How Does the Propagator Relate to the Heat Kernel on a Manifold?

Orbb
Messages
81
Reaction score
0
Hi,

I have a question about the relation between the propagator of a scalar field and the heat kernel. I'm not sure wether I should rather put this question into the math section: Given a Laplacian D on some manifold M, what I mean by heat kernel is just

K(x,y;s) = \langle x | \exp(-sD) | y \rangle

where x, y are distinct points on M and s is the diffusion time (in the sense that K obeys the heat eqn.). Now the propagator of a scalar field can be determined from K via

D^{-1}(x,y) = \int_0^{\infty} ds K(x,y;s).

What I want to ask now is wether there is a way to invert this expression such that given some propagator, I can determine the corresponding heat kernel. Can anyobdy help?

Cheers,
O
 
Physics news on Phys.org
Really no one with a hint? Is the question somehow ill-posed? Or should I add more detail about the specific calculation I am attempting?
 
Just a guess here:
The propagator is the inverse of the operator D. So if you were given the propagator, and invert it, you would get back the operator. From there, you should be able to construct your heat kernel.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Replies
3
Views
2K
Replies
13
Views
3K
Replies
5
Views
5K
Replies
3
Views
2K
Replies
6
Views
3K
Replies
5
Views
16K
Replies
1
Views
2K
Replies
5
Views
2K
Back
Top