What is the time evolution of the wave function?

[Panic Attack]

Ok, 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.


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)

 

[Panic Attack]

https://www.physicsforums.com/attachments/21538

Nobody wants to help Adam...

 

[criz.corral]

Hi Panic_Attack!

In the second boundary condition you made a mistake, because you
already know that B=0, You really have:

\Psi(x=L) = ASin(kL) = 0

And this condition say to you:

kL = n\pi

where

k^{2}\equiv\frac{2mE}{\hbar^{2}}

Then you got E_{n}

To find the wave function you don't know A yet, but try to normalize the wave function.

On (b) part.. Have you heard about the Evolution Operator? Maybe this simplify your problem.

(Sorry my english sucks)

 

[Panic Attack]

Hi Panic_Attack!

In the second boundary condition you made a mistake, because you
already know that B=0, You really have:

\Psi(x=L) = ASin(kL) = 0

And this condition say to you:

kL = n\pi

where

k^{2}\equiv\frac{2mE}{\hbar^{2}}

Then you got E_{n}

To find the wave function you don't know A yet, but try to normalize the wave function.

On (b) part.. Have you heard about the Evolution Operator? Maybe this simplify your problem.

(Sorry my english sucks)

Thanks so much for replying to my question. Fortunately I was able to find an answer without using the evolution operator. I basically went through solving with the schrodinger equation with setting up the solutions of the differential equations based on the regions. And had the same K value you got too. Then I normalised the wave function with it squared over the integral and found A too... I really apreciate your help, sorry I couldn't reply sooner.

Your english sounds better than mine! lol