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If I understand correctly, spin is an intrinsic property of particles, which follows the algebra of angular momentum, but has nothing to do with an "orbital angular momentum" in that the particle is not like a small sphere that rotates on itself, to which we could attribute an angular momentum as usual in classical mechanics, cf. this and this Phys.SE posts.

My question is: why does spin combine with orbital angular momentum to give total angular momentum? To me this is surprising, as spin, in fact, has nothing to do with a classical angular momentum. Any insight on that?

Community
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Frank
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2 Answers2

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That spin follows the angular momentum algebra is no accident - like angular momentum, it is part of the conserved quantity - the Noether charge - associated to rotations.

The reason why the $\mathfrak{so}(3)$ transformations of spin should be indeed those associated to the $\mathfrak{so}(3)$ of spatial rotations is not answerable in QM alone - you have to take it "on faith" or rather, as an experimental fact that spin is indeed (part of) the Noether charge associated to spatial rotations and not some other $\mathfrak{so}(3)$. But, when you enter QFT, you will find that every quantum field should transform in some representation of the spatial rotation group (or rather in relativistic QFT, in some representation of the Lorentz group, of which the rotations are a subgroup), and that is exactly what spin then is - the "label" of the representation the quantum field transforms in.

Since orbital angular momentum is what comes from the quantization of classical mechanics as the Noether charge of the spatial rotations, you then find that your total quantum Noether charge for the rotations will have become the sum of spin and angular momentum.

ACuriousMind
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    ACuriousMind - so where is the starting point then? Should we start from rotational invariance, and from that both orbital angular momentum and spin emerge via an irreducible representation of the Lorentz group? (spin being one of the spaces in the direct sum decomposition of a tensor product of spaces (I recently learned how to do that! :-)) – Frank May 16 '15 at 22:35
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    @Frank: For a single non-relativistic particle in 3D, you would have $L^2(\mathbb{R}^3)\otimes\mathcal{H}_s$ as its space of states, where $\mathcal{H}_s$ is the $\mathfrak{so}(3)$-rep associated to the spin value $s$. The orbital angular momentum now comes from decomposing the "wavefunction space" $L^2(\mathbb{R}^3) = \oplus_l\mathcal{H}_l$, where $l$ is orbital angular momentum. Then you have $(\oplus_l\mathcal{H}_l)\otimes\mathcal{H}_s$, and you can now decompose this again as $(\oplus_l\mathcal{H}_l)\otimes\mathcal{H}_s = \oplus_j \mathcal{H}_j$ with $j$ total angular momentum. – ACuriousMind May 16 '15 at 22:53
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Magnetic moment, in classical physics, is related to current in a loop, which in turn can be connected to angular momentum of a charged particle. Thus, in classical physics, magnetic moment and angular momentum are connected. In fact, they are proportional with the constant of proportionality being the gyromagnetic ratio.

Moving to quantum mechanics, some particles have an intrinsic magnetic moment. We can relate their magnetic moment to an "intrinsic angular momentum" that we call spin. It turns out that this is not just a mathematical construct. The spin angular momentum has to be added to the orbital angular momentum to get a conserved quantity.

Amey Joshi
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