Assume a photon has a wave function, what can be said about its Hamiltonian and eigenvalues? I understand a photon does not have an associated wave function, but what if we assumed a photon does in fact have a wave function. How would this look and how could its Hamiltonian be determined?
 A: In classical mechanics, a particle with 0 mass is not so clear, and this extends to quantum mechanics as well.
In relativity, however, light has a rigorous framework. In relativity, $E \neq \frac{p^2}{2m}$ for a free particle. It's instead:
$$E^2 = (mc^2)^2+(pc)^2$$
and so for light, which has 0 mass:
$$E^2=(pc)^2 \implies E=pc$$
So in some sense, you can define the photon hamiltonian (in 1 dimensional case) as:
$$H = c i \hbar \frac{\partial}{\partial x}$$
So The Schrodinger Equation becomes:
$$i\hbar \frac{\partial \psi}{\partial t} =i \hbar c \frac{\partial\psi}{\partial x} \implies \frac{\partial \psi}{\partial t} = c \frac{\partial \psi}{\partial x}$$

Note that for a complete hamiltonian of a photon, you actually need to learn about relativistic quantum mechanics. This is because even though for a photon:
$$E=pc$$
momentum is still a 3-vector while energy is a number. And the Schrodinger equation would have a gradient which is a vector.
To fix this, you need to work with relativistic equations which keeps energy squared and momentum squared:
$$-\hbar^2\frac{\partial^2 \psi^\mu}{\partial t^2}=-\hbar^2 c^2\nabla^2\psi^\mu \implies \frac{1}{c^2} \frac{\partial^2 \psi^\mu}{\partial t^2}=\nabla^2 \psi^\mu$$
Which you can say is Maxwell's equation describing light. But this also doesn't give the complete picture. Because I'm trying to keep the answer in the scope of your question, I cannot answer. For a complete picture, you need QFT. For more information check this.
