Redundant eigenfunctions of an operator In quantum mechanics, the eigenvalues and eigenfunctions of the operator $\hat{L}_z$ can be calculated by solving the differential equation
$$
-i\hbar\dfrac{\partial\Phi}{\partial \varphi} = \lambda\Phi(\varphi)\ ,
$$
where $\varphi \in [0,2\pi]$ is the azimuthal angle in spherical coordinates. If $\lambda = m\hbar$, then the eigenfunctions are
$$
\Phi_m(\varphi) = A e^{im\varphi}\ ,
$$
and applying the periodicity condition over $\varphi$, $\Phi(0)=\Phi(2\pi)$, we get that $m\in\mathbb{Z}$.
However, because of some compatibility between this operator and another quantum operator, $\hat{L}^2$, it should be possible to get the same result from the process of getting the eigenfunctions of $\hat{L}^2$. The problem of the eigenvalues of this operator is
$$
-\dfrac{\hbar^2}{\sin\theta}\dfrac{\partial}{\partial\theta}\left(\sin\theta\dfrac{\partial Y(\theta,\varphi)}{\partial\theta}\right) - \dfrac{\hbar^2}{\sin^2\theta}\dfrac{\partial^2 Y(\theta,\varphi)}{\partial\varphi^2} = \lambda Y(\theta,\varphi)\ ,
$$
where, after using the separation of variables $Y(\theta,\varphi)=\Theta(\theta)\Phi(\varphi)$, we can isolate the differential equation
$$
-\hbar^2\dfrac{\partial^2\Phi}{\partial\varphi^2} = \mu\Phi\ .
$$
Inspecting different possibilities, lets focus on $\mu > 0$, thus $\mu = m^2\hbar^2$. Being this the case, the general solution to the differential equation is
$$
\Phi(\varphi) = Ae^{im\varphi} + Be^{-im\varphi}\ .
$$
Applying the same condition as before on the periodicity over $\varphi$, we get that
$$
A + B = Ae^{i2\pi m} + Be^{-i2\pi m}\ ,
$$
which suggests that $m\in\mathbb{Z}-\{0\}$. The case where $\mu=0$ also works so, definitively, $m\in\mathbb{Z}$.
This result is similar to the previous one, but now there is an extra term, $B\exp(-im\varphi)$, which does not vanish. Some texts suggest that because $m$ takes values all over $\mathbb{Z}$, this second term is redundant, and therefore the eigenfunctions can be reduced to just $A\exp(im\varphi)$.
I would understand this redundancy if we were trying to make a representation of some function on the basis $\{\Phi_m\}$,
$$
\Psi(\varphi) = \sum_{m\in\mathbb{Z}} c_m\Phi_m(\varphi)\ ,
$$
where the coefficients $c_m$ could take into account the repetitions over the same exponential functions. However, I'm not able to see how this could be also true when working with a single eigenfunction. I hope that someone can give me a more detailed mathematical explanation on why it is ok to just remove the second exponential if $m\in\mathbb{Z}$, or any other explanation about how $B$ would vanish.
 A: The eigenspaces for a periodic problem can be 2-dimensional. That's not the case for 2nd order Sturm-Liouville problems with separated endpoint conditions. Here the eigenspace at $m=0$ is 1-dimensional. The other eigenvalues are $1^2, 2^2, 3^2,\cdots$, and the eigenspaces corresponding to the remaining eigenvalues $\{ \sin(m\pi x),\cos(m\pi x) \}$ are 2-dimensional. You're no longer dealing exclusively with 1-dimensional eigenspaces only.
By bringing complex numbers into the eigenvalue problem for a real equation, you have created confusion about the eigenspace dimensions, except at $\lambda=0$. This can be convenient, but it adds to the confusion, and will not generally work out to be so clean for the general periodic Sturm-Liouville case.
A: Eigenvectors for a fixed eigenvalue are not uniquely determined, as the set of eigenvectors for a fixed eigenvalue is a subspace. So what you usually want to do when choosing eigenvectors is to form an orthonormal basis.
If you say that the eigenvectors are "just" $Ae^{im\varphi}$ with $m\in\mathbb Z$, then $Ae^{im\varphi}$ and $Ae^{-im\varphi}$ are eigenvectors for the same eigenvalue. And linear combinations will again give you an eigenvectors for the same eigenvalue, so $Ae^{im\varphi}+Be^{-im\varphi}$ is still there.
If you want to keep considering eigenvectors of the form $Ae^{im\varphi}+Be^{-im\varphi}$, you get that $m$ and $-m$ give you eigenvectors for the same eigenvalue, so it makes sense to restrict to $m\geq0$.
Since in the end you want an orthonormal basis, $\{e^{im\varphi}\}_{m\in\mathbb Z}$ (with coefficients to normalize) is enough. Or, if you  restrict to $m\geq0$, you need to make choices for $A$ and $B$; typically one chooses $A=B=\tfrac12$ and $A=\tfrac12$, $B=-A$, to get the basis $\{\cos m\varphi\}_{m\in 0\cup\mathbb N}\cup\{\sin m\varphi\}_{m\in\mathbb N}$ (with appropriate coefficients to normalize).
