Finding the eigenfunctions and eigenvalues associated with an operator

[Orodruin]

What is the real part of $e^{i\sqrt q \phi}$?

 

[Gaussian97]

Consider the following operator

$$\hat Q = \frac{d^2}{d \phi^2}$$

Find its eigenfunctions and eigenvalues.

The eigenvalue equation is

$$\frac{d^2}{d \phi^2} f(\phi) = q f(\phi)$$

Well, since Orodruin didn't answer my questions I will post the last message, sorry if I'm being repetitive:
1) $f_1$ CAN BE a normalizable periodic function.
2) $f_2$ IS NOT a solution to the problem stated at the beginning.

Good luck.

 

[PeroK]

We have

$$f_2(\phi) = D e^{i \phi \sqrt{q}} + E e^{- i \phi \sqrt{q}}$$

Let's apply the given boundary condition $f(\phi +2\pi ) = f(\phi)$ to $f_2(\phi)$

$$D e^{i (\phi + 2\pi) \sqrt{q}} + E e^{- i (\phi + 2\pi) \sqrt{q}} = D e^{i \phi \sqrt{q}} + E e^{- i \phi \sqrt{q}}$$

$$D e^{-i \phi \sqrt{q}} + E e^{i \phi \sqrt{q}} = D e^{i \phi \sqrt{q}} + E e^{- i \phi \sqrt{q}}$$

$$D(e^{-i \phi \sqrt{q}} - e^{i \phi \sqrt{q}}) + E(e^{i \phi \sqrt{q}} - e^{- i \phi \sqrt{q}}) = 0$$

Mmm but this implies that $D=0$ and $E=0$, which leads to the trivial solution.

There must be something wrong in here... I am thinking.

I'm still not sure how we got here! To go back to the beginning. For the equation:
$$\frac{d^2 f}{d\phi^2} = q\phi$$
We can write down the solutions:
$$f_1 = e^{\sqrt{q}\phi}, \ \ f_2 = e^{-\sqrt{q}\phi}$$
Now, we know that an exponential of a real variable is not periodic. And, we know that the exponential of an imaginary variable is periodic, via Euler's equation:
$$e^{i\phi} = \cos \phi + i\sin \phi$$
We need, therefore, $\sqrt q$ to be imaginary. I.e. we need $q$ to be a negative real number.

Moreover, the required periodicity of $2\pi$ implies that $\sqrt q = in$, for $n = 0, 1, 2 \dots$.

We claim, therefore, that the eigenvalues of $\frac{d^2}{d\phi^2}$ (when restricted to $2\pi$-periodic functions) are $q_n = -n^2$, each with two eigenfunctions:
$$f_{n+}(\phi) = e^{in\phi} = \cos(n\phi) + i\sin(n\phi), \ \ f_{n-} = e^{-in\phi} = \cos(n\phi) - i\sin(n\phi) $$
Except the special case $n = 0$, where there is only one eigenfunction $f_0(\phi) = 1$.

And that's all that there was supposed to be to it.

Finally, note that linear combinations of these eigenfunctions are also eigenfunctions. For example:
$$g_{n_1} = \frac 1 2 (f_{n+} + f_{n-}) = \cos(n\phi), \ \ g_{n_2} = \frac 1 {2i} (f_{n+} - f_{n-}) = \sin(n\phi)$$
Gives an alternative pair of eigenfunctions with eigenvalue $-n^2$.

There is, therefore, no unique solution in terms of eigenfunctions. The solution is actually a 2D eigenspace (for $n \ne 0$).

 

[JD_PM]

I'm still not sure how we got here! To go back to the beginning. For the equation:
$$\frac{d^2 f}{d\phi^2} = q\phi$$
We can write down the solutions:
$$f_1 = e^{\sqrt{q}\phi}, \ \ f_2 = e^{-\sqrt{q}\phi}$$
Now, we know that an exponential of a real variable is not periodic. And, we know that the exponential of an imaginary variable is periodic, via Euler's equation:
$$e^{i\phi} = \cos \phi + i\sin \phi$$
We need, therefore, $\sqrt q$ to be imaginary. I.e. we need $q$ to be a negative real number.

Moreover, the required periodicity of $2\pi$ implies that $\sqrt q = in$, for $n = 0, 1, 2 \dots$.

We claim, therefore, that the eigenvalues of $\frac{d^2}{d\phi^2}$ (when restricted to $2\pi$-periodic functions) are $q_n = -n^2$, each with two eigenfunctions:
$$f_{n+}(\phi) = e^{in\phi} = \cos(n\phi) + i\sin(n\phi), \ \ f_{n-} = e^{-in\phi} = \cos(n\phi) - i\sin(n\phi) $$
Except the special case $n = 0$, where there is only one eigenfunction $f_0(\phi) = 1$.

And that's all that there was supposed to be to it.

Finally, note that linear combinations of these eigenfunctions are also eigenfunctions. For example:
$$g_{n_1} = \frac 1 2 (f_{n+} + f_{n-}) = \cos(n\phi), \ \ g_{n_2} = \frac 1 {2i} (f_{n+} - f_{n-}) = \sin(n\phi)$$
Gives an alternative pair of eigenfunctions with eigenvalue $-n^2$.

There is, therefore, no unique solution in terms of eigenfunctions. The solution is actually a 2D eigenspace (for $n \ne 0$).

I have to say that, to me, what you are saying here is correct.

However I want to also follow what Orodruin is saying, to see where we end up.If we are all correct, we should get the exact same answer right? :)

 

[PeroK]

I have to say that, to me, what you are saying here is correct.

However I want to also follow what Orodruin is saying, to see where we end up.If we are all correct, we should get the exact same answer right? :)

Yes, of course. It depends how long you want to spend on this.

 

[JD_PM]

What is the real part of $e^{i\sqrt q \phi}$?

$$\cos(\phi \sqrt q )$$

Ahhh

So using Euler’s formula we end up with

$$(2D - 2E)\sin(\phi \sqrt q ) = 0$$

Let’s label a new constant; $F=2D - 2E$

Thus we have

$$F\sin(\phi \sqrt q ) = 0$$

Let's assume $F \neq 0$. Thus

$$\sin(\phi \sqrt q ) = 0$$

$$\phi \sqrt q = 2\pi n$$

$$q = \frac{4\pi^2}{\phi^2} n^2$$

Where $\phi \neq 0$. OK to match the correct eigenvalue solution ($q_n = -n^2$ as PeroK showed; note that Griffiths states the same solution as PeroK) we need $\frac{4\pi^2}{\phi^2} =-1$. This looks awkward to me, there has to be something wrong in what I have done...

 

[JD_PM]

Yes, of course. It depends how long you want to spend on this.

I am afraid I really like trying different methods in the problems I solve... I cannot avoid to at least try and see what I get

 

[PeroK]

$$D e^{i (\phi + 2\pi) \sqrt{q}} + E e^{- i (\phi + 2\pi) \sqrt{q}} = D e^{i \phi \sqrt{q}} + E e^{- i \phi \sqrt{q}}$$

$$D e^{-i \phi \sqrt{q}} + E e^{i \phi \sqrt{q}} = D e^{i \phi \sqrt{q}} + E e^{- i \phi \sqrt{q}}$$

There must be something wrong in here... I am thinking.

I don't understand this step at all.

 

[JD_PM]

$$D e^{i (\phi + 2\pi) \sqrt{q}} = D e^{-i \phi \sqrt{q}}$$

Right?

 

[PeroK]

$$D e^{i (\phi + 2\pi) \sqrt{q}} = D e^{-i \phi \sqrt{q}}$$

Right?

Wrong! E.g. with $\sqrt q = 1$.

 

[JD_PM]

Mmm so I am stuck in here then

$$D e^{i (\phi + 2\pi) \sqrt{q}} + E e^{- i (\phi + 2\pi) \sqrt{q}} = D e^{i \phi \sqrt{q}} + E e^{- i \phi \sqrt{q}}$$

 

[JD_PM]

Let's see if Orodruin wants to go further.

I am willing to go for the extra method (I am quite curious) but as it is not strictly necessary (I find PeroK's answer satisfactory enough) I would perfectly understand if Orodruin feels it is not necessary.