Set-up and solve the quantum mechanical problem of a particle of mass m insidea one-dimensional infinite potential well of width a. Find the energy eigen values andthe eigen wave functions. Show that the energy states are discrete. Sketch the wave-function, probability density and energy for first and second excited states of theparticle

To set up and solve the quantum mechanical problem of a particle of mass m inside a one-dimensional infinite potential well of width a, we can use the time-independent Schrödinger equation. The Schrödinger equation for this system is given by:

-ħ²/2m * d²ψ(x)/dx² + V(x)ψ(x) = Eψ(x)

where ħ is the reduced Planck's constant, m is the mass of the particle, ψ(x) is the wave function, V(x) is the potential energy function, E is the energy eigenvalue, and x is the position coordinate.

In this case, the potential energy function V(x) is infinite within the well (0 ≤ x ≤ a) and zero outside the well. Therefore, V(x) can be represented as:

V(x) = 0, for 0 ≤ x ≤ a V(x) = ∞, for x < 0 or x > a

Inside the well, the Schrödinger equation becomes:

-ħ²/2m * d²ψ(x)/dx² = Eψ(x)

To solve this equation, we assume that the wave function ψ(x) can be written as a linear combination of sine and cosine functions:

ψ(x) = A sin(kx) + B cos(kx)

where A and B are constants to be determined, and k is the wave number.

Applying the boundary conditions, we find that the wave function must be zero at x = 0 and x = a. This gives us the following conditions:

ψ(0) = A sin(0) + B cos(0) = 0 ψ(a) = A sin(ka) + B cos(ka) = 0

From the first condition, we have B = 0, since cos(0) = 1. From the second condition, we have sin(ka) = 0, which implies ka = nπ, where n is an integer.

Therefore, the allowed values of k are given by:

k = nπ/a

Substituting this back into the wave function, we have:

ψ(x) = A sin(nπx/a)

To normalize the wave function, we integrate the absolute square of the wave function over the entire well and set it equal to 1:

∫[0,a] |ψ(x)|² dx = 1

∫[0,a] |A sin(nπx/a)|² dx = 1

Solving this integral, we find that A = √(2/a).

Therefore, the normalized wave function for the particle inside the well is:

ψ(x) = √(2/a) sin(nπx/a)

The energy eigenvalues E can be found by substituting the wave function into the Schrödinger equation:

-ħ²/2m * d²ψ(x)/dx² = Eψ(x)

Differentiating ψ(x) twice with respect to x, we have:

-ħ²/2m * (-n²π²/a²) √(2/a) sin(nπx/a) = E √(2/a) sin(nπx/a)

Simplifying, we find:

E = (n²π²ħ²)/(2ma²)

This shows that the energy eigenvalues are discrete and depend on the quantum number n.

To sketch the wave function, probability density, and energy for the first and second excited states of the particle, we can substitute the values of n = 1 and n = 2 into the wave function and energy equation.

For the first excited state (n = 1), the wave function is:

ψ(x) = √(2/a) sin(πx/a)

The probability density |ψ(x)|² is given by:

|ψ(x)|² = (2/a) sin²(πx/a)

The energy for the first excited state is:

E₁ = (π²ħ²)/(2ma²)

For the second excited state (n = 2), the wave function is:

ψ(x) = √(2/a) sin(2πx/a)

The probability density |ψ(x)|² is given by:

|ψ(x)|² = (2/a) sin²(2πx/a)

The energy for the second excited state is:

E₂ = (4π²ħ²)/(2ma²)

By sketching these wave functions, probability densities, and energies, we can visualize the behavior of the particle inside the infinite potential well for the first and second excited states.