Finding the eigenfunctions and eigenvalues associated with an operator

[JD_PM]

Homework Statement

Problem 3.6 Introduction to QM by Griffiths; second edition (I have rewritten it, as my issue is finding the eigenvalues and eigenvalues only)

Consider the following operator

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

Find its eigenfunctions and eigenvalues.

Hint: Note we are using functions $f(\phi)$ on the finite interval $0 \leq \phi \leq 2 \pi$

Relevant Equations

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

$f(\phi +2\pi ) = f(\phi)$

The eigenvalue equation is

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

This is a second order linear homogeneous differential equation. The second order polynomial associated to it is

$$\lambda ^2 - q = 0 \rightarrow \lambda = \pm \sqrt{q}$$

As both roots are real and distinct, the solution to the differential equation is

$$f(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$$

Once here I have two issues:

a) My solution does not match the provided one: $f_{\pm} (\phi) = C e^{\pm \phi \sqrt{q}}$ (maybe they are equivalent and I do not see it XD)

b) Then the solution says 'The periodicity condition $f(\phi +2\pi ) = f(\phi)$ implies that $\sqrt{q} 2\pi = 2in \pi$ but I do not see why.

Thanks

 

[PeroK]

Homework Statement:: Problem 3.6 Introduction to QM by Griffiths; second edition (I have rewritten it, as my issue is finding the eigenvalues and eigenvalues only)

Consider the following operator

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

Find its eigenfunctions and eigenvalues.

Hint: Note we are using functions $f(\phi)$ on the finite interval $0 \leq \phi \leq 2 \pi$
Relevant Equations:: $\frac{d^2}{d \phi^2} f(\phi) = q f(\phi)$

$f(\phi +2\pi ) = f(\phi)$

The eigenvalue equation is

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

This is a second order linear homogeneous differential equation. The second order polynomial associated to it is

$$\lambda ^2 - q = 0 \rightarrow \lambda = \pm \sqrt{q}$$

As both roots are real and distinct, the solution to the differential equation is

$$f(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$$

Once here I have two issues:

a) My solution does not match the provided one: $f_{\pm} (\phi) = C e^{\pm \phi \sqrt{q}}$ (maybe they are equivalent and I do not see it XD)

b) Then the solution says 'The periodicity condition $f(\phi +2\pi ) = f(\phi)$ implies that $\sqrt{q} 2\pi = 2in \pi$ but I do not see why.

Thanks

For a) you are looking for two linearly independent eigenfunctions. The book solution is your solution with $A= 0$ and $B = 0$. You could choose many other combinations of $A, B$.

For b), why not? Try $q = 1$, say. Is that a periodic function?

 

[Orodruin]

As both roots are real and distinct

This is not true, the roots will not be real if $q$ is negative.

 

[JD_PM]

This is not true, the roots will not be real if $q$ is negative.

Ahhhh

So we indeed have

$$\lambda ^2 - q = 0$$

We have three possible cases now:

1) $q > 0$

In this case the solution to the differential equation is:

$$f(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$$

2) $q < 0$

In this case the solution to the differential equation is:

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

3) $q = 0$

In this case the solution to the differential equation is (literally) trivial.

The sum of the solutions is also a solution. Thus:

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

For a) you are looking for two linearly independent eigenfunctions.

Now I have them.

Also, as you suggested, the book just sets $D=E=0$. This to me is not OK.

If they want us to get rid of the constants they should provide us with suitable boundary conditions.

Your following comment suggests that I am wrong on the above sentence though (ie. the condition $f(\phi +2\pi ) = f(\phi)$ is enough to get the provided answer $f_{\pm} (\phi) = C e^{\pm \phi \sqrt{q}}$)

For b), why not? Try $q = 1$, say. Is that a periodic function?

It is a periodic function but I am afraid I do not see your point yet.

 

[Orodruin]

If they want us to get rid of the constants they should provide us with suitable boundary conditions.

There are suitable boundary conditions: periodic boundary conditions. The constants are not ”just set” to zero, it is the only solution compatible with periodicity.

It is a periodic function but I am afraid I do not see your point yet.

so you are essentially saying here that $e^{\phi} = e^{(\phi + 2\pi)}$.

 

[PeroK]

The sum of the solutions is also a solution. Thus:

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

You were looking for eigenfunctions where $q$ was you eigenvalue. That $f$ is not an eigenfunction. It's does not correspond to any given eigenvalue: neither $q$ nor $-q$. Unl;ess $q = 0$, of course, when you have a constant function.

There must be two linearly independent eigenfunctions here. You are asked to find them. Eigenfunctions have an arbitrary constant. If $f$ is an eigenfunction, then so is $Af$ for any $A \ne 0$. So, let's leave any arbitrary constants out. Two linearly independent eigenfunctions are:
$$f_1(\phi) = e^{\phi \sqrt{q}}, \ \ f_2(\phi) = e^{-\phi \sqrt{q}}$$
That's one option. You can check those by plugging them into your equation.

But, because the eigenspace is two dimensional you have an infinite choice of pairs of linearly independent eigenfunctions. You could, instead, choose:
$$g_1(\phi) = e^{\phi \sqrt{q}} + e^{-\phi \sqrt{q}} , \ \ g_2(\phi) = e^{\phi \sqrt{q}} - e^{-\phi \sqrt{q}}$$
Or, equivalently:
$$g_1(\phi) = \cosh(\phi \sqrt{q}), \ \ g_2(\phi) =\sinh(\phi \sqrt{q})$$
Note that at this level of textbook not everything is going to be spelled out every time. Griffiths could have explained all this, but he assumes that either earlier in his book or otherwise you will have encountered this mathematics before.

 

[Orodruin]

Two linearly independent eigenfunctions are:

f1(ϕ)=eϕ√q, f2(ϕ)=e−ϕ√qf1(ϕ)=eϕq, f2(ϕ)=e−ϕq​

f_1(\phi) = e^{\phi \sqrt{q}}, \ \ f_2(\phi) = e^{-\phi \sqrt{q}} That's one option. You can check those by plugging them into your equation.

An operator comes with a domain. Eigenfunctions need to be in that domain. In this case the domain is $2\pi$-periodic functions and those functions are not in the domain because they are not $2\pi$-periodic.

While you could a priori extend the domain to just twice differentiable functions, this then leaves you without the necessary boundary conditions to apply Sturm-Liouville’s theorem, which is fundamental for the expansion in eigenfunctions as used in, for example, QM.

 

[PeroK]

An operator comes with a domain. Eigenfunctions need to be in that domain. In this case the domain is $2\pi$-periodic functions and those functions are not in the domain because they are not $2\pi$-periodic.

While you could a priori extend the domain to just twice differentiable functions, this then leaves you without the necessary boundary conditions to apply Sturm-Liouville’s theorem, which is fundamental for the expansion in eigenfunctions as used in ä, for example, QM.

In the above, I haven't restricted the values of $q$ yet using the boundary conditions.

 

[PeroK]

We have three possible cases now:

1) $q > 0$

2) $q < 0$

3) $q = 0$

An eigenvalue $q$ is in general a complex number.

 

[Orodruin]

In the above, I haven't restricted the values of $q$ yet using the boundary conditions.

This is irrelevant. The definition of the vector space is clear. Eigenvectors need to be in the vector space or they are not eigenvectors.

 

[Orodruin]

An eigenvalue $q$ is in general a complex number.

Not if the operator is self-adjoint.

 

[JD_PM]

OK So we have the following eigenfunctions

$$f_1(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$$

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

Applying the periodic boundary condition we see that the terms with complex exponentials are not periodic. Thus $D=0$ and $E=0$.

The issue now is that I do not know how to explicitly get the values of $A$ and $B$.

My guess is that we should somehow get A=1 and B=1. Thus we would end up with

$$f_1(\phi) = \cosh(\phi \sqrt{q})$$

Which is one of the eigenfunctions PeroK suggested.

 

[PeroK]

OK So we have the following eigenfunctions

$$f_1(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$$

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

Applying the periodic boundary condition we see that the terms with complex exponentials are not periodic. Thus $D=0$ and $E=0$.

The issue now is that I do not know how to explicitly get the values of $A$ and $B$.

My guess is that we should somehow get A=1 and B=1. Thus we would end up with

$$f_1(\phi) = \cosh(\phi \sqrt{q})$$

Which is one of the eigenfunctions PeroK suggested.

You're not quite getting the point of all this. First: only $e^{\pm \phi \sqrt{q}}$ are potential eigenfunctions with eigenvalue $q$. $e^{\pm i \phi \sqrt{q}}$ would have eigenvalue $-q$.

Second:
$$f_1(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$$
Represents an arbitrary linear combination of two linearly independent functions. This represents a 2D eigenspace. For some values of $q$ this is a subspace of the space of periodic functions. For other values of $q$ this is a subspace of a more general space of non-periodic functions.

As with any 2D linear space, you have a choice of basis. One obvious choice of basis is $e^{\pm \phi \sqrt{q}}$. You are not proving that $A = 1, B = 0$ and $A = 0, B = 1$ here. You are choosing these values for $A, B$ to get the basis you want. You could equally choose $A = 1, B = 1$ and $A = 1, B = -1$ etc. This is exactly the same as for any 2D linear space of any vectors.

 

[Orodruin]

OK So we have the following eigenfunctions

$$f_1(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$$

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

Applying the periodic boundary condition we see that the terms with complex exponentials are not periodic. Thus $D=0$ and $E=0$.

Do you know what ”periodic” means?

 

[JD_PM]

You're not quite getting the point of all this. First: only $e^{\pm \phi \sqrt{q}}$ are potential eigenfunctions with eigenvalue $q$. $e^{\pm i \phi \sqrt{q}}$ would have eigenvalue $-q$.

It looks to me like you are assuming that, intrinsically, $q$ is positive always.

Orodruin suggested I should take into account the three possible intrinsic cases: $q>0, q<0$ and $q=0$ and that is what I did in #4. That is a way of solving differential equations I have seen before (for instance in An introduction to PDE by Strauss).

I think we should come to an agreement on how to approach the problem though.

What do you think about it?

 

[JD_PM]

Do you know what ”periodic” means?

Yes. It is when a function yields the same values after complete revolutions. Two famous examples are the sine and cosine functions:

$$\sin(\phi + 2\pi) = \sin(\phi)$$

$$\cos(\phi + 2\pi) = \cos(\phi)$$

 

[JD_PM]

I am at this point now:

Following the reasoning at #4, I got the eigenfunctions

$$f_1(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$$

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

My issue is that I do not know how to get $f_{\pm} (\phi) = C e^{\pm \phi \sqrt{q}}$ out of them. I know I have to use the periodic boundary condition $f(\phi +2\pi ) = f(\phi)$ to get it.

What follows at #12 is simply wrong, so let's show what I have tried so far:

Applying the periodic boundary condition $f(\phi +2\pi ) = f(\phi)$ on $f_1(\phi)$ we get

$$A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}} = A e^{(\phi + 2\pi) \sqrt{q}} + B e^{-(\phi + 2\pi) \sqrt{q}} \ \ \ \ (1)$$

Applying the periodic boundary condition $f(\phi +2\pi ) = f(\phi)$ on $f_2(\phi)$ we get

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

As $e^{\phi} = e^{(\phi + 2\pi)}$, equation $(1)$ becomes trivial and tells us nothing useful.

Plugging Euler's formula $e^{ix} = \cos(x) + i\sin(x)$ into $(2)$ we get

$$D \cos(\phi \sqrt{q}) + D i\sin(\phi \sqrt{q}) + E \cos(\phi \sqrt{q}) - E i\sin(\phi \sqrt{q}) = D \cos((\phi +2\pi) \sqrt{q}) + D i\sin ((\phi +2\pi) \sqrt{q}) + E \cos((\phi +2\pi) \sqrt{q}) - E i\sin ((\phi +2\pi) \sqrt{q}) \ \ \ \ (3)$$

We know that sin and cos functions are periodic, which means that

$$\sin(\phi + 2\pi) = \sin(\phi)$$

$$\cos(\phi + 2\pi) = \cos(\phi)$$

Thus equation $(3)$ becomes also trivial and tells us nothing useful either.

There has to be something wrong in here...

 

[Gaussian97]

Following the reasoning at #4, I got the eigenfunctions

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

Applying the periodic boundary condition $f(\phi +2\pi ) = f(\phi)$ on $f_1(\phi)$ we get

$$A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}} = A e^{(\phi + 2\pi) \sqrt{q}} + B e^{-(\phi + 2\pi) \sqrt{q}} \ \ \ \ (1)$$

As $e^{\phi} = e^{(\phi + 2\pi)}$, equation $(1)$ becomes trivial and tells us nothing useful.

Well, as other comments pointed out before, $f_2$ is NOT an eigenfunction of your operator with eigenvalue $q$ (please, take the time to apply the operator to this function and see that what this is indeed true as many other said to you before).

Also, $e^{\phi}=e^{\phi+2\pi}$ is simply not true at all, and you have $e^{\phi \sqrt{q}}$ so, even if your property was true, doesn't imply that $e^{\phi \sqrt{q}}$ is periodic.

 

[Orodruin]

Yes. It is when a function yields the same values after complete revolutions. Two famous examples are the sine and cosine functions:

$$\sin(\phi + 2\pi) = \sin(\phi)$$

$$\cos(\phi + 2\pi) = \cos(\phi)$$

So please explain why you think that $e^{\phi\sqrt{q}}$ is periodic.

 

[PeroK]

It looks to me like you are assuming that, intrinsically, $q$ is positive always.

No, no, no. This seems to be a common mistake that $q$ means positive and $-q$ means negative.

$q$ can be any real number. So can $-q$. But, by definition, you are looking for eigenfunctions corresponding to $q$. Not $q$ when $q$ is positive and $-q$ when $q$ is negative.

 

[JD_PM]

OK so let's forget about $f_2(\phi)$ for the moment.

We all agree that $f_1(\phi)$ is an eigenfunction with eigenvalue $q$.

At this point it is time to apply the given periodic boundary condition.

But $f_1(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$ is not a periodic function so if we apply $f(\phi +2\pi ) = f(\phi)$ we are not going to get a correct equation.

How to proceed then?

Mmm there is still something I am missing in all of this...

 

[Orodruin]

OK so let's forget about $f_2(\phi)$ for the moment.

We all agree that $f_1(\phi)$ is an eigenfunction with eigenvalue $q$.

At this point it is time to apply the given periodic boundary condition.

But $f_1(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$ is not a periodic function so if we apply $f(\phi +2\pi ) = f(\phi)$ we are not going to get a correct equation.

How to proceed then?

Mmm there is still something I am missing in all of this...

So, you have shownthat the only possible eigenfunction of the form $f_1$ that is periodic is the trivial zero function. Thus, there are no such eigenfunctions. Why have you ignored $f_2$?

 

[Gaussian97]

We all agree that $f_1(\phi)$ is an eigenfunction with eigenvalue $q$.

At this point it is time to apply the given periodic boundary condition.

But $f_1(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$ is not a periodic function so if we apply $f(\phi +2\pi ) = f(\phi)$ we are not going to get a correct equation.

How to proceed then?

$f_1$ can be a periodic function, try to work out the example for specific cases, for example, try to solve it for $q= -4, 0, 4$.

So, you have shownthat the only possible eigenfunction of the form $f_1$ that is periodic is the trivial zero function. Thus, there are no such eigenfunctions. Why have you ignored $f_2$?

No, he/she hasn't shown what you say of $f_1$. And he/she ignored $f_2$ because many of us pointed out that it's not an eigenfunction of our problem, so we can ignore it.
EDIT: Of course $f_2$ is an eigenfunction of the operator we are interested in, but with eigenvalue $-q$, so it's not a solution of the equation we are studying, i.e. $$\frac{d^2f(\phi)}{d\phi^2}=qf(\phi)$$

 

[Orodruin]

$f_1$ can be a periodic function, try to work out the example for specific cases, for example, try to solve it for $q= -4, 0, 4$.No, he/she hasn't shown what you say of $f_1$. And he/she ignored $f_2$ because many of us pointed out that it's not an eigenfunction of our problem, so we can ignore it.

If you had actually read the thread, you would have seen that the OP is assuming $q > 0$.

 

[Orodruin]

No, he/she hasn't shown what you say of $f_1$. And he/she ignored $f_2$ because many of us pointed out that it's not an eigenfunction of our problem, so we can ignore it.

Also, thisis just wrong. $f_2$ is an eigenfunction of the differential operator, this much is obvious, it just has eigenvalue $-q$, not $q$. This does not mean it is not an eigenfunction!

 

[Gaussian97]

If you had actually read the thread, you would have seen that the OP is assuming $q > 0$.

Yes, I have read it, and I wasn't aware that we should assume $q>0$, where did he/she tell that? I'm not able to find such information (perhaps I'm not understanding correctly some posts, I'm not native English speaker). My apologies

Also, thisis just wrong. $f_2$ is an eigenfunction of the differential operator, this much is obvious, it just has eigenvalue $-q$, not $q$. This does not mean it is not an eigenfunction!

Of course, you're right, $f_2$ is an eigenfunction of the operator with eigenvalue $-q$, but we are interested in finding solutions with eigenvalue $q$, not function with eigenvalue $-q$ (at least as far as I have understood). So, when I said that $f_2$ is not an eigenfunction of our problem I mean that it hasn't the correct eigenvalue (which is $q$).
Sorry for the confusion.

 

[JD_PM]

Why have you ignored $f_2$?

Sorry, what I meant is that I wanted to first focus on $f_1$ and then go for $f_2$.

So, you have shown that the only possible eigenfunction of the form $f_1$ that is periodic is the trivial zero function. Thus, there are no such eigenfunctions.

Alright so what you mean, if I am not mistaken, is that as $f_1$ does not satisfy the periodic boundary condition (well only the trivial case satisfies it but, as we know by definition, the eigenfunction cannot be zero), we can claim it is not a valid eigenfunction, don't you?

Having a look at $f_2$, we also see that the periodic condition doesn't apply...

So now I am wondering: if the periodic condition does not apply to either case, how can we get an answer in the first place?

 

[Orodruin]

Having a look at $f_2$, we also see that the periodic condition doesn't apply...

Are you sure? Have you checked explicitly?

 

[JD_PM]

Are you sure? Have you checked explicitly?

Honestly no, I had a quick look.

That was a bad idea, let's do it the right way.

 

[JD_PM]

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.

 

[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.