- #1
AwesomeTrains
- 116
- 3
Hey!
1. Homework Statement
I've been given the time dependent schrödinger equation in momentum space and have to calculate the force, as a function of the position, acting on a particle with mass m.
[itex]\vec{F}(\vec{r})=-\nabla{V(\vec{r})}[/itex]
[itex](\frac{\vec{p}^2}{2m}-a\nabla_\vec{p}^2)\psi(\vec{p},t)=i\frac{\partial}{\partial t}\psi(\vec{p},t)[/itex] and [itex]a∈ℝ[/itex]
2. Homework Equations
Fourier transform:
[itex]\psi(\vec{p},t)=\frac{1}{\sqrt{2\pi}}\int \psi(\vec{r},t)e^{i\vec{p}\vec{r}} \, dr^3[/itex]
[itex]\psi(\vec{r},t)=\frac{1}{\sqrt{2\pi}}\int \psi(\vec{p},t)e^{-i\vec{p}\vec{r}} \, dp^3[/itex]
Dirac delta function:
[itex]\delta(\vec{p}-\vec{p}´)\frac{1}{2\pi}\int e^{i\vec{r}(\vec{p}-\vec{p}´)} \, dr[/itex]
ħ=1
3. The Attempt at a Solution
I tried plugging in the expression for [itex]\psi(\vec{p},t)[/itex] into the given equation, to get it transformed into position space.
I got this far:
[itex]\int (\frac{\vec{p}^2}{2m}-i\frac{\partial}{\partial t}) \int \psi(\vec{r}^{'},t) e^{i\vec{p}(\vec{r}^{'}-\vec{r})} \, dr´^3 \, dp^3=-a\psi(\vec{r},t) [/itex]
I wanted to use the property of the delta function again but there's that [itex]\vec{p}^2[/itex] which is messing things up. How do I continue from here? Or is this a wrong approach/ has I made a mistake?
Any hints are appreciated
Kind regards
Alex
1. Homework Statement
I've been given the time dependent schrödinger equation in momentum space and have to calculate the force, as a function of the position, acting on a particle with mass m.
[itex]\vec{F}(\vec{r})=-\nabla{V(\vec{r})}[/itex]
[itex](\frac{\vec{p}^2}{2m}-a\nabla_\vec{p}^2)\psi(\vec{p},t)=i\frac{\partial}{\partial t}\psi(\vec{p},t)[/itex] and [itex]a∈ℝ[/itex]
2. Homework Equations
Fourier transform:
[itex]\psi(\vec{p},t)=\frac{1}{\sqrt{2\pi}}\int \psi(\vec{r},t)e^{i\vec{p}\vec{r}} \, dr^3[/itex]
[itex]\psi(\vec{r},t)=\frac{1}{\sqrt{2\pi}}\int \psi(\vec{p},t)e^{-i\vec{p}\vec{r}} \, dp^3[/itex]
Dirac delta function:
[itex]\delta(\vec{p}-\vec{p}´)\frac{1}{2\pi}\int e^{i\vec{r}(\vec{p}-\vec{p}´)} \, dr[/itex]
ħ=1
3. The Attempt at a Solution
I tried plugging in the expression for [itex]\psi(\vec{p},t)[/itex] into the given equation, to get it transformed into position space.
I got this far:
[itex]\int (\frac{\vec{p}^2}{2m}-i\frac{\partial}{\partial t}) \int \psi(\vec{r}^{'},t) e^{i\vec{p}(\vec{r}^{'}-\vec{r})} \, dr´^3 \, dp^3=-a\psi(\vec{r},t) [/itex]
I wanted to use the property of the delta function again but there's that [itex]\vec{p}^2[/itex] which is messing things up. How do I continue from here? Or is this a wrong approach/ has I made a mistake?
Any hints are appreciated
Kind regards
Alex