## Friday, 31 January 2014

### The SWAP operator and separable measurements

Let $\Sigma$ be an alphabet, let $n = \abs{\Sigma}$, and assume $n\geq 2$. Also let $\X^{A} = \mathbb{C}^{\Sigma}$ and $\X^{B} = \mathbb{C}^{\Sigma}$,  and recall that the swap operator $W\in\Lin(\X^{A}\otimes\X^{B})$ may be defined as
$W = \sum_{a,b\in\Sigma} E_{a,b} \otimes E_{b,a}.$
Define $\Pi_0,\,\Pi_1\in\Proj(\X^{A}\otimes\X^{B})$ and $\sigma_0,\sigma_1\in\Density(\X^{A}\otimes\X^{B})$ as follows:
$\Pi_0 = \frac{1}{2} \I\otimes\I + \frac{1}{2} W,\qquad \Pi_1 = \frac{1}{2} \I\otimes\I - \frac{1}{2} W,\qquad \sigma_0 = \frac{1}{\binom{n+1}{2}}\Pi_0,\qquad \sigma_1 = \frac{1}{\binom{n}{2}}\Pi_1.$

Theorem:

If $\mu:\{0,1\}\rightarrow\Pos(\X^{A}\otimes\X^{B})$ is a separable measurement, then
$\frac{1}{2} \ip{\mu(0)}{\sigma_0} + \frac{1}{2} \ip{\mu(1)}{\sigma_1} \leq \frac{1}{2} + \frac{1}{n+1}.$

Proof:

Assuming that $\mu$ is a separable measurement allows $\mu(0)$ to be expressed as
$\mu(0)=\sum_{a\in\Gamma}P_a\otimes Q_a,$
where $\{P_a : a\in \Gamma\}\subset \Pos(\X^A)$ and $\{Q_a : a\in \Gamma\}\subset \Pos(\X^B)$. Moreover, since $\mu$ is a measurement it must satisfy the completeness condition that $\mu(0)+\mu(1)=\I\otimes\I$ implying that $\mu(1)$ can be expressed in terms of $\mu(0)$ as
$\mu(1)=\I\otimes\I-\mu(0)=\I\otimes\I-\sum_{a\in\Gamma}P_a\otimes Q_a.$
Write $\sigma_0$ and $\sigma_1$ more explicitly as
\begin{align*} \sigma_0 &= \frac{1}{\binom{n+1}{2}}\Pi_0=\frac{1}{(n+1)n}(\I\otimes\I + W), \\ \sigma_1 &= \frac{1}{\binom{n+1}{2}}\Pi_0=\frac{1}{(n-1)n}(\I\otimes\I - W). \end{align*}
Then,
\begin{align*} \ip{\mu(0)}{\sigma_0} + \ip{\mu(1)}{\sigma_1} &= \ip{\mu(0)}{\sigma_0} + \ip{\I\otimes\I-\mu(0)}{\sigma_1} \\ &=\frac{1}{(n+1)n}\ip{\mu(0)}{\I\otimes\I + W} + \frac{1}{(n-1)n}\ip{\I\otimes\I-\mu(0)}{\I\otimes\I - W} \\ &=\frac{1}{(n+1)n}(\ip{\mu(0)}{\I\otimes\I }+\ip{\mu(0)}{ W}) \\ & \ \ \ + \frac{1}{(n-1)n}(\ip{\I\otimes\I}{\I\otimes\I} -\ip{\I\otimes\I}{W} ) \\ & \ \ \ + \frac{1}{(n-1)n}(\ip{\mu(0)}{W}-\ip{\mu(0)}{\I\otimes\I} ). \end{align*}
Now observe that
$\ip{\mu(0)}{\I\otimes\I }=\tr\left(\mu(0)^\ast\I\otimes\I\right)=\tr(\sum_{a\in\Gamma}(P_a\otimes Q_a)(\I\otimes\I))=\tr(\sum_{a\in\Gamma}P_a\otimes Q_a)=\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a),$
$\ip{\I\otimes\I}{\I\otimes\I}=\tr((\I\otimes\I)(\I\otimes\I))=\tr(\I\otimes\I)=n^2,$
\begin{align*} \ip{\mu(0)}{ W}=\tr(\sum_{a\in\Gamma}(P_a\otimes Q_a)W)&=\sum_{a\in\Gamma}\sum_{i,j\in\Sigma}(e_i^\ast\otimes e_j^\ast)(P_a\otimes Q_a)W(e_i\otimes e_j) \\ &=\sum_{a\in\Gamma}\sum_{i,j\in\Sigma}(e_i^\ast\otimes e_j^\ast)(P_a\otimes Q_a)(e_j\otimes e_i) \\ &=\sum_{a\in\Gamma}\sum_{i,j\in\Sigma}(e_i^\ast P_a e_j)\otimes(e_j^\ast Q_a e_i) \\ &=\sum_{a\in\Gamma}\sum_{i,j\in\Sigma}e_i^\ast P_a e_je_j^\ast Q_a e_i \\ &=\sum_{a\in\Gamma}\sum_{i\in\Sigma}e_i^\ast P_a Q_a e_i \\ &=\sum_{a\in\Gamma}\tr(P_a Q_a), \\ \end{align*}
and by similar arguments used in the previous calculation $\ip{\I\otimes\I}{W}=\tr(\I\I)=\tr(\I)=n$.
Therefore, the original expression of interest can be simplified as
\begin{align*} \ip{\mu(0)}{\sigma_0} + \ip{\mu(1)}{\sigma_1}&=\frac{1}{(n+1)n}(\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a)+\sum_{a\in\Gamma}\tr(P_a Q_a)) \\ & \ \ \ + \frac{1}{(n-1)n}(n^2 -n ) \\ & \ \ \ + \frac{1}{(n-1)n}(\sum_{a\in\Gamma}\tr(P_a Q_a)-\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a) ) \\ &=1+\frac{2}{(n+1)(n-1)n}\left(n\sum_{a\in\Gamma}\tr(P_a Q_a)-\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a)\right) \\ &\leq 1+\frac{2}{(n+1)(n-1)n}(n^2-n) \\ &=1+\frac{2}{(n+1)}. \end{align*}
Here, the inequality follows from the fact that the quantity $\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a)$ is minimized when the projectors $P_a$ and $Q_a$ both have rank $1$ so that $\tr(P_a)\tr(Q_a)=1$ implying that $\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a)=n$.
Furthermore, in this case,  $\sum_{a\in\Gamma}\tr(P_a Q_a)=n$.

Therefore, dividing both sides of the inequality by $2$ gives
$\frac{1}{2} \ip{\mu(0)}{\sigma_0} + \frac{1}{2} \ip{\mu(1)}{\sigma_1} \leq \frac{1}{2} + \frac{1}{n+1}.$