Reason for the generation of London Dispersion Forces
You are correct that simply stating that the Schrodinger equation is time 
dependent does not really answer the question.
In fact, time-dependence is not a component of the derivation of the dispersion forces. At a surface level, they just represent how much the ground state energy of a system of polarizable molecules is lowered by their interactions. 
It's the same idea as any of the other intermolecular forces (coulomb, dipole-dipole, etc.) where some interaction between particles leads to a lower energy state than the free particles. 

Generally, the London Dispersion interaction potential is not derived in a general chemistry class because it is purely quantum mechanical and rather involved. These slides provide a clear and thorough (albeit advanced) derivation, but I will explain the main points below.

Derivation Explanation

Let's imagine for this model that a polarizable atoms can be thought of as a simple harmonic oscillator with a positive charge ($+q$) on one end of a spring and a negative charge ($-q$) on the other end. The spring has an equilibrium length of 0, so that the atom has zero average dipole.
This model of an polarizable atom captures the physics we want because applying a force of some kind to the spring will induce a dipole in this atom  ($\vec{p} = q \vec{d}$).

So far we have only considered one of these atoms. What if we move to consider multiple of them together?
Remember we're really interested in the energy difference between two non-interacting atoms and two interacting atoms, so let's start by finding the non-interacting ground state energy of two atoms. We'll let the two atoms have two different spring frequencies $\omega_1$ and $\omega_2$ respectively. 

Setting $z_1$ and $z_2$
as the displacements of the springs from equilibrium for atoms 1 and 2 respectively and $m$ as the mass of an atom,
we can write the total potential energy of the system as:
$$
U_{\text{non-interact}} = \frac{1}{2}\left(m \omega_1^2z_1^2 + m \omega_2^2z_2^2\right)
$$

Solving the Schrodinger equation for this potential yields the ground state energy for the two non-interacting harmonic oscillators:
$$
E = \frac{\hbar}{2} (\omega_1 + \omega_2)
$$

Now, we turn on the interaction and the problem becomes harder more interesting because the positions of the springs affect the magnitude of the interaction. Physically, because the dipole-dipole interaction can lower the energy, we expect that the interacting ground state of the two atoms will be a balance between minimizing the stretching of the springs (which increases the energy) and maximizing the amount of dipole-dipole interaction (which lowers the energy). 

Setting $z$ as the distance between the two atoms, with some algebra one can show that the interaction energy between these two dipoles is given by 
$$
U_{\text{interact}}(z,z_1,z_2) = \frac{q^2}{2\pi\epsilon_0}\frac{z_1 z_2}{z^3}
$$

Now, our total potential energy for the system is given by:
$$
U_{\text{total}}(z,z_1,z_2) = U_{\text{non-interact}}(z_1,z_2) + U_{\text{interact}}(z,z_1,z_2)
$$
While intimidating looking, this potential can also be solved exactly by changing variables to $z_+ = (z_1 + z_2)$ and $z_- = (z_1 - z_2$) and rearranging terms to make it look like the non-interacting potential energy. Once that is done the ground state energy can be found just as easily as in the non-interacting case.

With this done, we can find that the (one dimensional) energy stabilization is given by:
$$
U(z) = \frac{-\alpha^2 I}{2}\frac{q^2}{(4\pi\epsilon_0)^2}\frac{1}{z^6}
$$
Where the final step was to replace some of our less measurable parameters with their atomic counterparts: polarizability ($\alpha$) and ionization potential ($I$)