we know that the wavefunction for free particle having momentum and energy is a plane wave solution. so it makes sense to guess that particle moving in a circle has a wavefunction of similar form because it essentially resembles the same quantum system, the difference being the constraint. hope this helps.
In the earlier lectures he defined the plane wave (Psi (x,t) = exp(ikx-iwt)) which represented a particle with energy E and a momentum p. Then he wanted to find an equation (i.e. The Schroedinger Equation for free particle) which has as solution this plane wave. As he wanted to derive this equation he introduced the momentum operator (i call it P). The idea to introduce this momentum operator P was that it suffices the following condition: P*Psi(x,t)=p*Psi (x,t) (with psi (x,t) = plane wave of a particle and p = the momentum of this particle) (You can watch the video number 21 where he is introducing the momentum operator) And this condition is exactly the eigenvalue equation with the eigenfunctions Psi (x,t= = exp(ikx-iwt)) and the eigenvalues p. In other words: In Video 21 he already derived the momentum eigenfunctions.
Old question, but just in case someones is looking for it. He used the boundary condition, Ψ(x) = Ψ(x+L), try this and you'll find that e^ikL = 1, and therefore kL = 2πm -> k = 2πm/L. The momentum p_m is discrete and given by p_m = ħk_m = ħ2πm/L. Those are the eigenvalues of momentum operator ^p: ^p ψ_m = p_m ψ_m. Substitute the operator ^p = ħ/i ∂/∂x above and you'll find a first order differential equation which the solutions are given by ψ_m (x) = e^iħ2πmx/L. Remember that solution is not normalized yet, and you should normalize it before continuing the calculation.
Yes. I think the advantage of this method is that it generalizes to finding the eigenfunctions of other quantities. The Fourier Transform method only works for p, but as long as you can write down the general form of eigenfunctions for an arbitrary observable in position space you can do this kind of analysis.
