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Suppose an operator $O$ has eigenfunction normalized $f$ corresponding to eigenvalue $n.$ Of course, any function $cf$, with $c$ on the unit circle, is also a normalized eigenfunction. Thus, if a measurement of $O$ on some system returns a value of $n$, we have the state collapse to some $cf$. How is the constant $c$ determined? Does $c$ end up being physically irrelevant for the later time evolution of the system? (I can see it being physically irrelevant for future observables, but what about interference with other waves?)

Similarly, suppose an operator $O$ has degenerate spectrum at eigenvalue $n$, with orthonormal eigenfunctions $f_1$ and $f_2$. When a measurement of $O$ returns a value of $n$, can we in general determine what linear combination $c_1f_1+c_2f_2$ the collapsed state is in? Is the ratio $\frac{c_2}{c_1}$ perhaps given by the ratio $\frac{\langle f_2|S\rangle}{\langle f_1 | S\rangle}$, where $S$ is the state at the time of collapse?

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  • $\begingroup$ Upon more thought, I'm embarrassed that I didn't come up with the simplest of all possibilities: upon collapse, you simply disregard all eigenfunctions that don't permit the observed measurement, and renormalize what remains. Thus, for example, in the first case the constant $c$ would have the same phase as $\langle f|S\rangle$. Is this indeed the case? $\endgroup$ – Arjun Puri Mar 7 at 6:40
  • $\begingroup$ (Pure) quantum states are not (normalized) functions. They are equivalence classes of functions that differ by phase. In other words, yes the phase is not observable. In any experiment you will always measure some probability and these are invariant with respect to a phase change in the wavefunction. $\endgroup$ – lcv Mar 7 at 9:21
  • $\begingroup$ @ArjunPuri Yes your guess is correct, just dispose of all other terms and renormalize, that is the standard way. Projections are a bit untrustworthy, really, for any continuous space, but any time the eigenvalues are discrete this is correct and it is probably correct to some approximation in every case. $\endgroup$ – doublefelix Mar 7 at 13:39
  • $\begingroup$ @doublefelix I will accept an answer stating the same $\endgroup$ – Arjun Puri Mar 9 at 0:16
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As you guessed in your comment, the usual formalism is to just discard all terms inconsistent with the measurement outcome and then re-normalize. Nothing is changed about the phases of existing terms consistent with the measurement.

A concrete example: If an operator $O$ has two eigenfunctions with eigenvalue 1, say $|1a\rangle$ and $|1b\rangle$, and the state before measurement is

$|\psi\rangle=\alpha_1|1a\rangle+ \alpha_2 |1b\rangle + \alpha_3|2\rangle$

then if the eigenvalue 1 is found, the state after measurement is

$|\psi'\rangle=N(\alpha_1|1a\rangle+ \alpha_2 |1b\rangle)$

where $N=1/||\alpha_1|1a\rangle+ \alpha_2 |1b\rangle||$.

I feel it is my duty to warn you that this works nicely for discrete states as above, but I consider it highly untrustworthy for continuous variables like position. Yet there is no well-accepted alternative. This is just one manifestation of what is known as the measurement problem.

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Consider the wavefunction of the form $\psi=e^{i\theta_1}f_1+ e^{i\theta_2}f_2$. Since we can represent any unimodular complex number $c$ in this way as $e^{i\theta}$. Let us see what the probability density of this state looks like: $$|\psi|^2=|f_1|^2+|f_2|^2+ e^{i(\theta_2-\theta_1)}f_1^*f_2+ e^{i(\theta_1-\theta_2)}f_1f_2^*$$

Notice that the interference effect only depends on the difference between the phases. Or as you rightly say, the ratio between the coefficients. This is because the phases cancel out in the direct terms and only the cross terms give the interference.

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  • $\begingroup$ I have a feeling my question wasn't clear enough to understand. I'm wondering if we can determine which $e^{i\theta}$ a wave function "chooses" as its constant upon collapse. $\endgroup$ – Arjun Puri Mar 7 at 6:36
  • $\begingroup$ @ArjunPuri, In order to determine anything we need to be able to observe it. So under observation the phase of a single state is not seen because it’s mod squared out. But the phase difference between different states can be observed. So we can’t determine $e^{i\theta}$ but can determine $e^{i\Delta\theta}$ $\endgroup$ – Superfast Jellyfish Mar 7 at 6:41
  • $\begingroup$ @ArjunPuri "a wave function "chooses" " completely contradicts the probabilistic nature of quantum mechanics, which is given by $Ψ^*Ψ$. The " wavefunction weights the probabilities, but a single outcome is random. Also "collapse " after measurement means a new wavefunction based on the boundary conditions of the measurement has resulted, and most probably very much different than the operator's eigenfunctions. $\endgroup$ – anna v Mar 7 at 8:18
  • $\begingroup$ in the formula in this answer, the probability of getting either f1 or f2 for a single mesurement is given, whoever is larger will have a larger probability for the single measurement to follow. $\endgroup$ – anna v Mar 7 at 8:49

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