A particle is confined to a one-dimensional box (an infinite well) on the x-axis between x = 0 and x = L. The potential height of the walls of the box is infinite. The normalized wave function of the particle, which is in the ground state, is given by with 0

The wave functions of a particle confined to an infinite potential well between x = 0 and x = L are
Psi_n (x) = sqrt(2/L)sin((npix)/L) , where n is an integer (n = 1, 2, 3...). These wave functions are normalized so that the probability of finding the particle in the well is 1 and the probability of finding the particle outside of the well is 0.
Since the particle in this problem is in the ground state, n = 1 and its wave function is
Psi_1(x) = sqrt(2/L)sin(pix/L) .
b) The probability of finding the particle between x = 0 and x = L/3 is then
P = int_0 ^ (L/3) |Psi_1|^2 dx
Let's work with the integrand first and rewrite it using a trigonometric half-angle identity:
|Psi_1|^2 = 2/Lsin^2(pix/L) = 2/L*1/2*(1 - cos(2pix/L)) = 1/L - 1/Lcos(2pix/L) .
Then, the original probability integral breaks up into the two integrals. The first one is
int_0 ^ (L/3) 1/L dx = 1/L*L/3 = 1/3
and the second one is
int_0 ^(L/3) 1/Lcos(2pix/L)dx = 1/L *L/(2pi) (sin(2pix/L) |_0 ^(L/3) =1/(2pi)sin(2pi/3) = 1/(2pi)*sqrt(3)/2 = sqrt(3)/(4pi)
So the probability will be P = 1/3 - sqrt(3)/(4pi) = 0.196 , which confirms your result.
 
Part a seems to be less straightforward. The probability density function |Psi(x)|^2 describes the probability of finding a particle at a given point. I am not sure what is meant by probability per unit length. Just dividing the total probability (1) by the length L would result in 1/L, not 2/L.
 
 
 

Hello!
Unfortunately, you still haven't given the wave function and I have to guess it. If the probability per unit length of finding the particle is a constant, then the probability density function is also a constant on [0, L]. This constant must be 1/L for the total probability be 1. For a non-constant wave function Psi(x) (and therefore a non-constant probability density function p_d(x) = |Psi(x)|^2 ) the results would be different.
In general, it must be int_(-oo)^(+oo) p_d(x) dx = 1, here it is int_0^L p_d(x) dx.
a) the probability per unit length is 1/L (it might be 2/L for some other probability density function). It is true only for intervals inside [0, L].
b) it is int_0^(L/3) p_d(x) dx and it is equal to 1/L*L/3 = 1/3 for the "guessed" function.