# Normalization of a wave function

1. Oct 8, 2014

### leehufford

1. The problem statement, all variables and given/known data
A particle is described by the wave function psi(x) = b(a2-x2) for -a < x < +a and psi(x) = 0 for x < -a and x > +a, where a and b are positive real number constants.

a) Using the normalization condition, find b in terms of a.
b) What is the probability to find the particle at x = +a/2 in a small interval of width 0.010a?
c) What is the probability for the particle to be found between x = +a/2 and x = +a?

2. Relevant equations

Psi(x) = b(a2-x2)
The integral of Psi(x)2 from negative infinity to infinity must equal one, so that way it can exist.
P(x) dx = (Psi(x))2

3. The attempt at a solution

I have an answer, but I was hoping someone could confirm I did this right, this concept is brand new to me and my answer from A looks weird to me.

Part A:

b2* Integral of (a2-x2)(a2-x2) dx = 1 (From -a to a, since everywhere else Psi is zero).

b2[a4x - (2/3)a2x2 + (1/5)x5] dx = 1 evaluated from -a to a...

b2[(a5-(2/3)a5+(1/5)a5) - (-a5 + (2/3)a5-(1/3)a5) = 1

b2(16a5/15) = 1, so b = sqrt(15/16a5).

Part B:

P(x)dx = (Psi(x))2 dx

= (15/16a5)(a2-x2)2 dx
= (15/16a5)(a2 - (1/4)a2)2(0.010a)
= (15/16a5)(3/4)a4(0.010a)
=0.007, or 0.7 percent chance of finding the particle there.

Part C: I didn't attempt part C because I wanted to hopefully get some feedback that I was on the right track before I started part C. Looks like I will actually have to integrate for part C since the interval is bigger. Thanks for reading.

Lee

Last edited: Oct 8, 2014
2. Oct 8, 2014

### Orodruin

Staff Emeritus
I did not check your computations in detail, but your approach is correct and assuming you did the math correctly your results for (a) and (b) should therefore be correct as well. Just watch out for (what I assume is) the typo in aa in your problem formulation.

3. Oct 8, 2014

### leehufford

Thank you Orodruin for your reply and noticing the typo. It's been corrected. I'm pleased to know the approach is correct since I have already double checked the math.

4. Oct 8, 2014

### Staff: Mentor

There are indeed many typos, but there is also an error in part b.

That equation (and the number you get from it) is incorrect.

5. Oct 8, 2014

### leehufford

DrClaude - thank you for the reply. I noticed i forgot to square the (3/4), so my final equation is now

(15/16a5)(9/16)a4(0.010a) = 0.005 = 0.5 percent chance.