1
$\begingroup$

The question is related to both this question about the one photon wave functional, and this one about the QED Hamiltonian.

What does the free electromagnetism vacuum state functional look like? My best guess based on the above links and Weinberg's "The Quantum Theory of Fields, Vol. I" sections 8.2 and 8.3 is $$\Psi[A_\perp,\Pi_{||}] \propto \exp\left(-\frac{1}{2} \int \operatorname{d}^3k \left[\left(\delta_{ij} - \frac{k_ik_j}{k^2}\right)kA_i A_j\right]\right)\, \delta\left[i\frac{k_ik_j}{k^2}\Pi_j + \frac{k_i}{k^2}J^0\right],$$ where I've included the $J^0$ term for convenience in later usage where it may be minimally coupled to a matter field. The first factor is your standard Gaussian functional for a free field, and the second is a Dirac delta functional that quantum states tend to be when the Hamiltonian doesn't contain both canonically conjugate variables. Since $\Pi_i=-E_i$, the delta functional is equivalent to saying that the irrotational part of the electric field is fixed by the charge density, which is just the Gauss's law/Coulomb's law part of Maxwell's equations.

Comparing this to equation 2.16 from Greensite's "Calculation of the Yang-Mills Vacuum Wave Functional" (1979) (which the paper cites as matching one from a textbook by Wheeler), it matches except for the extra Dirac delta functional (which, in the case of the paper, would have $J^0=0$).

In particular, why is the constraint functional typically excluded? In my mind, it's no different than if part of a Hamiltonian were free, $H=\frac{p^2}{2m}$, and with no dependence on $x$ the ground state just becomes a momentum eigenstate with $p=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.