A mathematical issue raised from perturbation theory

[kof9595995]

Take the usual time-independent perturbation theory in QM for example,H'=H_0+V, a basic assumption is we can expand the new states of H' in terms of the old ones of H_0, most of the textbooks justify this assumption by reasoning that the set of eigenfunctions of Hamiltonian is complete, regardless of the old H_0 or the new H'. But I guess this assumption can't be always true, for example, you can't possibly write a free-particle wavepacket in terms of eigenfuntions of infinite square well. So when can this assumption be applied exactly?

 

[Chopin]

I asked pretty much this same question https://www.physicsforums.com/showthread.php?t=432504". I never completely got the answer I was looking for, but after then looking around a little bit more, it seems like the answer is that you can apply this assumption whenever the wavefunctions have support over the region you're interested in--that is, they are nonzero in that region. So an infinite square well has wavefunctions with vanishing support outside of the well, meaning they can't be a complete basis for plane waves, but they are a complete basis for anything else with support over the same region.

I may not have been 100% correct on all the mathematical details of that, but I think the basic argument is correct.

 

[tom.stoer]

The problem with the infinite square well is that the Hilbert spaces on which H_0 and H' are defined are different. Mathematically you do not have an infinite potential outside the well but you set the wave functions to zero restricting the Hilbert space to L²[a,b]where a and b are the boundaries of the well.

But of course coming from the free particle Hilbert space (which is larger) you can write the eigenfunctions of H' as a Fourier integral which is equivalent to "summing" over the (generalized) eigenfunctions of the free particle.

If you consider only perturbtations V which do not change the Hilbert space then expanding the full solutions in terms of the free solutions is fine. In case V restricts the Hilbert space it should work as well.

 

[A. Neumaier]

Take the usual time-independent perturbation theory in QM for example,H'=H_0+V, a basic assumption is we can expand the new states of H' in terms of the old ones of H_0, most of the textbooks justify this assumption by reasoning that the set of eigenfunctions of Hamiltonian is complete, regardless of the old H_0 or the new H'. But I guess this assumption can't be always true, for example, you can't possibly write a free-particle wavepacket in terms of eigenfuntions of infinite square well. So when can this assumption be applied exactly?

The right assumption is that H_0 and V are defined on the same dense domain of the Hilbert space, and that V is relatively compact with respect to H_0. These terms are explained in functional analysis texts, or in the math. physics books by Reed and Simon (Vol.1) or Thirring (Vol.3).

 

[kof9595995]

Thanks, I'll try to look at it.

 

[strangerep]

The right assumption is that H_0 and V are defined on the same dense domain of the Hilbert space, and that V is relatively compact with respect to H_0. These terms are explained in functional analysis texts, or in the math. physics books by Reed and Simon (Vol.1) or Thirring (Vol.3).

What does "with respect to" mean in this context? The meanings of "compact",
"relatively compact subset", etc, are easy enough to find, but I didn't find the
"with respect to" bit. I'm guess it has something to do with range(V) being
relatively compact in range(H_0) ?

 

[A. Neumaier]

What does "with respect to" mean in this context? The meanings of "compact",
"relatively compact subset", etc, are easy enough to find, but I didn't find the
"with respect to" bit. I'm guess it has something to do with range(V) being
relatively compact in range(H_0) ?

The linear operator V is compact relatively to H if for every bounded set B in the
Hilbert space H, the set {V psi | H psi in B} has a compact closure.