Probability of finding a free particle inside the left half of a 1-D box of length L

In quantum mechanics, the probability of finding a particle in a certain region is given by the square of the wavefunction. For a free particle in a 1-D box, the wavefunction can be represented by a sine or cosine function. Let's assume that the particle is in the ground state, which corresponds to the lowest energy state of the system.

1. Ground state wavefunction

The ground state wavefunction for a particle in a 1-D box of length L can be represented as:

ψ(x) = √(2/L) * sin((π/L) * x)

where x is the position of the particle inside the box, and L is the length of the box.

2. Probability density

The probability density, |ψ(x)|^2, gives the probability of finding the particle at a particular position x. For a free particle in the ground state, the probability density is:

|ψ(x)|^2 = (2/L) * sin^2((π/L) * x)

3. Finding the probability within the left half of the box

To find the probability of finding the particle inside the left half of the box, we need to integrate the probability density function from x = 0 to x = L/2. The integral is given by:

P = ∫[0 to L/2] |ψ(x)|^2 dx

P = ∫[0 to L/2] (2/L) * sin^2((π/L) * x) dx

Simplifying the integral, we have:

P = (2/L) * ∫[0 to L/2] sin^2((π/L) * x) dx

4. Solving the integral

Using a trigonometric identity, sin^2(θ) = (1/2) * (1 - cos(2θ)), we can rewrite the integral as:

P = (2/L) * ∫[0 to L/2] (1/2) * (1 - cos((2π/L) * x)) dx

Simplifying further, we have:

P = (1/L) * [x - (L/(2π)) * sin((2π/L) * x)] evaluated from 0 to L/2

P = (1/L) * (L/2 - (L/(2π)) * sin(π)) - (1/L) * (0 - (L/(2π)) * sin(0))

P = (1/2) - (1/π) * sin(π)

Since sin(π) = 0, the probability simplifies to:

P = (1/2) - 0

P = 1/2

5. Conclusion

Therefore, the probability of finding a free particle inside the left half of a 1-D box of length L is 1/2. This means that there is an equal chance of finding the particle on either side of the box.