# My paper to the gravitational research foundation 2018 (in a nut shell)

Discussion in 'Physics & Math' started by curvature, Aug 11, 2018.

1. ### curvatureRegistered Member

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Decay Rate for Quantum Superposition

From Flat Space To A Curved Hilbert Space: All For A Gravitational Schrodinger Equation

Yes, it is true the Hibert space is a vector space, but that doesn't mean us clever mathematicians cannot do some magic on the space so that we can get a definition or some clue behind The role of gravity in the phase space.

The Fubini-Study metric was,

$d(\psi,\psi) = \arccos\ (|<\psi|\psi>|)$

If $|\psi>$ is separable, then it can be written as

$|\psi>\ = |\psi_A> \otimes\ |\psi_B>$

Then the metric is

$ds^2 = ds^2_A + ds^2_B$

Then a curve in the metric may be taken as

$\frac{ds}{dt} = \sqrt{ \dot{s}^2_A + \dot{s}^2_B}$

Keep in mind, a curve in a Hilbert space may take on the following appearance, with an understanding of geometry - the following equation also makes use of the Wigner function which allowed me to write it as an inequality (as was shown previously). The main idea to construct the phase space is also thereby including uncertaintities which naturally come form the upper bounds of a spaceime equation written in terms of its geometry - spacetime geometry is known to exist in both string and LQG theories.

$R_{\mu \nu} = [\nabla_x\nabla_0 - \nabla_0 \nabla_x] \geq \frac{1}{\ell^2}$

And there may be some indication (see http://cgpg.gravity.psu.edu/people/Ashtekar/articles/qgfinal.pdf) that there are eigenvalues attached to the definition of the Planck length in the phase space, and it is, simpy:

$R_{\mu \nu} = [\nabla_x\nabla_0 - \nabla_0 \nabla_x]$

Which as you will see, has boiled down to three commutation components, which when you write them out, will become the Riemann tensor. In order to give the vacuum an intrinsic property of uncertainty, (something which some literature) has hinted at as requiring if vacuum fluctuations are to exist, we adopt a geometric interpretation of the uncertainty principle from the Cauchy Schwarz $L^2$ space in which the expectation of the uncertainty is found as the mean deviation of the curvature of the system:

$\sqrt{|<\nabla^2_i> <\nabla^2_j>|} \geq \frac{1}{2}i(<\psi|\nabla_i \nabla_j|\psi> + <\psi|\nabla_j,\nabla_i|\psi>) = \frac{1}{2}<\psi|\nabla_i, \nabla_j|\psi> = \frac{1}{2} <\psi|R_{ij}|\psi>$

It is also noted here, but will not be worked on, that such a spacetime uncertainty identification as the antisymmetric indices of a curvature tensor will have implications with the Bianchi identities, which will be further true up to three cyclic Christoffel symbols.

Is the Spacetime Uncertainty Consistent in this Theory?

The whole point of this, exists on whether in principle the physics holds up.

I want people to consider geometry as an observable (which it is under the treatment of general relativity) and for a full transition into quantum theory would require that geometry be described by Hermitian matrices.

There appears to be slight change in notation when considering the Hermitian Ricci Curvature and you can follow that in the first reference. You don't need to do anything fancy, we just impose there exists a Hermitian manifold - which is the complex definition of a Riemannian manifold and so you can also have the complex definition of the Ricci curvature. This means at least in principle, the space time non-commutivity can still remain since it is known that two Hermitian operators may not commute. It also means geometry can in principle be described as an observable which I feel is important for the unification theories that involve ''measurable lengths.''

Moving on, we now have a possible application of the spacetime uncertainty principle, in a new kind of form. We showed at the very start of these investigations, how you might interpret it as two connections of the gravitational field, one with spatial derivatives and another with time. It also exists that there can be non-commuting Hermitian operators (since space and time are treated as observables [x, ct] ≠ 0) - also keep in mind, the fourth dimension is manifested in observable three dimensions as the curvature path of a moving infinitesimal (test) particle): In gravitationally-warped spacetime the motion through time manifests as motion through space.

To move on, I started to investigate different ways to look at the theory - I came across a paper by I. Białynicki-Birula, J. Mycielski, on ''Uncertainty relations for information entropy in wave mechanics'' (Comm. Math. Phys. 1975) contains a derivation of an uncertainty principle based on an information entropy and features here as

$- \int |\psi|^2\ln[ |\psi(q)|^2] - \int |\hat{\psi}|^2 \ln[|\hat{\psi}(p)|^2]$

which can in fact formally satisfy the difference of the energy expectation of geometry equation we derived not long ago. Notice also, the first term is the entropy related to the state $(q)$ and the difference is taken from the entropy related to the state $(p)$. The investigation then led to the Bure metric, which is

$F(\sigma, \rho) = \min_{A^{*}_{i}A_i} \sum_i \sqrt{Tr(\sigma A^{*}_{i}A_i)} \sqrt{Tr(\rho A^{*}_{i}A_i)}$

As shown from Venn diagrams, it is possible to show a distinct difference between the classical and quantum entropy by noting that inequalities arise relating that entropies are weaker in the quantum case.

Last edited: Aug 11, 2018
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3. ### curvatureRegistered Member

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The inequality that already exists for the quantum theory is larger by a factor of two,

$S(A:B) \leq 2\min[S(A),S(B)]$

What is this strange notation? We will come back to this in a moment, because we have discovered something about our theory! I take you back to the equation/inequality we derived near the beginning of this work:

$\sqrt{|<\nabla^2_i> <\nabla^2_j>|} \geq \frac{1}{2}i(<\psi|\nabla_i \nabla_j|\psi> + <\psi|\nabla_j,\nabla_i|\psi>) = \frac{1}{2}<\psi|\nabla_i, \nabla_j|\psi> = \frac{1}{2} <\psi||R_{ij}|\psi>$

This inequality satisfies the quantum inequality bound

$<\psi|\nabla_i, \nabla_j|\psi>\ =\ <\psi|R_{ij}|\psi>\ \leq\ 2\sqrt{|<\nabla^2_i> <\nabla^2_j>|}$

This means the gravitational entropy of our system can reach twice the classical upper bound. The factor of two can further be interpreted as qubits. The same kind of situation arises in the Von Neumann entropy, in which the base log of 2 in information theory is almost always calculated using the base 2 log. Now, the strange notation located in the equation a few moments ago, can be understood say, in similar context to the Shannon entropy when written as an equation for information entropy in terms of quantum correlation,

$H(A) = H(A|B) + H(A:B)$

Here, $H(A|B)$ is the entropy of $A$ (after) having measured the system that become correlated in $B$. And, further, $H(A:B)$ is the information grained about $A$ by measuring $B$. As seems well known in literature, these two quantities complement each other so that $H(A)$ is unchanged overall to satisfy the second law.

In a similar context to the investigation into I. Białynicki-Birula, J. Mycielski paper on the ''Uncertainty relations for information entropy in wave mechanics,'' we will now investigate the relative entropy and the probability distributions between them (which can also be seen as the difference taken between two states) and so related to the difference of geometry equations:

$D(p|q) = \sum_lp_l \log_2(\frac{p_l}{q_i}) = \sum_l p_l(\log_2 p_l - \log_2 q_l)$

The way to view this equation, is as an information gain between two distributions (under observation), so is an expectation difference of two states, which is exactly what the difference in geometry equation is all about.

For the interesting single particle wave function case, you could argue that it is possible in the absence of external forces and other particle dynamics, a single wave function could be capable of collapsing under its own gravitational weight by assuming there is an analogue of the centre of mass for a wave function, which can be interpreted as fluctuating around the absolute square of its wave function. For a single particle, the entropy can be seen as

$-\sum^{n}_{i = 1} p_i \log q_i = \sum^{n}_{i = 1} p_i \log n = \log n$

which is known as the entropy of $q$. The inequality exists $h(p) \leq h(q)$ with equality $\iff$ $p$ is uniform (a uniform probability distribution). Obviously, in the gravitationally-induced collapse model, we must rely on a non-equilibrium in the system to cause a mechanical and deterministic collapse.

We also now, expand on the definition of the expectation value in information theory. When you treat geometry like an observable, like we have, the expectation value can be related to the density matrix in the following way:

$Tr(R_{ij} \rho) = Tr(R_{ij} \sum_i p_i|\psi_i><\psi_i|) = \sum_i p_i\ Tr(R_{ij}|\psi_i><\psi_i|) = \sum_i p_i\ Tr(<\psi|R_{ij}|\psi>) = \sum_i p_i<\psi_i|R_{ij}|\psi>$

This retrives the expectation value of the geometry of the systems for any collection of states $\rho$. This is a very simple solution and helps the continuation of the toy model for future development. When the systems are correlated we mean there exists a quantum entanglement - an entangled state is a linear superposition of separable states; the quantum divergence is

$S(A:B) \leq 2\min[S(A),S(B)]$

And we have shown that this still remains true for the expectation value for gravity. This has allowed quantum entanglement to enter this theory in a natural way, but since we are closing in on the investigation, it may be worthwhile saying a few things about it. Unlike the classical conditional entropy,

$S(a|b) = S(a,b) - S(b)$

which remains positive always, while the quantum mechanical equivalent form

$S(\rho_A|\rho_B) = S(\rho_{AB}) - S(\rho_B)$

is not. The state is entangled if $S(\rho_A|\rho_B) < 0$ where

$S(\rho_A \otimes \rho_B) = S(\rho_A) + S(\rho_B)$

Note also that

$S(\rho_A \otimes \rho_B|\rho_{AB}) = S(\rho_A) + S(\rho_B) - S(\rho_{AB}) \geq \frac{1}{2}| \rho_A \otimes \rho_B - \rho_{AB}|^2$

$\geq \frac{(Tr(\mathcal{O}_A \mathcal{O}_B)( \rho_A \otimes \rho_B - \rho_{AB} ))^2}{2|\mathcal{O}_A|^2|\mathcal{O}_B|^2}$

$= \frac{( <\mathcal{O}_A><\mathcal{O}_B> - <\mathcal{O}_A\mathcal{O}_B>)^2}{2|\mathcal{O}_A|^2|\mathcal{O}_B|^2}$

Which is known as the upper bound of correlation. In which the mutual information is

$I(A:B) = I(\rho_A) + I(\rho_B) - I(\rho_{AB})$

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

and

$\rho_{AB} = \sum_i \rho_A \otimes \rho_B$

Both these equations are always separable, so if

$S(\rho_A|\rho_B) = S(\rho_A)$

then it is a separable state no correlations. If instead we have

$0 < S(\rho_A|\rho_B) < S(\rho_A)$

then it is said to have ''classical correlations'' and if

$S(\rho_A|\rho_B) < 0$

means we have the quantum correlations. Going back now to the energy difference equation I derived from Anandan's hypothesis using the squared Christoffel symbol,

$\Delta E = \frac{c^4}{8 \pi G} \int (\Delta R_{ij})\ dV =\frac{c^4}{8 \pi G} \int <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi>\ dV$

While the wave function may not be time dependent in the Heisenberg picture, we can use a variational principle to vary the wave function - the total variation will split this part up

$<\psi|(R_{ij} - <\psi| R_{ij}|\psi>)|\psi>$

into

$<\delta_{ab} R_{ij}>\ =\ <\psi| R_{ij}| \delta_{ab}\psi> + <\delta_{ab} \psi |R_{ij}|\psi>$

for both geometries, where the subscript of $\delta_{ab}$ denotes a ''two particle system.''

$\frac{ds}{dt} \equiv \sqrt{<\dot{\psi}|\dot{\psi}>} = \int \int\ |W(q,p)^2|\ \sqrt{<\psi|R_{ij}^2|\psi>}\ dqdp\ \geq \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}$

When does entanglement happen?

A separable state is one that can be written like

$|\psi>\ = |\psi_A> \otimes\ |\psi_B>$

for any states $|\psi_A>$ and $|\psi_B>$ in which (and now here comes the important part) then the trace over the system will pick the terms in the system such that

$|\psi_{AB}>$

Then the system is entangled. Likewise, if you ever encounter notation like $|00> + |11>$ then the system is entangled.

Last edited: Aug 11, 2018
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5. ### curvatureRegistered Member

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Minimum Distance in Relativity and the Non-Commutation of the Phase Space

First... a blow by blow account of what led up to the proposal of the curve equation: Anandan proposes an equation

$E = \frac{k}{G} (\Delta \Gamma)^2$

and I offer also true

$E = \frac{c^4}{G} \int \Delta \Gamma^2\ dV = \frac{c^4}{G} \int \frac{1}{R^2} \frac{d\phi}{dR}(R^2 \frac{d\phi}{dR})\ dV$

That was after correcting the constant of proportionality, and after I derived a following inequality by making use of the Wigner function - The Mandelstam-Tamm inequality for instance I have shown can be written in the following way, with $c = 8 \pi G = 1$ (as usual), we can construct the relationship: (changing notation only slightly)

$|<\psi(0)|\psi(t)>|^2 \geq \cos^2(\frac{[<\Gamma^2> - <\psi|\Gamma^2|\psi> ]\Delta t}{\hbar}) = \cos^2(\frac{[<H> - <\psi|H|\psi> ]\Delta t}{\hbar})$

$= \cos^2(\frac{\Delta H \Delta t}{\hbar})$

This linking of geometry to the energy of the system can be understood through a curve equation I derived using the same principles

$\frac{ds}{dt} \equiv \sqrt{<\dot{\psi}|\dot{\psi}>}\ = |W(q,p)| \sqrt{<\psi|\Gamma^4|\psi>} \geq \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}$

By understanding that the curve equation when squared provides solutions to the time-dependent Schrodinger equation (in the following way) ~

$\frac{1}{ i \hbar}H|\psi>\ = |\dot{\psi}>$

Then you can construct a more serious equation that may be seen as a gravitational analogue to the Schrodinger equation when the covariant derivative acts on the tensor components and I calculate it as:

$\nabla_n|\dot{\psi}>\ = \frac{c^4}{8 \pi G} \int \int\ |W(q,p)^2|\ (\frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i})|\psi>\ dqdp \geq \int\ \frac{1}{\hbar}(\frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i})dV|\psi>$

Also keep in mind, when you take the bra solution $<\dot{\psi}| \nabla$ and it;s product with its conjugate above, you could understand the product as producing an object like $<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>$ which shows the non-commutation between covariant derivatives - the wave functions, in its most simplest form can be understood as

$|\psi>\ = e^{iHt}|q>$

$<\psi| =\ <q| e^{-iHt}$

When we take the product of the bra and ket solutions, we get an analogous identity found in General relativity - we simply take the tangent vector $\frac{dx^{\mu}}{d\tau}$ and allow the covariant derivative to act on this (and further set it to zero) and defines the minimum curve, or better yet, as a geodesic for a minimum distance,

$\nabla_n\frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>}$

We know how the covariant derivative acts on the curve from the following equation

$T_{nm}(y) = \nabla_n V_m = \frac{\partial V_m}{\partial y^{n}} + \Gamma_{nm} V_{r}(x)$

The interesting thing about setting this to zero and using this as the definition of the minimum distance is that non-commutation between the covariant derivatives in a phase space is generally not zero $[\nabla_i \nabla_j] \ne 0$! Except this time around there is no non-commutation because the order of two connections with derivatives in time commute $[\nabla_j,\nabla_j] = 0$.

The covariant derivative also acts on rank 2 tensors in the following way

The covariant derivative acting on our connection must obey a rank 2 tensor,

$\nabla_n\Gamma^{ij} = \frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i}$

It also follows then the stress energy tensor responds in much the same way

$\nabla_nT^{ij} = \frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i}$

which were the primary tools we used to construct our form of the non-linear Schrodinger equation.

Temperature in the Hilbert Space

A npn-linear Schrodinger equation was proposed:

$\nabla_n|\dot{\psi}>\ = \frac{c^4}{8 \pi G} \int \int\ |W(q,p)^2|\ (\frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i})|\psi>\ dqdp \geq \int\ \frac{1}{\hbar}(\frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i})dV|\psi>$

When this solution is taken as the product with its conjugate, you end up with an acceleration term (which relies) on the commutative properties of the covariant derivative on the RHS (by ensuring it satisfies the geodesic equation of GR) by taking the square root

$\nabla \frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>} = 0$

Notice then that squared solution actually consists of

$\frac{ds}{dt} \cdot [\nabla, \nabla] \cdot \frac{ds}{dt}$

We have the square of a curve in this expression and can be seen as related to the temperature (when dealing with a constant mass, we simply set it to 1), then the possible ''linking'' of temperature in the Hilbert space (and a concept of geometry) can be offered as:

$k_BT = \frac{1}{2}(\frac{ds}{dt} \cdot \frac{ds}{dt})$

https://en.wikipedia.org/wiki/Maupertuis'_principle

It was curious to me that my model predicted:for the geodesic equation:

$\nabla_n\frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>}$

The reason why it was curious because initially I made a mistake, I thought there was a non-commutative property going on the RHS... this is because covariant derivatives do not generally commute, but if I had taken relativity more seriously, I would have seen these objects must commute to satisfy the zero value of the geodesic equation in GR otherwise, it was permitting something strange. If it has been a true non-commutative phase space, then the geodesic equation naturally would have deviated from zero, but this didn't make sense.

In this formulation, the way to understand why the covariant derivatives commute is because they share the same derivative, so it doesn't matter what order the operators are in they commute classically $[\nabla_j,\nabla_j] = 0$ and so the following identity holds

$\nabla_n \dot{\gamma}(t) = \nabla_n\frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>} = 0$

$\nabla_n\frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>}$

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7. ### Q-reeusValued Senior Member

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Recalling your post here: http://www.sciforums.com/threads/ne...vity-challenges-gr.160900/page-2#post-3535472
Amusing that, a short while later, here there is a clear attempt to marry gravity/spacetime with QM.
Once again: These long serial posts have all the fingerprints suggesting this 'equation': curvature = SimonsCat = theorist. I think the mods should do some careful checking.

8. ### curvatureRegistered Member

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whose theorist?

9. ### curvatureRegistered Member

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Yes a unification in a phase space can happen regarding gravity: The concept of phase space and field quantization are actually two different things.

10. ### Q-reeusValued Senior Member

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Before my time here, but others have fingered SimonsCat as a sock of theorist. It then evidently follows.....

11. ### curvatureRegistered Member

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A phase space is different to a quantiation of a field - the way we think of a phase space, is when ordinary classical mechanics tends to breakdown and the wave function becomes pervasive. So a phase space is like smearing the classical into the quantum realm.

12. ### Q-reeusValued Senior Member

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Actually a phase space is simply a representation of all possible parameter values of a system and can be entirely classical in concept:
https://en.wikipedia.org/wiki/Phase_space
The QM application mentioned there has nothing to say about modifying an underlying spacetime background along QM lines.
The fact that you talk of spacetime uncertainty and reference to LQG articles that are for sure about spacetime quantization tells me there is a contradiction of statements.

13. ### curvatureRegistered Member

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Yes and it represents all those possible parameters through a wave function. It is called a ''smearing effect'' from classical to quantum Though, a wave function can act classically and deterministically thought of, the appearance of waves at the micro level is more than often a quantum effect. These quantum waves do exist aornd us, but its not a classical concept. It was things like wave functions that destroyed classical models of atoms to the ground, six feet under. Yet still, a type of quantization in phase is NOT to be taken to mean a field quantization.

14. ### curvatureRegistered Member

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No, the fact both string theory and LQG came to the same formula for scattering suggests pretty strongly this is not a coincidence.. in other words, it may in fact be a non-trivial equation..

15. ### Q-reeusValued Senior Member

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Let's digress a bit. Has your paper as per OP title been accepted for publication? Still pending a decision? If neither of those, are you prepared to post in full referee(s) detailed reasons for rejection?

16. ### NotEinsteinValued Senior Member

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curvature: Hello Reiku's sockpuppet. How are you today?

17. ### Q-reeusValued Senior Member

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My bad for not including Reiku in my 'equation' in #4. Which presumably should now read: curvature = SimonsCat = theorist = Reiku.

18. ### NotEinsteinValued Senior Member

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How long do you think it'll take before Reiku gets banned again, now that he's been outed?

19. ### Q-reeusValued Senior Member

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Taking into account advanced quantized (or not) spacetime uncertainty......hard to say. Hopefully not long.

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20. ### NotEinsteinValued Senior Member

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Is he Reiku, is he curvature, or both? It's a superposition! Luckily, with his posted math we can calculate his "Decay Rate for Quantum Superposition".

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21. ### exchemistValued Senior Member

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Yes this is obviously Reiku. Ban requested.

22. ### RainbowSingularityValued Senior Member

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it sounds very exciting !

Last edited: Aug 11, 2018
23. ### curvatureRegistered Member

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That's not how the process worked.

They evaluate papers quickly to see if they make sense and the probably have teams of scientists just to review it. Hundreds and hundreds participated and only a single scientist, out of several winners, won, because the rest of the winners where in groups, often with high calibre names. I thanked them for giving me more time as I had to edit the paper twice and I offered to forfit my paper, but they extended it. I told them I was happy for the experience and that I would come back next year. Nice people really.