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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.

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I didn't get that. Could you add some details and notation? What does the dual space act on? On what space do we end up getting a hermitian inner product? –  Gunnar Magnusson Jun 17 '12 at 18:04
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@GunnarMagnusson: sure, done! Hope it makes more sense now! –  Edward Hughes Jun 17 '12 at 18:19
    
Also - should I split this into 2 questions as a result of my edit? Any advice from a moderator would be appreciated! –  Edward Hughes Jun 17 '12 at 19:07
    
@hgbreton: Perhaps you'd have better chances of having your question answered at physics.stackexchange.com. –  William Jun 17 '12 at 19:10
    
Have a look at the very first pages of Folland's Quantum Field Theory: a tourist guide for mathematicians. The author addresses concisely the physicist's notations from a mathematician's point of view. –  Giuseppe Negro Jun 17 '12 at 19:12
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2 Answers 2

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The question might be old, but I feel as though I might be able to shed some light on this problem: I had a similar qualm when studying QM.

First, we must realize what we mean by an inner-product, and a bilinear pairing.

  1. If $V$ is a $\mathbb C$-vector space, an inner product is a nondegenerate, (sesqui/bi)linear, positive definite map $\langle \cdot,\cdot\rangle: V \times V \to \mathbb C$ (sometimes written as $V\otimes V \to \mathbb C$ to account for bilinearity).
  2. If $V$ is a complex vector space, let $V^*$ be its dual space. By definition, we have a non-degenerate bilinear pairing $\langle \cdot,\cdot\rangle : V^* \times V \to \mathbb C$ which acts by taking $\gamma \in V^*$ and $v \in V$ to $\gamma(v) \in \mathbb C$. (again, this is sometimes written as $V^* \otimes V \to \mathbb C$).

Aside: In more general contexts there are even issues with degeneracy, and positive definiteness is a meaningless property. One can extend these ideas to bilinear pairings of arbitrary modules, where if $M$ is an $R$-module, we define $M^* = \operatorname{Hom}_{R\text{-mod}}(M,R)$, and examine the maps in $\operatorname{Hom}_{R\text{-mod}}(M\otimes_R M^*,R)$. There is no reason to suspect such maps should be non-degenerate, and without an ordering on $R$ there is no notion of positivity.

While the two notions above look similar, there is no reason they should be related. For sufficiently grotesque vector spaces, they won't be. The power of the Riesz representation theorem is that it tells us that when $(V, \langle \cdot, \cdot \rangle)$ is a Hilbert space, then the bilinear pairing is a bona-fide inner product and there is an isometric (complex anti-)isomorphism $\Phi: V \to V^*$ which makes them equivalent.

The Dirac notation is exploiting this equivalence, by saying that the element $\langle \psi | \phi \rangle$, thought of as the linear functional $\langle \psi|$ acting on $|\phi\rangle$, is precisely the same thing as the inner product $\langle \psi |\phi\rangle$.

Now, the motivation for (separable) Hilbert spaces is, to my knowledge, just that they are very nice. There is some corroboration coming from the fact that wave functions $L^2(\mathbb R^n,\mathbb C)$ and discrete state representations $\mathbb C^n$ (or $SU(n)$ if you prefer the operator viewpoint) all have natural Hilbert space structures compatible with the quantum mechanics.

It actually turns out that Hilbert spaces are not good enough for quantum mechanics. The reason is that the momentum operator $i\frac\partial{\partial x}$ is not a bounded operator: alternatively, $x \to e^{ix}$ (the eigenfunction of the momentum operator) is not a square integrable map. Hence the correct place to do things in quantum mechanics is the collection of Schwarz distributions, also known as rigged Hilbert space.

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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?).

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