quantum field theory with differential forms ed

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

Table of Contents

prelude ed

In conventional quantum mechanics, there is a distinction between time and space. Space degrees of freedom are "observables", while time is an axis along which these observables change. This is even true for qft, where the whole state of a field along a 3d-slice is encoded in a state vector that changes along the 4th axis. (At least during quantization)

Here we have change in all 4 directions and end up with outer derivatives \[df\] instead of \[\dot f\].

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.

Schrödinger equation?

Heisenberg picture ed

Classically, the dynamics come from the Poisson brackets of \[{A, \mathcal H} = dA\] for an observable form function \[A(\omega, \pi)\]. This should correspond to the Heisenberg equation

\[ [A, \mathcal H] = i \, dA \]

So, let's try for the (space-time-dependent) operator \[\omega\]...

\[ [\omega, \mathcal H] = \frac{1}{2} [\omega, \pi \wedge \star \pi] \]
But what about \[\star \pi\]? Paranoidly checking the classical brackets gives \[\{\omega, \star \pi\} = \frac{\partial \star \pi}{\partial \pi} \]... this is not defined in the strong sense. But if we allow more general derivatives, this gives a linear operator, that maps each \[\mu\] onto \[\star \mu\]... so
\[\{\omega, \star \pi\} = \star \qquad \] or \[ \qquad [\omega, \star \pi] = i \star\]
Ok, back to the Heisenberg equation:
\[ [\omega, \mathcal H] = \frac{1}{2} (\omega \wedge \pi \wedge \star \pi - \pi \wedge \star \pi \wedge \omega) \]
\[ \qquad = \frac{1}{2} (\omega \wedge \pi \wedge \star \pi - \pi \wedge \omega \wedge \star \pi + \pi i \star) \]
\[ \qquad = \frac{1}{2} (\omega \wedge \pi \wedge \star \pi - \omega \wedge \pi \wedge \star \pi + i \pi \star + i \star \pi) \]
Put together:
\[ d\omega = \frac{1}{2}(\pi \star + \star \pi) \]

degrees
If we take the degrees of operators and state functions serious, then the consistency of this equation enforces \[ (n-k-1) + (n-s) = (k+1) + s \] or
\[ s = n - k - 1 \]
(so the state vectors have the same degree as the momentum)

Maxwell ed

Categories: Ideen, Physik