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Nonlinear second order differential equation for quantum mechanical system?

GROUPS:

Background: I am working on a model to describe materia as excited empty curved space time. The core idea is old. John Archibald Wheeler is one of the well known inventors of this idea.

https://en.wikipedia.org/wiki/John_Archibald_Wheeler

Here is another paper, unfortunately behind a pay wall:

Curved Empty Space-Time As the Building Material of the Physical World: An Assessment

http://www.sciencedirect.com/science/article/pii/S0049237X09706044

The following article can be accessed:

GRAVITATIONAL GEONS REVISITED

https://arxiv.org/pdf/gr-qc/9610074.pdf

I will work on an alternative method to describe the problem. I am using the path integral method provided from Richard Feynman. To describe the method in detail would be a very long journey, too long for this forum. But I can describe my problem as follows.

I have two differential equation systems. Each system contains of two equations. One equation provides a rule for the other equation, which is a wave equation. The first set delivers a Schödinger type equation and the second set delivers a Dirac type equation. The equation delivering the rule is the same for both sets of differential equation systems.

My problem is, that I cannot evaluate the solution of the equation, which provides the rule for the wave equations. I tried it with mathematica, but unfortunately I could not find a solution.

The path integral method is well known to derive e.g. the Schrödinger equation. The path integral method will start with a Lagrange function describing the system which leads finally to a set of differential equations in dependency of a Taylor expansion. The following document shows an example how to do this (look at equations 41, 42, 46 and 47):

http://www-f1.ijs.si/~ramsak/seminarji/susic.pdf

The Lagrange related to my system provides for the constant term of the expansion the following equation, which is the above mentioned equation providing the rule for both differential equation sets:

$\psi \left(r,t\right){\rm \; }={\rm \; }\frac{{\rm 1}}{A} {\rm \; }\left(\sqrt{\frac{{\rm 2}\pi i\hbar }{m} } \right)^{{\rm 3}} {\rm \; }\left(u\left(r\right)v\left(r\right)\right)^{-\frac{{\rm 3}}{{\rm 2}} } {\rm \; }\psi \left(r,t\right)$

The norm factor $A$ can be canceled with the root factor. The wave function $\psi \left(r,t\right)$ will also be canceled. For this, the product of the functions must be

$u\left(r\right)v\left(r\right)={\rm 1}$    (1)

The following definition specifies in detail the equation to be solved:

Definitions

The variable $r$ is defined as the radius in spherical coordinates.

Let

$g\left(r\right){\rm \; :}={\rm \; 1\; }-{\rm \; }py\left(r\right){\rm \; }+{\rm \; }q^{{\rm 2}} y\left(r\right)^{{\rm 2}} {\rm \; }={\rm \; }g\left(y\left(r\right)\right)$    (2)

with $p$ and $q$ elements of real numbers

Let $\mathop{\lim \; }\limits_{r{\rm \; }\to {\rm \; }\infty } y\left(r\right){\rm \; }={\rm \; 0}$

Define $u\left(r\right){\rm \; :}={\rm \; }\frac{{\rm 1}}{{\rm 4}r} \frac{\partial }{\partial r} \left(\frac{{\rm 1}}{r} \frac{\partial }{\partial r} \sqrt{-g\left(y\left(r\right)\right)} \right)$    (3)

Define $v\left(r\right){\rm \; :}={\rm \; }\left[\frac{\partial }{\partial r} \left(\frac{{\rm 1}}{y} \right)\right]^{{\rm 2}} {\rm \; }\left(-g\left(y\left(r\right)\right)\right)^{-\frac{{\rm 3}}{{\rm 2}} } $    (4)

inserting (2), (3) and (4) into (1) leads to the following differential equation which has to be solved:

$\left(\left(y'-ry''\right)\left(-p+\left(p^{{\rm 2}} +{\rm 4}q^{{\rm 2}} \right)y-{\rm 3}pq^{{\rm 2}} y^{{\rm 2}} +{\rm 2}q^{{\rm 4}} y^{{\rm 3}} \right)+r\left(y'\right)^{{\rm 2}} \left(p^{{\rm 2}} -{\rm 4}q^{{\rm 2}} \right)\right)\left(\frac{y'}{y^{{\rm 2}} } \right)^{{\rm 2}} \frac{{\rm 1}}{{\rm 16}r^{{\rm 3}} } \frac{{\rm 1}}{\left({\rm 1}-py+q^{{\rm 2}} y^{{\rm 2}} \right)^{{\rm 3}} } {\rm \; }={\rm \; 1}$

with abbreviation $y'=\frac{\partial y}{\partial r} $

POSTED BY: Guido Nierhauve
Answer
2 months ago

I've attached a notebook which I put into the Library Archive some years ago. It derives the Schrodinger equation for the simple harmonic oscillator using the path Integral approach using Mathematica. Your problem is much more complex, but this may be of some use.

Attachments:
POSTED BY: Frank Kampas
Answer
2 months ago

Dear Mr. Frank Kampas. Thank you very much for this informative mathematica example. The reason why this problem is a little bit more complex is the Lagrange function I use. The derivation of the Schrödinger equation with path integral method leads to an equation for the constant term in the Taylor expansion which defines only a norm factor. The system I am using needs to solve an equation in the constant term of the Taylor expansion, which is related to the curved space time.

POSTED BY: Guido Nierhauve
Answer
2 months ago

There was a typo in eq. (4). I had to add a "-" sign. The minus below the root is due to the requirement, that the line element related to the metric needs to be greater than zero.

POSTED BY: Guido Nierhauve
Answer
2 months ago

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POSTED BY: Moderation Team
Answer
2 months ago

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