Hydrogen atom with discrete nonlinear Schrödinger equation

  • Thread starter cryptist
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  • #1
cryptist
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Hi everyone,
How can I solve hydrogen atom with discrete nonlinear schrödinger equation? Could you help me with the mathematics of that, please?
 

Answers and Replies

  • #2
fzero
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What physics do you propose is responsible for nonlinearity? Why do you believe that the standard linear equation is insufficient?
 
  • #3
cryptist
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I want to consider all the possible forces and potential fields even gravitational forces between proton and electron, which is always neglected. Isn't there will be nonlinearity in equations in that case?
 
  • #4
Nicodemus
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I want to consider all the possible forces and potential fields even gravitational forces between proton and electron, which is always neglected. Isn't there will be nonlinearity in equations in that case?

Gravitational forces between a proton and electron? I realize that both do exert such a field, but at those distances it's a non-issue. Sounds like you want to screw in a nail.
 
  • #5
tom.stoer
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Why should the nice linear Schrödinger equation become non-linear? If the proton is located at r=0 you simply add V=GmM/r as a potential term for the electron, that's all. This results in a tiny correction to the usual Coulomb term.
 
  • #6
cryptist
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So there is no need for nonlinearity? Then, how should be the discrete formula?
 
  • #7
tom.stoer
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Which discrete formula? Usually you have

[tex]U_C(r) = -\frac{e^2}{4\pi\epsilon_0}\frac{1}{r}[/tex]

Now you add the potential according to Newton's law of gravitation

[tex]U_N(r) = -Gm_em_p\frac{1}{r}[/tex]

The total potential energy then reads

[tex]U(r) = -\left(\frac{e^2}{4\pi\epsilon_0} + Gm_em_p\right)\frac{1}{r} = -\frac{e^2}{4\pi\epsilon_0} \left(1 + \epsilon\right)\frac{1}{r}[/tex]

This results in a rescaling of the energy levels due to the term

[tex]\epsilon = \frac{4\pi\epsilon_0Gm_em_p}{e^2}[/tex]
 
  • #8
alxm
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This is all fairly pointless. To begin with, special-relativistic corrections are on the order of millionths of the electronic energy. A few orders-of-magnitude later, QED corrections come into play. So the Schrödinger equation itself ceases to be a useful description of electrons in an atom many orders of magnitude before any effects of gravity come into play.

Even if you did this correction, what would you compare it to? There are no measurements of electronic states that are anywhere near that accurate. As far as I know, the electronic levels of hydrogen have already been calculated to well within today's experimental accuracy without taking gravity into account. There isn't much point in improving the model if you don't have anything to test it against.
 
  • #9
tom.stoer
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Of course the corrections are tiny, but that is not the point. cryptist talked about a non-linear equation in order to take gravity into account. All what I did is to show that Newtonian gravity does not change the linear structure of the equation.
 

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