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Disclaimer: I am not a physicist; I am a geometer (and a student!) trying to learn some physics. Please be gentle. Thanks!

When solving the Schrödinger equation for a particle in a spherical potential, it seems common to separate variables into angular and radial components. The angular evolution can then be expressed in terms of eigenfunctions of the Laplace-Beltrami operator $\Delta$ on the sphere, i.e., the spherical harmonics. (It is my understanding that these eigenfunctions or eigenstates also have some physical significance, namely that eigenfunctions with the same eigenvalue correspond to states of equal energy.)

When solving the Dirac equation (again with a spherical potential) you'd expect a similar story: separate into angular and radial components and write the angular evolution in terms of the eigenfunctions of the Riemannian Dirac operator $D$ on the sphere. And, you'd expect these eigenfunctions would have a similar physical interpretation to the non-relativistic case (after all, the only thing we changed was the energy-momentum relationship).

However, the references I'm finding on the Dirac equation with central potential write solutions in terms of the spherical spinors $\Omega$, which are themselves simple functions of the spherical harmonics $Y_l^m$. This situtation seems odd to me because, although eigenfunctions of $D$ are also eigenfunctions of $\Delta$, the opposite is not true. In particular, $D$ will have both positive and negative eigenvalues, and so eigenspaces with equal value but opposite sign get "mixed" when we square $D$ (recall that on, say, Euclidean $R^3$, $D^2=\Delta$). I'm not sure about the physical interpretation, though, because I don't understand the physical meaning of eigenfunctions of the Dirac operator.

Here are some more concrete questions:

  • what do eigenfunctions of the $D$ represent physically?
  • why are the spherical harmonics used for separation of variables rather than eigenfunctions of $D$?
  • alternatively, are there cases where eigenfunctions of $D$ are used to solve Dirac's equation?

Pedagogical references are appreciated. Thanks!

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I think the [special-relativity] tag could be dropped from this, since although the theory of Dirac fermions requires SR "behind the scenes" in order to be consistent, we're not talking about relativity directly. Did you have a particular reason for including the tag? –  David Z Dec 2 '10 at 6:20
    
Done. I guess I added the tag because I saw the Lorentz group coming up a lot in this context... and because I'm not a physicist! ;) –  fuzzytron Dec 2 '10 at 18:42

2 Answers 2

up vote 4 down vote accepted

There might be a small bit of confusion in the question here.

[Skip all this stuff below until the next bracketed comment.]

First recall that $\mathcal{so}(3)$ has representations that integrate to representations of $SU(2)$ rather than $SO(3)$, and these are the "half-integral" spin. Likewise for $\mathcal{so}(3,1)$ and $SU(2)\times SU(2)$ and $SO(3,1)$, respectively.

Now for a manifold with Lorentzian metric, a representation of $\mathcal{so}(3,1)$ indicates what kind of particle/object/tensor/spinor you're talking about. The representation tells you how to transform the object/section when you change coordinates. (N.B.: since we're talking about bundles associated to the frame bundle, changing coordinates induces a change of trivialization of the bundle. Ordinarily, the two concepts are indpendent.) More specifically, tensors correspond to "integral spin" representations, e.g. the 4-dimensional representation is vectors, while the 6-dimensional anti-symmetric representation describes two-forms (such as field strengths).

So spinors are sections of bundles which correspond to representations of $\mathcal{so}(3,1)$ which integrate to a representation of $SU(2)\times SU(2)$. Practically speaking, in order to covariantly differentiate such an object, you can do the following. The Levi-Civita connection already gives you an element of $\mathcal{so}(3,1)$ (a matrix) for each tangent index. Plus you have a representation of $\mathcal{so}(3,1)$, so you act with this element of $\mathcal{so}(3,1)$ by your representation. This is what the gamma matrices are about.

[Oy, this is getting too long.]

Now here's the thing: the spinor representations decompose into different irreducible components, and the Dirac operator maps one representation (positive chirality) into another (negative chirality)! That is, it maps sections of one spinor bundle into sections of another. You can of course look at the Dirac operator on the sum of these bundles, but eigenvectors do not have an evident physical interpretation (they are of mixed chirality). In flat space, the square of the Dirac operator is a multiple of the identity endomorphism on the bundle, so eigenvalues make perfect sense and can be written in terms of functions times the (global) basis elements for spinors.

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+1: very nice answer. And by the way, I wouldn't mind the bracket part being even longer (for example some mention of Clifford algebras might be useful, seeing as the OP is a geometer). –  Marek Dec 2 '10 at 11:01
    
I'm ok with this; I think the irreducible representations you're considering (the Weyl spinors?) correspond to what I'd call even and odd functions (i.e., functions taking values in the even or odd part of the Clifford algebra, resp.). And the spinor Dirac operator maps from even to odd and vice versa. I'm trying to understand the physical meaning though: what do these even and odd parts correspond to? You mentioned chirality -- does chirality have a physical meaning here? Maybe that would help explain why one wouldn't want to consider eigenvectors of mixed chirality. –  fuzzytron Dec 2 '10 at 18:29
    
@fuzzytron: yes, chirality does have a physical meaning: left-handed and right-handed particles. If you reflected the space (this is called P transformation) you would swap such particles. As for the mixed chirality: when actually trying to get a classical approximation it so happens that there exists a basis (mixing left and right chiralities) where the top component spinor is much larger than the bottom one (which can thus be neglegted). The top spinor then corresponds to the classical two-component wavefunction modeling electron with spin. –  Marek Dec 2 '10 at 19:40

I think the essential physics is simply this: The conserved quantity associated with the spherical symmetry of the system is the angular momentum vector. In the case of the scalar particle described by the Schrödinger equation, this is unambiguous.

When one progresses to the Dirac operator, however, you are now describing a particle with intrinsic spin. Now, the conserved quantity is $\vec J = \vec S + \vec L$. In general, this means that you would expect some sort of spin-orbit coupling in the system, breaking the degeneracy between different values of the z component of spin seen in the Schrödinger-equation based solution for the central force problem.

I've never seen anyone do this by a direct application of the Dirac equation. Typically, the approach is to solve this using the Schrödinger equation, and then treat the spin-orbit coupling as a small perturbation on top of that, and spin as another quantum number that shows up whose explanation is just a handwavey QFT. But all of this information would have to be embedded in the Dirac equation.

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1  
"I've never seen anyone do this by a direct application of the Dirac equation." -> Really? I thought it was pretty standard exercise to derive all the basic results (described by Pauli equation) directly from the Dirac equation. –  Marek Dec 2 '10 at 11:05
    
Your QFT book breaks the Dirac equation into spherical harmonics and solves the hydrogen atom problem directly from the Dirac equation? –  Jerry Schirmer Dec 2 '10 at 14:05
1  
"Typically, the approach is to solve this using the Schrödinger equation, and then treat the spin-orbit coupling as a small perturbation on top of that" Yes, I've come across this approach as well and suspect it's the "right" thing to do in practice, but since my interest is in understanding the connection with geometry I'm willing to consider less practical stuff. ;) –  fuzzytron Dec 2 '10 at 18:44
1  
yes, I happened to have QM and QFT courses that solved hydrogen atom in numerous ways: Schroedinger, Pauli, Klein-Gordon, Dirac. We also did all those Stark and Zeeman effects and also solved the problem algebraically using Laplace-Runge-Lenz vector. I can say by the end I came to quite hate hydrogen atom. Thank god we at least didn't solve it from the point of view of QED, Standard Model and string theory :-) –  Marek Dec 2 '10 at 19:27
    
That's actually pretty neat, though I'd actually be interested in a first-principles derivation of the Lamb shift, now that you mention it. :) –  Jerry Schirmer Dec 2 '10 at 23:18

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