This statement is an excerpt from Quantum Mechanics (Cohen-Tannoudji), but I don't quite understand why it holds: given all our kets live in a certain Hilbert space $\mathcal H$, then all the bras will live in the dual Hilbert space $\mathcal H^*$, and both of them will be equal in dimension and therefore they will be isomorphic. By definition, if they are isomorphic then there exists an isomorphism between the two spaces, which means for every ket there is one and only one bra, and viceversa. Nonetheless, this goes directly against the statement of the Cohen-Tannoudji, because an isomorphism is by definition surjective, whilst the aforementioned statement only requires injectivity and actively rejects surjectivity. Why is my reasoning wrong? I'm thinking it might be failing because I'm using concepts of linear algebra specific for when the space has a finite dimension, but it makes no sense either because nothing is stopping a Hilbert space from having a finite dimension as well.
1 Answers
The kets and bras do not always live in a Hilbert space, but when you write something like the "position eigenbras" $\langle x\rvert$, this lives in a larger space that is part of the Gel'fand triple of a rigged Hilbert space, see also this answer by user1504 for more details. Such a triple consists of a Hilbert space $H$, a subspace $S\subset H$ - and this subspace is not a Hilbert space, so its dual $S^\ast$ is not isomorphic via Riesz representation but we have a sequence of inclusions $S\subset H\subset S^\ast$. The bras live in $S^\ast$ and the kets in $S$ (not that most physics texts are careful about that, mind you, since they write both $\lvert x\rangle$ and $\langle x\rvert$ with abandon) and so there are, in this sense, "more" bras than kets.
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