Asymptotic Expansion of the Schrödinger kernel?

2013-01-14T20:24:56Z

My stackexchange post [http://math.stackexchange.com/questions/275830/schrodinger-kernels-on-manifolds] was somewhat unsatisfactory (also because I may not have stated clear enough what my interest was). So here it goes!

Let $M$ be a compact Riemannian manifold and $\Delta$ be the Laplace-Beltrami operator. It is well-known that the solution operator to the heat equation $e^{t \Delta}$ is smoothing for $t>0$ and has a smooth integral kernel $k_t(x, y) \in C^\infty(M \times M)$. Furthermore, $k_t$ has an asymptotic expansion $$ k_t(x, y) \sim \underbrace{(4 \pi t)^{-n/2} \exp \left( -\frac{1}{4t} \mathrm{dist}(x, y)^2 \right)}_{:= e_t(x, y)} \sum_{j=0}^\infty t^j \Phi_j(x, y) $$ meaning that $$ \left| k_t(x, y) - e_t(x, y) \sum_{j=0}^N t^j \Phi_j(x, y) \right| \leq C t^{N+1}$$ uniformly in $x$ and $y$ in a neighborhood of the diagonal.

Now by by formally substituting $t \rightarrow it$, one gets the formal asymptotic series $$ e_{it}(x, y) \sum_{j=0}^\infty (it)^j \Phi_j(x, y),$$ which has the property that it formally (i.e. termwise, as asymptotic series in $t$) solves the Schrödinger equation $ \left(i \frac{\partial}{\partial t} + \Delta\right)k_t = 0.$

Now my question is the following: Does this asymptotic series have any relation to the solution operator $e^{it\Delta}$ of the Schrödinger equation, or to its distribution kernel?

Answer by Carlo Beenakker for Asymptotic Expansion of the Schrödinger kernel?

2013-01-15T13:10:53Z

In this 2006 paper you can find the long-time asymptotics of the Schrödinger kernel on Riemannian manifolds. My understanding is that the analytic continuation to imaginary time gives the correct answer provided there are no zero-energy resonant states.

Asymptotic Expansion of the Schrödinger kernel? - MathOverflow

Asymptotic Expansion of the Schrödinger kernel?

Answer by Carlo Beenakker for Asymptotic Expansion of the Schrödinger kernel?