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I've seen this statement in many papers. However the function $e^{-iHt}$ satisfies the Cauchy-Riemann condition for all t. So why do we say $e^{-iHt}$ is holomorphic in the lower half plane specifically?

Does this have something to do with infinity or does "holomorphic" imply "bounded"?

For the statement, see for example:

https://arxiv.org/abs/1803.04993 below eq.(2.7)

https://arxiv.org/abs/0710.5373 below eq.(2.62)

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    $\begingroup$ Quite clearly $\exp(-iHt)$ is holomorphic everywhere in the complex plane (but not on the Riemann sphere). So the relevance of the statement you see has to depend on the context (presumably something to do with contour integrals). Without that context, this question is unanswerable. $\endgroup$
    – TimRias
    Commented Nov 4, 2021 at 9:30

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