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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?

user246795
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    Why is it useful to have both row and column vectors in linear algebra? – Andrew Feb 16 '22 at 21:43
  • @Andrew But why do we need to consider a separate "dual space" that has a isomorphism to be able to use the idea of both row and column vectors? – user246795 Feb 16 '22 at 21:45
  • @JasonFunderberker I agree that the notation is convenient, however I don't understand where the concept of dual spaces becomes significant. It seems to me (though I have very little experience in QM yet) that we could redefine the notation simply using the vector space and the inner product, without needing the ideas of dual spaces? – user246795 Feb 16 '22 at 21:56
  • The column vectors live in the dual space of the row vectors. So whether you know it or not, you are using the concept of a dual space in regular linear algebra. There are clearly uses for having both, since $u^T v$ is not the same thing as $u v^T$ (it's not even the same type of mathematical object). The analog in quantum mechanics is that $\langle \chi | \psi \rangle$ and $| \chi \rangle | \psi \rangle$ are two different kinds of objects, but both are useful. On top of this, since $\langle \chi | \phi \rangle = \langle \phi | \chi \rangle^\star$, there is a numerical difference (...) – Andrew Feb 16 '22 at 22:02
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    (...) you can think of kets as being like row vectors, and bras as being like column vectors (for systems with a finite number of states, like a single spin, this is actually true), but with the twist that to go from a row to a column vector, you also complex conjugate the values. It's just all part of what makes the formalism work; if you didn't introduce the concept explicitly, you would still be using it implicitly (since you need to compute inner products between states, for example), but you would be making your work harder by not taking advantage of a nice mathematical structure. – Andrew Feb 16 '22 at 22:04
  • I don't think that's a necessity for physicist's (contrary to mathematician's) version of QM. In Sakurai's modern quantum mechanics revised edition, the author(s) do(es) not mention dual space.. – AlphaF20 Feb 20 '22 at 09:11
  • @AlphaF20 Not only do I disagree with your statement, because to use a framework correctly it is necessary to understand its structure. But also Sakurai mentions the dual space in chapter 1.2 (In my edition on pages 12-13). Also later in the book he speaks about "dual correspondence", where he (laxly) uses that the dual space is isomorphic to the space itself. Of course this is more complicated for infinite dimensional vector spaces. – AlmostClueless Feb 20 '22 at 09:47
  • @AlmostClueless I mean the formal concept 'dual space'. In my version, Sakurai used 'dual to' and 'dual correspondence'. I think Sakurai's approach is informal without introducing the mathematical definition. And I think op refers to the formal approach, since the Riesz-Representation theorem was mentioned. It might be that I misunderstood the op/Sakurai at some point (op's last sentence about only ket space is used, maybe op wants to get rid of bra space entirely) – AlphaF20 Feb 20 '22 at 14:16

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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.

ACuriousMind
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  • So there are contexts where the notion of dual spaces become useful, but for most basic applications of the bra-ket notation they needn't be considered? – user246795 Feb 17 '22 at 09:27
  • Does the bra notation simply point us to consider bra vectors as functionals, instead of seeing them as "inner product ket vectors" and the dual space allows us to do this because there is an isomorphism that makes both methods equivalent? And that there are benefits to considering them as functionals in more advanced applications? – user246795 Feb 17 '22 at 10:05
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    @user246795 I don't really understand what you're asking - you sound as if there was some additional burden to "considering" duals spaces, but there isn't. If you write down bras, you're writing down dual vectors, regardless of whether you're consciously aware of it. For an elementary comparison, I can view $3\cdot 4$ as "multiplication" in its own right or as the repeated addition $4+4+4$. That's just two equivalent views. Would you ask when someone "needs" to consider one or the other? – ACuriousMind Feb 17 '22 at 10:26
  • Ok that makes sense, but could I just clarify that it is the (anti-)isomorphism that makes them equivalent mathematically? – user246795 Feb 17 '22 at 12:46
  • @user246795 Yes. In mathematics we often consider isomorphic objects "the same" for all practical purposes. – ACuriousMind Feb 17 '22 at 13:35