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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.

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

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tom.stoer

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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.

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A. Neumaier

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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).

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Thanks, I'll try to look at it.

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strangerep

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What does "with respect to" mean in this context? The meanings of "compact",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).

"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) ?

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A. Neumaier

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The linear operator V is compact relatively to H if for every bounded set B in theWhat 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) ?

Hilbert space H, the set {V psi | H psi in B} has a compact closure.

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