A particle is initially in its ground state in an infinite one - dimen...
Introduction:
In this problem, we are given a particle initially in its ground state in an infinite one-dimensional potential box with sides at x = 0 and x = a. We are then asked to calculate the probability of finding the particle in the ground state of the new box when the wall at x = a is suddenly moved to x = 3a.
Understanding the problem:
To solve this problem, we need to calculate the probability of finding the particle in the ground state after the change in the potential box. The ground state wavefunction for an infinite one-dimensional potential box is given by:
ψ(x) = √(2/a) * sin(nπx/a)
Where n is the quantum number for the state of the particle and a is the length of the box.
Calculating the probability:
To calculate the probability of finding the particle in the ground state of the new box, we need to normalize the wavefunction and then find the probability density.
Normalizing the wavefunction:
To normalize the wavefunction, we need to calculate the normalization constant A:
∫|ψ(x)|² dx = 1
Integrating the modulus squared of the wavefunction gives:
∫(2/a) * sin²(nπx/a) dx = 1
Using the trigonometric identity sin²θ = (1/2)(1 - cos(2θ)), we can rewrite the integral as:
(2/a) * ∫(1/2)(1 - cos(2nπx/a)) dx = 1
Simplifying the integral, we have:
(1/a) * [(1/2)x - (a/4nπ)sin(2nπx/a)]|₀ᵃ = 1
Evaluating the integral at the limits of integration, we get:
(1/a) * [(1/2)a - (a/4nπ)sin(2nπa/a)] - (1/a) * [(1/2) * 0 - (a/4nπ)sin(2nπ * 0/a)] = 1
Simplifying further, we have:
(1/2) - (1/2nπ)sin(2nπ) = 1
The term sin(2nπ) is equal to zero since sin(2nπ) = sin(0) = 0. Therefore, the equation becomes:
(1/2) = 1
This implies that the normalization constant A is equal to √2.
Calculating the probability density:
To find the probability density, we need to calculate |ψ(x)|²:
|ψ(x)|² = (2/a) * sin²(nπx/a)
Substituting the value of the normalization constant A, we have:
|ψ(x)|² = (2/a) * sin²(nπx/a)
Since we are interested in the ground state (n = 1), the probability density becomes:
|ψ(x)|² = (2/a) * sin²(πx/a)
Calculating the probability:
To find the probability of finding the particle in the ground state of the new box, we need to integrate the probability density over the range x = a to x = 3