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

ReplyOptionsLet's try to express this more formally.

Following the Hermitian operator formalism, the observable can be written as

(1)The measured state $\ket{\phi}$ by definition belongs to some $m$-dimensional $\lambda_k$-eigenspace of $O$, $1 \leq k \leq n$, with the projector $P_k$. The state can be expressed as

(2)where $\ket{\alpha_j}$ are basis vectors of the $\lambda_k$-eigenspace. Because $\ket{\phi}$ is a $\lambda_k$-eigenvector of $O$, the result of a measurement of $\ket{\phi}$ according to $O$ is by definition a $\lambda_k$-eigenvector of $O$. Alternatively, and perhaps more correctly in light of part b), we could say that $\bra{\phi} P_i \ket{\phi} = 0$ for all $i \neq k$ so the measurement will be performed by projector $P_k$ with probability 1. Because

(3)the result is

(4)ReplyOptionsIn the above explanation, we intentionally said "the result of a measurement of $\ket{\phi}$ according to $O$" instead of "the result of applying $O$ to $\ket {\phi}$". An observable only describes how a measurement is performed, but is not by itself applied to a state vector to produce the measured state. In other words, it would be incorrect to attempt to express that result as

(1)That would suggest that the second measurement would produce $\ket{\phi'} = \lambda_k \ket{\phi}$, which in the general case wouldn't even be a unit vector, and thus not a valid state. Thus, generally $O^2 \neq O$, and that's okay because we don't apply it to states. $O$ is simply a notation allowing to summarize an observation in a single expression. The actual conversion of a state into the result is performed by one of the projectors $P_i$ (sans the associated eigenvalue), selected with probability $\bra{\psi} P_i \ket{\psi}$.

ReplyOptions