I have been reading that for a wave function to be in the Schwartz function space (highly desirable for physical systems in the Rigged Hilbert Space (RHS) description of QM), among other criteria, it should be representable by a finite, or countably infinite, basis. For instance, a particle in a box, or a harmonic oscillator, can be represented by its countably infinite, orthogonal, energy eigenfunctions as its basis. Note, the wave function for a harmonic oscillator is often described by a Gaussian function, which is relevant to the following question.

A free particle (no potential energy or boundary conditions) can also be described by a Gaussian wave packet (which spreads in space over time). This wave packet is an integral over plane waves, each times a scaling factor which is a function of that plane wave's momentum. In that case, due to the above integral, there is a continuum infinity of energy eigenfunctions; they are not countably infinite. So, the energy eigenfunctions cannot form a countably infinite basis for that Gaussian wave packet. However, I think that the wave packet can still be in Schwartz space by representing it by a countably infinite number of orthogonal Hermite polynomials, which do occupy Schwartz space.

So, my question: is the free particle Gaussian wave packet (with a continuum of energy eigenvalues and eigenfunctions) in the Schwartz space, or in the "dual Schwartz space" of RHS?

  • $\begingroup$ Yes, it is. You can check it pretty easily from the definition. $\endgroup$ – Javier Oct 11 '16 at 23:31
  • $\begingroup$ @ Javier. Thank you. Do the Hermite polynomials, in fact, form an orthogonal countably infinite basis for the free particle Gaussian wave packet, and that is why you say yes? $\endgroup$ – David Oct 11 '16 at 23:36
  • $\begingroup$ Yes, they are, because they're the eigenstates of a self-adjoint operator, but you can just go to the definition of Schwartz space and verify that the Gaussian packet satisfies the conditions. $\endgroup$ – Javier Oct 12 '16 at 1:18
  • $\begingroup$ @javier The definition of Schwartz space does not explicetly require its functions to be expressible using a countably infinite basis. However, the Schwarz space is also used as a space that is isomorphic to the dense subspace that is a subset of the Hilbert space of the so-called Rigged Hilbert Space. That subspace requies the functions represented as vectors (or rays) in it to be expressible by a countably infinite orthogonal basis. So, in addition to the free particle wave packet satisfying Schwartz criteria, it also must have a countably infinite basis. $\endgroup$ – David Oct 12 '16 at 1:46
  • $\begingroup$ A vector can't have a basis, that's a property of the vector space. If you want to know if a specific function is in Schwartz space or not, you just check the definition. You're asking about a specific function, not the whole space. $\endgroup$ – Javier Oct 12 '16 at 2:02

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