Wave Packet Momentum

Given a Wave Packet, how do we find the expected value of momentum, given the definition of wave packet? We know that

⟨p^⟩=∫−∞∞ψ∗(x)[−iℏddx]ψ(x) dx\langle \hat{p} \rangle = \int_{-\infty}^{\infty} \psi^*(x)\left[-i\hbar \frac{d}{dx}\right]\psi(x)\, dx

Apply ddx\frac{d}{dx} to ϕ(x)eikx\phi(x)e^{ikx} via product rule

ddxψ(x)=ddx[ϕ(x)eikx]=ddxϕ(x)eikx+ik ϕ(x) eikx\frac{d}{dx}\psi(x) = \frac{d}{dx}\left[ \phi(x)e^{ikx}\right]= \frac{d}{dx} \phi(x)e^{ikx} + ik\,\phi(x)\, e^{ikx}

To get we integrate

⟨p^⟩=−iℏ∫−∞∞ϕ(x)e−ikx(dϕ(x)dxeikx+ikϕ(x)eikx)dx\braket{\hat{p}}= -i\hbar \int_{-\infty}^{\infty} \phi(x)e^{-ikx}\left(\frac{d\phi(x)}{dx}e^{ikx} + ik\phi(x)e^{ikx}\right)dx =−iℏ∫ϕ(x),dϕ(x)dxe−ikxeikx dx+(−iℏ)(ik)∫ϕ2(x)e−ikxeikx dx=-i\hbar\int \phi(x),\frac{d\phi(x)}{dx}e^{-ikx}e^{ikx}\,dx + (-i\hbar)(ik)\int \phi^2(x)e^{-ikx}e^{ikx}\,dx =−iℏ∫−∞∞ϕ(x)dϕ(x)dx dx+ℏk∫−∞∞ϕ2(x) dx= -i\hbar \int_{-\infty}^{\infty} \phi(x)\frac{d\phi(x)}{dx}\,dx + \hbar k \int_{-\infty}^{\infty} \phi^2(x)\,dx =−iℏ∫−∞∞ϕ(x)dϕ(x)dx dx+ℏk(1)= -i\hbar \int_{-\infty}^{\infty} \phi(x)\frac{d\phi(x)}{dx}\,dx + \hbar k\quad(1)

Note

ddx[ϕ2(x)]=2 ϕ(x)dϕdx  ⟹  ϕ(x)dϕdx=12ddx[ϕ2(x)]\frac{d}{dx}[\phi^2(x)] = 2\,\phi(x)\frac{d\phi}{dx} \implies \phi(x)\frac{d\phi}{dx} = \frac{1}{2}\frac{d}{dx}[\phi^2(x)]

Note

∫−∞∞ϕ(x)dϕdx dx=12∫−∞∞ddx[ϕ2(x)] dx=12[ϕ2(x)]−∞∞\int_{-\infty}^{\infty} \phi(x)\frac{d\phi}{dx}\,dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{d}{dx}[\phi^2(x)]\,dx = \frac{1}{2}\Big[\phi^2(x)\Big]_{-\infty}^{\infty} =12(0−0)=0= \frac{1}{2}(0-0) = 0

So sub into (1)

⟨p^⟩=−iℏ(0)+ℏk\braket{\hat{p}}=-i\hbar(0)+\hbar k =ℏk=\hbar k