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Furthermore, neither of these functions has eigenvectors in $\mathcal{H}$: if $X\,f(x) = \lambda f(x) = x f(x)\,\forall x\in\mathbb{R}$ then $f(x) = 0$ for $x\neq\lambda$ and the eigenfunction $e^{-i\,k\,x}$ of $P$ is not normalisable.

For QM we take the dense subset $S$ to be the "smooth" functions that still belong to $\mathcal{H}$ when mapped by any member of the algebra of operators generated by $X$ and $P$. That is, $S$ is invariant under this algebra and comprises precisely the Schwartz space of functions than can be multiplied by any polynomial and differentiated any number of times and still belong to $\mathcal{H}$. Any function in $\mathcal{H}$ can be arbitrarily well (with respect to the Hilbert space norm) by some function in $S$.

At the same time, we kit the dense subset $S$ out with a stronger topology than the original Hilbert space one. Why do we do this? One of the basic problems with $\mathcal{H}$ is that the Dirac delta $\delta:\mathbf{L}^2(\mathbb{R})\to \mathbb{C};\;\delta\;f(x) = f(0)$, which can be construed as an eigenvector of $X$, is not a continuous linear functional on $\mathcal{H}$ even though of course it is a linear functional. To see this, consider the image of $f(x) + \exp(-x^2/(2 \sigma^2)$ under the delta funcional: we can choose a $\sigma$ to make this function arbitrarily near to $f(x)$ as measured by the $\mathbf{L}^2$ norm, but with images $f(0)$ and $f(0)+1$, respectively, under the Dirac $delta$. So we kit the dense subset $S$ out a topology that is strong enough to "ferret out" all useful linear functionals and make them continuous. We now have a topological dual (space of all linear functionals continuous with respect to the stronger topology) $S^*$ of $S$ such that $S\subset\mathcal{H} = \mathcal{H}^*\subset S^*$.

Furthermore, neither of these functions has eigenvectors in $\mathcal{H}$: if $X\,f(x) = \lambda f(x) = x f(x)\,\forall x\in\mathbb{R}$ then $f(x) = 0$ for $x\neq\lambda$ and the eigenfunction $e^{i\,k\,x}$ of $P$ is not normalisable.

For QM we take the dense subset $S$ to be the "smooth" functions that still belong to $\mathcal{H}$ when mapped by any member of the algebra of operators generated by $X$ and $P$. That is, $S$ is invariant under this algebra and comprises precisely the Schwartz space of functions than can be multiplied by any polynomial and differentiated any number of times and still belong to $\mathcal{H}$. Any function in $\mathcal{H}$ can be arbitrarily well approximated (with respect to the Hilbert space norm) by some function in $S$.

At the same time, we kit the dense subset $S$ out with a stronger topology than the original Hilbert space one. Why do we do this? One of the basic problems with $\mathcal{H}$ is that the Dirac delta $\delta:\mathbf{L}^2(\mathbb{R})\to \mathbb{C};\;\delta\;f(x) = f(0)$, which can be construed as an eigenvector of $X$, is not a continuous linear functional on $\mathcal{H}$ even though of course it is a linear functional. To see this, consider the image of $f(x) + \exp(-x^2/(2 \sigma^2)$ under the delta funcional: we can choose a $\sigma$ to make this function arbitrarily near to $f(x)$ as measured by the $\mathbf{L}^2$ norm, but with images $f(0)$ and $f(0)+1$, respectively, under the Dirac $\delta$. So we kit the dense subset $S$ out a topology that is strong enough to "ferret out" all useful linear functionals and make them continuous. We now have a topological dual (space of all linear functionals continuous with respect to the stronger topology) $S^*$ of $S$ such that $S\subset\mathcal{H} = \mathcal{H}^*\subset S^*$.

1. This answer to the Physics Stack Exchange question "Rigged Hilbert space and QM" and also
2. The discussions under the Math Overflow Question "Good references for Rigged Hilbert spaces?"

1. This answer to the Physics Stack Exchange question "Rigged Hilbert space and QM" and also
2. The discussions under the Math Overflow Question "Good references for Rigged Hilbert spaces?"

Less flippantly, here is where we get practical. In quantum mechanics, we need to implement the Heisenberg uncertainty principle, so we need Hermitian observables $\hat{X}$ and $\hat{P}$ fulfilling the canonical commutation relationship (CCR) $[\hat{X},\,\hat{P}]=i\,\hbar\,I$ (see my answer here and here). It's not too hard to show that a quantum space truly implementing the HUP cannot be finite dimensional - if it were, then $\hat{X}$ and $\hat{P}$ would have square matrix representations and the Lie bracket $[\hat{X}, \hat{P}]$ between any pair of finite square matrices has a trace of nought, whereas the right hand side of the CCR certainly does not have zero trace. So we consider them to be operators on the Hilbert space $\mathbf{L}^2(\mathbb{R}^N)$, which is a Hilbert space with dimensionality $\aleph_0$, i.e. it has countably infinite basis vectors, for example, the eigenfunctions of the $N$-dimensional harmonic oscillator. Vectors in this Hilbert space are "everyday wavefunctions" $\psi:\mathbb{R}^N\to\mathbb{R}^N$ as conceived by Schrödinger with the crucial normalisability property:

Now, for convenience, we want to work in co-ordinates wherein one of $\hat{X}$ and $\hat{P}$ is the simple multiplication operator $X \psi(x) = x\,\psi(x)$. In my answer here I show that this means that there are co-ordinates where $X \psi(x) = x\,\psi(x)$ and, needfully $\hat{P} \psi(x) = -i\,\hbar \,{\rm d}_x \psi(x)$.

$S^∗$ is the space of tempered distributions as discussed in my answer here. $S^∗$ includes the Dirac delta, $e^{i\,k\,x}$ and is bijectively, isometrically mapped onto itself by the Fourier transform. Intuitively, functions and their Fourier transforms are precisely the same information for the tempered distributions. This ties in with the fact that position and momentum co-ordinate are mapped into each other by the Fourier transform and its inverse.

1. This answer to the Physics Stack Exchange question "Rigged Hilbert space and QM" and also
2. The discussions under the Math Overflow Question "Good references for Rigged Hilbert spaces?"

In the latter, Todd Trimble's suspicions are correct that the usual Gel'Fand triple is $S\subset H = \mathbf{L}^2(\mathbb{R}^N)\subset S^*$ with $S$ , $S^∗$ being the Schwartz space and tempered distributions as discussed in my answer here. The Wikipedia article on rigged Hilbert space is a little light on here: there's a great deal of detail about nuclear spaces that's glossed over so at the first reading I'd suggest you should take a specific example $S$ = Schwartz space and $S^∗$ = Tempered Distributions and keep this relatively simple (and, for QM most relevant) example exclusively in mind - for QM you won't need anything else. The Schwarz space and space of tempered distributions are automatically nuclear, so you don't need to worry too much about this idea at first reading.

Less flippantly, here is where we get practical. In quantum mechanics, we need to implement the Heisenberg uncertainty principle, so we need Hermitian observables $\hat{X}$ and $\hat{P}$ fulfilling the canonical commutation relationship (CCR) $[\hat{X},\,\hat{P}]=i\,\hbar\,I$ (see my answer here and here). It's not too hard to show that a quantum space truly implementing the HUP cannot be finite dimensional - if it were, then $\hat{X}$ and $\hat{P}$ would have square matrix representations and the Lie bracket $[\hat{X}, \hat{P}]$ between any pair of finite square matrices has a trace of nought, whereas the right hand side of the CCR certainly does not have zero trace. So we consider them to be operators on the Hilbert space $\mathbf{L}^2(\mathbb{R}^N)$, which is a Hilbert space with dimensionality $\aleph_0$, i.e. it has countably infinite basis vectors, for example, the eigenfunctions of the $N$-dimensional harmonic oscillator. Vectors in this Hilbert space are "everyday wavefunctions" $\psi:\mathbb{R}^N\to\mathbb{R}^N$ as conceived by Schrödinger with the crucial normalisability property:

Now, for convenience, we want to work in co-ordinates wherein one of $\hat{X}$ and $\hat{P}$ is the simple multiplication operator $X \psi(x) = x\,\psi(x)$. In my answer here I show that this means that there are co-ordinates where $X \psi(x) = x\,\psi(x)$ and, needfully $\hat{P} \psi(x) = -i\,\hbar \,{\rm d}_x \psi(x)$.

$S^∗$ is the space of tempered distributions as discussed in my answer here. $S^∗$ includes the Dirac delta, $e^{i\,k\,x}$ and is bijectively, isometrically mapped onto itself by the Fourier transform. Intuitively, functions and their Fourier transforms are precisely the same information for the tempered distributions. This ties in with the fact that position and momentum co-ordinate are mapped into each other by the Fourier transform and its inverse.

1. This answer to the Physics Stack Exchange question "Rigged Hilbert space and QM" and also
2. The discussions under the Math Overflow Question "Good references for Rigged Hilbert spaces?"

In the latter, Todd Trimble's suspicions are correct that the usual Gel'Fand triple is $S\subset H = \mathbf{L}^2(\mathbb{R}^N)\subset S^*$ with $S$ , $S^∗$ being the Schwartz space and tempered distributions as discussed in my answer here. The Wikipedia article on rigged Hilbert space is a little light on here: there's a great deal of detail about nuclear spaces that's glossed over so at the first reading I'd suggest you should take a specific example $S$ = Schwartz space and $S^∗$ = Tempered Distributions and keep this relatively simple (and, for QM most relevant) example exclusively in mind - for QM you won't need anything else. The Schwarz space and space of tempered distributions are automatically nuclear, so you don't need to worry too much about this idea at first reading.