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There is a vast literature available for the heat kernel. Nevertheless, I haven't been able to find almost anything useful about the kernel of the equation $\frac{1}{\mathbb{i}} \frac{\partial u}{\partial t} - \Delta u =0$. So far, I have three questions about it:

1) if $e(t,x,y)$ is the above kernel, is it bounded in $x$ (in $\mathbb{R}^n$ it is)?

2) since $e$ is not real, it is impossible to construct a Gaussian probability starting from it; what, then, can still be done? I assume one can still construct a complex measure, but what are its "nice" properties?

3) finally, do you know of any rich monography on the subject? I have found plenty of titles about the "usual" heat kernel, yet none regarding my question - which I find strange given how important the Schrödinger equation is.

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As far as I know, the term "Mehler Kernel" is used for the integral kernel of the heat equation corresponding the the harmonic oscillator, $$ \partial_tu + \Delta u + x^2 u = 0.$$ The equation you are talking about is the Schrödinger equation. The kernel on $\mathbb{R}^n$ is $$ e_t(x, y) = (4\pi i t)^{-n/2} e^{-\frac{1}{4 i t} |x-y|^2},$$ that is the heat kernel with imaginary values plugged in for $t$. Despite the the formal similarity to the heat kernel, it has vastly different properties: For example, the integral $$u(t, x) := \int_{\mathbb{R}^n} e_t(x, y) u(y) \mathrm{d} y$$ does not converge absolutely for arbitrary functions in $L^2$, and it is not smoothing in the sense that it maps distributions to smooth functions. However, it preserves smoothness.

In particular, it does not define a complex $\sigma$-aditive measure!

On manifolds, this gets way worse: For example, I know from some corner of my memory that on the sphere, it is not smooth at all; instead, it is a singular distribution at every point at most times.

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  • $\begingroup$ Are you sure about the non-smoothness of $e$? Isn't the operator $\frac{1} {\mathbb{i}} \frac{\partial} {\partial t} - \Delta$ hypoelliptic? $\endgroup$
    – Alex M.
    Oct 20, 2014 at 14:06
  • $\begingroup$ No, why would it be? $\endgroup$ Oct 21, 2014 at 7:24
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    $\begingroup$ Depending on the sign conventions, $z$ may be in either the left or the right half plane to make the operator still hypoelliptic. For $z$ on the imaginary axis (the "boundary case"), it never hypoelliptic. You can easily verify this on $S^1$ by Fourier transform for example. $\endgroup$ Oct 22, 2014 at 15:39
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    $\begingroup$ I don't know any "real" references, just posts like mathoverflow.net/questions/114492/… and mathoverflow.net/questions/114492/… $\endgroup$ May 3, 2015 at 20:05
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    $\begingroup$ math.stackexchange.com/questions/275830/… $\endgroup$ May 4, 2015 at 8:40

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