# Can quantum measurement process be thought of as a sieve?

Consider an observable represented by the Hermitian operator

$$A=\sum_{a'}a' |a'\rangle \langle a'|.$$

As I read on Sakurai's textbook, the process of measuring $A$ throws a system initially into the state $|u\rangle$ into one of the eigenstates of $A$. Mathematically this amounts to applying exactly one of the projections $|a'\rangle \langle a'|$. So I'm asking:

1) Is it correct to think at the observable $A$ as if it were a sieve:

so that the state $|u\rangle$ ends up randomly (but with a prescribed probability distribution) into one of the eigenstates $|a'\rangle$?

2) If this is the case, then an observable is nothing more than a collection of orthoprojections, one for each output of the sieve. But then what is the physical meaning of $A|u\rangle$?

Thank you.

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1) After the measurement the state $|u\rangle$ ends up randomly into one of the eigenstates of $\hat{A}$ (Born rule). However before the measurement the state $|u\rangle$ is not a collection of eigenstates of $\hat{A}$ (mixed state), it's a superposition of them (pure state), so you can't really call it a sieve. But when you have a mixed state of eigenstates of $\hat{A}$ then the measurement is indeed like a sieve.

The action of the operator $\hat{A}$ on the state $|u\rangle$ is not a measurement, generally it gives you an another ket which is a superposition of the eigenstates of $\hat{A}$.

2) Observables are hermitian operators and they are weighted sums of projection operators, as you wrote, not just a collection of them.

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Very well! Thank you. –  Giuseppe Negro Dec 13 '11 at 19:09
I was thinking about your answer and decided to ask you. You said: "when you have a mixed state of eigenstates of $\hat{A}$ then the measurement is indeed like a sieve". Isn't this always the case? Indeed, $\hat{A}$ is Hermitian, so (disregarding mathematical difficulties such as continuous spectrum and the like) it has a complete orthonormal system of eigenstates. Then every state is a superposition of eigenstates and the measuring process throws it into exactly one of them . Am I wrong? –  Giuseppe Negro Dec 15 '11 at 11:39
Mixed state is a statistical ensemble of pure states (kets) and is different from a superposition (pure state - one ket or the sum of many kets). So the case of a mixed state of eigenstates of A and the measuring of A is like selecting randomly one of the eigenstates of A. In the case of superposition the observer becomes entangled with the system $\to$ some kind of superposition. There is no selection of eigenstates. However we perceive only one value of A after the measurement. This is a so called measurement problem. The best way of thinking this is many worlds. –  Andyk Dec 15 '11 at 20:21

Oh, boy! More standard brainwashing by textbooks. No, what actually happened is the measuring apparatus has at least a pointer indicating the value of the observable, and the process of "measurement" induces an entanglement of the system with the pointer of the apparatus, which in turn becomes entangled with the environment and the mental states of the observer.

$\sum_i c_i |a_i \rangle \otimes | 0 \rangle \rightarrow \sum_i c_i | a_i \rangle \otimes | p_i \rangle$

Besides, the correct more general treatment is to use positive operator valued measures (POVMs), not projector valued measures (PVMs).

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This is something I cannot truly understand right now, both physically (entanglement = ?) and mathematically (POVM = ?). But I will return here when it's time. Thank you! –  Giuseppe Negro Dec 13 '11 at 19:10