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Homework Help: Calculate the potential

  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data

    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](\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. Relevant 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]

    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 :smile:
    Kind regards
  2. jcsd
  3. May 8, 2015 #2
    Hi, I am not sure how you are expected to use the Dirac Delta function, but my best guess at finding an answer would be to solve for the wave function in momentum space, then do the furier transform, and then solve for the potential in the Scroedinger Equation for position space.
  4. May 9, 2015 #3
    Thanks for the reply.
    I tried doing that but got stuck already in the first step. I couldn't find the solution in momentumspace therefore I tried transforming the whole equation.
    Can you explain how you solve this equation? Do I have to guess a solution?
  5. May 9, 2015 #4
    Hi AwesomeTrains, I see what you mean I started the solution and ran into the same problem. I tried another approach that I am not 100% sure is correct although it does give me an answer for the force. Basically try substititing in the equation for the furier transform instead of you wave function. You can then express your formula as
    [itex]\int(\frac{\vec{p}^2}{2m}-a\nabla_\vec{p}^2)[\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}]d\vec{r}^3=\int i\frac{\partial}{\partial t}\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}d\vec{r}^3[/itex]
    then substitute in the hamiltonian for ##i\frac{\partial}{\partial t}\psi(\vec{r},t)## and compare like therms in your integral.
  6. May 9, 2015 #5
    Thanks alot for helping me out :biggrin:
    Okay I'll try and continue from where you stopped:
    [itex]\int(\frac{\vec{p}^2}{2m}-a\nabla_\vec{p}^2)[\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}]d\vec{r}^3=\int i\frac{\partial}{\partial t}\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}d\vec{r}^3[/itex]
    Then from the first equation we have that (Dividing by [itex]\psi(\vec{r},t)[/itex]):
    [itex](\frac{\vec{p}^2}{2m}-a\nabla_\vec{p}^2)=i\frac{\partial}{\partial t}[/itex]
    Plugging it in:
    [itex]\int(\frac{\vec{p}^2}{2m}-a\nabla_\vec{p}^2)[\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}]d\vec{r}^3=\int (\frac{\vec{p}^2}{2m}-a\nabla_\vec{p}^2)\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}d\vec{r}^3[/itex]

    How do I continue from here? Did I plug in the wrong hamiltonian? :-p

    If I would plug in the general one [itex]H=T+V[/itex] I would by comparison, like you said, get that [itex]V(\vec{p})=a\nabla_\vec{p}^2[/itex] since the first term looks like the kinetic energy.

    Do I then transform the potential into position space and then do the derivative to get the force? Is this how you got your result?

    Kind regards
  7. May 9, 2015 #6
    $$\int(\frac{\vec{p}^2}{2m}-a\nabla_\vec{p}^2)[\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}]d\vec{r}^3=\int (\frac{\vec{p}^2}{2m}+a\vec{r}^2)\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}d\vec{r}^3$$
    This is because the laplacian of our momentum coordiantes acts on ##\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}##, where the only term with ##\vec{p}## dependence is the exponential function. This then means
    $$\int \left(\frac{\vec{p}^2}{2m}+a\vec{r}^2\right)\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}d\vec{r}^3=\int i\frac{\partial}{\partial t}\psi(\vec{r},t)e^{i\vec{p}\cdotp\vec{r}}d\vec{r}^3$$
    Also, note that in position space, the hamiltonian becomes
    $$\left(\frac{\vec{p}^2}{2m}+V(\vec{r})\right)\psi(\vec{r},t)=i\frac{\partial}{\partial t}\psi(\vec{r},t)$$
    This is the nesessary equation to use here, as we cannot use the hamiltonian for momentum space when the operators act on the wavefunction for position space.
  8. May 9, 2015 #7
    Ah I see, then [itex]V(\vec{r})=a\vec{r}^2[/itex] and it follows that [itex]\vec{F}(\vec{r})=-2a\vec{r}[/itex] is that what you got aswell? If it is then thanks for the help :)
    Last edited: May 9, 2015
  9. May 10, 2015 #8
    Yeah that is what I got as well, glad I could help.
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