What do we miss if we consider the semi-normed vector space ${\cal L}^{2}(\mathbb{R})$ of square integrable functions on $\mathbb{R}$ as the state space in one-dimensional quantum mechanics instead of Hilbert space $L^{2}(\mathbb{R})={\cal L}^{2}(\mathbb{R})/\!\sim$ which consists not of functions but of equivalence classes of functions?
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$\begingroup$ This is a very good issue proper of Mathematical Physics concerning the mathematical description of quantum states. I cannot see a reason to close it. (Otherwise also my reasearch field should be considered improper here and I should move to math stack exchange.) $\endgroup$– Valter MorettiCommented Oct 3 at 7:27
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2 Answers
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The point is that, dealing with the foundation of Quantum Theory as suggested, the completeness property of the vector space whose elements represent pure states would cease to hold. Fundamental theoretical results at the basis of Quantum Theory, as the spectral theorem itself, use that hypothesis pervasively.
In a sense, it would be as using only rational numbers to formulate classical mechanics. Perhaps it is still possible: after all there is no chance to distinguish rational numbers from real numbers with physical instruments, but it would turn out very cumbersome.
From a philosophical perspective, but this is not my area of investigation, the issue concerns the existence of theoretical entities.
ADDENDUM. Above, I am referring to completeness referred to the (Hilbert space) norm structure. Contrarily, the complete seminormed space ${\cal L}^2(\mathbb{R})$ is not Hausdorff (just in view of the definition of seminorm when it is not a norm, as in the considered case). In particular, the limit of a sequence is not unique, with several drawbacks in the physical interpretation. Taking the obvious quotient, to get rid of the problem, is just equivalent to pass to the usual Hilbert space structure.
ADDENDUM2. There are in fact Hilbert spaces of proper functions to describe quantum states, as the Segal-Bargmann space. Since some of these spaces are separable as $L^2(\mathbb{R})$ is, there exist infinitely many Hilbert space isomorphisms connecting them. However one should search for physical intepretations. In a sense, the domain of the "wavefunctions" of SB space can be interpreted as "a sort of" phase space. See the discussion in the above wiki page.
answered Oct 1 at 12:43
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$\begingroup$ I am not sure that completeness is a problem here. Riesz-Fisher theorem (as in Riesz and Nagy Functional Analysis) establishes completeness in seminorm for ${\cal L}^{2}(\mathbb{R})$ and completess in norm for $L^{2}(\mathbb{R})$. Completeness is a problem for the space of continuous functions on $\mathbb{R}$ and the completion of this space is $L^{2}(\mathbb{R})$. Am I wrong? $\endgroup$– val 72Commented Oct 1 at 21:29
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$\begingroup$ Yes I was not precise. I meant completeness in the norm generated by the inner product. The Hilbert space structure is strongly exploited in the proof of the spectral theorem. $\endgroup$ Commented Oct 2 at 7:16
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$\begingroup$ Furthermore, the complete seminormed space you consider is not Hausdorff (just in view of the definition od seminorm which is not a norm as in this case). This implies in particular that the limit of sequences is not unique with several drawbacks inthe physical intepretation. Taking the quotient to get rid of the problem is equivalent to pass to the Hilbert space structure. $\endgroup$ Commented Oct 2 at 7:46
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$\begingroup$ Consider the following construction: starting with $\mathcal{L}^{2}(\mathbb{R})$ equipped with the usual seminorm $\left\Vert \centerdot\right\Vert$ , let $\mathcal{N}=\left\{ f:\left\Vert f\right\Vert =0\right\}$ and $\mathcal{M}=\left\{ f:\forall g\in\mathcal{L}^{2}(\mathbb{R}),\,f-g\notin\mathcal{N}\textrm{ or }f=0\right\}$ . Is the space $\left(\mathcal{M},\left\Vert \centerdot\right\Vert _{\mathcal{M}}\right)$ a complete normed space isometrically isomorphic with $L^{2}(\mathbb{R})$? $\endgroup$– val 72Commented Oct 2 at 23:01
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$\begingroup$ Apologies, I mean $\mathcal{M}=\left\{ f:f\notin\mathcal{N}\textrm{ or }f=0\right\}$ $\endgroup$– val 72Commented Oct 3 at 4:40
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In quantum mechanics, we generally consider the state space to be some hilbert space $\mathcal{H}$ over $\mathbb{C}$ and (pure) states to be one-dimensional orthogonal projectors on $\mathcal{H}$. A part of this is especially, that for $\lambda \in \mathbb{C}$ with $|\lambda| = 1$, the two vectors $\phi, \lambda \phi$ are the same state. The reason would be that no measurement would be able to distinguish between them. In the same notion, if we dont consider the hilbert space but the semi normed vector space, we would distinguish between things that measurement could not distinguish between.
