# Finite Hilbert Space v.s Infinite Hilbert Space in Perturbation Theory

Hi all,
I have a question about the concept of complete set when I apply the perturbation theory in two situations -Finite Hilbert Space and Infinite Hilbert Space.
Consider a Hamiltonian H=H0+H', where H0 is the unperturbed Hamiltonian and H' is the perturbed Hamiltonian. Let ψ_n be the complete set of unperturbed Hamiltonian H0.

Do ψ_n still constitute a complete set of the overall Hamiltonian H for either finite Hilbert space or infinite Hilbert space if the perturbed Hamiltonian is turned on?

Ck

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Hi all,
I have a question about the concept of complete set when I apply the perturbation theory in two situations -Finite Hilbert Space and Infinite Hilbert Space.
Consider a Hamiltonian H=H0+H', where H0 is the unperturbed Hamiltonian and H' is the perturbed Hamiltonian. Let ψ_n be the complete set of unperturbed Hamiltonian H0.

Do ψ_n still constitute a complete set of the overall Hamiltonian H for either finite Hilbert space or infinite Hilbert space if the perturbed Hamiltonian is turned on?

Ck
You can think of a complete set of states as a basis in Hilbert space, much like an ordinary (geometrical) vector space. A state of the Hamiltonian H0 can be seen as a vector in this Hilbert space, which may be one of the basis states, or a linear combination of several of them. So if you change your state by perturbing it, you are basically just rotating your vector into a new one, which can still be described using the same basis. This works both for finite and infinite dimensional Hilbert spaces. (Strictly speaking, infinite dimensional spaces require some additional care mathematically, but for physics intuition one can use the same picture.)

I assume you could run into some problems if H0 and H' live in Hilbert spaces of different dimensionality though (eg. H0 has a discrete spectrum, while H' has a continuous one), but I don't think I've ever seen that happen in practice. I'm speculating now, but to me it seems you should be able to use an overcomplete set of states then.

strangerep
Do ψ_n still constitute a complete set of the overall Hamiltonian H for either finite Hilbert space or infinite Hilbert space if the perturbed Hamiltonian is turned on?
In the case of QFT, you could look up "Haag's theorem" which implies that, in general, the eigenstates of the free Hamiltonian fail to span the Hilbert space of the full interacting theory.

tom.stoer
The only problem I am aware of is when the interaction forces us to change the Hilbert space. But I am not aware of any reasonable example in perturbation theory.

You can think of a complete set of states as a basis in Hilbert space, much like an ordinary (geometrical) vector space. A state of the Hamiltonian H0 can be seen as a vector in this Hilbert space, which may be one of the basis states, or a linear combination of several of them. So if you change your state by perturbing it, you are basically just rotating your vector into a new one, which can still be described using the same basis. This works both for finite and infinite dimensional Hilbert spaces. (Strictly speaking, infinite dimensional spaces require some additional care mathematically, but for physics intuition one can use the same picture.)

I assume you could run into some problems if H0 and H' live in Hilbert spaces of different dimensionality though (eg. H0 has a discrete spectrum, while H' has a continuous one), but I don't think I've ever seen that happen in practice. I'm speculating now, but to me it seems you should be able to use an overcomplete set of states then.
Consider 2x2 Hamiltonian H=H0+H'.
Let Φ_a and Φ_b be two eigenstates of H0 and ψ_a and ψ_b be two eigenstates of H.
What is the relation between {Φ_a, Φ_b} and {ψ_a, ψ_b}? Are they just linear combination of each other?

The only problem I am aware of is when the interaction forces us to change the Hilbert space. But I am not aware of any reasonable example in perturbation theory.
What is the meaning of change of Hilbert space?

tom.stoer
Consider the infinite square well with
V(x) = 0 for a<x<b (I)
V(x) = +∞ for x<a or x>b (II)
The associated Hilbert space is L²[a,b]; the region (II) is excluded.

Now consider a (large!) perturbation which changes the potential to
V(x) = 0 for a<x<b (I)
V(x) = V0 for x<a or x>b (II)
The associated Hilbert space is now L²[-∞,+∞]; the region (II) can no longer be excluded; the solutions for the original case do not span this new Hilbert space. Of course this change of the Hilbert space must not and cannot be treated perturbatively.

This may seem to be rather artificial, but there are physically relevant problems were similar problems occure. Consider an atom with an 1/r potential. It has well-known bound states plus a continuum of scattering states. Now add a constant electric field which results from a linear potential U(x) = u0x. The resulting Hamiltonian is no longer bounded from below; at first glance the problem is not even well-defined mathematically.

DrDu
Hi all,
I have a question about the concept of complete set when I apply the perturbation theory in two situations -Finite Hilbert Space and Infinite Hilbert Space.
Consider a Hamiltonian H=H0+H', where H0 is the unperturbed Hamiltonian and H' is the perturbed Hamiltonian. Let ψ_n be the complete set of unperturbed Hamiltonian H0.

Do ψ_n still constitute a complete set of the overall Hamiltonian H for either finite Hilbert space or infinite Hilbert space if the perturbed Hamiltonian is turned on?

Ck
The problem is usually that H and H0 are unbound operators and are not defined on the total Hilbert space but only on dense subsets. These ranges of definition do not coincide in general, e.g. when H' is a delta function. They do coincide iff H' is relatively bounded wrt H0 with bound <1 (Kato-Rellich theorem).
I found references where these terms are explained here
http://www.math.pku.edu.cn/teachers/fanhj/courses/fl5.pdf [Broken]
or here
http://www.fuw.edu.pl/~derezins/mat-u.pdf

The reference of the expert on perturbation theory is
Tosio Kato,
Perturbation Theory for Linear Operators

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