Question

Calculate d

Answer 1

Where
<p>=m(d<x>/dt)=−ih/2π∫Ψ∗(∂Ψ/∂x)dx

Answer 2

\[<p>=m<v>=m(d<x>/dt)=−ih/2π∫Ψ^{∗}(∂Ψ/∂x)dx\]

Answer 3

if you know the hamiltonian of the system, you can use the Ehrenfest theorem:
\[\frac{d\langle A \rangle }{dt} = \frac{i}{\hbar}\langle [H,A] \rangle + \left \langle \frac{\partial A}{\partial t} \right \rangle \], where A is an operator
but if you only have this equation, you just have to derivate it:
\[\frac{d\langle p \rangle }{dt} = -i\hbar \int \frac{d}{dt} \left(\Psi^{*} \frac{\partial \Psi}{\partial x} \right) dx \]

Answer 4

well i tried deriving it but i got stuck

Answer 5

\[{d⟨p⟩\over dt}=−iℏ∫({d \over dt}(Ψ^∗{∂Ψ\over∂x}))dx \]
\[=−iℏ∫({d{\Psi^*} \over dt}{∂{\Psi} \over dx}+Ψ^∗{∂^2Ψ\over∂x∂t})dx \]
\[=−iℏ∫(({-iℏ \over 2m}{∂^2 Ψ^* \over ∂x^2}+{i \over ℏ}VΨ^*){∂Ψ\over∂x} +Ψ^∗{∂\over∂x}({iℏ\over2m}{∂^2\Psi \over ∂x^2}+{i \over ℏ}V \Psi))dx\]

Answer 6

\[=-iℏ \int\limits ({iℏ\over2m} ( \Psi^{*}{{∂^3 \Psi \over ∂x^3} - {∂^2 \Psi^* \over ∂x^2}{∂ \Psi \over ∂x})+ {i \over ℏ} (V \Psi^*} {∂ \Psi \over ∂x} -\Psi^* {∂V \over ∂x} \Psi))dx\]
\[=\int\limits( {-iℏ^{2} \over 2m} (\Psi^* {∂^{3}\Psi \over ∂x^{3}}-{∂\Psi^{*} \over ∂x^{2}}{∂\Psi \over ∂x})+(V \Psi^{*}{∂\Psi \over ∂x}-\Psi^{*} {∂ \over ∂x}V \Psi))dx\]