Question

A. Normalize the wave function Ψ=Ae^(-ax^2) where A is the normalization constant and a is an integer.

A= ?

B. What is the expected value of the momentum? <p> = ?

Answer 1

Answer 2

To normalize the wave function Ψ=Ae^(-ax^2), we need to find the value of A that makes the integral of the absolute square of the wave function equal to 1 over the entire domain:

1 = ∫|Ψ|^2 dx from -∞ to +∞

|Ψ|^2 = |Ae^(-ax^2)|^2 = A^2e^(-2ax^2)

1 = ∫A^2e^(-2ax^2) dx from -∞ to +∞

1 = A^2 * ∫e^(-2ax^2) dx from -∞ to +∞

The integral on the right-hand side can be evaluated using a Gaussian integral:

∫e^(-ax^2) dx = sqrt(π/a)

So,

1 = A^2 * ∫e^(-2ax^2) dx from -∞ to +∞

1 = A^2 * √(π/a)

Solving for A:

A = 1 / √(√(π/a))

Thus, A = (a/π)^(1/4)

To find the expected value of momentum, we use the formula:

<p> = ∫Ψ* (-iħ ∂/∂x) Ψ dx from -∞ to +∞

Taking the complex conjugate of Ψ:

Ψ* = Ae^(ax^2)

Differentiating Ψ with respect to x:

∂Ψ/∂x = -2axAe^(-ax^2)

Substituting into the formula for <p>:

<p> = -iħ A^2 ∫(∂Ψ/∂x) e^(-ax^2) dx from -∞ to +∞<p> = -iħ A^2 ∫(-2axAe^(-ax^2))e^(-ax^2) dx from -∞ to +∞<p> = 2ħaA^2 ∫x e^(-2ax^2) dx from -∞ to +∞

The integral on the right-hand side can be evaluated using integration by parts:

u = x, dv = e^(-2ax^2) dx du = dx, v = -1/2a e^(-2ax^2)

<p> = 2ħaA^2 [(-1/2a)xe^(-2ax^2)| from -∞ to +∞ - ∫(-1/2a)e^(-2ax^2) dx from -∞ to +∞]

The first term evaluates to 0 because e^(-2ax^2) goes to 0 faster than x goes to ±∞. The integral on the right-hand side is just the same Gaussian integral we evaluated earlier:

<p> = 2ħaA^2 * (-1/2a) * √(π/2a)

Substituting in the expression for A:

<p> = -iħ (aπ)^(1/2) * 2a/π * (-1/2a) * √(π/2a)<p> = iħ * √(a/2π)

Therefore, the expected value of momentum is <p> = iħ * √(a/2π).