Ex10 10
$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 10.10}$
For $\ket\psi \in A\otimes B$, show that $\ket\psi$ is unentangled if and only if $S(\tr_B\rho) = 0$, where $\rho = \ket\psi\bra\psi$.
page revision: 0, last edited: 18 Nov 2012 19:48
Let $\ket{\psi} \in A \otimes B$ be a pure state with density operator $\rho = \ket{\psi}\bra{\psi}$.
(1)The von Neumann entropy over the partial trace is
where $\lambda_i$ are the Schmidt coefficients in the Schmidt decomposition $\ket{\psi} = \sum_{i=0}^{K-1} \lambda_i \ket{\psi_i^A}\otimes \ket{\psi_i^B}$.
If $\ket{\psi}$ is unentangled, then the Schmidt decomposition only has one term. That is, one $\lambda_i$ is 1 and the rest are 0. With the convention $0\log_2(0) = 0$, this implies $S = 0$.
The Schmidt coefficients $\lambda_i$ are positive and satisfies $\sum \lambda_i^2 = 1$. Hence, if more than one coefficient is non-zero, then the non-zero coefficients are less than 1 and $S > 0$.
So, if $S=0$, then only one coefficient is non-zero and must equal 1. This means, the Schmidt sum only has one term and $\ket{\psi}$ is unentangled.