Riemann vs. Lebesgue integral in QM

[AxiomOfChoice]

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
$$\int_{-\infty}^{\infty} |\psi(\vec r,t)|^2 d^3x.$$
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 $-\infty$ to $\infty$ 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.

 

[martinbn]

They are talking about Lebesgue-integrable functions. No matter what functions show up in practice you need the Hilbert space. There is a completeness requirement in the definition of Hilbert space, which is used in many theorems and cannot be avoided, that you don't have if only Riemann integration is considered.

 

[A. Neumaier]

Can someone give me an example of a wavefunction that arises in practice whose modulus squared is Lebesgue integrable but NOT Riemann integrable?

Wave functions that arise in practice are typically real analytic. But if one wants to prove general results about quantum mechanics, one needs to be able to take limits in the topology defined by the 2-norm. But the space of functions whose modulus squared is Riemann integrable is not a Hilbert space.

 

[AxiomOfChoice]

Wave functions that arise in practice are typically real analytic. But if one wants to prove general results about quantum mechanics, one needs to be able to take limits in the topology defined by the 2-norm. But the space of functions whose modulus squared is Riemann integrable is not a Hilbert space.

By "2-norm", do you mean the norm given by

$$\| \psi \| = \sqrt{\int_{-\infty}^\infty \psi(\vec x,t) \psi^*(\vec x,t) d^3 x}$$

 

[A. Neumaier]

By "2-norm", do you mean the norm given by

$$\| \psi \| = \sqrt{\int_{-\infty}^\infty \psi(\vec x,t) \psi^*(\vec x,t) d^3 x}$$

Yes.

 

[Fredrik]

AoC, you should either drop the time dependence on the right-hand side, or introduce a t on the left as well. (The first option is better, if all you want to do is to define a norm on $L^2(\mathbb R^3)$).

 

[homology]

perhaps then this norm is over $$R^4$$ since our wave functions are time dependent

 

[Fredrik]

perhaps then this norm is over $$R^4$$ since our wave functions are time dependent

No, the relevant Hilbert space is $L^2(\mathbb R^3)$. You should think of the time-dependent wavefunctions as curves in that space. More precisely, if we define $\psi_t$ by $\psi_t(\vec x)=\psi(\vec x,t)$ for all $\vec x$, then $\psi_t$ is a member of $L^2(\mathbb R^3)$ and $t\mapsto\psi_t$ is a curve in $L^2(\mathbb R^3)$.

 

[homology]

Fredrick,

why is this the case? that is, what would be the mistake with using R^4?

though it seems that what you're saying, resembles the case in mechanics where you gave an inner product at each point on the configuration manifold and the parameter t moves you along this surface through different tangent spaces, us that what you're after?

 

[A. Neumaier]

why is this the case? that is, what would be the mistake with using R^4?

though it seems that what you're saying, resembles the case in mechanics where you gave an inner product at each point on the configuration manifold and the parameter t moves you along this surface through different tangent spaces, us that what you're after?

The configuration space of a single particle is R^3, not R^4, and the Schroedinger equation describes the dynamics how a wave function in L^2(R^3) changes with time.

If you were to integrate over R^4, all integrals would diverge.

 

[homology]

oh of course, doh...the inner product is only over space...ack...shouldn't respond to these things in the morning

 

[weejee]

If we considered the infinite square well problem, where the domain is compact, is the Riemann integral sufficient, or do we still need the Lebesgue integral?

 

[Fredrik]

You still need the Lebesgue integral to get a Hilbert space. I think you can e.g. find a sequence of Riemann integrable functions (all with the same compact domain) that converges pointwise to a function that isn't Riemann integrable.