In quantum mechanics, a particle in a box has discrete energy levels, while a particle in infinite space has a continuum of energy levels. The discreteness comes from the "compactness" of the box. Can we use this idea to explain why spin is quantized? Yes! The "box" in this case, however, is the group SU(2)!
What's SU(2)? Well, it's the group of 2x2 unitary matrices with determinant 1. It's this group, rather than the group SO(3) of rotations in 3 dimensions, that is used to describe rotation symmetry in quantum mechanics. SU(2) is the "double cover" of SO(3), meaning that two different elements of SU(2) correspond to one element of SO(3). The point is that in quantum mechanics, if we rotate something by 360 degrees, its wavefunction need not return to its original value. It can either return to its original value or get multiplied by -1. So in quantum mechanics, we need to distinguish between a 360 degree rotation and no rotation. This is nicely described using SU(2).
Geometrically, SU(2) is the same as what we mathematicians call S^{3}, or the 3-sphere - which is the unit sphere in R^{4}. We get the usual rotation group SO(3) by taking SU(2) and identifying each element x with the corresponding element -x. Here "identifying" means "treating as if it were the same" or, more geometrically speaking, "gluing together". Thus SO(3) is geometrically the same as the space you get by taking S^{3} and identifying each point x with the antipodal point -x.
The orientation of a body is described by a point in SO(3) so the study of a spinning body is mathematically the same as that of a free particle moving on SO(3). When we do quantum mechanics we merely need to remember to work with SU(2) instead of SO(3). So the quantum-theoretic study of spin is pretty much the same as the study of an abstract free particle moving around on S^{3}, satisfying Schroedinger's equation. The energy levels of this abstract particle correspond to the spin of the rotating body. The allowed spins are therefore quantized, and it turns out that they are 0, 1/2, 1, 3/2, etc.. If a wavefunction has integer spin, it will be unchanged by a 360 degree rotation. If it has half-integer spin (that is, 1/2, 3/2, or so on), it will be multiplied by -1 when we do a 360 degree rotation.
Michael Weiss, reading stuff like the above, asked:
OK, another "box" question. The standard prescription for second quantization (the creation/destruction operator stuff) yields a discrete spectrum. Where's the box this time?
And I replied:
I presume you're referring to the fact that when you quantize the harmonic oscillator, the spectrum of the Hamiltonian is discrete. The stuff about second quantization and the discrete spectrum of the number operator is just a special case of this - if you let your harmonic oscillator have lots of degrees of freedom.
But let me consider the 1-dimensional harmonic oscillator, to avoid extraneous complications. Here the "box" that makes its spectrum discrete is the circle.
Let me toss out a buzzword or two. Pontryagin duality. Compact <=> Discrete. Lie groups.
Ah, what a nice buzzing sound those words make. Yes, this is basically it: if we think of the circle as, say, the unit circle in the complex plane, we immediately see that it's a compact group. In this guise it's called U(1). Think of how the classical harmonic oscillator works: letting p denote momentum and q denote position, time evolution is given by
p cos(t) p - sin(t) q |-> q sin(t) p + cos(t) qIn other words, in phase space the harmonic oscillator just goes round in circles... or in other, buzzier words, we have an action of U(1) as symplectomorphisms of phase space. So when we quantize, since nothing much goes wrong, we get an action of U(1) on a Hilbert space; the self-adjoint generator - the Hamiltonian - thus has integer spectrum.
Integer spectrum? Here I have renormalized the harmonic oscillator Hamiltonian so its eigenvalues are 0,1,2,3,.... Often one sees the eigenvalues written as 1/2,3/2,5/2, and so on. As in the case of spin, these half-integers suggest that a double cover may be lurking about somewhere, and indeed there is! If we try to quantize, not just this U(1) group, but the whole symplectic group Sp(2), we are forced to work with its double cover, the metaplectic group Mp(2).
Indeed, the relation between the Spin groups Spin(n) and the rotation (or "special orthogonal") groups SO(n) is very analogous to the relation between the metaplectic groups Mp(n) and the symplectic groups Sp(n). The difference is really just one of signs: rotations preserve a nondegenerate bilinear form with
<v,w> = <w,v>
while symplectomorphisms preserve one with
<v,w> = - <w,v>
This issue of signs also makes all the difference between fermions and bosons. The bosonic and fermionic worlds are finally reconciled by supersymmetry; and the orthogonal groups and symplectic groups are unified in the orthosymplectic supergroup, about which I am shockingly, depressingly ignorant. But I know Michael Weiss is fond of some of Shlomo Sternberg's writings, so I mention this anyway - because Sternberg has written a bunch about the orthosymplectic supergroup.
Michael then wrote:
(If I knew how these buzzwords answered the question, I would write more. Hopefully someone will be sufficiently irritated by this post to fill in the blanks.)
It pissed me off so much that I wrote the above epistle in retaliation.
But then he replied:
O sly one, the original question was what's the compact group that's responsible for the discrete spectrum (of the number operator)? You answered U(1), and a nicer compact group never touched ground while walking. But then you went on to talk about Sp(2), a group I have sadly neglected in my education.Enchanted by your pleasant line of patter, I failed to notice at the time that Sp(2) IS NOT COMPACT! (OK, so I'm shouting.) Are you pulling a fast one? Will the real symmetry group please stand up?
And so I said:
Consider the classical phase space of a particle on the real line. This phase space is just R^{2}, with coordinates p and q standing for momentum and position.
Sp(2) is the group of linear transformations of this phase space that preserve the basic geometrical structure underlying classical physics - the symplectic structure. Concretely, Sp(2) is just the group of 2x2 real matrices with determinant 1. It's a 3-dimensional group, so we can understand it pretty well by pondering three different 1-dimensional subgroups of it.
One of them I alluded to already: it's the group of matrices like
cos t -sin t sin t cos tThese are just rotations. This subgroup is shaped like a circle, and called U(1). It's compact. Physically they correspond to the time evolution of a harmonic oscillator. In other words, if the momentum and position of a harmonic oscillator starts out at (p,q), after time t it will be (p cos t - q sin t, p sin t + q cos t), at least if the frequency of the oscillator is chosen right. It's this U(1) subgroup that explains the discrete spectrum of the quantized harmonic oscillator Hamiltonian.
Another subgroup is
1 t 0 1These are just shears. Physically they correspond to the time evolution of a free particle: if the particle starts out with momentum and position (p,q), after time t the momentum and position will be (p, q + tp), at least if the mass of the particle is chosen right. Since this subgroup is not compact, when we go to quantum mechanics we don't expect to get a discrete spectrum for the Hamiltonian of a free particle - and we don't.
A third subgroup is
1/r 0 0 rThis stretch-squashing corresponds to a dilation in configuration space: the position gets multiplied by r, while the momentum gets multiplied by 1/r. (Remember that short distance scales correspond to big momenta - that's what's going on here.) This subgroup isn't compact either and in quantum theory we don't get a discrete spectrum for the operator generating it.
Okay, what about quantum mechanics? Well, it's actually the double cover of the symplectic group Sp(2), the so-called "metaplectic group" Mp(2), that has a nice unitary representation on the Hilbert space L^{2}(R), which is the Hilbert space for a single quantum particle on the real line. This representation is nicely described using the quantum versions of p and q, which are operators on L^{2}(R). In particular, the harmonic oscillator Hamiltonian is
(p^{2} + q^{2})/2,
the free particle Hamiltonian is
p^{2}/2,
and the generator of dilations in configuration space is
qp.
Only the first one has a discrete spectrum, and the reason it must is that U(1) is compact. Note: since in quantum theory we work with the double cover Mp(2) instead of Sp(2), the harmonic oscillator Hamiltonian really generates the action of the double cover of the original U(1) group described above. The (nontrivial) double cover of a circle is again a circle, so even in quantum theory the time evolution of a harmonic oscillator is described by a group like U(1), but it's a U(1) that's "twice as big". This is why the eigenvalues of the harmonic oscillator Hamiltonian get to be half-integers: 1/2, 3/2, 5/2.... In other words, by the time the classical oscillator first gets back to its original state, the quantum one is back but with its phase changed by a factor of -1! The second time the classical oscillator gets back to its original state, the quantum one really gets back to its original state.
As I hinted above, we can "renormalize" the harmonic oscillator Hamiltonian, subtracting off the zero-point energy to get
(p^{2} + q^{2} - 1)/2,
to make its eigenvalues integral. This gives a representation of the U(1) subgroup of Sp(2), rather than its double cover. However, we cannot consistently renormalize all the generators of Sp(2) in this way to get a representation of Sp(2) on L^{2}(R); all we can get is a representation of its double cover, Mp(2). This is called the Segal-Shale-Weil representation.
This funny extra 1/2 in the eigenvalues of the harmonic oscillator Hamiltonian can be thought of as the "zero-point energy" or "vacuum energy" due to the uncertainty principle. However, we've seen that the fact that it's exactly 1/2 is no coincidence! Just as you need to give a particle of half-integer spin two rotations of 360 degrees for it to get back to the way it was, with no funny phase factor of -1, so you need to let the harmonic oscillator wait twoclassical periods for it to get back to exactly the way it was. In the first case we are using the fact that SO(3) has a double cover SU(2) - or more generally, SO(n) has a double cover Spin(n). In the second case we are using the fact that Sp(2) has a double cover Mp(2) - or more generally, Sp(2n) has a double cover Mp(2n).
As I've said before, SO(n) is to fermions as Sp(2n) is to bosons. The first has to do with the canonical anticommutation relations, and Clifford algebras, while the second has to do with canonical commutation relations, and Weyl algebras. So there is a big beautiful pattern here.
This could certainly use a more detailed explanation, but I'll leave that for some other time, some other place. If anyone is desperately eager to see the big picture, you can read the book I helped write about this stuff: