If we take the free particle Hamiltonian, the eigenvectors (or eigenfunctions), say in position representation, are like $e^{ikx}$. Now these eigenfunctions are non-normalisable,so they don't belong the normal $L^2(\mathbb R^d)$ but the rigged Hilbert space.
My question hereto is that any unitary operator defined as a map in Hilbert space preserves the norm. But in the case of free particle, although the operator $e^{iHt}$ is unitary (since $H$ is hermitian), there is (atleast) no (direct) condition of norm preserval, as the norm cannot be defined for these eigenfunctions.
Now, how can one connect the unitarity of $e^{iHt}$ and norm preserval in this context ?
PS : I know one can use box-normalised wavefunctions and do away with the calculations and then take $L \rightarrow \infty$ limit. But I am rather interested in the actual question of unitarity and norm in rigged Hilbert spaces.
Answer
The so-called rigged spaces are made with a triple $(S,\mathscr{H},S')$; where $\mathscr{H}$ is the usual Hilbert space, $S$ is a dense vector subspace of $\mathscr{H}$, and $S'$ the dual of $S$.
Usually when $\mathscr{H}=L^2(\mathbb{R}^d$, then $S$ is taken to be the rapidly decrasing functions, and $S'$ the tempered distributions. If this is the case, then there is no notion of norm for the "extended" eigenvectors in $S'$, since the latter is not a Banach (or metrizable) space.
So even if $e^{-itH}e^{ikx}=e^{-itk^2/2m}e^{ikx}$, and therefore the evolution indeed acts as a phase, there is no norm to be preserved.
The point is that rigged Hilbert spaces are, as far as I know, simply a mathematical convenience to justify the emergence of "generalized eigenvectors" for some (very special) self-adjoint operators that have purely continuous spectrum. If you want to do (meaningful) quantum mechanics, you have to consider states of the Hilbert space, where the evolution is indeed unitary and everything works.
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