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