# Infinite Square Well

Tags:
1. Oct 25, 2016

### danmel413

1. The problem statement, all variables and given/known data
ISW walls at 0 and L, wavefunction ψ(x) = { A for x<L/2; -A for x>L/2. Find the lowest possible energy and the probability to measure it?

2. Relevant equations
Schrodinger equation

ψ(x)=(√2/L)*(sin(nπx/L)

cn=√(2/a)∫sin(nπx/L)dx {0<x<a}

En=n2π2ħ2/2ma2

3. The attempt at a solution
First I normalized and found A= 1/√L

Lowest energy = E12ħ2/2mL2

But I'm going wrong on finding the probability.

P1=|c1|2

When finding cn, I split the integral into two parts, one for ψ(x) = A for x<L/2 and one for ψ(x) = -A for x>L/2 and I get as a result:

(-2/(√L)nπ)(2cos(nπ/2-cosnπ-1) which equals 0 for odd n's and 8/(√L)nπ for even ones. Problem arises when I put that into my probability I'm left with my probability dependent on L - how is it possible that the probability of a particle having a certain energy be dependent on the size of our made up box? I've checked the integral over a thousand times and can't find a mistake. Help?

2. Oct 25, 2016

### Staff: Mentor

Is this equation correct?

Not sure what that represents.

Are these two statements compatible?

3. Oct 25, 2016

### drvrm

the probability of a particle confined in a box of infinite depth is a confinement and its energy goes higher as the width is lower.
the position probability is never a constant inside a box .pl. check your calculation and the limits . it may be zero at certain points.and maximum at other points.
one should see a text book for plot of probability for different states .

4. Oct 25, 2016

### TSny

All of your work looks good, except I don't agree with your numerical factor out front in your result for the integration. As you say, L should not appear. I also get √2 in the numerator instead of 2. So, check your integration again (makes a thousand and 1 times).

When writing fractions with the symbol /, make sure you put the entire denominator in parentheses to agree with "order of operations".

5. Oct 25, 2016

### TSny

That's true if the particle is in one of the energy eigenstates. But here, it is assumed that the particle is in the state given (which would be a superposition of energy eigenstates). The boundary conditions require that the wavefunction goes to zero at the endpoints of the well. So, I guess you need to assume the wavefunction goes from ±A to zero over a negligible interval near the endpoints.

6. Oct 28, 2016

### danmel413

Thanks everyone for your help. I realized my entire problem came from me not copying down the coefficient equation correctly when I went to solve it.