Bounds on operator 2-norms on partial traces of linearly related operators - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T01:40:24Z http://mathoverflow.net/feeds/question/43673 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43673/bounds-on-operator-2-norms-on-partial-traces-of-linearly-related-operators Bounds on operator 2-norms on partial traces of linearly related operators Niel de Beaudrap 2010-10-26T14:54:47Z 2010-10-27T14:10:46Z <p>Consider an arbitary positive semidefinite operator &rho;, acting on &#8450;<sup>A</sup>&nbsp;&otimes;&nbsp;&#8450;<sup>B</sup>&nbsp;&otimes;&nbsp;&#8450;<sup>C</sup>, for A,B,C finite. Also, let P be an orthogonal projector on &#8450;<sup>B</sup>&nbsp;&otimes;&nbsp;&#8450;<sup>C</sup>&nbsp;. For the sake of concision, I will write R&nbsp;=&nbsp;1<sub>&#8450;<sup>A</sup></sub>&nbsp;&otimes;&nbsp;P&nbsp;; this of course is also an orthogonal projector. Consider the completely positive transformation</p> <blockquote> <p>M(&rho;)&ensp;=&ensp;(1&thinsp;&minus;&thinsp;R)&thinsp;&rho;&thinsp;(1&thinsp;&minus;&thinsp;R)&ensp;+&ensp;R&thinsp;&rho;&thinsp;R .</p> </blockquote> <p>As R is an orthogonal projector, it is easy to show that ||&nbsp;M(&rho;)&nbsp;||<sub>2</sub>&ensp;&le;&ensp;||&thinsp;&rho;&thinsp;||<sub>2</sub>&nbsp;. This is because we may represent &rho; as matrix in a basis consisting of the eigenvectors of R; if we divide &rho; 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.</p> <p>I am interested in how the map M may similarly affect the operator 2 norm of reduced operators on &#8450;<sup>A</sup>&nbsp;&otimes;&nbsp;&#8450;<sup>B</sup>. So I would like to know:</p> <p>Is it also true that ||&nbsp;tr<sub>C</sub>(&nbsp;M(&rho;)&nbsp;)&nbsp;||<sub>2</sub>&ensp;&le;&ensp;||&nbsp;tr<sub>C</sub>(&rho;)&nbsp;||<sub>2</sub> &nbsp;&mdash;&nbsp; where tr<sub>C</sub> is the trace operator acting on &#8450;<sup>C</sup>, taken in tensor product with 1<sub>&#8450;<sup>A</sup></sub>&nbsp;&otimes;&nbsp;1<sub>&#8450;<sup>B</sup></sub>&nbsp;?</p> http://mathoverflow.net/questions/43673/bounds-on-operator-2-norms-on-partial-traces-of-linearly-related-operators/43783#43783 Answer by Niel de Beaudrap for Bounds on operator 2-norms on partial traces of linearly related operators Niel de Beaudrap 2010-10-27T10:35:31Z 2010-10-27T14:10:46Z <p>Okay, it turns out in retrospect that the problem is trivial. The answer is "no": such a bound does not hold in general.</p> <p>A simple counterexample is yielded by taking A=1 (so that we effectively deal with ℂ<sup>B</sup>&nbsp;⊗&nbsp;ℂ<sup>C</sup> throughout), B=C=2, and taking</p> <blockquote> <p>P = &frac12;&thinsp;<strong>&psi;</strong><strong>&psi;*</strong> &emsp;where <strong>&psi;</strong> = <strong>e</strong><sub>1</sub>&thinsp;&otimes;&thinsp;<strong>e</strong><sub>2</sub>&ensp;&minus;&ensp;<strong>e</strong><sub>2</sub>&thinsp;&otimes;&thinsp;<strong>e</strong><sub>1</sub></p> </blockquote> <p>for standard basis vectors <strong>e</strong><sub>j</sub> for ℂ<sup>2</sup>. Note that P is the projector onto the antisymmetric subspace of ℂ<sup>B</sup>&nbsp;⊗&nbsp;ℂ<sup>C</sup>. The map M may then be re-presented as</p> <blockquote> <p>M(&rho;)&ensp;=&ensp;&frac12;&thinsp;&rho;&ensp;+&ensp;&frac12;&thinsp;U&rho;U* &emsp; where U&ensp;=&ensp;1<sub>ℂ<sup>B</sup></sub> ⊗ 1<sub>ℂ<sup>C</sup></sub>&ensp;&minus;&ensp;2P.</p> </blockquote> <p>The operator U is unitary, and has the effect of 'swapping' the two spaces B and C; that is, for all tensor products <strong>&alpha;</strong>&nbsp;&otimes;&nbsp;<strong>&beta;</strong>&nbsp;, we have U(<strong>&alpha;</strong>&nbsp;&otimes;&nbsp;<strong>&beta;</strong>)&nbsp;=&nbsp;<strong>&beta;</strong>&nbsp;&otimes;&nbsp;<strong>&alpha;</strong>&nbsp;. We may then construct an operator &rho; 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.</p> <blockquote> <p>&rho;&ensp;=&ensp;1<sub>ℂ<sup>B</sup></sub> &otimes; <strong>e</strong><sub>1</sub><strong>e</strong><sub>1</sub>*.</p> </blockquote> <p>We then have tr<sub>C</sub>(&rho;) = 1<sub>ℂ<sup>B</sup></sub>&nbsp;, which has a 2-norm of $\sqrt 2$&thinsp;; and tr<sub>C</sub>(U&rho;U*)&nbsp;=&nbsp;2&thinsp;<strong>e</strong><sub>1</sub><strong>e</strong><sub>1</sub>*, which has a 2-norm of $2$. By the convexity of the 2-norm, we may then show that ||&nbsp;tr<sub>C</sub>(&nbsp;M(ρ)&nbsp;)&nbsp;||<sub>2</sub>&nbsp;&gt;&nbsp;||&nbsp;tr<sub>C</sub>(ρ)&nbsp;||<sub>2</sub> for this choice of P and &rho;. A similar construction can be made for any B=C>1, and letting P be the projector onto the antisymmetric space of ℂ<sup>B</sup>&nbsp;⊗&nbsp;ℂ<sup>C</sup>&nbsp;.</p> <p>I'm interested now in what upper bounds may be obtained for ||&nbsp;tr<sub>C</sub>(&nbsp;M(ρ)&nbsp;)&nbsp;||<sub>2</sub>&nbsp;&minus;&nbsp;||&nbsp;tr<sub>C</sub>(ρ)&nbsp;||<sub>2</sub>&nbsp;, or related quantities, in the case that P is a rank-1 projector on ℂ<sup>B</sup>&nbsp;⊗&nbsp;ℂ<sup>C</sup>&nbsp;. If anyone can show such an interesting such bound, I may 'accept' it; but for the meantime, this answers my original question.</p>