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

We want to find operators \(\hat\omega, \hat\pi\), so that the Poisson bracket \(\{\omega, \pi\} = 1\) turns into a commutator relation \([\hat\omega, \hat\pi] = i\).

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

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\[ \{\omega, \star \pi\} = \star \qquad \text{ or } \qquad [\omega, \star \pi] = i \star \]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)

Schrödinger picture ed

momentum operator
An "obvious" guess for the operators 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?
From the regular Schrödinger equation \( H |\Psi\rangle = i \dot |\Psi\rangle \), we would suspect \( \mathcal H |\Psi\rangle = i \, d |\Psi\rangle \)...

space-time evolution
We start by demanding that states propagate unitarily. That means, there should be a unitary operator field \(U(x)\) so that the state field satisfies\[ |\Psi(x)\rangle = U(x) |\Psi(0)\rangle\](...??...) To mimic regular qm, we write\[ i \, dU = \Omega U \]and want to show that \(\Omega = \mathcal H\).

Let's look at the Heisenberg equation \([A(x), \mathcal H(x)] = i \, dA(x)\). The right side using \( A(x) = U^\dagger A_0 U\) gives\[ i \, dA = i \, dU^\dagger \, A_0 \, U + i \, U^\dagger \, A_0 \, dU = - U^\dagger \Omega A_0 U + U^\dagger A_0 \Omega U = U^\dagger [A_0, \Omega] U \]Comparing this with the left side\[U^\dagger [A_0, \mathcal H] U\]we can see that \(\Omega = \mathcal H\) ...(hmmm, PROVE?)

Schrödinger equation
Now that we have \(U(x)\):\[ i\,d|\Psi_{x}\rangle = i\,dU |\Psi_0\rangle = \mathcal H U(x)|\Psi_0\rangle = \mathcal H |\Psi_{x}\rangle \]

Maxwell ed

Categories: Ideen, Physik