Our choice of basis surely cannot effect possible outcomes of a measurement? Common sense says that, of course, the outcome of a measurement on a quantum system cannot be affected by what base we choose to represent it in. However, while studying QM text, it seems like they sometimes just almost suggest it... My question is quite vague, so let me instead give you an example which I think contains most of my confusion.
Consider a system in a basis of simultaneous eigenstates of $L^2$ and $L_z$, with respective eigenvalues $l(l+1)\hbar^2$ and $m\hbar$. Let $\mathbf{\hat{n}}$ be a unit vector in a direction specified by polar angles ($\theta,\phi$). 
Clearly
$L_n = \sin\theta\cos\phi L_x+\sin\theta\sin\phi L_y +\cos\theta L_z$
Now I have two slightly different questions:


*

*What are the possible results of a (precise) measurement of
    $L_n$, $L_n^2$? 

*What are the possible expectation values of
    $L_n$, $L_n^2$


I would tell you


*

*$m\hbar$, $l(l+1)\hbar^2$ or maybe $m \hbar \cos\theta$, but this would mean that my choice of direction in space affected the outcome of the measurement?

*No idea and makes me question my answer to 1) again...

 A: Phenomena in quantum mechanics may be expressed using any basis (that's the English word for the set of vectors, not a "base"). It doesn't mean that all bases are equally useful for a given situation. In particular, a fundamental postulate of quantum mechanics says that right after every measurement, the system is found in one of the eigenstates of the observable that was just measured. 
That's why the basis of the eigenstates of $K$ is obviously more useful to describe the measurement of $K$ than other bases. Note that which basis is useful – or which basis describes possible post-measurement state – does depend on the kind of a measurement we decide to perform. This dependence of the "right analysis of the physical system" on the chosen way of observing it is really a main point of quantum mechanics. 
If the physical analysis could be made independently of the nature of observations, the theory would be by definition classical physics, not quantum mechanics. Such independence on the observations could be called "common sense" by someone – but that changes nothing about the fact that Nature contradicts this assumption.


*

*$L_n$ always has eigenvalues $m\hbar$ where $m$ is an integer. $L_n^2$ has eigenvalues $m^2\hbar^2$ – because it's just the square of the operator from the previous sentence. This shouldn't be confused with $L^2$ which has the eigenvalues $\ell(\ell+1)\hbar^2$ where $\ell=0,1,2,3,\dots$ $L^2$ may always be measured simultaneously with any $L_n$ and in that case, $m\in\{-\ell, -\ell+1,\dots, \ell-1,\ell\}$.

*The expectation value of any operator may be any number from the interval between the lowest and highest eigenvalue. So if we measure $L^2$ and $L_n$ at the same moment, then $\langle L_n\rangle$ may be any real number between $-\ell$ and $+\ell$.
Note that for every $\vec n$, the spectrum of $L_n$ is the same. For all choices of $\vec n$, the operators $L_n$ are conjugate to each other i.e. 
$$\exists U\in U({\mathcal H}):\quad L_{n'} = U L_n U^\dagger $$
Here, the operator $U$ is an operator on the Hilbert space that represents a rotation that turns the $\vec n$ axis to $\vec n'$ (passively or actively, one would have to be careful).
The corresponding eigenstates of $L_n$ and $L_{n'}$ are also conjugate to each other – but the detailed sets of eigenvectors are different. So when we measure $L_n$, we bring the system to one of the basis vectors of $L_n$ by the measurement, and if we measure $L_{n'}$, the candidate post-measurement states are elements of the basis of different eigenvectors.
