I'm starting to learn about Dirac notation in Quantum Mechanics, and am coming from a pure background. The notes I'm reading states that we assume that the action of the dual space on the state space $V$ induces a Hermitian inner product.
More explicitly if we identify $<\psi|\in V $ with $|\psi>\in V^*$ we define an inner product on $V$ by $(|\psi>,|\phi>)\rightarrow <\psi|\phi>$. This is basically what is written in the notes, and seems to be the way of doing things in QM, but coming from a pure background prompted me to consider its rigour.
Thinking more rigorously I was wondering under what conditions this is true in general?
It feels like I should be doing something like a reverse engineered Riesz Representation Theorem, but I can't work out exactly how. Sorry if this is a stupid/obvious question - I may be a little rusty after a week or so away from maths!
Edit: After a bit more reading and thinking it seems that the more rigorous way to go is the following. Assume that $V$ a Hilbert space, w.r.t. to an inner product which we'll notate by $<\psi|\phi>$. By the Riesz Repn Theorem we know that we may consider $<\psi|$ to be an element of the dual space $V^*$. My question now becomes, why can we assume $V$ a Hilbert space? What's the motivation for that in QM?
Thanks in advance.
I don't understand the question. Let $V$ be a vector space over an arbitrary field $k$ equipped with a function $\langle -, - \rangle : V \times V \to k$ which is linear in the second argument. Then every $v \in V$ gives rise to a dual vector $\langle v, - \rangle : V \to k$ which the Dirac notation denotes by $\langle v |$. (For inner product spaces this assignment is itself conjugate-linear.)
There is no need to appeal to Riesz representation (which is about the converse of this claim) and this has nothing to do with the completeness of the underlying space (although completeness is necessary and sufficient for Riesz representation to hold; is this what you're asking about?).