Ex4 13

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 4.13}$

a) Let $O$ be any observable specifying a measurement of an $n$-qubit system. Suppose that after measuring $\ket\psi$ according to $O$ we obtain $\ket\phi$. Show that if we now measure $\ket\phi$ according to $O$ we simply obtain $\ket\phi$ again, with certainty.

b) Reconcile the result of part a) with the fact that for most observables $O$ it is not true that $O^2 = O$.

page revision: 0, last edited: 18 Nov 2012 18:25

SOooooo I'm not sure how to answer this. I have trouble using the correct language but basically I'm thinking this

If after a measurement of $|\psi>$ according to $O$ we obtain $|\phi>$ this means that one of the subspaces of the measurement is generated by $|\phi>$ with projector $|\phi><\phi|$. This means the operator for the measurment of observable $O$ contains $|\phi><\phi|$. Since the subspaces of any observable are orthogonal when $O$ is applied to $|\phi>$ the only non orthonal projector in the operator is $|\phi><\phi|$.

ie.

(1)where $|i><i|$ is the projection operator of the ith subspace that $O$

(2)this is true because the subspaces defined by an observable are orthogonal ie $|i><\phi|=0$ for all i.

Now for $O$ vs $O^2$ I'm really confused. Unless $O^2$ is a different measurement than two individual $O$ measurements I can't think of how they aren't the same…..thoughts anyone

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