Solve Infinite Square Well Homework: Find Energy, Probability

AI Thread Summary
The discussion focuses on solving a homework problem related to an infinite square well (ISW) where the wavefunction is defined in two parts. The lowest energy is correctly calculated as E1=π²ħ²/2mL², but confusion arises regarding the probability calculation, particularly how it seems dependent on the box size L. Participants suggest re-evaluating the integration process and checking the numerical factors involved, as the probability should not depend on L. It is emphasized that the wavefunction must adhere to boundary conditions, and the probability distribution varies within the box. Ultimately, the original poster identifies a mistake in copying the coefficient equation, which was the source of the confusion.
danmel413
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Homework Statement


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?

Homework Equations


Schrodinger equation

ψ(x)=(√2/L)*(sin(nπx/L)

cn=√(2/a)∫sin(nπx/L)dx {0<x<a}

En=n2π2ħ2/2ma2

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?
 
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danmel413 said:
ψ(x)=(√2/L)*(sin(nπx/L)
Is this equation correct?

danmel413 said:
cn=√(2/a)∫sin(nπx/L)dx {0<x<a}
Not sure what that represents.

danmel413 said:
Lowest energy = E12ħ2/2mL2

(-2/(√L)nπ)(2cos(nπ/2-cosnπ-1) which equals 0 for odd n's
Are these two statements compatible?
 
danmel413 said:
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?

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 textbook for plot of probability for different states .
 
danmel413 said:
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)

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".
 
drvrm said:
the position probability is never a constant inside a box .
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.
 
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.
 
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