# Inverse wavefunction incorporates lower Planck gravitational bound

1. Oct 7, 2004

### Loren Booda

The wavefunction for a hypothetical quantum box of size Planck length (L), when inverted through L, models the universe with this lower bound required by quantum gravitational constraints. The initial quantum box solutions are given by:

$$\phi_n=\sqrt(2/L)\\sin(n \pi x/L)$$

However, having inverted the scale (so L-->1/L), the now "inverse quantum box" generates permitted states, all above L, i. e. those which obey this "ultraviolet" limit to quantum gravity:

$$\phi_n^-^1=\sqrt(L/2)\\sin(L/n \pi x)=\sqrt(1/2P)\\sin(1/Pn \pi x)$$.

This new wavefunction represents an envelope that modifies other wavefunctions when considering gravitational effects. It applies the correspondence principle, using the reciprocal Planck limit in terms of probability P=1/L, to find a particle outside the Planck region. The correspondence principle is, in effect, a manifest transformation of such forbidden quantum gravitational states (those <L) inverted through P.

2. Oct 7, 2004

### ZapperZ

Staff Emeritus
Since WHEN is

$$\frac{1}{sin(A/B)} = sin(B/A)$$ ???!!!

Zz.

3. Oct 7, 2004

### Loren Booda

ZapperZ,

My notation $$\phi_n^-^1$$ should read $$\phi_n^-$$, indicating the inverse wavefunction, not the multiplicative reciprocal of $$\phi_n$$. Thanks for pointing this out. Sorry for the confusion. Please read on.

4. Oct 7, 2004

### ZapperZ

Staff Emeritus
That makes even less sense, because ALL of the rest of the variables in there WERE done as a reciprocal. Unless you're willing to show how you rederive the "inverse wavefunction", then simply doing what you did is meaningless.

You can't just do this kind of "inverse" operation to the wavefunction when you make the shift from L to 1/L. This is because the reciprocal space is the fourier transform of the real space. It isn't simply an "inverse" space. Refer to a solid state physics text on what a reciprocal space is.

Zz.

5. Oct 7, 2004

### Tom Mattson

Staff Emeritus
Isn't the inverse of sin(u), arcsin(u)?

6. Oct 7, 2004

Staff Emeritus
Arcsin is the functional inverse of sine, to be sure, just as cosecant is the algebraic inverse, but that has nothing to do with this problem which is a about the reciprocal space in optics, not reciprocal trig functions.

7. Oct 8, 2004

### Loren Booda

Sorry - it seems I had not fully thought out my idea. A universal "particle in a box" wavefunction that obeys the bounds of quantum gravitation might better (and more simply) be derived as

$$\phi_n(r)=\sum_{n=1}^{10^{61}} \ sin(nr/L)$$

Reciprocal lattices do not seem to address problems in radial coordinates, unless that coordinate system is considered isotropic, like the above radial relation, or like the one-dimensional solution in x which I originally attempted to deduce.

Last edited: Oct 8, 2004
8. Oct 8, 2004

### ZapperZ

Staff Emeritus
Unless that coordinate system is considered ISOTROPIC?

How does a "coordinate system" dictate the isotropy or anisotropy of anything? Does that mean that if I have an isotropic system, simply by changing coordinate system, I can make it anisotropic? I didn't realize something like the Noether Theorem hinges on what coordinate system I use!

I hate to say this, but this is not getting any better each time. If you look at the original post in this thread, you were not asking a question, or seeking clarification. You were making a statement out of nowhere. I criticize you for not doing your homework before you make such bold statements, especially when you made what I consider to be elementary errors that resulted in a rather silly conclusion. I'm just surprised this didn't get dumped automatically into the TD section.

Zz.

9. Oct 8, 2004

### Loren Booda

ZapperZ,

You're right, the forces at PF have been more than tolerant with this, my speculation. I will cease pursuing this thread unless prompted otherwise.

By the way, I meant a physical system with isotropic symmetry. Thanks again for your correction.