Infinitely many degeneracy of Landau level: Countable or Uncountable? Description of Landau levels can be found in many standard textbooks of quantum mechanics and here. Two ubiquitous solutions apply either the symmetric gauge $\vec{A}=(-\frac{1}{2}By,\frac{1}{2}Bx,0)$ or the Landau gauge $\vec{A}=(-By,0,0)$ to the Schrödinger equation.
(1) Taking $\{\hat{H},\hat{L}_z\}$ as the complete set of conserved quantities, the symmetric gauge solution often calls for confluent hypergeometric equation in the end, which gives $E=(2n_\rho+m+|m|+1)\hbar\omega_L$, in which $n_\rho=0,1,2,\cdots,\,m=0,\pm1,\pm2,\cdots$.
(2) The above Landau gauge solution diagonalizes $\hat{p}_x$ and reduces the problem to a 1D harmonic oscillator in $y$-direction with completely not fixed equilibrium point $y_0=-\frac{p_x}{qB}$.
We know if the 2D plane in which the electron moves is finite, the degeneracy is the number of flux quanta in the plane. However, if the plane is infinitely large, then the degeneracy of each landau level should be infinity. Intriguingly, this degeneracy appears to be a countable infinity for (1) since $m$ takes discrete values, while being uncountable for (2) since $y_0$ is in continuum. How to resolve this?
One might say this is probably not physically observable, but I guess it is hard to deny this is a well-defined question in its own right.
 A: Expanding my comments: Quantum mechanics is formulated in separable Hilbert spaces, i.e. Hilbert spaces with a countable orthonormal basis. The usual space describing a three dimensional particle is the space of square integrable functions $L^2(\mathbb{R}^3)$. This is separable, as well.
On separable Hilbert spaces, the spectrum of a self-adjoint operator has a continuous and discrete part. Eigenvectors that belong to the Hilbert space are associated only with the eigenvalues of the discrete spectrum. It is possible to formally associate eigenvectors with the continuous spectrum, but they do not belong to the Hilbert space.
Given an eigenvalue (of the discrete spectrum), it has a multiplicity determined by the dimension of the Hilbert subspace spanned by its eigenvectors. This subspace is always contained in the Hilbert space itself, so it is also at most separable, i.e. it admits also at most a countable orthonormal basis.
To get a concrete example (borrowing from another answer): an harmonic oscillator operator $H_x=-\Delta_x + x^2$, with $x\in\mathbb{R}$, on $L^2(\mathbb{R})$ has a purely discrete spectrum, with each eigenvalue of multiplicity 1. If we consider the same operator but acting on $L^2(\mathbb{R}^2)=L^2_x(\mathbb{R})\otimes L^2_y(\mathbb{R})$, i.e. it acts only on the $x$ variable but not on the $y$, it has still the same discrete spectrum but with infinite multiplicity. This is because if $\phi_\lambda(x)$ is an eigenfunction of $H$ (on $L^2(\mathbb{R})$) with eigenvalue $\lambda$, then for any $\psi\in L^2(\mathbb{R})$ we have that on $L^2(\mathbb{R}^2)$:
$$H_x(\phi_\lambda(x)\psi(y))=\lambda\phi_\lambda(x)\psi(y)$$
i.e. $\phi_\lambda(x)\psi(y)$ is an eigenvector of $H$. The multiplicity has become infinite, but its cardinality is still countable since it is the same cardinality of $L^2(\mathbb{R})$, which is countable.
The fact that an operator acts like the multiplication by an "uncountable" variable (i.e. defined for any point of $\mathbb{R}$ which is uncountable) does not mean the associated Hilbert space where it acts on is uncountable, because the space itself is a space of functions.
A: The Hamiltonian as given in
http://en.wikipedia.org/wiki/Landau_quantization is\ (note that $x$ and $y$
are interchanged) is
\begin{equation*}
H=\frac{p_{x}^{2}}{2m}+\frac{1}{2}m\omega _{c}^{2}(x-\frac{\hbar p_{y}}{%
m\omega _{c}})^{2},
\end{equation*}
acting in $\mathcal{H}=L^{2}(\mathbb{R}^{2},dxdy)$.Then, since
\begin{equation*}
x-\frac{\hbar p_{y}}{m\omega _{c}}=\exp [-ip_{x}\frac{\hbar p_{y}}{m\omega
_{c}}]x\exp [+ip_{x}\frac{\hbar p_{y}}{m\omega _{c}}]=U(p_{y})xU^{-1}(p_{y}),
\end{equation*}
we have
\begin{equation*}
H=U(p_{y})\hat{H}U^{-1}(p_{y}),\;\hat{H}=\{\frac{p_{x}^{2}}{2m}+\frac{1}{2}%
m\omega _{c}^{2}x^{2}\}\otimes \mathsf{I}_{y}=\{\sum_{n}E_{n}\mid \varphi
_{n}><\varphi _{n}\mid \}\otimes \mathsf{I}_{y},
\end{equation*}
where $\mathsf{I}_{y}$ is the unit operator acting in $L^{2}(\mathbb{R},dy)$
.Thus $H$ and $\hat{H}$ are unitarily related and hence have, as operators
in $\mathcal{H}$, the same spectrum. But, although
\begin{equation*}
h=\frac{p_{x}^{2}}{2m}+\frac{1}{2}m\omega _{c}^{2}x^{2}
\end{equation*}
acting in $L^{2}(\mathbb{R},dx)$ has discrete spectrum, this is no longer
true for $\hat{H}$ acting in $\mathcal{H}$. Thus, as an operator in $
\mathcal{H}$, $\hat{H}$ has no discrete spectrum at all. Only if we make a
direct integral decomposition (possible since $y$ does not occur in $H$ or $
\hat{H}$), here a Fourier transformation in the $y$-variable, $L^{2}(\mathbb{
R},dy)\rightarrow L^{2}(\mathbb{R},dk)$, $p_{y}\rightarrow k$, we obtain on
each fiber, labelled by $k$, the operator $h$, which has discrete spectrum,
\begin{equation*}
\hat{H}\rightarrow \int^{\oplus }dk\hat{H}(k),\;\hat{H}(k)=h.
\end{equation*}
Thus $h$, acting in $L^{2}(\mathbb{R},dx)$, has discrete spectrum but  $\hat{%
H}$, acting in $\mathcal{H}$ does not.
