How does the quantum path integral relate to the quantization of energy? So, the quantum path integral is a generalization of the classical principle of least action- but here we know that all paths contribute something finite to the probability density. What confuses me is that this doesn't seem to involve the quantization of energy (action) at all. How does this come into play? Is it a separate assumption, only coming into play when we worry about the non-commutativity of certain operators?
 A: The path integral in quantum mechanics computes the evolution kernel, which is the matrix element of the evolution operator: 
$\mathrm{exp}(iH t) $, ($H$ is the Hamiltonian), between two position eigenstates.
The path integral expresses the evolution kernel as a sum over paths:
$U(x,t, x_0) = \int_{x(0)=x_0}^{x(t)=x} \mathrm{exp}(\frac{iS}{\hbar}) \mathcal{D}x$.
where $U$ is the evolution kernel, $S$ is the classical action.
On the other hand, the evolution operator has an expansion as a sum over the energy eigenstates:
$U(x,t, x_0) =\sum_n \mathrm{exp}(\frac{-iE_n t}{\hbar})\psi_n(x) \psi_n^{*}(x_0)$
where, $\psi_n(x)$ are the energy eigenstates.
From this expression, it is clear that the evolution kernel has a discrete spectrum, whenever the energies are quantized.
In other words, in the case of quantized values of the energy, the Fourier transform of the evolution kernel:
$\hat{U}(x,\omega, x_0)\equiv \int_{-\infty}^{\infty} U(x,t, x_0) \mathrm{exp}(i\omega t) dt$
will be a sum of Dirac delta functions centered at the frequencies:
$\omega_n = \frac{E_n}{\hbar}$
Please observe that the weight of each Dirac delta function is just the projection operator on the corresponding discrete eigenstate.
A: In general, path integral does not imply quantization of anything. First, you can use path integrals for classical wave optics. Second, even in quantum mechanics there are non-bonding potentials (take the free particle potential $V(x)=0$ for the simplest case).
However, sometimes from the path integral point of view it can be argued that some things are quantized. Perhaps the best known example is the Aharonov-Bohm effect where to have non-ambiguous solution phase $e^{i S/\hbar}$ can not change when a particle makes a full circle around a flux of magnetic field.
A: The quantization is not necessary in quantum mechanics. It is essential, but not necessary. Energy of a free particle is not quantized. Even for oscillator you are free to construct any state and energy. 
Quantization appears at the next step. If you want for instance study stable states which do not change with time. It is not separate assumption. It is different problem.
