3
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ -@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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.