My textbook says:

"To determine the state of a system at a given instant, it suffices to perform on the system a set of measurements corresponding to a complete set of commuting observables (CSCO)"

This is my understanding. Please correct me if I'm wrong:

Let's say $H$ is an observable and CSCO on its own. Also its eigenstates ( $|\psi_i\rangle$ ) are discrete and non-degenerate and its eigenvalues are $E_i$.
I take $N$ systems that are in the same state $|\psi\rangle$ and make measurements for $H$. From the $N$ numbers I get, I can assign a probability $p_i$ to the result of each measurement, $E_i$. Then the state of the system will be: $$|\psi\rangle=\sum_i \sqrt{p_i} |\psi_i\rangle$$

Is this right or something else is intended?

Ref: Cohen-Tannoudji - vol. 1 - page 295

  1. A short comment: If $H$ is an observable and a CSCO, then its eigenstates are necessarily non-degenerate. You wrote it as if those were separate requirements.

  2. This is not correct. If you repeat your procedure with a very large number $N$ of measurements, then you can indeed figure out that the system state before the measurement was $$ |\psi\rangle = \sum_i \mathrm e^{\mathrm i \varphi_i} \sqrt{p_i}\, |\psi_i\rangle , $$ but you do not get any information about the phases $\varphi_i$.

  3. What your textbook is trying to say is much simpler, I think: you know the state of the system after the measurement for certain. (If you measure the energy $E_i$, then the state afterwards is $|\psi_i\rangle$.)

  • $\begingroup$ In addition to point 3 above I'd say that it is possible to do what @Ali suggests, which is figure out an arbitrary state by performing many measurements, but for that one needs to perform a protocol called "quantum state tomography" in which the measurements are more complicated than a CSCO. $\endgroup$
    – oleg
    Jul 30 '19 at 0:18

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