Question 1.7: Calculate d(p)/dt . Answer. d(p)/dt = (-∂V/∂x) . (1.38) This...

From Eq. 1.33,

$\left\langle p\right\rangle = m \frac{d \left\langle x\right\rangle }{dt} = i\hbar \int{\Bigl(\Psi ^*\frac{\partial \Psi }{\partial x} \Bigr) } dx$.    (1.33).

$\frac{d\left\langle p\right\rangle }{dt} = – i\hbar \int{\frac{\partial  }{\partial t} \bigl(\Psi ^*\frac{\partial \Psi }{\partial x}\bigr) }dx$.

But, noting that $\frac{\partial ^2 Ψ}{\partial x \partial t} = \frac{\partial ^2 Ψ}{\partial t \partial x}$.

and using Eqs. 1.23-1.24:

$\frac{\partial  Ψ}{ \partial t} = \frac{i\hbar}{2m} \frac{\partial ^2 Ψ}{\partial x ^2} – \frac{i}{\hbar}V Ψ$.        (1.23).

$\frac{\partial  Ψ^*}{ \partial t} = -\frac{i\hbar}{2m} \frac{\partial ^2 Ψ^*}{\partial x ^2} + \frac{i}{\hbar}V Ψ^*$.    (1.24).

$= \frac{ i\hbar}{2m} \biggl[ Ψ^* \frac{\partial ^3 Ψ}{\partial x ^3} – \frac{\partial ^2 Ψ^*}{\partial x ^2} \frac{\partial  Ψ}{\partial x } \biggr] + \frac{ i}{\hbar}\biggl[ V Ψ^* \frac{\partial  Ψ}{\partial x} – Ψ^* \frac{\partial  }{\partial x } (V Ψ)\biggr]$.

The first term integrates to zero, using integration by parts twice, and the second term can be simplified to $V Ψ^* \frac{\partial  Ψ}{\partial x} – Ψ^* V \frac{\partial  Ψ}{\partial x} – Ψ^* \frac{\partial  V}{\partial x} Ψ = – \left|\Psi \right|^2 \frac{\partial  V}{\partial x}$ .

So

$\frac{d\left\langle p\right\rangle }{dt} = – i\hbar \biggl(\frac{ i}{\hbar}\biggr)  \int { – \left|\Psi \right|^2 \frac{\partial  V}{\partial x}}dx = \left\langle – \frac{\partial V}{\partial x} \right\rangle$ .    QED