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...

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...

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.