I have been studying bra-ket notation for QM and have come across the concept of dual spaces and their relation to bra vectors. I can appreciate that the definition of a dual space is such that, given a vector space $\mathcal{V}$ on a field $\mathbb{F}$ the dual space is the set of all linear functionals $\phi: \mathcal{V} \rightarrow \mathbb{F}$. I also accept that for finite dimensional vector spaces there is an isomorphism between the space and its dual (and that for infinite dimensional spaces one can use the Riesz-Representation theorem guaranteeing a correspondence to the topological dual of $\mathcal{V}$).
Clearly, however I don't appreciate these topics enough to be able to justify why any of the ideas are required in QM. Why, for example, does the dual space (or topological dual space) even need to be considered for QM? Why does there have to be a 1-1 correspondence between bra and ket vectors for the formalism to work? Surely some alternate mathematical framework could be used just on the vector space itself (e.g. $\langle a|b \rangle$ might just be defined as being an operation between two ket vectors $\in \mathcal{V}$) where the idea of a dual space isn't needed?
Edit
Using the definition here where for a ket vector $|\phi \rangle$ we define the functional $f_{\phi}= \langle\phi|$ by $(|\phi \rangle, |\psi \rangle)$ where $( , )$ denotes the inner product on the Hilbert space, what is the use of considering the functionals as a dual space? Surely, all the useful information is contained in the above definition and only the ket vector space is needed?
Best Answer
Let $H$ be a (separable) Hilbert space, such as we have in quantum mechanics. You can say that its inner product is a map $H\times H\to\mathbb{C}, (\lvert v\rangle,\lvert w\rangle)\mapsto \langle v\vert w\rangle$, and for $\lvert v\rangle\in H$ we can define $\langle v\vert \in H^\ast$ by the map $H\to \mathbb{C}, \lvert w\rangle\mapsto \langle v \vert w\rangle$. This is how the idea of dual spaces arises naturally for inner product spaces - the inner product gives you a natural map $H\to H^\ast, \lvert v\rangle \to \langle v\rvert$ from the space to its dual and the Riesz representation theorem establishes that this map is an (anti-)isomorphism.
Therefore, it is mathematically completely equivalent to view $\langle v\vert w\rangle$ either as the inner product of two vectors in $H$, or as the action of an element of $H^\ast$ on $H$. There is rarely any "need" to consider a particular of these viewpoints - since they are equivalent, you can rephrase every statement about the inner product as a statement about the dual and vice versa - but some things might be easier to express.
Note that the dual space and the map from kets to bras is completely fixed by $H$ and its inner product - there is no choice here, there isn't any additional information needed to talk about the dual space, so it's a bit strange to ask whether we "need" the dual space - it's just something that exists.
The notion of the dual becomes more subtle and more relevant as something distinct from the inner product when you try to formalize what certain objects like the "eigenstates" $\lvert x_0\rangle$ of the position operator are. They aren't elements of the Hilbert space, which is easiest to see when people write them in the position basis as $f(x) = \delta(x - x_0)$ - this isn't even a proper function, let alone an element of the Hilbert space of square-integrable functions $L^2$. However, the $\delta$-function is a tempered distribution $S^\ast$, which is the dual space to the space of Schwartz functions $S$. That is, if you want to be careful about it, $\langle x\rvert$ exists as a bra on the space of "Schwartz kets", but not as a ket, nor as a bra on the full Hilbert space. The sequence of spaces $S\subset L^2 \subset S^\ast$ is known as a Gel'fand triple and leads one to the notion of a rigged Hilbert space.