torsdag 27 mars 2014

How to Make Schrödinger's Equation Physically Meaningful + Computable

The derivation of Schrödinger's equation as the basic mathematical model of quantum mechanics is hidden in mystery: The idea is somehow to start considering a classical Hamiltonian $H(q,p)$ as the total energy equal to the sum of kinetic and potential energy:
  • $H(q,p)=\frac{p^2}{2m} + V(q)$,
where $q(t)$ is position and $p=m\dot q= m\frac{dq}{dt}$ momentum of a moving particle of mass $m$, and make the formal ad hoc substitution with $\bar h =\frac{h}{2\pi}$ and $h$ Planck's constant:
  • $p = -i\bar h\nabla$ with formally $\frac{p^2}{2m} = - \frac{\bar h^2}{2m}\nabla^2 = - \frac{\bar h^2} {2m}\Delta$, 
to get Schrödinger's equation in time dependent form 
  • $ i\bar h\frac{\partial\psi}{\partial t}=H\psi $,
with now $H$ a differential operator acting on a wave function $\psi (x,t)$ with $x$ a space coordinate and $t$ time, given by 
  • $H\psi \equiv -\frac{\bar h^2}{2m}\Delta \psi + V\psi$,
where now $V(x)$ acts as a given potential function. As a time independent eigenvalue problem Schrödinger's equation then takes the form:
  •  $-\frac{\bar h^2}{2m}\Delta \psi + V\psi = E\psi$,
with $E$ an eigenvalue, as a stationary value for the total energy
  • $K(\psi ) + W(\psi )\equiv\frac{\bar h^2}{2m}\int\vert\nabla\psi\vert^2\, dx +\int V\psi^2\, dx$, 
as the sum of kinetic energy $K(\psi )$ and potential energy $W(\psi )$, under the normalization $\int\psi^2\, dx = 1$.  The ground state then corresponds to minimal total energy,

We see that the total energy $K(\psi ) + W(\psi)$ can be seen as smoothed version of $H(q,p)$ with
  • $V(q)$ replaced by $\int V\psi^2\, dx$,
  • $\frac{p^2}{2m}=\frac{m\dot q^2}{2}$ replaced by  $\frac{\bar h^2}{2m}\int\vert\nabla\psi\vert^2\, dx$,
and Schrödinger's equation as expressing stationarity of the total energy as an analog the classical equations of motion expressing stationarity of the Hamiltonian $H(p,q)$ under variations of the path $q(t)$. 

We conclude that Schrödinger's equation for a one electron system can be seen as a smoothed version of the equation of motion for a classical particle acted upon by a potential force, with Planck's constant serving as a smoothing parameter. 

Similarly it is natural to consider smoothed versions of classical many-particle systems as quantum mechanical models resembling Hartree variants of Schrödinger's equation for many-electrons systems, that is quantum mechanics as smoothed particle mechanics, thereby (maybe) reducing some of the mystery of Schrödinger's equation and opening to computable quantum mechanical models.

We see Schrödinger's equation arising from a Hamiltonian as total energy kinetic energy + potential energy, rather that from a Lagrangian as kinetic energy - potential energy. The reason is a confusing terminology with $K(\psi )$ named kinetic energy even though it does not involve time differentiation, while it more naturally should occur in a Lagrangian as a form of the potential energy like elastic energy in classical mechanics.  

17 kommentarer:

  1. and make the formal ad hoc substitution

    QM has a lot deeper theoretical justification than you give it credit for.

    It is possible that the earlier derivations used an ad hoc substitution like this (I don't really know since Schroedingers original derivation differs from a more modern one).

    Introductory text books certainly use to include the operator ad hoc.

    If you read Sakurai's Modern Quantum Mechanics, ch 1.7, you will see how the operator p in a plane wave basis is derived including powers of p.

    Further, I can't really understand your point in the last paragraphs. Have you read the different formulations of QM and the Ehrenfest theorem? If so you should be aware of the connections between a Hamiltonian or Lagrangian approach and the connection to a classical picture. What is new in that you write here??

  2. The idea is to seek a formulation of QM which is not an ad hoc uncomputable linear multidim Schrödinger equation requiring a statistical interpretation.

  3. And you don't think anybody has tried that...? ;-)

  4. Of course, but if you insist to use statistics it means that you have given up physical interpretation.

  5. Would you say that electrons are distinguishable particles?

    If so, how come that all experiments including some numbers of electrons, without exceptions, show that electrons follows a Fermi-Dirac distribution and not a Maxwell-Boltzmann distribution?

  6. Of course electrons in different locations in space are distinguishable particles, as in different atoms. Electrons within atoms partly loose identity and mix into some form of electron clouds.

  7. You need to motivate that position a lot more.

    If you you have two electrons in a system, say el1 and el2, at location x1 and x2. Call this (x1,el1) and (x2,el2).

    If we change the positions of the electrons, that is (x1,el2) and (x2,el1), what difference, in a physical sense, would this exchange make?

  8. This is the same as rotating new identical tires on your car. What difference in a physical sense would that exchange make according to your view?

  9. You're missing the point. Your tires are not, of course, identical, other then as an idealized construction! You can also decide to label one of them with a magic marker or whatever.

    But say that you have two electrons in your system, you decide to name them Frank and Kurt.

    You look away for an instance and then examine the system again. Which electron is Frank and which is Kurt. Can you tell a difference? Are they distinguishable? Is there a physical difference if Frank does Kurts job and vice verse?

  10. And if you don't mark the tires, can you then tell which is which?

  11. That's a silly question.

    You can mark tires, but you can't mark electrons. So how do you distinguish them if they are distinguishable?

  12. Why should I mark electrons?
    Why should I mark new identical tires?

  13. Have you used your theory on molecular Hydrogen and showed that the theory manages to bind in the most simplest case?

    If so, what bond length did you get?

  14. I guess that you haven't done this then.

    Are you planning on doing it?

  15. What have you done so far, when it comes to real calculations?

    How does it compare to real data?