Multiple questions about eigenstates and eigenvalues

[Sebas4]

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
$\psi\left(\theta\right)$ = $\psi\left(\theta + 2\pi\right)$.
The solution is
$\psi\left(\theta\right) = \frac{1}{\sqrt{2\pi R}}e^{ik\theta}$ with k = $0, \pm1, \pm2, \pm 3 ...$.

The next question is to derive the $$L_{z}$$ operator and find the eigenvalues for this operator.
I have derived the operator for $$L_{z}$$ and it is
$$\hat{L}_{z} = i\hbar \frac{\partial }{\partial \theta}$$.
The answer book states that the eigenvector
of the angular momentum can be found by filling in the solution of the Schrödinger equation, so
$$\hat{L}_{z}\psi\left(\theta\right) = \lambda\psi\left(\theta\right)$$ or
$$i\hbar \frac{\partial }{\partial \theta} \left(\frac{1}{\sqrt{2\pi R}}e^{ik\theta} \right) = \lambda \frac{1}{\sqrt{2\pi R}}e^{ik\theta}$$.
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.

 

[anuttarasammyak]

Summary:: Solving time-independent Schrödinger equation and find eigenvalues for observable angular momentum.

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

How is it an eigenstate of energy ? What is your Hamiltonian ?

 

[PeroK]

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?

The Hilbert space depends on the system. In this case, it's the Hilbert space of square-integrable functions on the ring. The different observables (in this case energy and angular momentum), then have bases that span this Hilbert 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
$\psi\left(\theta\right)$ = $\psi\left(\theta + 2\pi\right)$.
The solution is
$\psi\left(\theta\right) = \frac{1}{\sqrt{2\pi R}}e^{ik\theta}$ with k = $0, \pm1, \pm2, \pm 3 ...$.

The next question is to derive the $$L_{z}$$ operator and find the eigenvalues for this operator.
I have derived the operator for $$L_{z}$$ and it is
$$\hat{L}_{z} = i\hbar \frac{\partial }{\partial \theta}$$.
The answer book states that the eigenvector
of the angular momentum can be found by filling in the solution of the Schrödinger equation, so
$$\hat{L}_{z}\psi\left(\theta\right) = \lambda\psi\left(\theta\right)$$ or
$$i\hbar \frac{\partial }{\partial \theta} \left(\frac{1}{\sqrt{2\pi R}}e^{ik\theta} \right) = \lambda \frac{1}{\sqrt{2\pi R}}e^{ik\theta}$$.
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.

In any system where there is zero potential, the energy and momentum (or angular momentum) operators are closely related. In this case:
$$\hat H = -\frac {\hbar^2} {2mR^2} \frac{\partial^2}{\partial \theta^2} \ \ \text{and} \ \ \hat{L_z} = i\hbar \frac{\partial}{\partial \theta}$$It's not surprising that they have similar eigenvectors.

 

[vanhees71]

It should be
$$\hat{H}=\frac{1}{2m a^2} \hat{L}_z^2=-\frac{\hbar^2}{2m R^2} \partial_{\theta}^2, \quad \hat{L}_z=-\mathrm{i} \hbar \partial_{\theta},$$
where $R$ is the radius of the ring. The sign of the angular-momentum operator is, of course, a convention, but I've never seen another convention than the one with the minus. It's analogous to define the usual momentum operator as $\hat{p}=-\mathrm{i} \hbar \partial_x$ for motion along a straight line (or $\hat{p}=-\mathrm{i} \hbar \vec{\nabla}$ for motion in 3D space).

It's clear that the $\hat{L}_z$ eigenbasis is also an eigenbasis for $\hat{H}$ in this case, and they are simpler in the sense that they are non-degenerate, i.e., for each eigenvalue there's only one linearly independent eigenvector. The calculation in #1 is correct:
$$u_k(\theta)=\frac{1}{\sqrt{2 \pi R}} \exp(\mathrm{i} k \theta), \quad k \in \mathbb{Z}$$
with the scalar product of wave functions defined as
$$\langle \psi_1|\psi_2 \rangle=\int_0^{2 \pi} \mathrm{d} \theta R \psi_1^*(\theta) \psi_2(\theta).$$
The eigenvalues of $\hat{L}_z$ are $l_z=\hbar k$.

The energy eigenvalues are
$$E_k=\frac{\hbar^2}{2m R^2} k^2.$$
As you see, only the ground state with $k=0$ is non-degenerate, while the energy eigenvalues for each $k \neq 0$ is two-fold degenerate.

The angular-momentum eigenstates are obviously a complete set of orthonormal wave functions since you can express any square-integrable $2 \pi$ periodic function in terms of these functions, which leads to the usual Fourier series,
$$\psi_k=\langle u_k|\psi \rangle = \int_0^{2 \pi} \mathrm{d} \theta \frac{\sqrt{R}}{\sqrt{2 \pi}} \exp(-\mathrm{i} k \theta) \psi(x), \quad \psi(x)=\sum_{k \in \mathbb{Z}} \frac{1}{\sqrt{2 \pi R}} \psi_k \exp(\mathrm{i} k x).$$