Q:

Can excitons be understood in terms of classical quantum physics?

From what I understand, an exciton is an electron-hole pair in a semiconductor that exists in a bound state (through the electrostatic potential). I have seen it stated that this pair behaves in a way analogous to a hydrogen atom, i.e. as two particles bound through some potential, and having a radius, etc. 
It is not at all clear to me how that could be made concrete. Can the hole somehow be treated as a particle? If so, is it possible to write down the Schrӧdinger equation (or Hamiltonian) for the system, or does the statement belong to quantum field theory or other more general theories? In particular it's not at all obvious to me what would be the mass of a hole.

A:

Hole as a particle
First, hole can really be treated as a particle. For electrons, there are Pauli exclusion principle, so there are only one electron per state(state can be described by momentum $\vec p$, band index and spin). 
In semiconductors, there are valence band and conduction band. In ground state, valence band is completely occupied by electrons, so bulk momentum is zero. For convinience, we can shift energy, such as $E = 0$ for ground state, so
Removing one electron (with momentum $\vec p_1 $) from valence band will make a system to have a positive energy (for valence band, we have $ \varepsilon \approx - \frac{p^2}{2 m_v^*}$, therefore resultative energy will be $E \approx \frac{p_1^2}{2 m_v^*})$. In addition, bulk momentum would be $\vec P = - \vec p_1$.
It is convinient to use hole-formalism, where such pertrubation on ground state is treated like creation of hole, particle with charge $+\vert e \vert$ and energy spectra $\varepsilon(\vec p) = \frac{p^2}{2 m_v^*}$. Here occurs that $m_v^*$, that comes from valence band structure, is a mass of the hole.
What is exciton?
Now suppose we have a system with one hole(one valence electron absent) and one conduction electron. They have charges $+1$ and $-1$ respective, so they can interact electromagnetically. From quantum mechanics, for two interacting particles, we can't have energy of interacting particles to be written as $E_{1-2} = \frac{p_1^2}{2 m_1} + \frac{p_2^2}{2 m_2}$. Rather it can be written in such way: $E_{1-2} = \frac{P^2}{2(m_1 + m_2)}  + E_{inter}$, where first term stands for center of mass movement, and second stands for interaction. 
It occurs that for attractive force E_{inter} can be less than zero, with wave-function such that two particles are localized near each other. It is exactly what is called bound state. In semiconductor system, bound state of electron-hole pair is called exciton.
What actually happens in semiconductor system? Or how can I imagine that electon attracts such a void?
Absense of one valence electron means that there is some $+1$ net charge. Recall that semiconductor is a system that have background ions with positive charge and electrons "flying" over such background. It is difficult to imagine $10^{23}$ number of particles, constantly scattering on each other, but we can imagine that conduction electron is attracted to the places, where valence electron is absent. There are net charge distribution is positive-sign. In quantum-mechanical manner, it can be localized near such places, with gain in energy described approximately by hydrogen states(but with effective $Z, m$).
What formalism should be used to obtain this picture?
Well, exciton is indeed collective excitation, and strictly speaking must be considered from field point of view. However, with some physical intuition, we can approximate such system by some simple solvable models. For example, there we neglect interaction of conducting electron with valence electrons. It permits us to use two-particle wave-function formalism and write a Schrodinger equation for it.

