Why do we choose to use $L_z$ and $L^2$ to define a quantum state with angular momentum?

Question

Classically, we need to specify all three $(l_x, l_y, l_z)$ to define the vector $\vec{l}$. I understand that in Quantum mechanics we cannot define states corresponding to the operators $\hat{{L}}_x$, $\hat{{L}}_y$, $\hat{{L}}_z$ simultaneously due to commutator issues. So we choose $\hat{{L}}_z$ and $L^2$. But aren't we missing one? Angular momentum is a vector and we should be needing three quantities to specify it completely. What am I missing?

This is probably very basic and I would appreciate it if someone could shed light on this.

Answer 1

You are perhaps missing here the departure of quantum mechanics from classical mechanics.

Because components of angular momentum do not commute, we know via the generalized uncertainty principal that knowing $\vec{L} = (L_x, L_y, L_z)$ is impossible.

There is no question about "missing one". It is necessary to be "missing one" in order to satisfy the uncertainty principal (if I am understanding your confusion correctly).

Instead, we can only know angular momentum along one axis, e.g., $\vec{L_z}$ and the total angular momentum squared $\vec{L}^2$ at a given time (along with other observables that all mutually commute).

Answer 2

Angular momentum is a vector and we should be needing three quantities to specify it completely. What am I missing?

This is a very interesting feature of quantum theories. Remember that, classically, you would need the doublet $(x, p) $ to specify the state of a system. But in QM, $\psi (x) $ specifies the full state. $\psi (p) $ can be determined using a Fourier transform. Even the energy spectrum is sufficient for specifying the state.

This is because $X$, $P$ or $H$ span a complete eigenbasis on their own, because of the axiom $[X, P]=i\hbar$, which makes $X$ and $P$ dependent. Classically, (in the KvN formulation), this axiom would be $[X, P]=0$, so the information about both $X$ and $P$ would be needed to be specified independently in the simultaneous wavefunction $\psi (x, p) $.

So the answer to this question is that $L^2$ and $L_z$ form a complete eigenbasis. Actually, $L^2$ or $L_z$ alone form a complete eigenbasis, just like $X$, $P$ and $H$ do. But we use their simultaneous eigenbasis so we can bypass their degeneracy and uniquely label the states. The spectrum of $L_x$ and $L_y$ can be obtained through a change of basis.

Answer 3

You cannot use the components because they do not commute, and thus do not have common eigenvectors.

A better(?) answer goes back to classical mechanics, where an important concept for Hamilton-Jacobi theory is quantities in involution, i.e. they Poisson-commute. The choice of three quantities in involution is precisely $L_z$, $L^2$ and $H$. Thus the leap to quantum is strongly inspired by classical mechanics,

Answer 4

$L_z$ and $L^2$ form a complete set of commuting observables(C.S.C.O). This means that no two eigenstates of both of these operators may have the same eigenvalues. Thus, we don't really need the the other components to specify a state. To confirm that $L_z$ and $L^2$ form a C.S.C.O, look at their eigenvalues, $m\hslash$ and $l(l+1)\hslash$. With both of these eigenvalues, a Spherical Harmonic is fully defined, which is the wavefunction.