# Probability Density in an infinite 1D square well

## Homework Statement

The wave function of a particle of mass m confined in an infinite one-dimensional square well of width L = 0.23 nm, is:
ψ(x) = (2/L)1/2 sin(3πx/L) for 0 < x < L

ψ(x) = 0 everywhere else. The energy of the particle in this state is E = 63.974 eV.

1) What is the rest energy (mc2) of the particle?
ANS: 511133.6 ev

2) P(x) is the probability density.
That is, P(x)dx is the probability that the particle is between x and x+dx when dx is small.
For how many values of x does P(x) = 5 nm-1?
ANS: 6 values of x

3)What is the largest value of x for which P(x) = 5 nm-1?
ANS: ?

P(x)=∫|ψ(x)|2dx

E=h2n2/(8mL2)

## The Attempt at a Solution

For question 2 i only know its 6 because the question was multiple attempts. i don't know how to get that answer mathematically. I thought that by integrating P(x) squared and then solving for the x's would give me the 6 values of x but its not making sense to me. After i have the 6 values, then the answer to question 3 is just the largest value of x.

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kuruman
Homework Helper
Gold Member
For how many values of x does P(x) = 5 nm-1?
You have all you need.
P(x)dx= ψ*ψ dx = (2/L) sin2(3πx/L)dx
Solve
P(x) = 5 nm-1

You have all you need.
P(x)dx= ψ*ψ dx = (2/L) sin2(3πx/L)dx
Solve
P(x) = 5 nm-1
so set the integral equal to 5nm-1 right?

After integrating i get 5=(x/L)-(1/6π)sin(6πx/L) How do you even solve for x?

kuruman
Homework Helper
Gold Member
You don't integrate.. If you integrate x drops out and you get just a number, i.e. the probability that the particle be found within the limits of integration. You need to find at what values of x the probability density has the given value. It looks like you are confusing "probability" with "probability density". They are not the same.

gv3
You don't integrate.. If you integrate x drops out and you get just a number, i.e. the probability that the particle be found within the limits of integration. You need to find at what values of x the probability density has the given value. It looks like you are confusing "probability" with "probability density". They are not the same.
Ok I got it. Thank you. The answer was .21 nm. Originally when I first solved it I got .021nm and the hw said I was off by a power of ten. Do you know why it could be off? Also could you explain to me why question 2 is 6 values of x

kuruman
Homework Helper
Gold Member
Do you know why it could be off?
Not unless you reproduce and show me what you did.
Also could you explain to me why question 2 is 6 values of x
Plot P(x) as a function of x from 0 to L and you will see.

You have all you need.
P(x)dx= ψ*ψ dx = (2/L) sin2(3πx/L)dx
Solve
P(x) = 5 nm-1
I solved this and couldn't get 6. i tried with my calculator in radians and degrees.

kuruman
Homework Helper
Gold Member
Here is the plot of P(x) vs. x. Note that the 5 nm-1 line crosses the plot at 6 points.

gv3
Here is the plot of P(x) vs. x. Note that the 5 nm-1 line crosses the plot at 6 points.

View attachment 196861
Ah ok. I was graphing the wrong thing. Thanks for helping me! Sorry for the confusion.