Spin One-Half, and the Mysterious Factor 2

Lie group and Lie algebra representation theory fit flawlessly into the structure of quantum mechanics, and the most satisfying explanation of the mysteries of spin one-half lie in this connection. Careful exposition of this material belongs to standard textbooks. But I must at least justify my remark that half-integer spin is fundamentally non-classical.

Let us start with a quantum mechanical system which has $n$ states. The Hilbert space for this system is then isomorphic to ${\bf C}^n$. Suppose the system ``inhabits'' ordinary physical Euclidean 3-space, that is, we can picture the physical processes of the system as taking place in 3-space. Vague though this statement is, something precise will come out of it. Any classical physical system (e.g., a spinning ball) ``inhabits'' 3-space in this (as yet) fuzzy sense.

Rotate the system in 3-space. Or if you prefer, rotate the coordinate axes used to describe the system. Either way, we have a transformation from one state of the system to another. (Physicists like to distinguish ``active'' from ``passive'' transformations, along the lines of preference just mentioned, but we won't need to be so exact.) We will assume this transformation can be represented by a linear operator on the Hilbert space of the system.

In fact, no generality is lost by assuming the linear operator is unitary - one can prove this. Even more: since our Hilbert space is isomorphic to ${\bf C}^n$, the operator has a matrix representation, and it has been proved that one can arrange for the determinant of the matrix to be 1. In short, we have a unitary unimodular representation of the Lie group $SO(3)$ on ${\bf C}^n$:

$$\sigma: SO(3) \rightarrow SU(n)$$

This is the precise statement that emerges from our vague notion of ``inhabiting 3-space''.

Group representation theory now grinds away. It tells us first, that $\sigma$ is a direct sum of irreducible representations, and next, that an irreducible representation exists if and only if the dimension of the target space is odd (in which case the representation is essentially unique).

So we now have:

$$SO(3) \rightarrow SU(H_1) \oplus\ldots\oplus SU(H_r)$$

where $\Sigma_i H_i$ is a direct sum decomposition of the Hilbert space of our system, and $SU(H_i)$ is the group of unitary unimodular transformations on $H_i$; also, each $H_i$ has odd dimension.

Any state in subspace $H_i$ can be rotated into any other state in subspace $H_i$. But rotation never mixes different subspaces together: a state in subspace $H_i$ remains in that state. So it seems natural to concentrate on the irreducible representations.

Associated with the Lie group $SO(3)$ is the Lie algebra $so(3)$, generated by three Lie algebra elements; we can picture these generators as ``infinitesimal'' rotations about the $x$, $y$, and $z$ axes (as did Lie), or as angular velocities about the coordinate axes.

Suppose $\sigma:SO(3) \rightarrow SU(H)$ is an irreducible representation, and suppose $H$ has dimension $2j+1$. Then $\sigma$ induces a map from $so(3)$ to the Lie algebra $su(H)$, which just happens to consist of all the anti-Hermitian traceless operators on $H$. The three generators of $so(3)$ map to three operators I will label $iJ_x$, $iJ_y$, and $iJ_z$-- that factor of $i$ makes $J_x$, $J_y$, and $J_z$ Hermitian.

The generators of $so(3)$ ``look like'' angular velocities, so $J_x$, $J_y$, and $J_z$ are good candiates for the angular momentum operators. Ultimately this comes down to a matter of definition. Let us make this identification without further ado.

Finally, the operator $J_xJ_x+J_yJ_y+J_zJ_z$ commutes with all of $SU(H)$ (as can be shown by direct calculation, or more cleverly), and so by Schur's lemma, is a constant times the identity matrix. And now the punchline: it turns out that this operator is $j(j+1)I$, where $j$ (you may recall) is defined by the relation $\dim H = 2j+1$.

In one sense we should be well satisfied. Our original direct sum decomposition of the $n$-dimensional Hilbert space of the system simply decomposed that space into states with a definite magnitude for the angular momentum-- that is, with definite quantum number $j$. On the other hand, group representation theory has told us that $j$ must be an integer and $\dim H$ is odd. And we want $j=\frac{1}{2}$ for an electron.

But now something curious appears. $SO(3)$ has a double-cover, the Lie group $SU(2)$. Their Lie algebras are of course the same. Perhaps we pick up some additional representations by starting with $SU(2)$?

Indeed we do. There is a unique irreducible representation $SU(2) \rightarrow SU(k)$ for every integer $k$; for reasons that should be obvious by now, we set $2j+1=k$, and use $j$ in place of $k$ in all the formulas. The $SU(2)$ representation factors through the $SO(3)$ representation precisely when $j$ is an integer: $SU(2) \rightarrow SO(3) \rightarrow SU(2j+1)$.

This is as far as I will pursue the mathematics. Our Hilbert spaces are still rather bloodless. Schrödinger conjured up lovely images of wavefunctions spreading through 3-space, which is to say he picked a representation for the Hilbert space which ``inhabits 3-space'' in our sense. Naturally the Schrödinger wavefunctions will not serve to represent a spin one-half particle like an electron. Pauli solved this puzzle, scant months after the invention of quantum mechanics. Relativity led to further paradoxes. To Dirac belongs the glory of resolving these. But thereby hangs another tangled tale of history...

I will ramble a bit further though on classical imagery versus quantum reality, just with regard to this question of spin. The groups $SU(2)$ and $SO(3)$ have the same Lie algebra, and as we've seen, the Lie algebra begets the angular momentum operators. This above all is why classical pictures can carry us so far, even for half-integer spin. Push it far enough though, and any classical picture will finally break down, for $SU(2)$ and $SO(3)$ are different groups.

In some sense, ``the mysterious factor 2'' stems from this difference. Because $SU(2)$ is the double cover of $SO(3)$, factors of 2 appear in the formula for the covering projection. These factors ultimately find their way into the formula for the ratio of the electron's magnetic moment to its spin angular momentum. The orbital motion of an electron however ``inhabits 3-space'', and so no factors of 2 appear.

© 2001 Michael Weiss

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