Is it possible to determine which combination of three degenerate states an atom is in, not destroying the state? Suppose we have somehow determined that an atom's outer shell electron is in a $p$ state (i.e. with $l=1$). For example, waited enough for a cold boron atom to almost certainly come into electronic ground state.
Since the $p$ shell has threefold degeneracy, the atom now is in some state like
$$\newcommand\ket[1]{\left|#1\right\rangle}
\alpha\ket{m=-1}+\beta\ket{m=0}+\gamma\ket{m=1}.$$
Having chosen some $z$ axis, can we find differences of phases and magnitudes between $\alpha$, $\beta$ and $\gamma$ without destroying or only slightly changing the atom's state? If yes, then how? If no, why?
EDIT OK, it seems that by no-cloning theorem this can't be done perfectly. But what about approximate measurement of $\alpha,\beta,\gamma$?
 A: What you are trying to do is an instance of state estimation for a qutrit in some state 
$$|\psi⟩=\alpha|-1⟩+\beta|0⟩+\gamma|1⟩$$
using a single copy of the state. This is essentially impossible, due mostly to the no-cloning theorem. The most you can do is choose some orthogonal basis $\{|a⟩,|b⟩,|c⟩\}$, and measure along it, which will get you a hit on only one of the states, say $|a⟩$. The only information you will get from that is that 
$$⟨a|\psi⟩\neq0.$$
That is, you can conclude that the component of $|\psi⟩$ along some chosen vector is nonzero. However, this does not (and cannot) give you any information about the magnitude of this component, or about the existence or not of components along vectors orthogonal to $|a⟩$.
It's important to note that this procedure will completely destroy the state, and this is pretty much a necessary condition for gaining any information about it.
The case of a qubit is slightly easier to understand geometrically. There, if you measure on a basis $\{|a⟩,|b⟩\}$ and get $|a⟩$, then you know that the state is not on the state $|b⟩$; that is, you know a single point on the Bloch sphere that the state vector doesn't point along. However, this does not rule out points very close to that forbidden point (for which $⟨a|\psi⟩$ is small but nonzero).
Quantum states really are very dark black boxes. If you have axes to the box factory - that is, if you have access to multiple copies of states produced by the same preparation - then you can perform tomography on the boxes and get a fairly good idea of what the state is (including whether it's pure or mixed). The theory of quantum state tomography tells you how many different measurements you need to perform to estimate the state, depending on its dimensionality. However, it's important to note that each 'measurement' is itself performed on an ensemble of boxes and aims to estimate a given probability. If you only have access to a finite number of boxes, this will bound the accuracy of your tomography, in well-studied ways.
If you only have a single box, though, there's very little indeed that you can say.
