How to build an intuition regarding QM angular momentum?

Question

I am trying to build an intuition on how angular momentum algebra works.

From what I currently understand there is a set of rules we must know to deal with angular momentum:

1. The commutator of angular momentum components is non zero, so we can't diagonalize them simultaneously, so no simultaneous eigenstates; but we can diagonalize the $L^2$ with one of the components.
2. The spin is an intrinsic property of particles, but it behaves exactly like angular momentum, it has the same algebra, so we can say that it is a form of angular momentum, that can be summed with the orbital angular momentum.
3. The standard notation to write eigenvector and eigenstates of angular momentum is the following one: $$L^2|l \ m \rangle = \hbar^2 l (l+1)|l \ m\rangle$$ $$L_z|l \ m \rangle = \hbar m |l \ m \rangle$$ the notation for the spin has of course the same structure, since it is an angular momentum, but the symbols change: $$S^2|s \ s_z \rangle = \hbar^2 s (s+1)|l \ m\rangle$$ $$S_z|s \ s_z \rangle = \hbar s_z |s \ s_z \rangle$$ $l,m,s,s_z$ are usually called quantum numbers. Since spin and orbital angular momentum have the same algebra the rules that I am about to state regarding $l$ and $m$ apply also to the couple: $s,s_z$.
4. $l$ must be positive.
5. The following relation between $l$ and $m$ (or $s,s_z$) must hold: $$-l\leq m \leq l$$
6. Both $l$ and $m$ are quantize values, that move by unit jumps. (In the sense that the possible values of $l$ (or $m$) must be spaced by 1, for example l=0,1,2,3)
7. The values of $m$ must be symmetric with respect to zero (so for example $m=-1/4,3/4$ is not allowed)
8. Both $l$ and $m$ can only have integer or semi-integer values, but in the case of orbital angular momentum $l$ and $m$ must be integers.
9. Angular momenta can be summed; the algebra of the sum is the same regardless if we are summing orbital angular momenta or spin or both together. The total angular momentum of a system is usually called $J$. Taking for example the case of a system of two particles with spin but no orbital angular momentum, the total angular momentum is: $$\vec{J}=\vec{S}_1+\vec{S}_2$$ of course $J$ is an angular momentum so the commutator of its components is nonzero as well. And its quantum numbers are usually called: $j,j_z$.
10. We usually choose to diagonalize symultaneously as many things as possible, in this case we have two choices: we can diagonalize $J^2,J_z,S^2_1,S^2_2$ or $S^2_1,S^z_1,S_2^2,S^z_2$. (in the case of the sum of spin with orbital we would have the analogous: $J^2,J_z,S^2,L^2$ or $S^2,S_z,L^2,L_z$). We can switch between this two basis with the Clebsch-Gordan coefficients that are non trivial to find.
11. To determine what are the possible values of $j$ we have the following rule: $$|s_1-s_2|\leq j \leq s_1+s_2$$ or in the case of spin plus orbital we have the analogous: $$|s-l|\leq j \leq s+l$$ then it's easy to find the possible values of $j_z$ with $-j\leq j \leq j$.

My question is: Is there a way to make this topic more intuitive or am I doomed to remember this eleven rules? Of course all this rules have proofs, but I cannot find a way to intuitively picture this topic in my head, it seems like a sudoku with strange rules. To put it in another way: is there a way to reduce the number of rules one must remember from eleven to like three or four? And then have a way to quickly deduce the other facts from them? I am also searching for a way to understand this topic in a more physical and less mathematical way. Especially the change of basis with the C-G coefficients is problematic to me. Is there some kind of shortcut?

Answer 1

As far as doing computations, no. You have all the rules you need, and there is no shortcut (Clebsh-Gordan tables are the shortcut).

As far as a deeper understanding, I have to agree @G. Smith's comment. You can look deeper into Lie algebras and their representation theory. Another avenue that helped me is the rotational symmetries of ordinary rank-$n$ cartesian tensors, which divvies up the $3^n$ cartesian components into things that rotate like $Y_l^m$ for various $0\le l \le n$, with a deep connection to Young diagrams, the permutation group on $n$ letters, and partitions of $n$.

There is perhaps one area where you can compare angular momentum with something simpler, which is linear momentum:

1. Why do we use linear momentum?

2. Because it is conserved.

3. Why is it conserved?

4. Because physics is invariant under translations.

5. So how do we use it?

6. via momentum eigenstates: $|k\rangle = e^{ikx}$

The same thing applies to angular momentum, rotations, and $|l, m\rangle = Y_l^m(\theta, \phi)$.

So how do you combine two momenta?:

$$ |k_1\rangle|k_2\rangle = |K\rangle $$

where the total momentum is the sum of the components:

$$ K = k_1 + k_2 $$

and the analogous Clebsch-Gordon coefficients are all trivial:

$$ c_{k_1, k_2;K} = \delta(K-(k_1 + k_2)) $$

So here I've taken the tensor product of two representations (individual particle momenta) and turned it into a sum over representations of total momentum eigenstates.

The sum has one term because translations commute. It's so trivial that we rarely even think of it in these terms.

When we write $|k\rangle$, or $\langle k|\psi\rangle$, we understand that in terms of Fourier analysis, and that representing an arbitrary (normed) function of $x$ in terms of $k$ makes it easier to understand and work with.

Likewise, atomic orbitals, which are complicated functions of $e^{i\phi}$, and $\cos{\theta}, \sin{\theta}$ can be simply understood as eigenstates of rotations about the $z$-axis: $|l, m\rangle$.

Even before we studied quantum mechanics, when doing waves on a string, we knew about beats:

$$ \cos(2\pi f_1 t)\cos(2\pi f_2 t) = \frac 1 2\big(\cos(2\pi(f_1+f_2)t) + \cos(2\pi(f_1-f_2)t)\big)$$

or:

$$|f_1\rangle|f_2\rangle = \frac 1 2 [|f_1+f_2\rangle + |f_1-f_2\rangle]$$

That is: the product of two frequencies eigenstates is not a frequency eigenstate, but it is the sum over eigenstates of related frequencies.

So there is nothing conceptually new about combining individual eigenstates and re-expressing them in terms of total eigenstates. What is new is using eigenstates of operations that don't commute. That's what leads to algebraic complexity...and that algebraic complexity is completly defined by the commutation relations.

Answer 2

My question is: Is there a way to make this topic more intuitive or am I doomed to remember this eleven rules?

To be honest, remembering the first 8 rules is not particularly hard, and up to that point it can be rather intuitive.

1. Is not surprising if you remember that noncommutativity of (not infinitesimal) rotations is already true classically. Since angular momentum is the generator of rotations, it should not be surprising that they do not commutate.

The fact that angular-momentum operator has discrete spectrum comes immidiately from the Stone-Neumann theorem, and from the periodicity of angles. Intuitively speaking, the "l" quantum number can be related to the "size" of the angular momentum (which has to be non-negative), while the "m" quantum number is related to the projection of the angular moment to a given axis. 4) 5) 7) can be remembered rather simply from this picture.

That ℏ is a quanta of angular momentum, can be "seen" (for spatial angular momentum) from a rather old concept before QM: Let us demand that the "particle-waves" have unique phases. That means that on a Bohr-trajectory, $2\pi R=k\lambda$ which means that $L=PR=k\hbar$. This can be generalized to non-circular paths, and at least gives intuition to 6) and partially to 8).

3. Needs to be memorized, the rest you will look up either way, when you are using them.

is there a way to reduce the number of rules one must remember from eleven to like three or four? And then have a way to quickly deduce the other facts from them?

Yes, expect it is anything but quick. Still, I hope I could be of help.