Ex4 9

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 4.9}$

Design a measurement on a $3$-qubit system that distinguishes between states in which all bit values are equal and those in which they are not, and gives no other information. Write all operators in bra/ket notation.

twhittle 22 Dec 2013 09:11

In order to do this, we just need to assign two distinct eigenvalues to the two distinct eigenspaces we are trying to measure (all bit values are equal & all bit values are not equal). The solution below uses 2 for the eigenvalue for the eigenspace where all bit values are equal and 6 for the eigenvalue for the eigenspace where all bit values are not equal.

(1)

\begin{align} M = 2\left(|000\rangle\langle 000| + |111\rangle\langle 111|\right) + 6\left(|001\rangle\langle 001| + |010\rangle\langle 010| + |011\rangle\langle 011| + |100\rangle\langle 100| + |101\rangle\langle 101| + |110\rangle\langle 110|\right) \end{align}

vb314 29 Jul 2017 05:42

It could be argued that the simple

(1)

\begin{align} M = \ket{000}\bra{000} + \ket{111}\bra{111} \end{align}

which represents the comparison outcome as $1$ for true and $0$ for false is a better choice of eigenvalues, in line with the remark on p.56.

Last edited on 29 Jul 2017 05:46 by vb314 Show more