# Self adjoint operators, eigenfunctions & eigenvalues

• Incand
In summary: The eigenvalues will be the functions ##e^{ikx}## for ##k=0, 1, 2, \dots , n## and the eigenvalues for for ##A, B, C, D## and ##C## are ##ik, -k## and ##-k^2## respectively.
Incand

## Homework Statement

Consider the space ##P_n = \text{Span}\{ e^{ik\theta};k=0,\pm 1, \dots , \pm n\}##,
with the hermitian ##L^2##-inner product
##\langle f,g\rangle = \int_{-\pi}^\pi f(\theta) \overline{g(\theta)}d\theta##.
Define operators ##A,B,C,D## as
##A = \frac{d}{d\theta}, \; \; B= i\frac{d}{d\theta}, \; \; C= \frac{d^2}{d\theta^2}, \; \; D: f\to D f(\theta) = f(\theta) + f(-\theta)##.
Which of the operators are self-adjoint? Find the eigenvalues and eigenfunctions for each operator.

## Homework Equations

The operator ##T## is self adjoint iff
##\langle Tf,g \rangle = \langle f,Tg\rangle## for all ##f,g## in the ##P_n##.

## The Attempt at a Solution

Mostly looking for some quick input if I'm getting this roughly right or not since there's no answer and I'm a bit unsure.
I first write ##f(\theta) = \sum_{-n}^n c_k e^{ik\theta}##. We're allowed to differentiate since ##f(\theta)## is in ##C^\infty##. Differentiating we have ##f'(\theta) = \sum_{-n}^n ikc_k e^{ik\theta}## and ##f''(\theta ) = \sum_{-n}^n -k^2c_k e^{ik\theta}##. every part of ##f(\theta )## is ##2\pi##-periodic so the only time we get a contribution to the integral is when the the exponential terms cancel each other exactly.
A) In this case when the index matches so we have##\langle Af, g\rangle = \int_{-\pi}^\pi \sum_{-n}^n \left( ika_kb_k \right) d\theta = 2\pi \sum_{-n}^n ika_k\overline{b_k}##.
and ##\langle f, Ag\rangle = \int_{-\pi}^\pi \left( \sum_{-n}^n -ika_k\overline{b_k} \right) d\theta = -2\pi \sum_{-n}^n ika_k\overline{b_k}## so not self-adjoint.
Similary we see that both B and C are self-adjoint.

The eigenvalues would be the functions ##e^{ikx}## for ##k = \pm 1, \pm 2,\dots , \pm n## and the eigenvalues for for ##A,B## and ##C## are ##ik, -k## and ##-k^2## respectively.

The case ##D## is slightly harder we have
##\langle Df,g \rangle = \langle f,g \rangle + \int_{-\pi}^\pi f(-\theta)\overline{g(\theta)}d\theta## and
##\langle f,Dg \rangle = \langle f,g \rangle + \int_{-\pi}^\pi f(\theta)\overline{g(-\theta)}d\theta##
So we need to prove that ##\int_{-\pi}^\pi f(-\theta)\overline{g(\theta)} = \int_{-\pi}^\pi f(\theta)\overline{g(-\theta)}d\theta##. Again matching the coefficients we have for the left side
##\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{n-k}} \right) d\theta##
and the right side
##\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{n-k}}\right) d\theta## which matches so self adjoint. The eigenvalues should be ##\frac{a_k+a_{n-k}}{a_k}## and the eigenfunctions the same as before. Did I understand this right?

You have omitted the case ##k=0##, which gives an additional eigenvalue of 0 for A, B, C, with the eigenvector being any constant.

For D, I also conclude it is Hermitian but I get the integrals being like this

$$\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{-k}} \right) d\theta$$

The eigenvalues won't be of the form you gave because the ##a_k## coefficients are properties of the function ##f##, not the operator D.

Given the revised integral, can you work out what eigenvalues and eigenfunctions of D will be?

Incand
andrewkirk said:
For D, I also conclude it is Hermitian but I get the integrals being like this

$$\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{-k}} \right) d\theta$$

The eigenvalues won't be of the form you gave because the ##a_k## coefficients are properties of the function ##f##, not the operator D.

Given the revised integral, can you work out what eigenvalues and eigenfunctions of D will be?
Thanks for taking the time replying! You're absolutely right about the index in that sum and thanks for clarifying that I can't have those constants from the function as an eigenvalue.

If I write out ##Df(\theta)## I have ##\sum_{-N}^N \left(c_k e^{ik\theta} + c_ke^{-ik\theta} \right) = \sum_{-N}^N (c_k + c_{-k})e^{ik\theta}##
I'm not sure I get all eigenfunctions here but I was thinking if I take eigenfunctions of the form ##f\in P_n## with the added constraint that ##c_k+c_{-k}=0## I have the eigenvalue zero?

I also possibly see other eigenvalues For example
if I choose the constrant that ##c_{-k} = ac_{k}## with ##a## being a complex number I have the eigenvalue ##(1+a)##. But not sure if I'm allowed too do this? The eigenvalue doesn't vary like in the last case(which was obviously wrong) but I do use properties from the function I guess.

Incand said:
I was thinking if I take eigenfunctions of the form ##f\in P_n## with the added constraint that ##c_k+c_{-k}=0## I have the eigenvalue zero?
That sounds right. What would be a neat basis for the eigenspace with eigenvalue zero?

Incand said:
I also possibly see other eigenvalues For example
if I choose the constrant that ##c_{-k} = ac_{k}## with ##a## being a complex number I have the eigenvalue ##(1+a)##.
To see if this works, apply D to ##f(\theta)=e^{-ik\theta}+ae^{ik\theta}## and see what you get. Is it a complex scalar multiple of ##f(\theta)##?

Incand
andrewkirk said:
That sounds right. What would be a neat basis for the eigenspace with eigenvalue zero?
Perhaps ##\{e^{ik\theta}-e^{-ik\theta} \}_0^n ## that is the basis ##\{ 1, e^{i\theta}-e^{-i\theta}, \dots , e^{in\theta}-e^{-in\theta} \}##.
The basis vector aint orthogonal nor normalized but if one wants we could get there with Gram-Schmidts method.

andrewkirk said:
To see if this works, apply D to ##f(\theta)=e^{-ik\theta}+ae^{ik\theta}## and see what you get. Is it a complex scalar multiple of ##f(\theta)##?

##D\left( e^{-ik\theta}+ae^{ik\theta} \right) = e^{-ik\theta} + ae^{ik\theta} + e^{ik\theta} + e^{-ik\theta} = (1+a) \left(e^{-ik\theta}+e^{ik\theta} \right)##. Right this doesn't work since I should have a ##a## in there as well.

So instead let's look at the case when ##c_{-k} = c_{k}## we have
##D(e^{ik\theta}+ e^{-ik\theta}) =e^{ik\theta}+ e^{-ik\theta} + e^{-ik\theta}+ e^{ik\theta} = 2(e^{ik\theta}+ e^{-ik\theta})##. So we have the eigenvalue two for the eigenfunctions as a linear combination of ##\{e^{ik\theta}+e^{-ik\theta} \}_0^n##. I don't think there's actually any more eigenvalues.

## What is a self adjoint operator?

A self adjoint operator is a linear operator on a vector space that is equal to its own adjoint. This means that the operator and its adjoint have the same action on every vector in the vector space.

## What are eigenfunctions and eigenvalues?

Eigenfunctions and eigenvalues are a pair of mathematical concepts that are closely related to each other. An eigenfunction is a function that, when acted upon by an operator, returns a scalar multiple of itself. The corresponding scalar multiple is known as the eigenvalue of that particular eigenfunction.

## What are the applications of self adjoint operators, eigenfunctions, and eigenvalues?

Self adjoint operators, eigenfunctions, and eigenvalues have many applications in mathematics and physics. They are particularly useful in quantum mechanics, where they are used to describe the behavior of physical systems and the properties of particles.

## What is the difference between self adjoint operators and Hermitian operators?

The terms self adjoint operator and Hermitian operator are often used interchangeably, but there is a subtle difference between them. A self adjoint operator is defined on a real vector space, while a Hermitian operator is defined on a complex vector space. However, in most cases, the properties and behavior of these operators are very similar.

## How do you find eigenfunctions and eigenvalues of a self adjoint operator?

The process of finding eigenfunctions and eigenvalues of a self adjoint operator involves solving a mathematical equation known as the eigenvalue equation. This equation involves the operator, the eigenfunction, and the eigenvalue, and can be solved using various mathematical methods such as diagonalization or spectral decomposition.

• Calculus and Beyond Homework Help
Replies
3
Views
818
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
959
• Calculus and Beyond Homework Help
Replies
3
Views
640
• Calculus and Beyond Homework Help
Replies
1
Views
739
• Calculus and Beyond Homework Help
Replies
1
Views
729
• Calculus and Beyond Homework Help
Replies
1
Views
558
• Calculus and Beyond Homework Help
Replies
3
Views
678
• Calculus and Beyond Homework Help
Replies
5
Views
684
• Calculus and Beyond Homework Help
Replies
16
Views
823