quantum field theory with differential forms ed

Feeble thoughts for creating a qft out of field theory with differential forms

Table of Contents

Klein-Gordon ed

classical theory ed

Hamiltonian
\[ \mathcal H(\omega, \pi) = \frac{1}{2} \pi \wedge \star \pi + \frac{m^2}{2} \omega \wedge \star \omega \]

Schrödinger picture ed

We want to find operators \[\hat\omega, \hat\pi\], so that the Poisson bracket \[\{\omega, \pi\} = 1\] turns into a commutator relation \[[\hat\omega, \hat\pi] = i\]. An "obvious" guess would be

\[ \hat\omega = \omega \quad \hat\pi = - i \frac{\partial}{\partial \omega} \]
This suggests, that these operate on the vector space of form functions \[f(\omega)\] or \[f(\omega, x)\]. Let's try for \[\omega\] a \[k\]-form and \[f\] an \[s\]-form:
\[ [\omega, \pi] f = - i [\omega, \frac{\partial}{\partial \omega}] f = - i \omega \wedge \frac{\partial}{\partial \omega} f + i \frac{\partial}{\partial \omega} (\omega \wedge f) \]
\[ \qquad = - i \omega \wedge f' + i \omega' \wedge f + (-1)^{ks} i \omega \wedge f' \]
This is exactly
\[[\omega, \pi] = i\]
if \[ks\] is even. For the Klein-Gordon field, \[\omega\] is a \[k=0\]-form, so it is always true.

Heisenberg picture ed

Maxwell ed

Categories: Ideen, Physik