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We have a measurement $M$ that distinguishs between $\rho_1$ and $\rho_0$, if it has three possible answers 1,2,3 and whenever it answers something different than 3 it's correct. $M$ succeeds with probability $p$ if for every $\rho s$ above it answer the correct answer with probability at least $p$.

Now I found the most efficient value of $p$ that $M$ succeeds to disntiguish between $|0\rangle\langle0|$ and $|+\rangle\langle+|$, but I am not sure how to contruct a three dimensional measurement that achieves this value of $p$.

Any hints, or references?


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up vote 1 down vote accepted

OK, so I understand correctly that you want for result $1$, you know to have $|0\rangle\langle0|$, for result $2$, you know to have $|+\rangle\langle+|$, and for range 3 you don't know?

Since for result $1$ you want to be sure that you don't have $|+\rangle\langle+|$, the only way to get that is if $M_1=\lambda|-\rangle\langle-|=\lambda\left(\begin{array}{cc}\frac12&-\frac12\\-\frac12&\frac12\end{array}\right)$ and analogously $M_2=\mu|1\rangle\langle1|=\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$ (of course $\lambda$ and $\mu$ must not be larger than $1$). Now $M_1+M_2+M_3=\mathbf{1}$, and therefore $M_3=\left(\begin{array}{cc}1-\frac\lambda2&\frac\lambda2\\\frac\lambda2&1-\frac\lambda2-\mu\end{array}\right)$. The eigenvalues of the latter are $\frac{1}{2} \left(\pm\sqrt{\lambda ^2+\mu ^2}-\lambda -\mu +2\right)$. This is non-negative if $\lambda\le (2-2\mu)/(2-\mu)$. Now you want to maximize $\min(\lambda,\mu)$, which is maximal for $\lambda=\mu=2-\sqrt{2}$.

Thus the optimal measure should be $M_1=(2-\sqrt{2})|-\rangle\langle-|$, $M_2=(2-\sqrt{2})|1\rangle\langle1|$ and $M_3=\mathbf{1}-M_1-M_2$.

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Ok, but how do I actually construct this measurement apparatus? – MathematicalPhysicist Apr 19 '12 at 4:02

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