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

In summary, the conversation discusses the concept of a complete set in perturbation theory, specifically in two situations: finite Hilbert space and infinite Hilbert space. It is explained that a complete set of states can be seen as a basis in Hilbert space and can still be used to describe the system after perturbation. However, it is noted that there may be problems if the unperturbed and perturbed Hamiltonians exist in Hilbert spaces of different dimensionality. The concept of changing the Hilbert space is also discussed, which can occur in certain physically relevant problems. The Kato-Rellich theorem is mentioned as a reference for perturbation theory.
  • #1
ck00
19
0
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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  • #2
ck00 said:
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.
 
  • #3
ck00 said:
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.
 
  • #4
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.
 
  • #5
Hypersphere said:
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.

Thank you for your answering. I have a follow-up question.
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?
 
  • #6
tom.stoer said:
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?
 
  • #7
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.
 
  • #8
ck00 said:
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
or here
http://www.fuw.edu.pl/~derezins/mat-u.pdfThe reference of the expert on perturbation theory is
Tosio Kato,
Perturbation Theory for Linear Operators
 
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1. What is the difference between finite and infinite Hilbert spaces in perturbation theory?

The main difference between finite and infinite Hilbert spaces in perturbation theory is the size of the space. A finite Hilbert space has a limited number of basis states, while an infinite Hilbert space has an infinite number of basis states. This affects the way perturbation theory is applied, as the calculations are more complex in an infinite Hilbert space.

2. Can perturbation theory be applied to both finite and infinite Hilbert spaces?

Yes, perturbation theory can be applied to both finite and infinite Hilbert spaces. However, the calculations are more straightforward in a finite Hilbert space, while an infinite Hilbert space requires more advanced techniques.

3. How does the size of the Hilbert space affect the accuracy of perturbation theory?

The size of the Hilbert space can affect the accuracy of perturbation theory. In a finite Hilbert space, the calculations are more accurate as the number of basis states is limited. In an infinite Hilbert space, the calculations may be less accurate due to the infinite number of possible states.

4. Are there any real-world applications of perturbation theory in finite and infinite Hilbert spaces?

Yes, perturbation theory has many real-world applications in both finite and infinite Hilbert spaces. In quantum mechanics, it is used to calculate the energy levels of atoms and molecules. In solid state physics, perturbation theory is used to study the effect of external forces on the electronic properties of materials.

5. What are some challenges in applying perturbation theory in an infinite Hilbert space?

One of the main challenges in applying perturbation theory in an infinite Hilbert space is the convergence of the calculations. As the number of basis states is infinite, the calculations can become more complex and may not converge to a solution. In addition, the choice of basis states can also affect the accuracy of the results in an infinite Hilbert space.

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