Energy levels and Hilbert Spaces

In summary: The only incomplete set that I can think of is the set of all eigenvectors of the Hamiltonian, but that would be impossible to use as an observable because it would be infinite.
  • #1
carllacan
274
3
Hi.

Is there a Hilbert Space for each energy level of a system? (And, in general, for every point in time?)

I read in some book that if a equation for a problem accepts two different sets of wavefunction solutions (the case in question was the free particle and the sets of solutions in Cartesian and spherical coordinates) then the functions in one of those sets could be expressed by a linear combination of solutions of the same energy of the other set because of the completenes of the sets of eigenvectors of Hermitian operators.

But a complete set of vectors should span the whole hilbert Space, hence my question.
 
Physics news on Phys.org
  • #2
There is one Hilbert space in quantum theory. The set of (generalized) eigenvectors of a self-adjoint operator (hermitian is not enough!) is complete, i.e., any Hilbert-space vector can be realized as a (generalized) linear combination of the set of orthonormal eigenvectors.

What you are referring to is the case of degeneracy, i.e., an self-adjoint operator can have eigenvalues with more then one linearly independent eigenvector. Then you need to specify one or more other compatible observable(s) (of course all used observables must be mutually compatible, i.e., the self-adjoint operators representing them must commute) to pin down a particle's state by determining all these mutually compatible observables. If a set of such compatible observables has only one-dimensional common eigenspaces, the set is called complete. It is of course sufficient to use a minimal such complete set, i.e., it doesn't make sense to use a compatible set, where one observable is a function of the others.

E.g., for a particle with spin 0, the momentum components are a (minimal) complete set of compatible observables. Their spectrum is [itex]\mathbb{R}^3[/itex], which is entirely continuous, and you thus have no true (normizable) eigenvectors but "generalized" ones, which can only be "normalized to a Dirac-[itex]\delta[/itex] distribution:
[tex]\langle \vec{p}|\vec{p}' \rangle=\delta^{(3)}(\vec{p}-\vec{p}').[/tex]
Now consider a free particle. Its Hamiltonian is
[tex]\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2.[/tex]
This implies that any (generalized) momentum eigenstate is also an energy eigenstate, but determining the energy of the particle only fixes [itex]\vec{p}^2[/itex] and not the three momentum components, i.e., for each energy eigenvalue [itex]E \geq 0[/itex] there are infinitely many (generalized) eigenvectors.

Another set of compatible observables are the energy, [itex]\vec{L}^2[/itex], and [itex]L_z[/itex], where [itex]\vec{L}[/itex] is the orbital angular momentum of the particle. Now, indeed you can express any eigenvector of the Hamiltonian with a given energy eigenvalue [itex]E[/itex] as a linear combination of a complete set of other eigenvectors, i.e., you have
[tex]|E,l,m \rangle=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \; \sin \vartheta \, A(\vartheta,\varphi) |\vec{p}(\vartheta,\varphi) \rangle,[/tex]
with some function [itex]A(\vartheta,\varphi)[/itex]. Here [itex]\vartheta[/itex] and [itex]\varphi[/itex] parametrize the spherical shell with radius [itex]|\vec{p}|=\sqrt{2 m E}[/itex] in the usual way of spherical coordinates:
[tex]\vec{p}=|\vec{p}| \begin{pmatrix}
\cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \sin \vartheta
\end{pmatrix}.
[/tex]
 
  • #3
Thanks for your answer, I think I grap what you mean, but I'm going to need to ask a few things.

vanhees71 said:
What you are referring to is the case of degeneracy, i.e., an self-adjoint operator can have eigenvalues with more then one linearly independent eigenvector.
In this case the self-adjoint operator would be the Hamiltonian, the eigenvalues the energy levels and the l.i. eigenvectors the solutions, right?
vanhees71 said:
[...] It is of course sufficient to use a minimal such complete set, [...]
So just with the op with degeneracy and a compatible one we would already have minimal such set?
vanhees71 said:
[...]no true (normizable) eigenvectors but "generalized" ones [...]
What do you exactly mean by "generalized" eigenvectors? Are those "generalized" vectors still in the HS?
vanhees71 said:
determining the energy of the particle only fixes p⃗ 2 and not the three momentum components, i.e., for each energy eigenvalue E≥0 there are infinitely many (generalized) eigenvectors.
Tell me if I understood this part: the [itex]P^2[/itex] operator is not useful (by itself) to pin down the [itex]E_n[/itex] state because it is also degenerate (i.e. one [itex]P^2[/itex] -> many ψ(x)), so we also use [itex]L^2[/itex] and [itex]L_z[/itex]. The indices in [itex]\left|E, l, m\right\rangle[/itex] mean, respectively, that it is an eigenket of a certain E (and of [itex]\vec{p^2}=\sqrt{2mE}[/itex]) of a certain eigenvalue of [itex]L^2[/itex] and of a certain eigenvalue of [itex]L_z[/itex] Is it so?

At the end of it all I still can't put together an answer to my question. Why can't I express [itex]\left|E, l, m\right\rangle[/itex] as a linear combination of any complete set of eigenvectors, like the ones correspondings to another state [itex]\left|E', l', m'\right\rangle[/itex] or the ones that consitute a solutions to another problem?
 
  • #4
@Vanhees71, What's the difference between self-adjoint and hermitian operator?
 
  • #5
ChrisVer said:
@Vanhees71, What's the difference between self-adjoint and hermitian operator?

I can answer this one. A self-adjoint operator is a Hermitian operator with the additional condition that the domains of the operator and its adjoint are the same so that it really IS "self-adjoint". Self-adjoint is a stronger condition.
 
  • Like
Likes 1 person

1. What are energy levels in physics?

Energy levels in physics refer to the specific quantized states that a system can occupy. These states are determined by the amount of energy that a system possesses and are represented by discrete values rather than a continuous range.

2. What is the significance of energy levels in quantum mechanics?

In quantum mechanics, energy levels play a crucial role in determining the behavior of particles. They dictate the allowed energy values that particles can have and are used to calculate the probability of a particle being in a certain state.

3. What are Hilbert Spaces in relation to energy levels?

Hilbert Spaces are mathematical structures used to represent the possible quantum states that a system can occupy. In the context of energy levels, Hilbert Spaces are used to describe the different allowed energy states of a system and the probability of a particle being in a specific energy level.

4. How are energy levels and Hilbert Spaces related to the Schrödinger equation?

The Schrödinger equation is a fundamental equation in quantum mechanics that describes the time evolution of a quantum system. Energy levels and Hilbert Spaces are used in the Schrödinger equation to determine the allowed energy states of a system and the probability of a particle being in a certain state at a given time.

5. Can energy levels and Hilbert Spaces be applied to systems other than particles?

Yes, energy levels and Hilbert Spaces can be applied to various systems, including atoms, molecules, and even larger structures such as crystals. They are also used in fields such as nuclear physics and chemistry to study the energy states of nuclei and molecules.

Similar threads

Replies
3
Views
818
Replies
9
Views
1K
Replies
13
Views
2K
  • Quantum Physics
Replies
8
Views
2K
Replies
13
Views
2K
Replies
67
Views
4K
  • Quantum Interpretations and Foundations
Replies
5
Views
360
Replies
2
Views
1K
  • Quantum Physics
Replies
8
Views
1K
  • Quantum Physics
Replies
2
Views
920
Back
Top