Convergence of $e^{itH}$ as a sum for $H$ and unbounded operator

Let $H$ be an unbounded operator on a Hilbert space $mathcal{H}$. We may define the propagator $U(t)=e^{itH}$ via the functional calculus. However, if $H$ were bounded, the series $sum_{k=0}^infty frac{(itH)^k}{k!}$ converges in the operator norm, so we do not need the functional calculus to make sense of propagator. It is claimed in Reed and Simon as well as elsewhere that for $H$ unbounded the above series does not make sense. I see that the above should not converge in the operator norm if $H$ is not bounded, but I am wondering if we can make sense of it as a strong limit. So I ask:

  1. Does $sum_{k=0}^infty frac{(itH)^k}{k!}$ always exist as a strong limit and if not can you give an example of an operator for which this occurs?
  2. Can it happen that $sum_{k=0}^infty frac{(itH)^k}{k!}$ exists as a strong limit but that it does not equal $U(t)$ as defined via the spectral calculus or has the incorrect domain?

In other words, I am asking for a "proof" of the necessity of constructing the propagator via the functional calculus, and not via the more natural route of power series.

Mathematics Asked by user293794 on November 12, 2021

1 Answers

One Answer

If $H$ is self adjoint on $mathcal H$, then you can write it as its spectral decomposition: $$ H = int_{mathbb R} lambda , dP_lambda $$ Then restricted to $$mathcal H_n = left(int_{[-n,n]} , dP_lambda right) (mathcal H) = (P_{n+} - P_{-n-})(mathcal H)$$ we have that $H$ is invariant and bounded in norm by by $n$. And $$ mathcal H_0 = bigcup_n mathcal H_n $$ is dense in $mathcal H$. So on the dense subspace $mathcal H_0$, the infinite series does converge strongly.

Of course, this is effectively the functional calculus written in a slightly different way. But your example is rather close to considering $ifrac d{dt}$ on the dense subspace of functions whose Fourier transform is compactly supported, and hence the example you give is not far from this process.

It all really boils down to the single example $H f(x) = x f(x)$ on $L^2(Omega)$ where $Omega subset mathbb R$. This is precisely what the spectral decomposition does. Your example is $f(x) = frac 1{1 + (lambda x)^2} $, which although not compactly supported, definitely converges to $0$ quite rapidly as $x to pm infty$.

Answered by Stephen Montgomery-Smith on November 12, 2021

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