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Visualising Potential Wells in Two Body Systems

  1. Sep 22, 2013 #1
    I am currently working through Griffiths' textbook on quantum mechanics. The hydrogen atom was first modelled as a one body system with the proton fixed at the origin. In this case the potential was given by Coulomb's law,
    [tex]V(r) = -\frac{e^2}{4\pi\epsilon_0}\frac{1}{r} \ ,[/tex]
    where r is the radial coordinate.

    This potential is easy to visualise as a "potential well" -- at least in two dimensions -- with the proton at the centre with V=-∞ and then the potential approaching 0 as r goes to ∞.

    However, the hydrogen atom is then reconsidered as a multi body problem with the motion of the proton now accounted for. The positions of the particles are given by r1 and r2, and I understand the change into new coordinates: the separation distance r and the centre of mass R.

    My question is this: how can the potential
    [tex]V(\vec{r}_1,\vec{r}_2) = V(\vec{r}_1-\vec{r}_2) = V(\vec{r}) = V(|\vec{r}|) = -\frac{e^2}{4\pi\epsilon_0}\frac{1}{r} [/tex]
    now be visualised as a potential well. Can this only be done in some kind of 2n-dimensional "configuration space" (where n is the number of space dimensions) of tuples
    [tex](\vec{r}_1,\vec{r}_2) \ ?[/tex]

    Also, what is the interpretation of a general multi body potential
    [tex]V = V(t,\vec{r}_1,\vec{r}_2,\ldots,\vec{r}_m) \ ?[/tex]
    Is this the total potential energy of the system at time t when particle 1 is at position r1, particle 2 is at position r2 and so on?
     
    Last edited: Sep 22, 2013
  2. jcsd
  3. Sep 22, 2013 #2
    Well, in the two-body case since the potential only depends on the separation vector ##\vec{r}##, you can visualize it in the 3D space that ##\vec{r}## lives in. In general though, you are right: when you have ##n## particles the potential is a function of ##3n## spatial coordinates, and so becomes much harder to visualize. Note that the wave function also depends on ##3n## spatial coordinates, so when you have many particles you can no longer visualize the wave function as a wave propagating in real space. Instead it is a wave in configuration space.

    Yes.
     
  4. Sep 26, 2013 #3
    Thanks for the help!
     
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