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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 ...


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.


6

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^\...


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-\...


2

Use the fact that both distance measures are invariant under left- as well as right-multiplication (independently!) with an arbitrary unitary. This way, you can map $U$ to $I$ and $V$ to $R_z(\phi)$. Now (i) the matrices are both diagonal, making the trace distance trivial to compute, and (ii) the problem is only characterized by a single angle. This will ...


2

Let's start with expanding the calculation of $E$: $$ E(U,V)=\max_{|\psi\rangle}\sqrt{\langle\psi|(U-V)^\dagger(U-V)|\psi\rangle}. $$ Clearly, we want $|\psi\rangle$ to be the eigenvector with maximum eigenvalue of $$ 2I-V^\dagger U-U^\dagger V. $$ Let's note that If $|\psi\rangle$ is an eigenvector of $U^\dagger V$, then the eigenvalue must be of the form $...


2

Firstly, just because an operator $M$ is defined on a composite space, it does not mean that the operator itself has a tensor product structure. You need $$ M=\sum_{i,j,k,l}M_{ij,kl}|ij\rangle\langle kl| $$ Now let $$ O=\sum_{i,k}O_{i,k}|i\rangle\langle k|. $$ With these in place, you just calculate the two sides of the equation and see if they're the same. (...


2

A bound on the total variation distance Rammus already provided a short answer, but I'd like to elaborate a bit on why this is the case. This is basically the proof of theorem $9.1$ on page $405$ of Nielsen & Chuang. Note that they call the total variation distance the (classical) trace distance, to draw the connection to the quantum trace distance. For ...


2

$\newcommand{\bra}[1]{\langle #1\rvert}\newcommand{\braket}[2]{\langle #1\rvert #2\rangle}\newcommand{\ket}[1]{\lvert #1\rangle}\newcommand{\on}[1]{\operatorname{#1}}\newcommand{\ketbra}[2]{\lvert #1\rangle\!\langle #2\rvert} $Define $A\equiv \ketbra\phi\phi - \ketbra\psi\psi$. Being $A$ hermitian, $\on{Tr}(\sqrt{A^2})$ equals the sum of its singular values. ...


2

A derivation of this is given in Mark Wilde's book https://arxiv.org/abs/1106.1445 equation 9.173, pages 274-275.


1

I am no longer confused about this, since now I see in this equation we are already restricting to a subspace $||S\mathcal{E}(|u\rangle\langle u|)-S\mathcal{E}(|v\rangle\langle v|)|| = |||u\rangle\langle u|-|v\rangle\langle v|||$, and the contracting map has to contract every subspace.


1

This is not a saturation of the the Fuchs-van de Graaf upper bound, $F(\rho,\sigma)^2 = 1-d(\rho,\sigma)^2$, but rather the lower bound, $1 - \sqrt{F(\rho,\sigma)} \leq d(\rho,\sigma)$ (see, for example, here). Consider, $\rho = | \psi \rangle \langle \psi |$, a pure state and $\sigma = \frac{\mathbb{I}}{d}$ is the maximally mixed state (for a $d$-...


1

A way that's perhaps more compact (if less terse), we can write $$ \mathrm{Tr}_B(O_AM)=\mathrm{Tr}_B((O_A\otimes \mathbb 1_B)M)=\sum_i (\mathbb 1_A\otimes \langle i|_B) (O_A\otimes \mathbb 1_B)M(\mathbb 1_A\otimes |i\rangle_B)=\\=\sum_i O_A(\mathbb 1_A\otimes \langle i|_B) M(\mathbb 1_A\otimes |i\rangle_B)=O_A\mathrm{Tr}_B(M)$$ The only nontrivial thing I ...


1

Unfortunately $D_{\max}$ is not a continuous function and so functions built from it tend not to be continuous. For example consider consider the two states $$ \rho_{AB} = |00 \rangle \langle 00|, $$ and $$ \tau_{AB}(\epsilon) = (1-\epsilon) |00 \rangle \langle 00 | + \epsilon | 11\rangle \langle 11 |. $$ A quick calculation gives $I_{\max}(\rho_{AB}) = 1$ ...


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