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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 ...

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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 ...

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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.

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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 ...

5

The answer is no, as the following counter-example reveals. Let $\varepsilon\in(0,1)$ and define $$\rho_0 = \begin{pmatrix} \frac{1+\varepsilon}{2} & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & \frac{1-\varepsilon}{2} \end{pmatrix},\quad \rho_1 = \begin{pmatrix} \frac{1-\varepsilon}{2} & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & \frac{1+... 5 For arbitrary linear super-operators U_j and V_j\def\D{\mathrm{Diamond}} \def\Dn#1{\lVert #1 \rVert_\diamond}\def\le{\leqslant} , we have$$\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 $\... 4 Both definitions are used and authors usually make it clear which one they mean. Wikipedia also points this out under the Alternative Defintion section. 4 Here's a concrete example for a single qubit. We can always change the basis to have$|\psi\rangle=|0\rangle$. Let us further suppose that$\langle0|\rho|0\rangle=0$, so that $$\rho=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$ The requirement$\operatorname{Tr}[(\sigma-\rho)|\psi\rangle\!\langle\psi|]=\langle\psi|\sigma-\rho|\psi\rangle=\epsilon$then ... 4 In general, it would seem no. The quantity $$\mathrm{Tr}[(\rho - \sigma)|\psi\rangle\langle\psi|]$$ is only concerned with the distance between$\rho$and$\sigma$on the subspace$\mathrm{span}(|\psi\rangle)$. For example, we know we can decompose the Hilbert space as$\mathcal{H} = \mathrm{span}(|\psi\rangle) \oplus \mathrm{span}(|\psi\rangle)^{\perp}$. ... 4 No. Just take two Bell states. They have identical reduced density matrices yet are orthogonal, that is, as distant from each other as it gets. 3 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^\...

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