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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 $\mu=2$.
• 4,423
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 ...
• 13k
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 ...
• 4,423

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 ...

• 46.4k
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 ...
• 18.3k
Accepted

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.
• 4,998

• 13k

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 ...
• 4,998
Accepted

• 46.4k
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 ...
• 18.3k
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 ...