- #1
AxiomOfChoice
- 533
- 1
When we talk about "Hilbert space" in (undergraduate) QM, we are typically talking about the space of square-integrable functions so that we can make sense out of
[tex]
\int_{-\infty}^{\infty} |\psi(\vec r,t)|^2 d^3x.
[/tex]
But are we talking about Riemann-integrable functions or Lebesgue-integrable functions? Can someone give me an example of a wavefunction that arises in practice whose modulus squared is Lebesgue integrable but NOT Riemann integrable? Don't wavefunctions (and hence their modulus squares) have to satisfy certain continuity conditions, implying that the Riemann integral is sufficient?
I realize that writing the limits of integration above from [itex]-\infty[/itex] to [itex]\infty[/itex] might, to some, imply that we're talking about a Lebesgue integral, since Riemann integrals are by definition defined only on closed, bounded intervals. But I suppose it's possible that we're just talking about an improper Riemann integral instead.
[tex]
\int_{-\infty}^{\infty} |\psi(\vec r,t)|^2 d^3x.
[/tex]
But are we talking about Riemann-integrable functions or Lebesgue-integrable functions? Can someone give me an example of a wavefunction that arises in practice whose modulus squared is Lebesgue integrable but NOT Riemann integrable? Don't wavefunctions (and hence their modulus squares) have to satisfy certain continuity conditions, implying that the Riemann integral is sufficient?
I realize that writing the limits of integration above from [itex]-\infty[/itex] to [itex]\infty[/itex] might, to some, imply that we're talking about a Lebesgue integral, since Riemann integrals are by definition defined only on closed, bounded intervals. But I suppose it's possible that we're just talking about an improper Riemann integral instead.