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.


1 Answer 1


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.

  • $\begingroup$ So you mean to say, there is no such connection between unitarity and norm preserval in a non metrizable space (like rigged Hilbert space). $\endgroup$
    – user35952
    Jun 15, 2016 at 5:44
  • 1
    $\begingroup$ Yes; unitarity is defined in Hilbert spaces, and norm preserving operators in Banach spaces are called isometries. The space of distributions is a space that can be endowed with one of the topologies induced by duality with $\mathscr{S}(\mathbb{R}^d)$. Anyways, it is not a metrizable space, so neither Banach nor Hilbert. Hence the notions of unitarity and isometry cannot be defined on $S'$. $\endgroup$
    – yuggib
    Jun 15, 2016 at 6:33
  • $\begingroup$ But if the notions of unitarity is not defined in $S'$, how can one do quantum mechanics with it and be reconciled with probability conservation. Shouldn't there be an analogous property that can help with the probability interpretation in QM, if yes, what is it ? $\endgroup$
    – user35952
    Jun 15, 2016 at 7:08
  • $\begingroup$ So that is why box normalisation comes in ?? $\endgroup$
    – user35952
    Jun 15, 2016 at 7:28
  • $\begingroup$ Of course if you put everything in a box, there is no problem. However removing the box can be very tricky, if you want to do things in a rigorous (and meaningful) fashion. $\endgroup$
    – yuggib
    Jun 15, 2016 at 7:30

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