# Tag Info

8

Marsl is correct, and his "hint" is really more a sketch of a solution than a hint. Rather than viewing the question or its solution as just formal algebra, you can also approach his same solution more conceptually. The conceptual reasoning is really identical to the algebra, just phrased differently. You can rely on the following two facts: 1) Trace ...

7

Hint: To make your induction work, write \eqalign{p^{\otimes n} - q^{\otimes n} & = & \left(p^{\otimes(n-1)}\otimes p \right)-\left(q^{\otimes (n-1)} \otimes q\right)\\ & = & \left(p^{\otimes(n-1)}-q^{\otimes (n-1)} \right)\otimes p+\left(q^{\otimes (n-1)} \right) \otimes (p-q)} Then, use triangle inequality and finally the fact that ...

6

Yes, since the trace norm is the sum of the absolute value of the singular values, and the singular values can be found for each of the $a$ blocks independently.

5

For arbitrary linear super-operators $U_j$ and $V_j\def\D{\mathrm{Diamond}} \def\Dn#1{\lVert #1 \rVert_\diamond}\def\le{\leqslant}$, we have \def\D{\mathrm{Diamond}} \def\Dn#1{\lVert #1 \rVert_\diamond}\def\le{\leqslant} \begin{aligned} \D(U_1 U_2, V_1 V_2) &= \Dn{U_1 U_2 - V_1 V_2} \\&\le \Dn{U_1 U_2 - V_1 U_2} + \Dn{V_1 U_2 - V_1 V_2} \\&= ... 5 There is a geometric interpretation that you certainly can take seriously, but the geometry that you get is not as clean as you might have hoped. Trace distance between operator states is an example of a Banach norm on a vector space V. The rules for such a norm are that ||v|| > 0 when 0 \ne v \in V, ||\lambda v|| = |\lambda|\cdot||v|| for \... 2 For the way round that you've got your inequalities, I don't think there's much that can be said. To see why, let's consider the first expression \|U-V\|_1=\text{Tr}(\sqrt{2I-VU^\dagger-UV^\dagger}). $$Now, VU^\dagger is a unitary, and hence as a spectral decomposition. Let the eigenvectors be |\lambda_i\rangle with eigenvalues e^{i\lambda_i}. UV^\... 2 Short answer. The trace distance between two states more or less determines how distinguishable they are by any operational means. A trace distance of 0 means that they are indistinguishable (because they're equal); a trace distance of 2 indicates that they can be perfectly distinguished in principle. Long answer. We will show how, from the objective of ... 2 One way to understand the trace distance is to notice that it equals the (classical) trace distance (also referred to as Kolmogorov distance, see this post for some information about it) maximised over all possible POVMs on the states. To see this, start from the following expression for the trace distance:$$D(\rho,\sigma)\equiv\frac{1}{2}\mathrm{Tr}|\rho-\...

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