# Self adjoint operators, eigenfunctions & eigenvalues

1. Oct 21, 2015

### Incand

1. The problem statement, all variables and given/known data
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.

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

3. 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?

2. Oct 21, 2015

### andrewkirk

Your work looks mostly sound.
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?

3. Oct 22, 2015

### Incand

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.

4. Oct 22, 2015

### andrewkirk

That sounds right. What would be a neat basis for the eigenspace with eigenvalue zero?

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 $f(\theta)$?

What about other nonzero eigenvalues?

5. Oct 22, 2015

### Incand

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.

$D\left( e^{-ik\theta}+ae^{ik\theta} \right) = e^{-ik\theta} + ae^{ik\theta} + e^{ik\theta} + e^{-ik\theta} = (1+a) \left(e^{-ik\theta}+e^{ik\theta} \right)$. 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.