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Before the measurement of an observable, the quantum state is

$$|\Phi\rangle = \sum_i c_i |\psi_i \rangle,$$ with $|\psi_i \rangle$ called "pure states".

Once the measurement is done, the quantum state $|\Phi\rangle$ is projected onto one of the pure states $|\psi_{i}\rangle$.

Questions:

  1. Is a pure state an eigenvector of the observable used for the measurement?

  2. Before measurement, is the quantum state $|\Phi\rangle$ a superposition of pure states $|\psi_{i}\rangle$?

  3. What is the relation between the coefficients $c_i$ above and the probability $p_i$ to get the system in a pure state $|\psi_i\rangle$ (once the quantum state is projected, i.e the measurement is performed)? Can we write $|c_{i}|^2 = p_{i}$?

From the normalization condition for the quantum state $|\Phi\rangle$ we have $$\langle\Phi|\Phi\rangle = 1 = \sum_{i} |c_{i}|^2 = \sum_{i} p_{i} = 1 \, ,$$

but I can only arrive at $\left|c_i \right|^2 = p_i$ if the pure state basis is orthogonal $\left(\text{i.e.,}~\langle\psi_i|\psi_j\rangle=\delta_{ij}\right) $, can't I?

Thanks all, just a last question:

So I should rather think that a pure state can also be a superposition of basis states which are assimilated to eigenvectors of an observable. A pure state may not be only one eigeinevctor, it can be a linear combination of eigeinvectors, can't it?

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    $\begingroup$ It is probably worth noting that "pure state" is a technical term which does not mean what what you seem to think it does in this question (for example the state $|\Phi\rangle $ in your first equation is also a pure state). A better term for the states you are describing would be basis states. $\endgroup$ – By Symmetry Jul 28 '18 at 10:05
  • $\begingroup$ A measurement of $|\Phi\rangle$ "finds" the system in one of states $|\psi_n\rangle$ with some probability, but it does not mean any "collapse". On needs to make many measurements to find out the probabilities $p_n$ from statistics and thus (re)construct the measured state $|\Phi\rangle$ from the experimental results. $\endgroup$ – Vladimir Kalitvianski Jul 28 '18 at 11:15
  • $\begingroup$ -@BySymmetry. thanks for your remark. So I should rather think that a pure state can also be a superposition of basis states which are assimilated to eigenvectors of an observable. : a pure state may not be only one eigeinevctor, it can be a linear combination of eigeinvectors, can't it ? $\endgroup$ – youpilat13 Jul 28 '18 at 12:49
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1) The decomposition you wrote down,

$$ |\Phi\rangle = \sum_{i} c_{i} |\psi_{i}\rangle, $$

can be performed using any complete basis $\psi$. So it would be useful to choose the basis of eigenvectors of the operator you want to investigate.

2) Yes, this is called a superposition.

3) If you want to arrive at the state $\psi_k$, you project it onto your state $\Phi$:

$$ \langle \psi_k |\Phi\rangle = \sum_i c_i \langle \psi_k|\psi_i\rangle = \sum_i c_i \delta_{ki} = c_k. $$

And then you take the absolute square of this amplitude to get the probability. This is the probability of arriving at the state $\psi_k$ after the measurement:

$$ |\langle \psi_k |\Phi\rangle|^2 = |c_k|^2. $$

So your equation is correct.

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  • $\begingroup$ -@Stephan. Thanks a lot ! Does your last equation $|\langle \psi_k |\Phi\rangle|^2 = |c_k|^2 = p_k$ come from the relation of $<\Phi_k|\Phi_k>=\Phi_k^{*}\Phi_k$ ? I mean how do you get the equality between $p_k$ and $|c_k|^2$ ? regards $\endgroup$ – youpilat13 Jul 28 '18 at 4:52
  • $\begingroup$ The probability of a wave function Phi collapsing to the wave function psi after a measurement is given by the absolute square of the transition amplitude. I calculated the transition amplitude <Phi|psi> to be c_k and the absolute square is therefore |c|^2. That‘s why $p_k=|c_k|^2$! $\endgroup$ – Stephan Jul 28 '18 at 5:42
  • $\begingroup$ It is not always happening that the measurement gives you a system in an eigenstate in the sense that the system "moves to" $|\psi_n\rangle$ and stays in it. Often the measurement (detector) gives the eigenvalue and destroys completely the original $|\Phi\rangle$. $\endgroup$ – Vladimir Kalitvianski Jul 28 '18 at 11:00

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