Does measuring an observable $\hat{\theta}$ for a QM system in a state $|\psi\rangle$ preserve the expansion coefficients of $|\psi\rangle$? Consider an idealised hydrogen atom in the state $|\psi\rangle=\frac{1}{\sqrt{6}}(2|1,0,0\rangle-|2,1,0\rangle+|2,1,1\rangle)$ where $|n,l,m\rangle$ are the normalised eigenstates of the Hamiltonian $\hat{H}$ and of $|\hat{\vec{L}}^{2}|, \hat{L}_z$ where $\hat{L}$ is the angular momentum operator.
Assume now that we measure $\hat{L}_z$ with outcome $m=0$. What state is the system in after the measurement?

solution from my lecture notes: By the form of $|\psi\rangle$, the state after the measurement is $|\phi\rangle=\lambda(2|1,0,0\rangle-|2,1,0\rangle)$ with $\lambda$ the normalisation constant (so $\lambda=1/\sqrt{5}$).
Question: Why are the $|n,l,m\rangle$ in the expansion of $|\phi\rangle$ only ones that appear in the expansion of $|\psi\rangle$ and why are their coefficients preserved? Why not the other $|n,l,m\rangle$ with $n=1,2,\ldots$ and $l=0,1,\ldots,n-1$? I can see that their dot product with $|\psi\rangle$ vanishes and hence adding such a state to $|\phi\rangle$ does not change the probability of measuring this state. How do I continue?
 A: Strictly speaking, yes. As mentioned in the comments, a measurement can be modeled by the application of a projection operator onto some subspace of the Hilbert space corresponding to the measurement result.  However, projection operators do not preserve normalization, so if $P$ projects onto the $m=0$ subspace, then
$$P|\psi\rangle =\frac{1}{\sqrt{6}}\big(2|1,0,0\rangle-|2,1,0\rangle\big)$$
which clearly is no longer normalized. This isn't a problem, because physical states are only defined up to a multiplicative constant, meaning that $|\phi\rangle$ and $\lambda |\phi\rangle$ represent exactly the same physical state.  In that sense, representing the state of your system as a normalized vector is just a useful convention which you don't need to follow.
If you do choose to follow that convention, then you can re-normalize your state after the projective measurement by multiplying by an overall normalization constant, in this case given by $\sqrt{6/5}$.  Obviously that changes the coefficients, but not the physical state.
