Calculate time evolution of Schrodinger wave equation

[natugnaro]

[SOLVED] Calculate time evolution of Schrödinger wave equation

Homework Statement

At time t=0 particle is in state:


$$\psi\left(x\right)=\sqrt{2}A\phi_{1}(x)+\frac{A}{\sqrt{2}}\phi_{2}(x)+A\phi_{3}(x)$$

where $$\phi_{n}(x)$$ are eigenfunctions of 1-D infinite potential well.
a) Normalize the state
b) calculate $$\psi(x,t)$$
d) Calculate <E>

Homework Equations

$$\psi\left(x,t\right)=\sum_{n}b_{n}*\phi_{n}(x)*e^{\frac{-iE_{n}t}{h/2Pi}}$$

$$b_{n}=<\phi_{n}(x)|\psi(x,0)>$$ $$\Rightarrow$$ $$b_{n}=\int^{L}_{0}\sqrt{\frac{2}{L}}*Sin(\frac{n*Pi*x}{L})*\psi(x,0)*dx$$

The Attempt at a Solution

From normalization I've got A = (2/7)^1/2.

To calculate $$\psi(x,t)$$ I have used

$$\psi\left(x,t\right)=\sum_{n}b_{n}*\phi_{n}(x)*e^{\frac{-iE_{n}t}{h/2Pi}}$$

To get to bn coefficients I need to calculate the integral:

$$b_{n}=\int^{L}_{0}\sqrt{\frac{2}{L}}*Sin(\frac{n*Pi*x}{L})*\psi(x,0)*dx$$

This integral gives Sin[n*Pi] multiplied by some expression,
but this means thath all bn coefficients are zero (since Sin[n*Pi]=0 for every integer n).
So my $$\psi(x,t)$$ is always zero except for t=0, is this possible ?

This also means that probability of finding particle later inside the well is zero,
but particle should be located shomewere inside well, potential walls are infinite so particle
can't escape !?

 

[natugnaro]

Question is really simple. I have some state function (which is not zero) at t=0.
Then when I calculated Psi(x,t) for t other than zero I've got Psi(x,t)=0 !
Is something missing in my problem formulation ?

Is this possible ? simple yes or no would be great !

 

[natugnaro]

I've done integral for bn coefficients again , and I've got:

b1=0/0 , b2=0/0 , b3=0/0 , and for n>3 bn=0 , now I'm totally puzzled, how should I interpret this result? where in space is my particle ?
I could post my mathematica notebook if that helps.

Please help !

 

[Dick]

You don't NEED to calculate the bn's. You already have them. E.g. b1=sqrt(2)*A. The integral of phi_n(x)*phi_m(x) is equal to zero only if m is not equal to n. It's equal to 1 if m=n. They are orthonormal.

 

[natugnaro]

Ah !, I understand it now, so:
b1=2/sqrt(7) , b2=1/sqrt(7) , b3=sqrt(2/7) , bn=0 for n>3

and
$$\psi(x,t)=b_{1}e^{-i\omega_{1}t}\phi_{1}(x) + b_{2}e^{-i\omega_{2}t}\phi_{2}(x) + b_{3}e^{-i\omega_{3}t}\phi_{3}(x)$$

d)

$$<E>=\sum|b_{n}|^{2}E_{n}$$

$$<E>=\frac{1}{7}(4E_{1}+4E_{1}+18E_{1})=\frac{26}{7}E_{1}$$

Simple explanation but it helped me. Thanks a lot Dick , I was so much in calculating integrals that I could not see the obvious.
I also understan why I got 0/0 for first three bn's before.