Ex4 12

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 4.12}$

Suppose $O$ is a measurement operator corresponding to a subspace decomposition $V = S_1 \oplus S_2 \oplus S_3 \oplus S_4$ with projection operators $P_1$, $P_2$, $P_3$, and $P_4$. Design a measurement operator for the subspace decomposition $V = S_5\oplus S_6$ where $S_5 = S_1 \oplus S_2$ and $S_6 = S_3 \oplus S_4$.

page revision: 0, last edited: 18 Nov 2012 18:24

Since a projector onto a direct sum of spaces is a sum of projectors onto the individual spaces, the projectors onto $S_5$ and $S_6$ are $P_5 = P_1 + P_2$ and $P_6 = P_3 + P_4$. Thus, a measurement operator for $S_5 \oplus S_6$ can be written as

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