7
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

No, in general, the expansion in small times of the heat kernel $k_t(x,y)$ does not tell much about the Schrodinger semigroup. You may think about the circle case for which the kernel of $e^{t \Delta}$ has an expansion at any order

$ k_t(x,y)=\frac{1}{\sqrt{4\pi t}} e^{-\frac{(y-x)^2}{4t}} (1 +O(t^m))$

but the Schrodinger semigroup $e^{it \Delta}$ has a rather complicated behavior. The Schrodinger kernel in that case is actually a distribution. If $t$ is a rational multiple of $2\pi$, it is a finite linear combination of delta functions which is interestingly connected to Gauss sums.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.