Well, the disappointment from prelims is fading, so I'm posting about a topic which I find very cool, and which has been used as a segue in my class to explaining topological effects in quantum field theories.
Classically, there's no problem with electrodynamics if we assume the existence of magnetic charge (even though, of course, no such thing has ever been observed). In fact, things are actually much nicer with magnetic charge, as Maxwell's equations are perfectly symmetric between E and B fields (I would write them out if I knew how to do equations in my blog, anybody know?), so the physics is much more beautiful if magnetic charge exists.
Well, you might ask: what about the vector potential A? We know that B= \del X A, and that if magnetic charges exist \del (dot) B=m \delta^3 (r) (up to a constant for a monopole at the origin, where m is the "magnetic charge", and the delta is the Dirac delta function). We also know, from our course on multivariable calculus, that the divergence of a curl is zero. So, on the one hand, we need the divergence of the curl of A to be m*\delta^3 (r), but on the other hand, we know it must be zero. What is the resolution? Well, classically, there is no real problem, because our creation of the vector potential A was built from the ansatz that there are no magnetic monopoles. So, we can put back the monopole if we just get rid of the vector potential (which, for classical electrodynamics, is not necessary), and just use E and B in our calculations.
What if we treat the system as quantum mechanical? Well, for quantum mechanics, we
need the vector potential A, as we can do QM with either the Hamiltonian or path integral formalism, both of which will involve A (in classical world, we could just use the force laws, Newton's laws, and calculate trajectories). Can we still salvage the monopole?
The answer is yes, and in a most satisfying way. If we take the modern viewpoint of A not so much as a physical field, but as the connection on a U(1) G-bundle, then A need not be defined everywhere. It can be defined on different trivializatons of the bundle, and different A's are related to each other via the usual gauge transformations.
For example, we could cover R^3 with two different A's.
A_1=m/(r*sin(\theta)) * (1-cos(\theta))
A_2-=m/(r*sin(\theta)) * (-1-cos(\theta))
A_1 is singular on the negative z-axis, while A_2 is singular on the positive z-axis. Using them together, we can cover the whole space (well, except the origin, but the monopole is an inherent singularity). You can check that B=\del \cross A gives the formula for a monopole field, using each A on the part where it is defined.
A_2 and A_1 are related by a gauge transformation with the function f=-2m*\phi. Now, we also know how the wave function transforms under a gauge transformation, so on the overlap, we have:
\psi_2=exp^(-i*m*e*phi/hbar) * \psi_1, where e is the electron's charge
Requiring the wavefunction to be single-valued, we get N=(2*e*m)/(hbar), where N is an integer. This is the famous Dirac quantization condition. It's quite beautiful. By imposing a monopole on quantum mechanics, we can "explain" the quantization of charge (well, at least if we assume the quantization of magnetic charge, or vice versa).
We can get this quantization condition in other ways too. For example, we can put an electric monopole at the origin, and a magnetic monopole at some other point in space. We can then calculate the angular momentum stored in the field over all space (by integrating r X S, where S is the Poynting vector). Then, by using well known quantization of angular momentum, we can also get the quantization condition. This way seems a bit shady to me, though, as, to get the correct factor one must equate the angular momentum stored in the field (which I think of as "orbital" angular momentum) with half-integer values of angular momentum (which, is "spin" angular momentum).
A third way to derive it, is to build a Hilbert space out of position and momentum operators in the usual way. We want H=p^2/2m (B fields do no work), and we also have the Heisenberg equations of motion:
p/m=r(dot)=i*[H,r]
ep/m \cross B=p(dot)=i*[H,p]
We get what we want if we impose the following commutation relations:
[r_i, r_j]=0
[r_i, p_j]= i * \delta_{ij}
[p_i, p_j]=i*e*\epsilon_{ijk} B_k
The first two are usual for the commutation relations between position and momentum, while the third one is quite unusual. Of course, this is because, in this particular case, we are considering kinematical momentum (which is usually the same as canonical momentum, but in this case, is not). We can then examine the generator for translations by a, U(a)=exp^(i*a(dot)p/(hbar)). It can be calculated that:
(U(a_1)*U(a_2))*U(a_3)=exp(-i*CHI) U(a_1)*(U(a_2)*U(a_3))
Now, non-associativity of these three operators on a Hilbert space makes no physical sense, so we require that CHI be an integer times 2 pi. This gives the quantization condition as well.
What is CHI? Well, apparently (I have not calculated it) it is the flux of \del (dot) B through a tetrahedron with sides of lengths a_1, a_2, a_1 + a_2, a_2 + a_3, and a_1 +a_2 +a_3. Somehow CHI is related to the overall topology of the space. This was not explained by the professor entirely, but it looks to me a whole like lot simplicial homology. Anybody know what the exact relation is between the physical system and the homology of the space?
