Suppose a particle has wavefunction ψ(x,t=0)=Aexp−(x^2/2l^2). What is the average value (expectation value) of p^, ⟨p^⟩, for this state at t=0?
To find the average value or expectation value of the momentum operator p̂ for a given state ψ(x, t=0), you need to calculate the expression ⟨p̂⟩ = ∫ψ*(x, t=0)p̂ψ(x, t=0)dx, where ψ* represents the complex conjugate of the wavefunction ψ.
In this case, the given wavefunction is ψ(x, t=0) = Aexp[-(x^2/2l^2)]. To find ⟨p̂⟩, you need to substitute this wavefunction into the expression ⟨p̂⟩ = ∫ψ*(x, t=0)p̂ψ(x, t=0)dx.
The momentum operator p̂ is defined as p̂ = -iħ(d/dx), where ħ is the reduced Planck constant.
Now, let's substitute the given wavefunction and the momentum operator into the expression ⟨p̂⟩ = ∫ψ*(x, t=0)p̂ψ(x, t=0)dx:
⟨p̂⟩ = ∫[Aexp(-x^2/2l^2)]*(-iħ)d/dx[Aexp(-x^2/2l^2)] dx
To simplify this expression, we need to apply the derivative operator to ψ(x, t=0). Let's expand the derivative:
d/dx[Aexp(-x^2/2l^2)] = -x/l^2 [Aexp(-x^2/2l^2)]
Now we can substitute this back into the expression for ⟨p̂⟩:
⟨p̂⟩ = ∫[Aexp(-x^2/2l^2)]*[-iħ]*[-x/l^2][Aexp(-x^2/2l^2)] dx
Simplifying further, we get:
⟨p̂⟩ = iħ/l^2 ∫xexp(-x^2/l^2) dx
To solve this integral, you can use the technique of integration by parts. By letting u = x and dv = exp(-x^2/l^2) dx, you can find du and v, and then apply the integration by parts formula:
∫u dv = uv - ∫v du
After performing the integration by parts, you will end up with an expression for ⟨p̂⟩.