity in $M_{|C|}(\mathbb{C})$ is 1? – Jon Bannon Oct 26 '10 at 16:14
@Jon Bannon: close... it maps any rank-1 projector to 1. That is to say, it is the first-year-university trace operation, concieved as a positive map $\mathop{tr}:M_{∣C∣}(\mathbb C) \to M_1(\mathbb C)$. – Niel de Beaudrap Oct 27 '10 at 5:58
up vote 1 down vote accepted

Okay, it turns out in retrospect that the problem is trivial. The answer is "no": such a bound does not hold in general.

A simple counterexample is yielded by taking A=1 (so that we effectively deal with ℂB ⊗ ℂC throughout), B=C=2, and taking

P = ½ ψψ*  where ψ = e1 ⊗ e2 − e2 ⊗ e1

for standard basis vectors ej for ℂ2. Note that P is the projector onto the antisymmetric subspace of ℂB ⊗ ℂC. The map M may then be re-presented as

M(ρ) = ½ ρ + ½ UρU*   where U = 1B ⊗ 1C − 2P.

The operator U is unitary, and has the effect of 'swapping' the two spaces B and C; that is, for all tensor products α ⊗ β , we have U(α ⊗ β) = β ⊗ α . We may then construct an operator ρ for which the desired bound does not hold, by taking a tensor product of an operator with low 2-norm with one of high 2-norm, e.g.

ρ = 1Be1e1*.

We then have trC(ρ) = 1B , which has a 2-norm of $\sqrt 2$ ; and trC(UρU*) = 2 e1e1*, which has a 2-norm of $2$. By the convexity of the 2-norm, we may then show that || trC( M(ρ) ) ||2 > || trC(ρ) ||2 for this choice of P and ρ. A similar construction can be made for any B=C>1, and letting P be the projector onto the antisymmetric space of ℂB ⊗ ℂC .

I'm interested now in what upper bounds may be obtained for || trC( M(ρ) ) ||2 − || trC(ρ) ||2 , or related quantities, in the case that P is a rank-1 projector on ℂB ⊗ ℂC . If anyone can show such an interesting such bound, I may 'accept' it; but for the meantime, this answers my original question.

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