Collapse of state vector for degenerate eigenvalues Consider a state vector given by
$$|\psi\rangle=a|w_1\rangle + b|w_2\rangle+c|w_3\rangle$$
where $|w_1\rangle,|w_2\rangle$ and $ |w_3\rangle$ are orthonormal eigenstates of an observable operator $\Omega$. The eigenvalues of $|w_1\rangle$ and $|w_2\rangle$ are degenerate, i.e $w_1=w_2$.
If we make a measurement of $\Omega$ that yielded the value $w_1 $,  the state vector will collapse in to the normalised state $$|\psi'\rangle= {a|w_1\rangle+b|w_2\rangle \over \sqrt{a^2+b^2}}.$$
Why doesn't it collapse to either $|w_1\rangle$ or $|w_2\rangle$, i.e.
$$|\psi'\rangle={|w_1\rangle } $$
or
$$|\psi'\rangle={|w_2\rangle}?$$
Why does it collapse to $|w_1\rangle$ and $|w_2\rangle$? Is this just a rule we observed from experiments?
 A: The postulates of quantum mechanics says that a measurement will return an eigenvalue and that there is some eigenvector with the according coefficient to match the probability of measuring that eigenvalue.
I am going to use a different normalization, let
$$
|\psi'\rangle = \frac{a|w_1 \rangle + b|w_2\rangle}{\sqrt{a^2+b^2}}\\ 
d = \sqrt{a^2+b^2}
$$
then we have $$|\psi\rangle=a|w_1\rangle + b|w_2\rangle+c|w_3\rangle| =d |\psi'\rangle +c|w_3\rangle $$
and assuming $$a^2+b^2+c^2=1=d^2+c^2$$
Now we have a state $|\psi'\rangle$ with eigenvalue $w_1=w_2$ with probability $d^2=a ^2 +b^2$ which is in agreement with the postulate of quantum mechanics about measurements and the measurements that we would make on a system prepared in your given initial state.
The states $|w_1\rangle$ and $|w_2\rangle$ are just a basis that span the degenerate space. You could use two different linearly independent basis vectors to describe the space. This would change the single coefficients but the total probability would remain the same. And that probability + eigenvalue are the physical information that is contained by $|\psi\rangle$. The basis in which we express that information doesn't affects the physics. The state $|\psi'\rangle$ is also a perfectly valid eigenstate that could be used as part of a basis to span the degenerate space.
A: Every self-adjoint operator $\Omega$ (on a finite-dimensional Hilbert space, for simplicity) can be uniquely decomposed as
$$\Omega = \sum_i \omega_i \Pi_i$$
where the eigenvalues $\omega_i \in \mathbb R$ are all distinct, and the $\Pi_i$'s are orthogonal projection operators such that $\sum_i \Pi_i = \mathbf 1$.  This is sometimes called the spectral decomposition of $\Omega$. The statement of the Born rule is that given a pure state represented by a unit vector $\psi$, the probability that $\Omega$ takes the value $\omega_i$ is given by $\mathrm{Prob}_\psi(\Omega,\omega_i) := \langle \psi,\Pi_i \psi\rangle. $  If this outcome occurs, then the (generally non-normalized) post-measurement state vector will be $\Pi_i \psi$, which reproduces the rule you quote.
Ultimately, yes - the Born rule is consistent with experiment, which is why it is accepted as the correct way to assign probabilities to various measurement outcomes. While one can motivate the Born rule as being in some sense the most natural way to associate a state vector to a set of probabilities, it is ultimately a postulate which can only be validated empirically.
