3
$\begingroup$

After reading about the hydrogen atom and understanding how Schrodinger's equation explains most part of the atomic spectrum of an hydrogen atom, and also came to know that, it explains most of the chemical reactions and a huge tool in chemistry.

I am now almost convinced, that it is wise to accept the Schrodinger equation as a law that govern's the motion of subatomic particles like electrons at quantum scales. Now I am a little curious about one problem. How does an electron (a distribution of charge) move under the influence of its own electrostatic Coulomb's field. I am interested only in strictly theoretical sense, but also like to know if there is any practical importance to it.

I'd like to consider this problem first in a 1-D setup, purely due to my lack of acquaintance with partial differential equations. So lets consider a 1-D electron, as a linear charge distribution of constant density $\rho$ and distributed over a length $2r_e$.

Now I am interested to setup the Schrodinger equation for it, in the case where there is no external field. I'd appreciate some help/comments on setting it up, solution and analysis/interpretation of the resulting wave function and what it actually means, at diefferent energy levels(very high, very low, etc).

Improve this question
$\endgroup$
11
  • $\begingroup$ Comment to the question (v1): Note that standard 1D/2D/3D quantum mechanical treatment of the hydrogen atom considers the electron in the proton's electrostatic field rather than the electron's own electrostatic field. $\endgroup$ – Qmechanic♦ Jul 11 '13 at 13:28
  • $\begingroup$ Electrons are considered point particles - though this statement is sidenoted by this answer (does anyone care to expand on this?) - so the radius $r_e$ would be zero. But that's actually just a pedantic point in this context. The electric field around an electron (on its own in a vacuum) is spherically symmetric (as far as we can tell), so this would not influence the electron's movement. $\endgroup$ – Wouter Jul 11 '13 at 13:30
  • $\begingroup$ @Qmechanic : I know this is a different problem from the Hydrogen atom. I am just curious about this special problem. $\endgroup$ – Rajesh Dachiraju Jul 11 '13 at 13:33
  • 1
    $\begingroup$ @RajeshD Yes and so the only term in the Schrödinger equation for a free electron would be the kinetic energy one, since it is not influenced by its own (spherically symmetric) electric field. $\endgroup$ – Wouter Jul 11 '13 at 13:44
  • 1
    $\begingroup$ Well there is such a thing as self-interaction in QFT, but not (as far as I know) in the context of regular QM - which is what I think you're currently looking into? And I'm not even sure this self-interaction is possible for a free electron. Could be wrong though. $\endgroup$ – Wouter Jul 11 '13 at 14:00
-1
$\begingroup$

schrodinger equation is for a single particle problem. if you want to use it for electron you need to go deep to represent the mass and charge distribution then express the the electromagnetic potential due to this charge distribution and its rotation beside you may need to reform schrodinger equation to be in terms of mass function and angular momentum or make some assumption to approximate the solution. I think this must ends with minimum energy state of electron mass and spin.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.