Recent content by RedPsi

beta = 3hbar pi^2 /(2mL^2) in the end. It's interesting how making both potentials equal to each other yields the same thing. Anywho, thank you kuruman for your help, it has been invaluable!

! I totally overlooked that! Thank's so much! For a generic infinite-square well where we set v(x) = 0 then, alpha = E1 alpha + beta = E2 => beta = E2 - E1 = 3(hbar)^2(pi)^2/(2mL^2) hmm, is that so?

Yes, it assumes the form 2sin(theta)cos(theta) which is exactly what we have. Thus, the second term is in the second eigenstate (n=2 as mentioned in the starting post). Although this provides me with a new way to look at it, I still don't know what the functional form of the first term is...

Sure thing, A particle of mass m is confined to the interval 0 < x < L in one dimension, and it has wavefunction Ψ(x, t) = Aexp(−iαt)sin(πx/L)[1 + exp(−iβt)cos(πx/L)] (a) What is functional form of the potential energy V (x) in the region 0 < x < L? (b) Determine the value of A and...

Hi Kuruman, thanks for the reply! Well, it's definitely a start, but I still don't see how I would go about finding the beta value. The question is clearly indicating that I find some connection between the two potentials in order to find this, but I just don't see it. hmm.

Homework Statement The problem involves a particle confined to an interval 0 < x < L (in one dimension). It asks to solve for the functional form of the potential V(x) given a wave function (to get to the relevant question, I won't bother providing this). In order to solve for the potential, it...