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9 votes
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Positive semidefinite relationship after partial trace

No, not necessarily. For example, take $\rho$ to be a GHZ state and let $\sigma$ be the completely mixed state of one qubit. We then have $\lambda=4$ and $\mu=2$.
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  • 4,423
9 votes
Accepted

What's the 'physical consistency' in the partial trace scenario?

Measurement average Measurement average $\langle M \rangle_\rho$ of observable (a Hermitian operator) $M$ on the state $\rho$ is the average of measurement outcomes $m$ in the limit of infinite number ...
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8 votes
Accepted

Partial trace over a product of matrices - one factor is in tensor product form

The equation at the top of the question is not correct: there is a missing factor of $1/d$ on the right-hand side. Let's eliminate this factor from the left-hand side to make it simpler, so that the ...
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  • 4,423
7 votes

Partial trace over a product of matrices - one factor is in tensor product form

Here the important fact is that the maximally mixed state is in fact an identity matrix. Let me rewrite the expression on the left in index notation (the summation sign is omitted according to the ...
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6 votes

How can we upper bound the norm of a partial trace?

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 \...
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  • 3,837
6 votes
Accepted

Is the trace distance between multipartite states invariant under permutations?

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.
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6 votes
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If the partial traces $\rho_A,\rho_B$ are pure, does it imply that $\rho$ is a product state?

Yes; in fact, $\rho$ is both separable and pure. We can start by writing any state $\rho$ in its eigenbasis $$\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|,$$ where $p_i$ are probabilities (i.e., ...
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6 votes

Prove that $\rho_{AB} \leq |B|(\rho_A\otimes I_B)$ for any bipartite state $\rho_{AB}$

There's a very quick proof if you can use the properties of the Choi-Jamiołkowski isomorphism. Define a map that acts on subsystem $B$ as $$\Lambda(\rho_B) = \mathrm{Tr}(\rho_B) |B| I_B - \rho_B.$$ ...
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  • 86
5 votes

Is the trace distance between multipartite states invariant under permutations?

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 ...
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5 votes
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Fidelity of extensions of states

Let's start with the second question. There is nothing special about an extension $\sigma_{AR}^{\ast}$ that allows it to be optimal for the right-hand side of (1); any extension $\sigma_{AR}$ of $\...
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  • 4,423
5 votes

What's the 'physical consistency' in the partial trace scenario?

The point of physical consistency is about how we can define the operator $M$ as acting on system $\rho^A$, or we can define the operator $M \otimes \mathbb{1}_B$ as acting on the system $\rho^{AB}$, ...
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5 votes
Accepted

Can we express $\mathrm{tr}_A((A\otimes B)\rho_{AB})$ in terms of $A$, $B$, $\rho_A$ and $\rho_B$?

In general, the knowledge of the marginals $\rho_A$ and $\rho_B$ and the operators $A$ and $B$ is insufficient to compute $\mathrm{tr}_A((A\otimes B)\rho_{AB})$. Indeed, we can find two different ...
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4 votes
Accepted

How does $\mathcal E(\rho)=\mathrm{Tr}_{env}[U(\rho\otimes\rho_{env})U^\dagger]$ turn into $P_0\rho P_0+P_1\rho P_1$?

Let's start with a general state $$ \rho\otimes\rho_0=\sum_{x,y\in\{0,1\}}\langle x|\rho|y\rangle|x\rangle\langle y|\otimes |0\rangle\langle 0|. $$ If we apply the controlled-not, we have $$ \...
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  • 46.4k
4 votes
Accepted

How is the partial trace related to the operator sum representation?

I think it helps here to write things explicitly. Suppose $\mathcal E(\rho)=\operatorname{Tr}_E[U(\rho\otimes|\mathbf e_0\rangle\!\langle\mathbf e_0|)U^\dagger]$. Pick a basis for the environment in ...
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  • 18.3k
4 votes
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Does the trace distance between marginals bound the distance between the overall states?

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.
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4 votes

Can SWAP operators change trace of a product state?

The SWAP operator has been widely used to ``linearize'' polynomial functions of the density matrix. To understand this carefully, consider the following setup: Let $\mathcal{H} = \mathcal{H}_{A} \...
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4 votes
Accepted

Is the partial trace $\mathrm{Tr}_B(\rho)$ equal to $\sum_k \mathrm{Tr}[(\sigma_k\otimes I)^\dagger \rho]\sigma_k$?

They are exactly the same. Remember that you can write $$ \rho=\sum_{i,j}\rho_{ij}\sigma_i\otimes\sigma_j. $$ If you take the partial trace, you have $$ \rho_A=\sum_{i,j}\rho_{ij}\sigma_i \text{Tr}(\...
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  • 46.4k
4 votes
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Non-lockability of quantum max-entropy

Let $D_{\alpha}(\rho\|\sigma):= \frac{1}{\alpha - 1} \log \mathrm{Tr}[\rho^\alpha \sigma^{1-\alpha}]$ be the Petz-Rényi divergence for $\alpha \in (0,1)\cup(1,\infty)$. Note that for $\alpha \in (0,1)\...
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  • 3,837
4 votes
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Bob applies a projector - what happens to eigenvalues of Alice's reduced state?

Just notice that $$ \text{Tr}(\rho'_{AB}) = \text{Tr}(\Pi_B\rho_{B}). $$ One way to see this is to consider any decomposition $$ \rho_{AB} = \sum_i A_i \otimes B_i, $$ where $A_i, B_i$ just some ...
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  • 5,453
4 votes
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Partial Trace of Werner State

$\frac{s}{2}|0\rangle\langle1|\otimes|0\rangle\langle1|+\frac{s}{2}|1\rangle\langle0|\otimes|1\rangle\langle0|$ should disappear when you take the trace over them, as $\langle0|1\rangle$ and $\langle1|...
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  • 1,113
4 votes
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If $\text{supp}(\rho_{AB})\subset \text{supp}(\sigma_{AB})$, is $\text{supp}(\rho_{A})\subset \text{supp}(\sigma_{A})$?

If $\rho_{AB}, \sigma_{AB} \ge 0$, i.e. they are positive semi-definite then yes (otherwise no). To prove it observe that if $\text{supp}(\rho_{AB})\subset \text{supp}(\sigma_{AB})$ then we can write ...
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  • 5,453
4 votes
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What is the physical intuition behind taking the partial trace of a state?

TL;DR Tracing out a subsystem corresponds to discarding it. Suppose Alice has subsystem $A$ and Bob has subsystem $B$ of the composite system $AB$ in state $\rho_{AB}$. Tracing out subsystem $B$ ...
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4 votes

Why does $\rho^A=\mathrm{tr}_B(\rho^{AB})$ guarantee that $\mathrm{tr}(M\rho^A)=\mathrm{tr}((M\otimes I_B)\rho^{AB})$?

The equality follows by hitting both sides of $$ X_A\mathrm{tr}_B(Y_{AB})=\mathrm{tr}_B((X_A\otimes I_B)Y_{AB})\tag1 $$ with the trace and setting $X_A=M$ and $Y_{AB}=\rho^{AB}$. We can establish $(1)...
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3 votes

Given $\rho,\sigma$ such that $F(\rho,\sigma)=0$, what can we say about $F(\text{Tr}_A(\rho),\text{Tr}_B(\sigma))$?

The Bell states are orthogonal and their partial trace is equal - the maximally mixed states - and thus indistinguishable. So yes, this is the maximum you can achieve. Another extremal example could ...
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3 votes
Accepted

Fidelity With Bell State Calculation

No, you wouldn't find $0.9$ again. To make the partial trace calculation simpler you can note that the state $|\psi'\rangle$ is separable under the bipartition $a_1b_1 | a_2 b_2$, i.e. $|\psi'\rangle =...
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  • 3,837
3 votes
Accepted

Which simple property of partial trace are we using here?

mixed-product property of the Kronecker product: $(A \otimes B)(C \otimes D) = (AC \otimes BD)$ definition of the partial trace: $\text{Tr}_1(A\otimes B) = \text{Tr}(A)B$ cyclic property of the ...
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  • 5,453
3 votes

How does $\mathcal E(\rho)=\mathrm{Tr}_{env}[U(\rho\otimes\rho_{env})U^\dagger]$ turn into $P_0\rho P_0+P_1\rho P_1$?

Let $\newcommand{\rmD}{\mathrm{D}}\newcommand{\rmU}{\mathrm{U}}\newcommand{\calU}{\mathcal{U}}\newcommand{\CC}{\mathbb{C}}\rho\in\rmD(\CC^n),\sigma\in\rmD(\CC^m)$ be two arbitrary finite-dimensional ...
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  • 18.3k
3 votes

Prove that the partial trace is equivalent to measuring and discarding

Let me give you the structure for how you go about answering this question. Let $|\Psi\rangle$ be the pure state shared between Alice and Bob. Bob measures his system in an orthonormal basis $\{|\...
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  • 46.4k
3 votes
Accepted

Tensor product properties used to obtain Kraus operator decomposition of a channel

As pointed out in a comment, what you wrote as $\rho$ should more precisely be written as $\rho\otimes\mathbb 1$ (although the Kraus operator decomposition can be obtained similarly with any initial ...
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  • 18.3k
3 votes

How many Kraus operators are required to characterise a channel with different start and end dimensions?

The answer is yes, you need $d_1 d_2$ operators, as already pointed out in the other answer. Here I'll show explicitly why this is the case. Let $\Phi\in\mathrm{T}(\mathcal X,\mathcal Y)$ be a CPTP ...
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