On the difference between orbital angular momentum $\hat L$ and spin angular momentum $\hat S$

Question

Intuitively, from here:

Spin is the angular momentum of a particle in the frame where the particle is at rest. If the particle is moving, then there are additional contributions to its angular momentum, namely the orbital angular momentum.

Ok but, from a mathematical perspective, I'm just very confused.

Algebraically, they behave in a similar way: $L ^2 |l, m_l\rangle = \hbar^2 l (l+1) |l, m_l\rangle \\ L_z |l, m_l\rangle = \hbar m_l |l, m_l\rangle \\ S ^2 |s, m_s\rangle = \hbar^2 s (s+1) |s, m_s\rangle \\ S_z |s, m_s\rangle = \hbar m_s |s, m_s\rangle$

So they're treated as if they were the same.

But, while $L := X \times P = -i \hbar x \times \nabla $, the same cannot be said for $S$ I think.. is there a similar formula for it?

Another difference is the Hilbert spaces the two operators act on, although I didn't understand why. ${\scr H}_{L} \equiv L^2 (\mathbb R^3)$, while ${\scr H}_{S} \equiv \mathbb C^{2s+1}$. Uhm, why is that anyway?

So .. are they the same? In which sense?

Answer 1

Spin is just an intrinsic properties of elementary particles, like mass and charge. The spin of a composite system is the sum of the spins of the components. There is actually nothing "spinning", there is not rotation involved as the electron is a point particle and you cannot mark a side.

It is called spin, because it behaves like an intrinsic angular momentum, that is independent of the trajectory in space (hence the lack of a formula similar to $\mathbf{r}\times\nabla$. It behaves like that in the sense that, experimentally, it gives you the same behaviour as orbital angular momentum: for example, it was known that the Zeeman effect splits the energy levels according to the projections of the (orbital) angular momentum $m_\ell$, which ranges from $-\ell$ to $\ell$. But they also found that an $\ell =0$ state would be split in two! This was compatible with the electron having an "additional" angular momentum $s=1$ so that $m_s = \pm 1$. And it obeys the same algebra-style as the orbital one, in terms of operators and states.

The above is a bit of jabber and experimental talk, but spin itself is a fundamental property of particles, as proven by the Wigner representation of the representation of the Poincare' group: these are labelled solely by mass and spin. So when you talk about "identical particles" in the context of bosons and fermions, "identical" means that they transform in the same representation of the Poincare' group, that is they have the same mass and spin.

Regarding the Hilbert spaces:
The orbital angular momentum ($\mathcal{H}_L$) is made from two real continuous operators ($x$ and $p$) so it makes sense it has the same Hilbert space.
The spin angular momentum ($\mathcal{H}_S$) is an internal degree of freedom, it does not depend on position or momentum, so it cannot be encoded in a spatial function, but it has to be a column vector (that is then multiplied by the spatial part of the wavefunction to give you the total one). The spin operators have Pauli matrices which can be imaginary, hence the $\mathbb{C}$ set. Angular momentum of magnitude $S$ has $2S + 1$ projections, hence why you should allow $2S+1$ rows in your column vector.

Answer 2

Fix an axis, such as the z-axis. The total angular momentum operator $J_z$ for that axis is what we call the "infinitesimal generator of rotations" for that axis. What this means basically is that if the system, in state $\varphi$, is rotated by a tiny angle $\mathrm{d}\theta$ around the z-axis, then the state goes to $\varphi - \frac{i}{\hbar} J_z(\varphi) \mathrm{d}\theta$. In other words $J_z$ tells you the difference between the system's state and a slightly rotated copy of the system's state.

In some simple systems, such as those that consist of discrete massive particles, it's possible to imagine that rotation of the system consists of two processes happening simultaneously: the particles revolve around the axis (the way the Earth revolves around the Sun) and the particles also themselves rotate as if around a parallel axis passing through themselves. The orbital angular momentum $L_z$ is the infinitesimal generator of the first type of rotation. The spin angular momentum $S_z$ is the infinitesimal generator of the second type of rotation. A "full" infinitesimal rotation involves both processes, so $J_z = L_z + S_z$. We can define $J$, $L$, and $S$ around other axes analogously.

Based on this analysis, we can conclude that $L = -i\hbar \frac{\partial}{\partial \theta}$ (where again, an axis is assumed to be fixed). In other words $L$ acts at each point of the wave function by comparing the value of the wave function at that point with the value at a point at an infinitesimal angular displacement from that point. If you convert to Cartesian coordinates, you can derive that $\mathbf{L} = \mathbf{r} \times \mathbf{p}$.

There's no analogous formula for $S$, because the process that $S$ corresponds to acts differently on different types of particles. Some particles don't change at all when rotated; such particles are "scalar". The Higgs boson is the only elementary particle that is scalar. The pions are examples of composite particles that are scalar. Because such particles retain exactly the same state under rotation, $S = 0$. Electrons, on the other hand, have two possible basis spin states and it is possible to convert one to the other by rotating the electron around an appropriate axis. So $S$ acts nontrivially on electrons and other non-scalar particles.

Both $L$ and $S$ satisfy similar algebraic relations like those given by the OP, as well as commutation relations such as $[L_x, L_y] = i\hbar L_z; [S_x, S_y] = i\hbar S_z$ because these relations are based on the mathematical structure of rotations in three dimensions. But orbital and spin angular momentum correspond to different physical processes and that is the difference between them. They also, at the most fundamental level, can be viewed as acting on different Hilbert spaces: $L$ must work with a spatial wave function, whereas $S$ only needs to know the particle's current spin state. Of course, when we add them to get $J$, we have to take appropriate tensor products with the part of the state that each operator doesn't care about.