# Does it make sense to speak of amplitudes of finite closed boundaries in QFT?

A example of amplitude in Relativistic Quantum Mechanics or specifically in QFT is the amplitude of a field configuration on a space-like hyper-surface of space-time to "lead" to another field configuration on another space-like hyper-surface of space-time. In the path-integral picture one simply integrates over all possible field configurations on the interior, giving each a weight in the normal way. Now if one wants to generalize this to finite closed boundaries, we would get an amplitude for each field configuration on a finite closed boundary of space-time. but how would we interpret this? This question relates to interpretations of quantum mechanics, has anyone investigated this line ?

-
Actually in QFT (at least in particle physics) one usually calculates the amplitude between two spatial slices which are infinitely separated in time, i.e. the S-matrix. The infinite separation is needed to make the asymptotic state be Fock-like particles. Real time-dependent QFT (sometimes used in condensed matter) requires some very imaginative formulation of the path integral (see Keldysh formalism). I think what you're asking about has been pondered mostly by quantum gravity people, who have this problem in spades. See works by Rovelli. – genneth Mar 27 '12 at 14:45
@genneth I think you should post your very informative comments as an answer. – Slaviks Mar 27 '12 at 16:42
@genneth: This is not completely true--- it is true for S-matrix calculations, but the original pure field theory calculations of Schwinger, which were based on Feynman's path integral (in Schwinger's action principle reformulation) were between two finite time hypersurfaces, and this is still the cleanest way theoretically. The S-matrix thing was only in response to the quest for a pure S-matrix theory, which quantum field theory isn't. – Ron Maimon Mar 28 '12 at 8:03
@RonMaimon: perhaps I'm misunderstanding the thrust of the question or your comment, but I think the OP wants to know whether it is possible and what it would mean to assign an amplitude to a field configuration defined on the boundary of a spacetime hypervolume, i.e. including time-like parts of the boundary. I wasn't aware that there was anything done on this formalism outside of quantum gravity circles? – genneth Apr 1 '12 at 11:38
@RonMaimon: but I think that's the whole point of the OP's question --- what to do about the timelike (I assume that's what you meant to write in the last sentence) parts of the boundary, and what the resulting amplitude means. Mohamed should correct me if I has mis-understood. – genneth Apr 1 '12 at 16:44

A wave functional of a field A is $\Psi[A]$. This is the amplitude that there is a certain field configuration A. (Compare to ordinary quantum mechanics where a wave function $\psi(x)$ is the amplitude for a particle to be at position x). If a wave function is highly peaked at a particular value it means that this value is most likely. Similary a wave functional of a field A can be highly peaked at a certain field configuration f. It may be a Gaussian such as

$$\Psi[A] = \exp\left(-\int (A(x)-f(x))^2 dx^3 \right)$$

(This only works for bosons. Fermions don't have anything corresponding to a classical field).

The universe at any time is described by the wave functional of the fields (which may be highly peaked at a particular field configuration... or not). The amplitude that the universe will have a different wave functional $\Psi_{out}$ at a later time is given by the path integral:

$$\Delta[\Psi_{in},\Psi_{out}] = \int \Psi_{in}[A]\Psi_{out}[A]e^{i S[A] } D[A]$$

One can calculate this by expanding the wave functional in terms of particle amplitudes:

$$\Psi_{in}[A] = a + \int\psi(x)A(x,t_{in})dx^3 + \int\psi(x,y)A(x,t_{in})A(y,t_{in})dx^3dy^3+...$$

Where $\psi(x,y)$ is the amplitude for particles being found at both positions x and y. In particular we have:

$$\Delta_F(x,y) = \int A(x) A(y) e^{i S[A] } D[A]$$

which is the amplitude for a particle to travel from x to y.

As for closed boundaries, the time slice for the incoming data and the time slice for the outgoing data can be joined at the boundaries forming a nut shape.

-