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When we say that an operator is Hermitian in QM, does it depend on the Hilbert space under consideration, or not? Are there operators that are Hermitian in one Hilbert space but not in another?

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    $\begingroup$ Have you illustrated this with finite-dimensional matrices and submatrices? $\endgroup$ – Cosmas Zachos Feb 28 at 17:53
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    $\begingroup$ An operator is a map from one Hilbert space to another. It doesn't make sense to talk about it acting on a different Hilbert space surely? $\endgroup$ – jacob1729 Feb 28 at 19:17
  • $\begingroup$ Do we have to have a basis to answer that question? $\endgroup$ – Frank Feb 28 at 19:17
  • $\begingroup$ @jacob1729 - let's go from $H$ to itself, so there is only one space. $\endgroup$ – Frank Feb 28 at 19:18
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    $\begingroup$ @Frank yes, that's what I meant. I can't edit my comment anymore. My point stands that if you have any operator acting on some vector space, then there isn't a notion of it acting on some other vector space. $\endgroup$ – jacob1729 Feb 28 at 19:24
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There is a unique Hilbert space upto isomorphism for any given dimension.

So, a Hermetian operator on one Hilbert space of dimension $d$ is also a Hermetian operator on any other Hilbert space of the same dimension $d$.

So the answer to your question is no.

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