# Infinite Well Potential : going from 0<x<L to 0<x<2L

1. Feb 24, 2009

### lstellyl

1. The problem statement, all variables and given/known data
A particle of mass m is in the lowest energy (ground) state of the infinite potential energy well (V(x)=0 for 0<x<L, and infinite elsewhere)

At time t=0 the wall located at x=L is suddenly pulled back to a position at x=2L. This change occurs so rapidly that instantaneously the wave function does not change. Calculate the probability that a measurement of the energy will yield the ground-state energy of the new well. What is the probability that a measurement of the energy will yield the first excited energy of the new well?

2. Relevant equations

So far I am using the following relevant equations/formulas:

(1) $$\Psi (x) = \sum c_n \sqrt{\frac{2}{L}} \sin{\frac{n \pi x}{L}} \: \: for \: 0<x<L$$

(2) $$\int {\psi_{n}}^{\ast} \Psi dx = c_n$$

and

(3) $$E_n = \frac{n^2 h^2 \pi^2}{2 m L^2}$$

3. The attempt at a solution

This problem is tripping me up a little bit... I am not really sure what is implied/asserted by the phrase "instantaneously the wave function does not change"

So far, I have come up with the following solution....

Obviously, in the new system E will be able to take on 2 values instead of just the ground state since $$E'_1=\frac{h^2 \pi^2}{8 m L^2}$$ and $$E'_2 = E_1 = \frac{h^2 \pi^2}{2 m L^2}$$

I try to use (1) and (2) to calculate $$c_1^'$$ and $$c_2^'$$ using :

$$c'_2 = \frac{\sqrt{2}}{L} \int \sin{\frac{\pi x}{L}} \sin{\frac{\pi x}{L}}$$
from 0 to L which comes out as 1/sqrt(2) which seems reasonable and is nice (P = 1/2) so i left it...

the reasoning i used behind this calculation is that i am multiplying the psi_2 prime with the first wave equation, which has a c_1 of 1, and 0 elsewhere.... which makes the limits 0 to L...

next, i find c'_1 by:

$$c'_1 = \frac{\sqrt{2}}{L} \int \sin{\frac{\pi x}{L}} \sin{\frac{\pi x}{2 L}}$$

from 0 to L which comes out as
$$\frac{4 \sqrt{2}}{3 \pi}$$

in which case P=.3602, and thus this must be wrong unless the particle can take the 2nd excited energy state, which doesn't seem to make sense to me (plus that answer doesn't look as clean or nice)

Can someone help me out or explain to me where it is I am going wrong????

also... if anyone can qualitatively explain to me how it is the wave function "does not change"???

any help appreciated!!!

2. Feb 24, 2009

### lanedance

When the potential instantly changes so do the basis functions for the wavefunction... ie the eigenfunctions

So what you're effectively doing is decomposing the wavefunction in terms of the new eignfunctions

I haven't checked the integrals, but at a quick glance the reasoning looks right. The decomposition is no longer an energy eigenstate & so will evolve with time.

Its an assumption in the problem that the wavefunction does not change so you can decompose in term of the new energy eigenstates. So by changing the potential you are changing the whole state of the system.

Note also that the energy levels in the new potential will have different (lower) values for a given n as L has effectively doubled, so each energy level will be reduced by 1/4

3. Feb 25, 2009

### lstellyl

right... I got that the energy levels would be reduced by 1/4.... thus E2 of the new system is equal to E1 of the old system....

i guess the question I have is this:

is it correct for me to find the constants for the new system c1 and c2 by using the psi_n formula of sqrt(2/L)sin(n*pi*x/L) with L=2L (ie, using the psi_n of the NEW system) and then multiplying that by the wavefunction of the old system which i have determined to be sqrt(2/L)sin(n*pi*x/L) with L=1L since the particle is in the ground energy state, and thus the c1 of the old state is 1...

i feel like there is a flaw in my reasoning here...
using this logic, i found an answer i am happy with for c2, but c1 looks a bit odd to me...

PLUS, adding up c1^2 and c2^2 i do not get 1, meaning there would be some probability that the particle could be in the 2nd excited state if my c1 and c2 are correct