6
votes
Why do completely positive maps satisfy ${\rm Tr}[\Psi(\rho)_++\Psi(-\rho)_+]\leq{\rm Tr}[\Psi(\rho_+)]+{\rm Tr}[\Psi((-\rho)_+)]$?
The inequality can be proved as follows.
First, let $P$ and $Q$ be any two positive semidefinite operators, and consider the operator $\Psi(P - Q)$. Because $\Psi$ is positive, this is a Hermitian ...
- 6,187
5
votes
Accepted
Is there an identity for the partial transpose of a product of operators?
Your suspicion is correct, even when $A=B$. Consider the Hilbert space of two qubits and let $^{T_A}$ denote the partial transpose with respect to one of them.
Suppose that
$$
A=B=\begin{pmatrix}
1 &...
- 24.9k
4
votes
In a bipartite system $AB$, why does the entanglement negativity $\mathcal{N}(\rho^{T_A})$ measure the entanglement between $A$ and $B$?
There is no good definition of what is an "amount of entanglement". We have some requirements, such as saying that a measure of entanglement must be convex and cannot increase under local ...
- 3,082
3
votes
Accepted
How is the expression $\frac{\|\rho^{T_B}\|-1}{2}$ obtained from the definition of negativity?
Firstly
$$
\mathrm{Tr}[\rho_{AB}^{T_B}] = 1
$$
as the transpose map is trace-preserving. As the trace of $\rho_{AB}^{T_B}$ is equal to the sum of its eigenvalues we have
$$
\sum_i \lambda_i = 1
$$
...
- 6,638
2
votes
Accepted
What is the logarithmic Negativity of the Werner state?
Hint 1: Logarithmic negativity is easily computed once we know the eigenvalues of $\rho_w^\Gamma$ - where $^\Gamma$ denotes the partial transpose - since then it becomes a matter of substitution into ...
- 24.9k
2
votes
How do I calculate Logarithmic Negativity for the given bipartite state?
If we defined the logarithmic negativity as $E_N(\rho)= \log_2 \|\rho^{\Gamma_A} \|_1$ then given that
$$\rho = \frac{1}{2} |0\rangle \langle0| \otimes |+\rangle \langle+| +\frac{1}{2} |+\rangle \...
- 14.1k
1
vote
Accepted
Why does the entanglement negativity equal (in magnitude) the sum of the negative eigenvalues?
This is also discussed in the paper linked above. The trace norm of $X$ is defined as the sum of the absolute values of the eigenvalues of $X$: $\|X\|_1=\sum_i \lvert\lambda_i\rvert$.
$\newcommand{\tr}...
glS♦
- 26.9k
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