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5 votes

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 &...
Adam Zalcman's user avatar
  • 22.9k
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 ...
Mateus Araújo's user avatar
3 votes

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 $$ \sum_i \lambda_i = 1 $$ ...
Rammus's user avatar
  • 5,843
2 votes

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 ...
Adam Zalcman's user avatar
  • 22.9k
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 \...
KAJ226's user avatar
  • 13.9k
1 vote

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's user avatar
  • 25.2k

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