8

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 ...


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 ...


8

No dimension-independent bound is possible. Consider states $\rho_A$ and $\sigma_A$ that are close in $p$-norm (for $p>1$) but have relatively low fidelity. Specifically, assume $$ \|\rho_A - \sigma_A\|_p = \varepsilon $$ and $$ \operatorname{F}(\rho_A,\sigma_A) = \bigl\|\sqrt{\rho_A}\sqrt{\sigma_A}\bigr\|_1 = \delta, $$ where $\varepsilon$ is small and $...


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 ...


7

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 $\...


7

We can bound the amount of information that can be retrieved from $|\psi\rangle$ using Holevo's bound. Alice and Bob Let us first reformulate the situation in the terms usually employed in the context of Holevo's bound. Suppose Alice chooses two $n$-bit strings $x_0$ and $x_1$ independently and uniformly at random. Let $X$ represent the random variable ...


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

The $1$-norm decreases under partial trace and so we have an upper bound of $1$ when the states are normalized, $$ \|\mathrm{Tr}_B[|\psi_1\rangle \langle \psi_2|]\|_1 \leq \||\psi_1\rangle \langle \psi_2|\|_1 = 1. $$ This bound cannot be improved upon without extra information about the states. Here is a counterexample. Take $|\psi_1 \rangle = |00\rangle$ ...


6

A permutation of the qubits is a unitary operation. The trace distance is invariant under unitaries (https://en.wikipedia.org/wiki/Trace_distance#Properties). Thus, statement 1 is true.


6

Observe that, for any collection of matrices $A_i$, we have $$\sqrt{\sum_i |i\rangle\!\langle i|\otimes A_i} = \sum_i |i\rangle\!\langle i|\otimes \sqrt{A_i}, \\ {\rm Tr}\left(\sum_i |i\rangle\!\langle i|\otimes A_i\right) = \sum_i {\rm Tr}(A_i).$$ It immediately follows that $\|\sqrt\rho\sqrt\sigma\|_1\equiv {\rm Tr}|\sqrt\rho\sqrt\sigma|$ can be written as ...


5

There are different ways to prove what you want to prove, including the solution tsgeorgios has suggested, but for the sake of gaining greater intuition I would suggest starting with the recognition that the trace norm of any matrix is equal to the sum of its singular values. Once you have this, the inequality you are trying to prove follows pretty easily. ...


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

I'd like to add a small addition to the answer of @DaftWullie about why you would expect this operationally to be true -- without knowing permutations correspond to unitary matrices. It boils down to the Holevo-Helstrom Theorem (HHT) that says the trace distance between two states characterizes operationally the probability that we can distinguish the two ...


5

Recall that for any Hermitian operator $A$ and any unit vector $|\psi\rangle$ the real number $\langle \psi|A|\psi\rangle$, known as the Rayleigh quotient, is bounded by the largest eigenvalue $\lambda_{max}$ of $A$ $$ \langle \psi|A|\psi\rangle \le \lambda_{max}. $$ Moreover, the maximum is achieved when $|\psi\rangle$ is the unit norm eigenvector of $A$ ...


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.


4

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 $...


3

After @Rammus answer explaining that the given inequality does not hold in general, i'll try to prove a weaker statement. Define $ \Delta = V - \tilde{V} $. The assumption is equivalent to $$ \text{Tr}[I - \tilde{V}^{\dagger} V] \leq \epsilon \cdot |D| \implies \text{Tr}[V^{\dagger}V - \tilde{V}^{\dagger} V] \leq \epsilon \cdot |D| \implies \text{Tr}[\...


3

I'll use $W$ instead of $\tilde V$. I believe there must be more constraints in the paper as to relation between $W$ and $V$ because as it stands the inequality is not true. For example take $V$ to be the identity matrix and let $W$ also be the identity matrix except we make the penultimate and final elements on the diagonal of $W$ be $e^{i \alpha}$ and $e^{-...


3

We still have $ \big| \langle B, A \rangle \big| = \big|\text{Tr}(AB^{\dagger}) \big| \leq \text{Tr}|A| $ for any operator $B$ with operator norm $ ||B|| \leq 1 $. First observe that $ ||B|| \leq 1 $ implies that for any positive semi-definite operator $ Q = \sum_i \lambda_i |\psi_i \rangle\langle \psi_i| \in L(X, X) $: $$ \text{Tr}(Q) = \sum_i \lambda_i \...


3

Yes, the trace distance can only decrease under partial trace. One can see this via the variational characterization of the trace norm $$ \|\rho\|_1 = \max_{-I \leq M \leq I} \mathrm{Tr}[M\rho] $$ where $M$ is some hermitian operator satisfying the two operator inequalities $M \leq I$ and $M \geq - I$. This is sometimes also known as the duality between ...


3

If perturbations are sufficiently small and $\rho_{AB}$ has sufficiently broad support then a desired global state $\rho_{AB}'$ exists. Define $$ \rho_{AB}' = \rho_{AB} + (\rho_A' - \rho_A) \otimes \rho_B + \rho_A \otimes (\rho_B' - \rho_B) \tag1. $$ Note that $\rho_{AB}'$ is Hermitian and trace one, but may not be positive. However, $\rho_{AB}'$ is positive ...


3

This follows from the cyclicity of trace. \begin{align} \text{tr}\left(\left(\sqrt{\sqrt{B}}A\sqrt{\sqrt{B}}\right)^2\right) &= \text{tr}\left(\sqrt{\sqrt{B}}A\sqrt{\sqrt{B}}\sqrt{\sqrt{B}}A\sqrt{\sqrt{B}}\right)\\ &= \text{tr}\left(\sqrt{\sqrt{B}}A\sqrt{B}A\sqrt{\sqrt{B}}\right)\\ &= \text{tr}\left(A\sqrt{B}A\sqrt{\sqrt{B}}\sqrt{\sqrt{B}}\right)...


3

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 ...


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^\...


3

Note that if $p_{i}q_{i} = 0\,\,\forall i$, then for all $i$ either $p_{i} = 0$, $q_{i} = 0$, or both are $0$. Divide $\{i\} = \{1,\ldots,N\}$ into those $i$ for which these three different things happen: $\{N_{p_{i}}\} = \{i|p_{i} = 0\}$, $\{N_{q_{i}}\} = \{i|q_{i} = 0\}$, $\{N_{pq_{i}}\} = \{i|p_{i} = q_{i} = 0\}$. Then \begin{equation} \begin{split} \|P-...


3

You could try solving this numerically using semidefinite programming. We know the trace norm of an operator $X$ can be formulated as $$ \begin{aligned} \|X\|_1 &= \min_{Y,Z}\quad \frac12\mathrm{Tr}[Y+Z] \\ &\quad \mathrm{s.t.} \quad \begin{pmatrix} Y & X \\ X^* & Z \end{pmatrix} \geq 0. \end{aligned} $$ Furthermore, we can write your problem ...


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