- #1

Sebas4

- 13

- 2

- TL;DR Summary
- Solving time-independent Schrödinger equation and find eigenvalues for observable angular momentum.

I have multiple questions about eigenstates and eigenvalues.

The Hilbert space is spanned by independent bases.

The textbook said that the eigenvectors of observable spans the Hilbert space.

Here comes the question.

Do the eigenvectors of multiple observables span the same Hilber space?

Here comes the other question.

I solved the time-independent Schrödinger equation for a ring with radius R, the potential is 0 and the boundary condition is that

[itex]\psi\left(\theta\right)[/itex] = [itex]\psi\left(\theta + 2\pi\right)[/itex].

The solution is

[itex]\psi\left(\theta\right) = \frac{1}{\sqrt{2\pi R}}e^{ik\theta}[/itex] with k = [itex]0, \pm1, \pm2, \pm 3 ...[/itex].

The next question is to derive the [tex]L_{z}[/tex] operator and find the eigenvalues for this operator.

I have derived the operator for [tex]L_{z}[/tex] and it is

[tex]\hat{L}_{z} = i\hbar \frac{\partial }{\partial \theta}[/tex].

The answer book states that the eigenvector

of the angular momentum can be found by filling in the solution of the Schrödinger equation, so

[tex]\hat{L}_{z}\psi\left(\theta\right) = \lambda\psi\left(\theta\right)[/tex] or

[tex]i\hbar \frac{\partial }{\partial \theta} \left(\frac{1}{\sqrt{2\pi R}}e^{ik\theta} \right) = \lambda \frac{1}{\sqrt{2\pi R}}e^{ik\theta}[/tex].

Why is this? (Why apply an operator of the observable angular momentum on an eigenstate vector of energy)?

The solution of the time-independent Schrödinger equation is an eigenstate of another observable and not angular momentum.

The Hilbert space is spanned by independent bases.

The textbook said that the eigenvectors of observable spans the Hilbert space.

Here comes the question.

Do the eigenvectors of multiple observables span the same Hilber space?

Here comes the other question.

I solved the time-independent Schrödinger equation for a ring with radius R, the potential is 0 and the boundary condition is that

[itex]\psi\left(\theta\right)[/itex] = [itex]\psi\left(\theta + 2\pi\right)[/itex].

The solution is

[itex]\psi\left(\theta\right) = \frac{1}{\sqrt{2\pi R}}e^{ik\theta}[/itex] with k = [itex]0, \pm1, \pm2, \pm 3 ...[/itex].

The next question is to derive the [tex]L_{z}[/tex] operator and find the eigenvalues for this operator.

I have derived the operator for [tex]L_{z}[/tex] and it is

[tex]\hat{L}_{z} = i\hbar \frac{\partial }{\partial \theta}[/tex].

The answer book states that the eigenvector

of the angular momentum can be found by filling in the solution of the Schrödinger equation, so

[tex]\hat{L}_{z}\psi\left(\theta\right) = \lambda\psi\left(\theta\right)[/tex] or

[tex]i\hbar \frac{\partial }{\partial \theta} \left(\frac{1}{\sqrt{2\pi R}}e^{ik\theta} \right) = \lambda \frac{1}{\sqrt{2\pi R}}e^{ik\theta}[/tex].

Why is this? (Why apply an operator of the observable angular momentum on an eigenstate vector of energy)?

The solution of the time-independent Schrödinger equation is an eigenstate of another observable and not angular momentum.

Last edited: