On 14 Aug 2004 07:58:04 -0400, "Flip Tomato" wrote:


What motivates the definition of the gauge covariant derivative, other than
that it gives a nice result?


There are many possible answers to this question. Let me propose one
that appeals to me.

Consider quantum particle in the external EM field. Compute "velocity
operator" dx(t)/dt - which should be an "observable". Indeed, we can
measure velocities of particles. But when you compute it, you see that
is proportional to id/dx - A. But A is an electromagnetic potential
that, in itself is not an observable, because it can be replaced by A+d
lambda with the same physical implications. Therefore i d/dx must not be
an observable either! The only solution out of the dilemma is:

whenever you replace A by A+ d lambda, replace your wave function psi by
exp(i lambda) psi, so that the two contributions from d/dx and A will
cancel and velocity will be unchanged.

ark
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Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
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