# Particle in a box question

1. Nov 22, 2007

### lozzyjay

1. The problem statement, all variables and given/known data

Hey all!
A particle in a box of side L has wavefunction $$\psi$$(x) = Asin([Pi*x/L) for the ground state.
a) Determine the value of A for a properly normalised wavefunction
b) Determine the probability that x>0.75L

2. Relevant equations

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3. The attempt at a solution

Ok so I started off by normalising the wavefunction

$$\int$$$$\psi$$*$$\psi$$dx with the limits of L and 0

= A$$^{2}$$$$\int$$sin$$^{2}$$(Pi*x/L)dx which would equal 1.

Integrating...

∫ sin^{2} pi x/L) dx

∫[1 - cos(2pi x/L)]/2 dx

∫ dx - 1/2(L/2pi) ∫ cos(2pi x/L) (2pi/L)dx

x/2 - (L/4 pi) sin(2pi x/L)

applying limits 0 and L

[L/2 - 0] - (L/4pi)[sin(2pi L/L) - 0

L/2 - L/4 pi (0)

L/2.

Ok so I need a value though. So because it is in the ground state, the value of L would be $$\pi$$ I think, or is that incorrect?

A$$^{2}$$*Pi/2 = 1
A = 1/(\sqrt{Pi/2})
A = 0.798

Is this correct?

For the second part, b)
I'm totally confused, should I do the same as I did in part a), and integrate, but put in the limits of L and 0.75L?

2. Nov 22, 2007

### nrqed

Do not use any value for L. Just leave it as L. It's ok to have your wavefunction containing L. In the second part, yes, integrate from 0.75 L to L. You will find out that all L dependence will cancel out in your result, leaving you with an actual number.

3. Nov 22, 2007

### lozzyjay

I got:

Let y = pix/L. So the limit becomes pi and 3pi/4
∫sin^2(y) dy
= 0.5∫(1-cos2y) dy
= 0.5(y - 0.5sin2y), y from 3pi/4 to pi
=0.5(pi/4-0.5)

1 = A^2*0.5(pi/4-0.5)

I'm not sure if I did that right though...

And how do I go from this to finding the probability that x>0.75L I'm a bit confused...