Finding the eigenfunctions and eigenvalues associated with an operator

[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.