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

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.

• why is there a vector over ${\hat{L}}_z$?
– JEB
Dec 13, 2022 at 3:18
• Thanks for identifying the typos, JEB! Dec 14, 2022 at 15:02

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).

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.

• Thanks for the very nice explanation. Indeed the analogy makes sense that we cannot define the $(x, p)$ of a quantum particle. So, is it correct to say that, the complete knowledge of $\vec{L}$ in quantum mechanics is unknown to us due to commutator issues just like the complete knowledge of $(x,p)$ is unknown to us? Coming to the second part of your answer, we use a simultaneous eigenbasis of $|L, L_z\rangle$ to uniquely label our states. We cannot extend this to additionally include $L_x$ and $L_y$ in our label because they cannot commute with $L_z$. Is this correct? Dec 14, 2022 at 15:13
• @rahman62 That is correct. Dec 14, 2022 at 15:18
• But then, why do we choose to consider the common eigenbasis of $L^2$ and $L_z$ (and not $L^2$ and $L_x$ for example)? The answer is that $L_z$ has a much simpler expression in spherical coordinates than $L_x$ and $L_y$: $L_z=-i\hbar{\partial\over\partial\varphi}$ while $L_x$ and $L_y$ involves also the angle $\theta$. Dec 14, 2022 at 16:28

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,

$$L_z$$ 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.