Question:

A particle is confined in a one-dimensional infinite potential well of width L. The ground state energy of the particle is given by the equation E₀ = (h²π²)/(8mL²), where h is the Planck's constant and m is the mass of the particle.

a) Calculate the de Broglie wavelength (λ) associated with the ground state energy of the particle in terms of L.

b) Consider that the particle is in its ground state and is observed at time t = 0. Determine the probability density function (P(x)) for finding the particle at a position x within the well.

c) The particle is then projected onto an excited state represented by the wavefunction Ψ(x,t) = Asin(3πx/L)exp(-iE₁t/ℏ), where A is a constant, E₁ is the energy associated with the excited state, and ℏ is the reduced Planck's constant. Calculate the expectation value of position (x) for this excited state.

d) Discuss the physical interpretation of the expectation value of position for the excited state.

Answer:

a) The de Broglie wavelength (λ) associated with the ground state energy of the particle can be calculated using the de Broglie relation: λ = h / p, where p is the momentum of the particle.

In the ground state, the momentum of the particle can be determined using the equation p = √(2mE₀), where E₀ is the ground state energy.

Substituting the given values in the equation above, we have:

p = √(2mE₀) = √(2m * (h²π²)/(8mL²)) = √(h²π²/4mL²) = hπ/(2mL)

Now, using the de Broglie relation, we can calculate the de Broglie wavelength:

λ = h / p = (h / (hπ/(2mL))) = 2mL / π

Therefore, the de Broglie wavelength (λ) associated with the ground state energy of the particle is given by λ = 2mL / π.

b) The probability density function (P(x)) for finding the particle at a position x within the well can be determined using the wavefunction Ψ(x,t) associated with the ground state.

The ground state wavefunction, Ψ(x,t), is given by Ψ₀(x,t) = √(2/L) * sin((πx)/L) * e^(-iE₀t/ℏ), where E₀ is the ground state energy.

To calculate the probability density function, we need to find the square modulus of the wavefunction: |Ψ₀(x,t)|² = |√(2/L) * sin((πx)/L) * e^(-iE₀t/ℏ)|²

Simplifying the equation above, we have:

|Ψ₀(x,t)|² = (2/L) * sin²((πx)/L)

Therefore, the probability density function (P(x)) for finding the particle at a position x within the well is given by P(x) = (2/L) * sin²((πx)/L).

c) The expectation value of position (x) for the excited state can be determined using the given wavefunction Ψ(x,t) = Asin(3πx/L)exp(-iE₁t/ℏ).

The expectation value of position is calculated using the equation: ⟨x⟩ = ∫x|Ψ(x,t)|² dx, where |Ψ(x,t)|² is the square modulus of the wavefunction.

Substituting the given wavefunction, we have:

⟨x⟩ = ∫x|Ψ(x,t)|² dx = ∫x|Asin(3πx/L)exp(-iE₁t/ℏ)|² dx

Expanding the equation above, we have:

⟨x⟩ = ∫x|(A * sin(3πx/L) * exp(-iE₁t/ℏ))|² dx = ∫x|A|² * sin²(3πx/L) * dx = |A|² * ∫x sin²(3πx/L) dx

Using the trigonometric identity sin²θ = (1/2) * (1 - cos(2θ)), we can simplify the equation as follows:

⟨x⟩ = |A|² * ∫x (1/2) * (1 - cos(2(3πx/L))) dx = |A|² * 1/2 * ∫x (1 - cos(6πx/L)) dx = |A|² * 1/2 * [x - (L/(6π)) * sin(6πx/L)] | from 0 to L = |A|² * 1/2 * [(L - (L/(6π)) * sin(6π)) - (0 - (0/(6π)) * sin(0))] = |A|² * 1/2 * (L - (L/(6π)) * sin(6π))

Since sin(6π) = 0, the equation simplifies to:

⟨x⟩ = |A|² * 1/2 * L

Therefore, the expectation value of position (x) for the excited state is given by ⟨x⟩ = |A|² * 1/2 * L.

d) The physical interpretation of the expectation value of position for the excited state is that it represents the average position at which the particle is most likely to be found when it is in the excited state. In other words, it indicates the most probable location of the particle within the potential well when it exists in the excited state.