Looking at the Spacetime Uncertainty Relation as an Approach to Unify Gravity

[Dubbelosix]

Using the geometry featured in a reference we have has, the geodesic length is

[math]|\psi_0 - \psi_1| = \sqrt{2 - 2\cos \alpha} = 2 sin\frac{\alpha}{2}[/math]

The [math]\cos \alpha[/math] is calculated the following way:

[math]\cos \alpha = \frac{<\psi_0|\psi_1> + <\psi_1|\psi_0>}{2}[/math]

And from this, it states in literature that the inequality holds

[math]\cos \alpha \leq |<\psi_0|\psi_1>|[/math]

And so the length of a curve on the unit sphere connecting 

[math]\arccos |<\psi_0|\psi_1>|[/math]

With it, you can actually set up an inequality of the form

[math](t_1 - t_0)\sqrt{<\psi|H^2|\psi> - <\psi|H|\psi>^2} \geq \hbar \arccos |<\psi|\psi>|[/math]

Notice, we have encountered this before and so is additional physics to a previous look at the object - we looked at it in two forms, we could write it in terms of the curvature tensor with some additional constants and/or we could write it as the difference of the Hamiltonian energies. They are known as binding energies when you take the difference like this, just as we saw in the Penrose model and in this model (early on). 

The statistical equivalent of taking the difference of these Hamiltonians was

[math]\sqrt{|\psi|^2\ln[|\psi|^2] - |\hat{\psi}(i)|^2\ln[|\hat{\psi}(i)|^2]}[/math]

and so for a two particle system we would have

[math]\sqrt{|\psi|^2\ln[|\psi|^2] - |\hat{\psi}(i)|^2\ln[|\hat{\psi}(i)|^2]} - \sqrt{|\psi|^2\ln[|\psi|^2] - |\hat{\psi}(j)|^2\ln[|\hat{\psi}(j)|^2]}[/math]

Since for two particles [math](i,j)[/math].Since we already established this relationship with geometry, it must also stand true for the curvature tensor that the inequality also holds ~ with [math]c = 8 \pi G = 1[/math]

[math](t_1 - t_0)\sqrt{<\psi|R_{ij}^2|\psi> - <\psi|R_{ij}|\psi>^2} \geq \hbar \arccos |<\psi|\psi>|[/math]

 

http://www.physik.uni-leipzig.de/~uhlmann/PDF/UC07.pdf

[koti]

That's it. I'm taking a math course.

 

[Mordred]

10 hours ago, Dubbelosix said:

If we identify 

12<ψ˙|M|ψ˙> (1)

with wave functions

|ψ>=eiHt|q>

<ψ|=<q|e−iHt

In these last two equations, though the wave function will give rise to a probability of the wave function with dimensions of either 1/L . 1/L2 or cubed density (volume) 1/L3 we will not renormalize it in this approach, we'll leave that to the reader to assume there is some differential attached in there to remove the density. ie. <ψ|ψ> dV for a wave function in three dimensions. In the two equations, we will also assume for the largest part there is encoded a generalized position q . You'll see why when we reach equations 5 and 6.

and identify it as ''being akin'' to the following equation

kBT=12mc2=12dqdt⋅M⋅dqdt (2)

where q is the generalised position and so q˙ is the generalized velocity and knowing that

kBT dt2=12mc2 dt2=12dq⋅M⋅dq (3)

When considering the Hertz principle of least curvature, the mass is often set equal to 1 without loss of generality in the theory - you may do such a thing if all the masses are equal, or if you wish to continue with a ''massless theory'' - whatever the motive, keep in mind that a massless theory would not have the mass tensor. This is why you will come to encounter some authors write it in the form:

ds2≡c2 dt2=12dq⋅M⋅dq (4)

According to the almost similar nature of the first expressions relationship with equation 2, we can theorize some relationships, one related to the kinetic energy and another related to the metric

kBT=12<ψ˙|M|ψ˙> (5)

featuring the mass tensor on the RHS but not the mass term on the LHS, it would also hold that

ds2=12<ψ|M|ψ> (6)

The mass tensor in this model, appears to be a metric tensor. I'll even propose the possible quantization:

ℏ=12t−1<ψ|M|ψ>  

The temperature was once again, related to the curve,

kBT=12m(dsdt⋅dsdt)  

and the curve in our Hilbert space was identified using the Wigner function, here as the square of the metric ds2 ,

<ψ˙|ψ˙> =|W(q,p)2| <ψ|R2ij|ψ> ≥1ℏ2 <ψ|H2|ψ>

and as noted, will satisfy a Schrodinger equation 

1iℏH|ψ>=|ψ˙>

remember, ds=cdt . From this under the Jacobi formulation there exists an action related to the temperature:

kBT=p⋅q˙2  

In which 12p⋅q˙ is the minimized action. Further under the Jacobi formulation, the action is related to the following relationships for the temperature 

kBT dt=∫p⋅dq

So the interesting thing now which is different to my last thoughts, I thought the mass tensor could be just a mass term acting on a matrix, but it is actually identified as a more complicated object which appears to be a [metric tensor]. I need more literature on this to understand it.

 

Been a bit busy lately, yes the stress tensor is far more complex than many realize.

It isnt the mass term though, Think of it as your fluid relations.

For example [latex]T^{00}[/latex] is a scalar field in a set region.( particles in a box).

Your next diagonal is the pressure, (particle rate crossing an axis or hitting the sides of the box) QM wavefunctions applicable,,, Equations of state ie radiation, matter, Lambda.

The other two diagonals involve flux and vorticity.

Yes the stress tensor requires a metric (the box.) However it is not a metric. That is the metric tensor. In order for your fluidic formulas to work requires a coordinate basis.

5 hours ago, Dubbelosix said:

Using the geometry featured in a reference we have has, the geodesic length is

|ψ0−ψ1|=2−2cosα−−−−−−−−−√=2sinα2

The cosα is calculated the following way:

cosα=<ψ0|ψ1>+<ψ1|ψ0>2

And from this, it states in literature that the inequality holds

cosα≤|<ψ0|ψ1>|

And so the length of a curve on the unit sphere connecting 

arccos|<ψ0|ψ1>|

With it, you can actually set up an inequality of the form

(t1−t0)<ψ|H2|ψ>−<ψ|H|ψ>2−−−−−−−−−−−−−−−−−−−−−−√≥ℏarccos|<ψ|ψ>|

Notice, we have encountered this before and so is additional physics to a previous look at the object - we looked at it in two forms, we could write it in terms of the curvature tensor with some additional constants and/or we could write it as the difference of the Hamiltonian energies. They are known as binding energies when you take the difference like this, just as we saw in the Penrose model and in this model (early on). 

The statistical equivalent of taking the difference of these Hamiltonians was

|ψ|2ln[|ψ|2]−|ψ^(i)|2ln[|ψ^(i)|2]−−−−−−−−−−−−−−−−−−−−−−−−√

and so for a two particle system we would have

|ψ|2ln[|ψ|2]−|ψ^(i)|2ln[|ψ^(i)|2]−−−−−−−−−−−−−−−−−−−−−−−−√−|ψ|2ln[|ψ|2]−|ψ^(j)|2ln[|ψ^(j)|2]−−−−−−−−−−−−−−−−−−−−−−−−√

Since for two particles (i,j) .Since we already established this relationship with geometry, it must also stand true for the curvature tensor that the inequality also holds ~ with c=8πG=1

(t1−t0)<ψ|R2ij|ψ>−<ψ|Rij|ψ>2−−−−−−−−−−−−−−−−−−−−−−−√≥ℏarccos|<ψ|ψ>|

 

http://www.physik.uni-leipzig.de/~uhlmann/PDF/UC07.pdf

I recommend you examine your explanation, with regards to the differences between conformal vs canonical treatments. Though what you described isn't incorrect. It lacks the mathematical precision in the descriptive of the above two treatments.

(Hamilton is canonical)

I will be rather short on time this week, so won't be as thorough in my responses as I normally am.

(That and very unreliable connections in the field lol)

2 hours ago, koti said:

That's it. I'm taking a math course.

 

For the notation used in this thread vector calculus which obviously includes differential geometry, though largely were also using linearizations via linear algebra relations as per Clifford and Lie algebra. (linear is symmetric) so when non linear you linear approximate. (The two methods mentioned above)

 Both the QM and QFT above require the above two.

edit more accurately the formulas of this thread involve Clifford and Lie as they establish the tensor relations of the Lie/Clifford group.( see previous summary on products)

Edited October 24, 2017 by Mordred

[Dubbelosix]

The inequalities recently, as I noted, where objects we encountered before, in the previous studies, they were related to ''survival probabilities.'' The inequalities a few posts back, turns out has a name, they are called Mandelstam-Tamm inequalities. They are it turns out (something more I have learned) related to decay models, just as the zeno effect depended on the survival probabilities. The physics is converging. 

http://iopscience.iop.org/article/10.1088/0305-4470/16/13/021/pdf

http://streaming.ictp.trieste.it/preprints/P/83/041.pdf

http://sci-hub.bz/10.1088/0305-4470/16/13/021

[Dubbelosix]

http://sci-hub.bz/10.1007/BF02819419

http://aapt.scitation.org/doi/10.1119/1.16940

The Mandelstam-Tamm inequality can be written in the following way, with [math]c = 8 \pi G = 1[/math] (as usual), we can construct the relationship: (changing notation only slightly)

 

[math]|<\psi(0)|\psi(t)>|^2 > \cos^2(\frac{[<R_{ij}> - <\psi|R_{ij}|\psi> ]\Delta t}{\hbar}) = \cos^2(\frac{[<H> - <\psi|H|\psi> ]\Delta t}{\hbar})[/math]

[math] = \cos^2(\frac{\Delta H \Delta t}{\hbar})[/math]

for

[math]0 < t < \frac{\pi \hbar}{2 \Delta H}[/math]

where [math]|\psi(0)>[/math] is the state at [math]t=0[/math]. 

So how long does it take for a particle probability to reach the half life? In the zeno effect models, when a particle reaches its half life, it is likely to decay and any measurements made on it after its half life could very well result in the anti-zeno effect.

In the decay probability model, we say the system has reached it's half life when the decay probability [math]|<\psi(0)|\psi(t)>|^2[/math] is [math]\frac{1}{2}[/math]. The inequality tells us that the half life is greater than 

[math]\frac{\hbar}{4 \Delta H}[/math]

Another implication is that for short times the non-decay probability will fall off more slowly than the parabola [math]1 - (\frac{\Delta H t}{\hbar})^2[/math].

 

Mandelstam and Tamm derived this type of  inequality from a better known relation of theirs

[math]\Delta H \Delta A \geq \frac{\hbar}{2}|\frac{d<A>}{dt}|[/math]

Here [math]A[/math] is taken to be the projector on the state [math]|\psi(0)>[/math] so that 

[math]A = |\psi(0)><\psi(0)|[/math]

It is possible to obtain the identity

[math]\Delta<A^2>\ =\ <A>(1 - <A>)[/math]

In which 

[math]<A>\ = |<\psi(0)|\psi(t)>|^2[/math]

Edited October 24, 2017 by Dubbelosix

[Dubbelosix]

''In quantum mechanics, the Fubini–Study metric is also known as the Bures metric.^[2] However, the Bures metric is typically defined in the notation of mixed states, whereas the exposition below is written in terms of a pure state. The real part of the metric is (four times) the Fisher information metric.^[2]''

 

^''In the context of quantum mechanics, CP¹ is called the Bloch sphere; the Fubini–Study metric is the natural metric for the geometrization of quantum mechanics. Much of the peculiar behaviour of quantum mechanics, including quantum entanglement and the Berry phase effect, can be attributed to the peculiarities of the Fubini–Study metric.''

 

Added  because it is educational and, it gives us a new terminology for the Bure's metric, something we looked at much earlier. 

 

https://en.wikipedia.org/wiki/Fubini–Study_metric

 

[Dubbelosix]

You can also write the inequality in terms of the Wigner function, 

[math]|<\psi(0)|\psi(t)>|^2 \geq  \cos^2(|W(q,p)| <R_{ij}> - <\psi|R_{ij}|\psi> )\Delta t \geq \cos^2(\frac{[<H> - <\psi|H|\psi> ]\Delta t}{\hbar})[/math]

(I missed the greater than or equal to sign before at the beginning when I featured the inequality, but if the Wigner function gives rise to an inequality below, then the above may be a true inequality.

remember that

[math]|W(q,p)| \geq \frac{1}{\pi \hbar}[/math]

[math]\dot{s} = |W(q,p)|\sqrt{<\psi|R_{ij}^2|\psi>}[/math]

[math]\dot{s} = \sqrt{<\dot{\psi}|\dot{\psi}>}[/math]

 

(I think I done this right?)

 

Edited October 24, 2017 by Dubbelosix

[Dubbelosix]

I wanted to check to see whether the Wigner function was defined correctly in the work. It shows, there are things possibly missing from the description ~ here is an excerpt from some material I am reading

 

''The Wigner function is defined as the transform of the density operator, multiplied by the normalization factor [math]\frac{1}{2 \pi \hbar}[/math].''

 

Early work I read, considered inequalities of the form [math]|W(q,p)| \geq \frac{1}{\pi \hbar}[/math] but the factor of 2 in the denominator should not be forgotten about. So yeah! I've been oversimplifying the physics accidently. The function is related to the Weyl transform. 

[math]\bar{A}(x,p) = \int\ e^{\frac{-ipy}{\hbar}}\ <\frac{x + y}{2}|A|\frac{x - y}{2}>\ dy[/math]

This is the Weyl transform [math]\bar{A}[/math] of some operator [math]A[/math] and has been expressed in the x-basis [math]<x|A|x>[/math]. The Weyl transform itself turns an operator into a function of x and p.

One property of the Weyl transfrom is that the trace of the product of two operators is given by the integral over the phase space of their Weyl transforms,

[math]Tr[AB] = \frac{1}{\hbar} \int \int \bar{A}(x,p)\bar{B}(x,p)\ dxdp[/math]

Of course, densities are given, in this case in the position basis

[math]<x|\rho|x>\ = \psi(x)\psi*(x)[/math]

The Wigner function can be generalized into mixed, in much the same way as the density operator. Even though there is more physics to consider from the Wigner function than I previously took into account, it looks like I have most of relationships correct concerning the probability distributions of the form

[math]\int W(q,p)\ dp = <q|\rho|q>[/math]

[math]\int W(q,p)\ dq = <p|\rho|p>[/math]

 

http://www.stat.physik.uni-potsdam.de/~pikovsky/teaching/stud_seminar/Wigner_function.pdf

 

In that link though, it does state an identity for the Wigner function and the Planck action as

[math]W(q,p)^2\ dq dp = \frac{1}{\pi \hbar}[/math]

In such a case, we can see that 

[math]W(q,p)^2\ dq dp =\ <q|\rho|q><p|\rho|p> = \frac{1}{\pi \hbar}[/math]

Though if this:

[math]W(q,p)^2\ dq dp =\ <q|\rho|q><p|\rho|p> = \frac{1}{\pi \hbar}[/math]

Is a true relationship, like the linked worked seems to suggest http://www.stat.physik.uni-potsdam.de/~pikovsky/teaching/stud_seminar/Wigner_function.pdf then I have applied it correctly, save notation differences. 

The Weyl transform concerning two probability distributions:

[math]Tr[\rho_A\rho_B] =\ <\psi_A|\psi_B>[/math]

is

[math]W_{AB}(q,p)^2\ dqdp = \frac{1}{\pi \hbar}<\psi_A|\psi_B>[/math]

This has strong correlations to previous identities we looked at recently as well. As you can see, the Wigner function is a probability measure itself.
 

Edited October 25, 2017 by Dubbelosix

[Dubbelosix]

So it seems like I have used the Wigner function correctly. I have been primarily using it on a Hamiltonian which can otherwise be described by the curvature tensor. Here is another example of how you use it - you calculate the lowest ground state energy in such a way -

 

[math]<H> =  \int \int W(q,p)(\frac{p^2}{2m} + \frac{m \omega^2 q^2}{2})\ dqdp = \frac{\hbar \omega}{2}[/math]

 

We have defined mean values for the function and inequalities which has given a much richer physics. Some of the inequalities are weaker than others.

Or more generally

[math]<H> =  \int \int W(q,p)H(q,p)\ dqdp [/math]

 

paper on curvature and Hilbert space. 

https://sci-hub.bz/https://doi.org/10.1142/S0217732393001148

Edited October 25, 2017 by Dubbelosix

[stephaneww]

16 hours ago, Dubbelosix said:

...

 

<H>=∫∫W(q,p)(p22m+mω2q22) dqdp=ℏω2

...

it's look like the  Zitterbewegung for [latex] \omega [/latex] in quantum mechanics relativistic no ? 

 

(I give the french link because the formula don't appear on English version of wikipedia)

 

Edited October 25, 2017 by stephaneww

[Dubbelosix]

Almost!

But you need some other formulae to argue it. I have taken the liberty of finding a relevant paper by an author I highly regard as a theorist. It's been some time since I read it, but if my memory is correct, is not actually a bad introductory into those ideas. 

http://fqxi.org/data/essay-contest-files/Hestenes_Electron_time_essa.pdf?phpMyAdmin=0c371ccdae9b5ff3071bae814fb4f9e9

[stephaneww]

I am not able to develop or argue. I just note the resemblance.

Thank you for the link.

Edited October 25, 2017 by stephaneww

[Dubbelosix]

No problem, of course there is resemblance. The zitter motion comes from the angular momentum interpretation of [math]\omega[/math] which has dimensions of an inverse time. It is encountered many times in physics. In the early theory, it was believed the electron could have been a photon traversing a zig zag motion through space, hence, why we called it zitter motion. This interpretation was Dirac's. But like many of Dirac's idea's... and even objections, they never caught on. This was because he was a revolutionary mind of his day - when I hear people of talking about ''we need the next Einstein,'' conditions where much different than what you will count today for many reasons:

 

1) That is, peer reviewers may not do the best job

2) More papers are created today and as a result, finding the nail in the haystack could be a true saying when comparing the find for the next Einstein

3) Radical idea's today are often met with negative criticism, often guised as ''honest criticism.'' This may be based on either a lack of understanding, education or even just hostility which let's admit it, is encountered in every stretch of life.

4) Maybe the idea is too bizarre: ie. the physics may pertain to remodelling the theory or relativity. These kind of claims should be taken with a pinch of salt. However, if there ever does exist a claim, which there very well could be, that extensions to relativity is required, then yes, we will need to consider the evidence. It is considered by some physicists, that dark matter may be an indication that relativity breaks down on global scales. 

5) Take into consideration also, that our physics is incomplete and no one can come to agree on everything. 

 

There are loads of examples of 5). A recent discussion of the wave function and more specifically the double slit experiment (involving the Afshar experiment) provided a friend of mine (with excellent education) enough reason to believe it contradicts the Copenhagen interpretation. It seems, that great names have also taken it to mean this, including Doctor John Cramer, but maybe for bias reasons, since he thinks it supports his own Transactional interpretation.

 

I cannot uphold or confirm with any integrity that this is why John has done this, but I will state it for the audience since obviously some people will be thinking this way, but I have no doubt personally of Cramers own integrity concerning the matter. Personally I am not educated enough yet to understand why the experiment may be flawed, only that I know of one example in which I have often taken as an '''incompleteness'' to quantum mechanics concerning the wave function. As I wrote on a forum:

 

''The strangest thing about the double slit experiment that I have never forgotten and I am still puzzled today, that even when you reduce the experiment to one particle at a time being shot through the slits, the interference pattern on the screen still emerges. The photons are totally unrelated, but somehow the particles know which region of the interference pattern is the most likely to land on. Note, those particles have no knowledge about where the other particles landed or where even future particles will land. Though somehow, each photon reaches the screen ''knowing'' which regions which are the most likely landing and most unlikely landing spots. To me quantum theory has to be incomplete, because our physics does not explain how these particles know where to land.''

 

The only conclusions from an early age, and even now, that I think could even explain the 'single interference anomaly'' is some notion of determinism within the theory. The pilot wave was considered shortly, but really, I only mentioned it as an example, since when later my physicist friend commented that it is experimentally out of date. Once again, a statement I cannot uphold myself, maybe due to my own education. 

 

Nevertheless, these are great examples of how broad of understanding, and how broad the difference of thinking can be. In my case, its just lack of reading enough material. But rest assured, I do not believe I am far from it. A few more years of home education, and I think I'll be able to understand more cases of the Lie algebra, not just the confined knowledge I have of it, but something more akin to Mordred, who has expressed the importance of it within the sets or subspaces. It's a subject I am still learning, along with set theory, which is why, you will rarely encounter me talking about them. 

Edited October 26, 2017 by Dubbelosix

[stephaneww]

Hard to find a place in consensual physics when new ideas emerge.

Full of articles were peer-reviewed and applauded for being "burned"

There is a good chance that relativity is complete. There is too much proof of its validity: especially the recent proof of gravitational waves. There may be something to do on the side of quantum mechanics to bring determinism to it. Even if the Copenhagen school has proven itself. I do not see how we can marry determinism with indeterminism ....
But here it is a beginning of off topic ....

Edited October 26, 2017 by stephaneww

[Dubbelosix]

Ah, but a good theory should be falsifiable, I am just questionable how certain people approach that, even on forums like this. The idea of creative thought should always be commended, but ignorant idea's based on old dated idea's need to be overthrown in the invent of new physics coming to our door, literally almost every year. 

The old standard model, for instance, has actually been remodelled, many times over, even when predicted physics has not officially entered the standard model. That is because it takes time to experimentally confirm what is being said, and that takes equally good experimental physicists, like A. Eddington once was, taking Einstein seriously to measure the bending of light around the earth/and effects of the moon. 

Since it was a full eclipse. 

[Dubbelosix]

I better make sure I speak in my first language - it has been commented I do not speak good English. Sorry if anyone has had this impression here, I have not had reason to think it. I thought up until tonight, I understood and other posters understood me in the process. 

[stephaneww]

My english is much worse than yours 

[Dubbelosix]

Well that's very kind but honestly, it won't be an issue. I can understand most posters I encounter, no matter how bad your English. 

[Dubbelosix]

Wrote up a lot more so if I find time tomorrow, will update

[Dubbelosix]

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 the requite 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 princple, the space time non-commutivity can still remain since it is known that two Hermitian operators may not do this. It also means geometry can in principle be described as an observable which I feel is important for the unification theories that involve ''measureable 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. For the non-commuting Hermitian operators (since space and time are treated as observables [math][x, ct] \ne 0[/math] - also keep in mind, the fourth dimension is manifested in observable three dimensions as the curvature path of a moving infinitesimal (test) particle), we showed how to calculate them. The mean square uncertainty of both the spatial and time connections are

[math]\Delta <\nabla_i>^2 =\ <\psi|(\nabla_i - <\nabla_i>)^2|\psi> =\ <\psi|A^2|\psi>[/math]

[math]\Delta <\nabla_j>^2 =\ <\psi|(\nabla_j - <\nabla_j>)^2|\psi> =\ <\psi|B^2|\psi>[/math]

The scalar product of [math]A|\psi> + i \lambda B|\psi>[/math] - as a modulus, the scalar product must be greater or equal to zero. Expanding you get

[math]<\psi|A^2|\psi> + \lambda^2<\psi|B^2|\psi> + i \lambda <A\psi|B\psi>\ \geq 0[/math]

After reorganizing the inequalitis in terms of uncertainties you can find the following identity:

[math]\Delta <\nabla_i>^2 \Delta <\nabla_j>^2\ \geq - \frac{1}{4}<\psi|[A,B]|\psi>[/math]

The operators [math]A[/math] and [math]B[/math] are given as

[math]A = (\nabla_i - <\nabla_i>)^2[/math]

[math]B = (\nabla_j - <\nabla_j>)^2[/math]

(This is all pretty standard) it does show an application of the physics we have encountered so far, in a neat way. 

Just for clarity, the connections follow the non-commutation:

[math][\nabla_i, \nabla_j] = (\partial_i + \Gamma_i)(\partial_j + \Gamma_j) - (\partial_j + \Gamma_j)(\partial_i + \Gamma_i) = (\partial_i \partial_j + \Gamma_i \partial_j + \partial_i \Gamma_j + \Gamma_i\Gamma_j) - (\partial_j \partial_i + \partial_j\Gamma_i + \Gamma_j \partial_i + \Gamma_j \Gamma_i)= -[\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j][/math]

Before this post is over, I want to draw our attentions back to what I called the Hilbert-Wigner Distribution Inequality:

[math]\sqrt{<\dot{\psi}|\dot{\psi}>} = |W(q,p)| \sqrt{<\psi|R_{ij}^2|\psi>} \geq \frac{1}{\hbar}\sqrt{<\psi|H^2|\psi>}[/math] 

The Schrodinger equation as I have already noted can satisfy this [math]H|\psi> = i \hbar |\dot{\psi}>[/math]. A new question arises, can we use a non-linear (takes into account gravity) Schrodinger equation? Yes, it seems we might be able to think this way: http://sci-hub.bz/10.1063/1.1301592

A simple way to form a non-linear wave equation is through using the Schrodinger-Newton equation:

[math]i \hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m}\nabla^2 \psi + V\psi + m\phi\psi[/math]

If applying a non-linear version of the Schrodinger equation, we might be able to talk about gravity in the Hilbert vector space.

http://sci-hub.bz/10.1063/1.1301592

http://sci-hub.bz/10.1142/S0219887814500662

https://arxiv.org/pdf/1011.0207.pdf

fixed typos

 

Edited October 29, 2017 by Dubbelosix

[Dubbelosix]

A while back I did find an author who wrote the curvature tensor in the phase space similar to mine, except they wrote it in a more appropriate form: I had the notes reference this https://arxiv.org/pdf/hep-th/0007181v2.pdf  but cannot find where it is, so thinking I have referenced the wrong paper.  They find for the equality term ~



And they identify an inequality



Which is a very useful construction to remember and I wish I could remember who the author is - maybe I can find it?

In the paper they identified [math]\theta^{ij}[/math] as an antisymmetric tensor. When you consider the tensor [math]R_{ij}[/math] which is antisymmetric in the two indices, then the spacetime relationship they gave in the paper is identical to mine. 

Edited October 29, 2017 by Dubbelosix

[Dubbelosix]

So, I had a look at what this non-linear Schrodinger equation may look like when considering the metric curve and the metric itself. The metric curve with the non-linear Scrodinger-Newton equation is (without any inequalities imposed),

 

[math]\dot{s} \equiv \sqrt{<\dot{\psi}|\dot{\psi}>} = \sqrt{<\psi|\frac{\hbar^2}{4m^2}\nabla^4|\psi>} + \frac{1}{\hbar} \sqrt{<\psi|V^2|\psi>} + \sqrt{<\psi|\frac{m^2\phi^2}{\hbar^2}|\psi>}[/math]

 

The metric of the theory is

 

[math]ds \equiv (t_1 - t_2)\sqrt{<\dot{\psi}|\dot{\psi}>} = (t_1 - t_2)\sqrt{<\psi|\frac{\hbar^2}{4m^2}\nabla^4|\psi>} + (t_1 - t_2)\frac{1}{\hbar} \sqrt{<\psi|V^2|\psi>} + (t_1 - t_2)\sqrt{<\psi|\frac{m^2\phi^2}{\hbar^2}|\psi>}[/math]

 

There is also the idea of the Langrangian encoded in all this

 

[math]\mathcal{L} = <\psi|\frac{\hbar^2}{2m}\nabla^2|\psi> - <\psi|V|\psi> + <\psi|m\phi|\psi>[/math]

 

where the potential energy is simply the sum of an ordinary potential and the mass of the system which in turn interacts with its own gravitational field [math]\phi[/math]

 

[math]PE = <\psi|V|\psi> + <\psi|m\phi|\psi>[/math]

Edited October 30, 2017 by Dubbelosix

[Dubbelosix]

I have noticed myself the term

 

[math]<\psi|\frac{m^2\phi^2}{\hbar^2}|\psi>[/math]

 

Is very similar to the term you find in the Klein-Gorden equation 

 

[math]\frac{m^2c^2}{\hbar^2}\psi[/math]

 

Remember the gravitational potential [math]\phi = -\frac{Gm}{R}[/math] has units of velocity squared - you can know this from the definition of the standard gravitational parameter

 

[math]c^2R = Gm \rightarrow \frac{Gm}{R}[/math]

 

This is why when calculating powers of the gravitational parameter you often encounter the form [math]\frac{\phi}{c^2}[/math] (and is almost the metric of GR save an additional two constants).

To make sure a Schrodinger equation can have appropriate boundary conditions so that the physical nature of the model can be studied, requires a model for this non-linear wave equation in terms of the Green function which is usually attached to the ordinary potential [math]\mathbf{G}_FV[/math]. It will create a propagator in the model. I will try and do this and write it up soon.

Edited October 30, 2017 by Dubbelosix

[Dubbelosix]

Also this equation

[math]\dot{s} \equiv \sqrt{<\dot{\psi}|\dot{\psi}>} = \sqrt{<\psi|\frac{\hbar^2}{4m^2}\nabla^4|\psi>} + \frac{1}{\hbar} \sqrt{<\psi|V^2|\psi>} + \sqrt{<\psi|\frac{m^2\phi^2}{\hbar^2}|\psi>}[/math]

might be confusing, it can be seen as, without any of the simplifications I made as

[math]\sqrt{<\dot{\psi}|\dot{\psi}>} = \sqrt{<\psi|\frac{1}{\hbar^2}\frac{\hbar^4}{4m^2}\nabla^4|\psi>} + \sqrt{<\psi|\frac{1}{\hbar^2}V^2|\psi>} + \sqrt{<\psi|\frac{1}{\hbar^2}m^2\phi^2|\psi>}[/math]

Edited October 30, 2017 by Dubbelosix

[Dubbelosix]

Been trying to understand my physics clearly before continuing making some Hamiltonian with an interaction which would result in a Lippmann Schwinger equation. Inside of it, I can create the propagator. I want the solution to satisfy not just outgoing waves but also those incoming wave solutions. Generally, such an equation is used in scattering theory - from such an equation though, I could describe the scattering, or S-matrix. It's certainly not impossible, I just redefine the Langrangian as a Hamiltonian and the Lippmann Schwinger equation becomes non-linear because of the extra interaction term of the mass with its own gravitational field. Mordred, does the Lippmann Schwinger equation, automatically have solutions for both outgoing and ingoing waves?

Edited October 31, 2017 by Dubbelosix