How Do Eigenfunctions and Eigenvalues Evolve in a 1-D Infinite Potential Well?

[Panic Attack]

Homework Statement

I didn't put this in the right format the first time. My question is, I have a 1-D box confined at at x = 0 and x = L. So, points between 0 and L distances are the continuum state and otherwise distances be discontinous.
a) I need to find the egien functs: Un(x) and related egien values: En ... n are the excited levels represented as postive whole numbers.

The wave funct is: φ(x, t = 0) = (1/(3^1/2))U_2(x) + ((2/3)^1/2)U_3(x)


b) As time progresses, what will the function look like?
c) What is the prob. density (φ squared) and P(x,t) = total probability.

Homework Equations

Schrodinger Eq.

The Attempt at a Solution

What I have so far...

(-(h/2pi)^2)/2m * (d^2/dx^2)Psi(x) = E*Psi(x)
Psi(x)|x=0 = Asin(0) + Bcos(0) = B = 0 ?
Psi(x)|x=L = Asin(kL) + Bcos(kL) = 0 ?

[0 0; sin(kL) cos(kL)] *[A;B] = [0 0]

set KnL/2 = n*pi
En = (h/2pi)^2 *k^2]/2m
= [(h/2pi)^2] /2m * (2n*pi/L)

 

[NateD]

^2= [h^2/8mL^2]*n^2Un(x) = A_n sin[K_n(x-L/2)] + B_n cos[k_n(x-L/2)] A_n = sqrt(2/L)B_n = 0b) As time progresses, what will the function look like?The wave function will have the same form as before, but with time-dependent coefficients:φ(x,t) = (1/(3^1/2)U_2(x))e^(-iE_2t/h) + (2/3^1/2)U_3(x)e^(-iE_3t/h) c) What is the prob. density (φ squared) and P(x,t) = total probability.The probability density is given by φ(x,t)^2:φ(x,t)^2 = (1/3)*U_2(x)^2e^(-2iE_2t/h) + (4/9)*U_3(x)^2e^(-2iE_3t/h) + (2/3)*U_2(x)*U_3(x)e^(-i(E_2+E_3)t/h) The total probability is given by P(x,t):P(x,t) = ∫φ(x,t)^2dx = ∫(1/3)*U_2(x)^2e^(-2iE_2t/h) + (4/9)*U_3(x)^2e^(-2iE_3t/h) + (2/3)*U_2(x)*U_3(x)e^(-i(E_2+E_3)t/h)dx = 1