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

[Mordred]

Sorry been very busy with work, I might have time to study what you have added this weekend

[Dubbelosix]

Based on the Langrangian, the Hamiltonian of the theory with an interaction would look like and allow the Hamiltonian to have eigenstates [math]H_0[/math] which gives

[math](\mathcal{H}_0 + V + m\phi)|\psi>\ =\frac{\hbar^2}{2m} \nabla^2|\psi>[/math]

and allow 

[math]\frac{\hbar^2}{2m} \nabla^2 = E[/math]

allows us to write it more simply as

[math](\mathcal{H}_0 + V + m\phi)|\psi>\ = E |\psi>[/math]

As noted before we must try and construct a Lippmann-Schwinger-like equation from this - the general idea is that the continuity of the eigenvalues would make [math]|\psi> \rightarrow |\phi>[/math] as the potential [math]V[/math] goes to zerp. 

But our situation is a bit more complicated, we don't just have one potential. We have two interaction terms in this theory (making it non-linear essentially).

(Presumably) the solution we seek would take the form

[math]|\psi>\ = |\phi> + \frac{1}{E - H_0}(V + m\phi)|\psi>[/math]

And since I know not much about this equation, I've had to read literature. [math]E - H_0[/math] though is singular of course, and it has been suggested that making the denominator slightly complex removes it -

[math]|\psi>\ = |\phi> + \frac{1}{E - H_0 \pm i \epsilon}(V + m\phi)|\psi>[/math]

I have a feeling the denominator does not need to be complex but will need to look into it. Also note, I've simplified the notation a bit, a popular notation is to write the wave function with ingoing and outgoing wave solutions [math]|\psi^{\pm}>[/math]. This approach above brings us into some new physics, because the Lippmann Schwinger equation is a free particle solution where the interaction terms do not vary, whereas it can be assumed when you have a non-linear equation, parameters will almost certainly vary. 

And all that is an indication this needs to be considered carefully.

I know you haven't replied Mordred, but this was kind of the idea I had in mind above.

Edited November 3, 2017 by Dubbelosix

[Dubbelosix]

On Misunderstandings of the Role of Observables in QM

It was stated to me that

''No, position is not an operator at the relativistic level; and Nature is relativistic. Time not an operator at any level.''

You could say if you wanted that the relativistic version of QM is by definition a QFT, where position and momentum are no longer operators on a Hilbert space. It's not that quantum mechanics does not describe them as operators, it is just that one treatment of melding relativity with quantum mechanics resulted in a modern QFT which did not treat them as operators fundamentally. Though, some have argued in literature that it ''may be the only way'' to go forwards, but I don't think this is the case. Again, classical gravity may extend into the phase space. Classical gravity may have a definition in the Hilbert space (albeit) you have to modify your model. For instance, the Hilbert space is a vector space, it doesn't naturally induce curvilinear coordinates. To do that, you have to literally invent notions of curvature in the Hilbert space - is it so strange we come to do this? Absolutely not, the Hilbert space is after all an abstract space, but one that seems to work very well with quantum mechanics. Equally in an abstract space you can invent abstract tools. 

Gravity is a pseudo force in relativity, its not a real force by definition - so field theories that have quantized it may also be wrong from first principles. This may also be a reason not to expect any quantum corrections to gravity. I have chosen a model in which classical gravity extends even to the phase space of quantum particles. This idea that an understanding of gravity at the fundamental level remains classical is not entirely a new idea and in similar idea's, was studied earlier - as wiki puts it: ''Later on it was proposed as a model to explain the quantum wave function collapse by Diósi[4] and Penrose,[5][6][7] from whom the name "Schrödinger–Newton equation" originates. In this context, matter has quantum properties while gravity remains classical even at the fundamental level. ''

Certainly, one main foundation of the QFT approach is a direct quantization of gravity - I have been vocal about this as well. Relativity was clear about gravity - it is like the centrifugal force, it doesn't need a mediator to be explained. It can be understood perfectly in terms of geometry only. Continued failure to quantize gravity has made scientists skeptical of the QFT approach. Some have taken the divergence problems as evidence that gravity cannot, and by default the theory of relativity, be quantized. 

And of course, attempts have been made to find relativistic versions of our phase space

 

https://en.wikipedia.org/wiki/Newton–Wigner_localization

[Mordred]

In essence the above is accurate, gravity is best understood via spacetime curvature. No quantization is needed to accurately model gravity and indeed may or may not require a boson as a mediator.

 The more deep you get into the calculations of other fields involved in the same region. The more one starts to question whether or not gravity is a force as per a mediated field or if its simply the sum of all field interactions. 

 Now if gravity does require a mediator, it makes sense why we cannot yet find the graviton, it would need to be a spin 2 boson and the heaviest boson at that. However one must be careful as to how understands what particles truly are...

Field excitations is a fundamental key. So the best way I found to relate to this is to study wave formulas and constructive and destructive wave interference.

 This actually helps understand how virtual and real particle production occurs. Two overlapping waves that are in phase and of the same frequency will combine giving an increased ie increased amplitude. 

Which you already have the formula posted as to how this amplitude increase will affect particle number density. The same applies to destructive interference.

The pointlike characteristics of a particle is in essence the Compton wavelength, this is sets the boundary for the confinement to the pointlike characteristic. 

 Now with the above "Field excitations" it is easy to see that QFT promotes the field itself as the operators. This is where it changes from the QM treatment. (one of the articles I wrote on this site and linked to your thread previous mentions this)

Look at the two slit experiment itself, and look at it specifically in terms of constructive and destructive wave interference. This will demonstrate the above, and indeed one can see that although wave particle duality still exists, " All particles exhibit pointlike and wavelike characteristics". The pointlike characteristic does not mean that particles are corpuscular. Indeed there is no corpuscular aspect to any particle.

Edited November 7, 2017 by Mordred

[Dubbelosix]

Well written. Good points as well.

[koti]

+1 Mordred. And a +1 for me for actually comprehending what you posted. Since I read the Art Hobson article „There are no particles, there are only fields” I was thinking about how gravity might not be a force at all in a classical sense like you mention above but was scared to express my thoughts not to be ridiculed. Im so glad you wrote what you wrote above in plain English for me to digest.

 

[Mordred]

I would also say +1 to Dubbelesix who through diligence is arriving at a similar understanding.

Glad to see how well your picking up on this as well Koti

[Dubbelosix]

The Fubini-Study metric was,

 

[math]d(\psi,\psi) = \arccos\ (|<\psi|\psi>|)[/math]

 

If [math]|\psi>[/math] is separable, then it can be written as

 

[math]|\psi> = |\psi_A> \otimes\ |\psi_B>[/math]

 

Then the metric is

 

[math]ds^2 = ds^2_A + ds^2_B[/math]

 

Then a curve in the metric may be taken as

 

[math]\frac{ds}{dt} = \sqrt{ \dot{s}^2_A + \dot{s}^2_B}[/math]

 

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)

 

[math]\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>}[/math]

 

If you are interested in the correlations (entanglement) then here is some notes I wrote down:

In terms of probability density, the separable state also looks like

[math]\rho_{AB} = \rho_A \otimes \rho_B[/math]

Correlated entropy is

[math]S(\rho_A \otimes \rho_B) = S(\rho_A) + S(\rho_B)[/math]

so long as 

[math]S(\rho_A|\rho_B) < 0[/math]

and

[math]S(\rho_A \otimes \rho_B) = S(\rho_A) + S(\rho_B)[/math]

Note also that the upper bound of correlation is found as (from Carrols paper)

[math]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[/math]

[math]\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}[/math]

[math]= \frac{( <\mathcal{O}_A><\mathcal{O}_B> - <\mathcal{O}_A\mathcal{O}_B>)^2}{2|\mathcal{O}_A|^2|\mathcal{O}_B|^2}[/math]

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

https://www.cv-foundation.org/openaccess/content_cvpr_2015/papers/Feragen_Geodesic_Exponential_Kernels_2015_CVPR_paper.pdf

Edited November 8, 2017 by Dubbelosix

[Mordred]

Excellent your getting the correct tensor products involved now. I don't see any errors in the above well done.

I've always enjoyed Caroll's articles regardless of topic. His papers are always incredibly well written. I'm also glad to see your applying the boundary cutoffs.

I would still like to see you involve the Levi-Cevita with the i,j,k summation. This will give you the complete orthogonal group SO(1.3) and giving the full group with the curvature. (Under relativity the Levi-Cevita affine connection) is the one that corresponds to [latex] G_{\mu\nu}[/latex] where as the Kronecker affine connection i,j is the Minkowskii Euclid. [latex]\eta_{\mu\nu}[/latex]

Mainly as this will complete your understanding to arrive at [latex]SO(1.3)=SU(2)\otimes SU(2)/\mathbb{Z}[/latex] where the Z denotes parity, helicity and chirality following the right hand rule under lie groups.

Note this is also applicable to the LQG treatments

In essence the antisymmetric [latex][L^i,L^j]=i\hbar\epsilon^{i,j}_kL^k[/latex] this is where you should arrive if your doing everything correctly

lol how the latex displays the superscript and subscript of the last looks a bit off, the k subscript should be after the i,j superscript but that is trivial. Wonder why it changes the superscript subscript sequence?

Anyways the total antisymnetric Levi -Civita is [latex]\epsilon^{i,j}_k[/latex]

for some reason I can't keep subsript k at the end of that statement following after the superscript i,j

The Kronecker connection being [latex]\epsilon^{i,j}[/latex] though more commonly its written as [latex]\delta_{i,j}[/latex] you will see variations on which symbols are used but thats trivial

I used the presentation as per Rovelli under covariant LQC

Edited November 8, 2017 by Mordred

[Mordred]

See equation 1.7 for the above

https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.cpt.univ-mrs.fr/~rovelli/IntroductionLQG.pdf&ved=0ahUKEwjYz5nkm6_XAhVO-GMKHcgKCnMQFggdMAA&usg=AOvVaw1gqghLG0TDxskcgiMdoID9

[Dubbelosix]

I will certainly look into it, familiar with the Levi cevita from general relativity. Thanks for the link.

[Mordred]

Exactly so there are numerous QFT treatments to handle Levi-civita that will be vital to your modelling.  Ie Newton approximation, Schwartzchild metric etc...

As regards to the article I found it highly infomative.

Edited November 9, 2017 by Mordred

[Dubbelosix]

Yes I have so far as well, quite a bit of reading. 

 

[Dubbelosix]

I am reading back and I spoke about say a correlated system as [math]S(\rho_A \otimes \rho_B)[/math] and there are some things I need to be clear about, I didn't write that very well before. Compared to the object above, a separable state is one that can be written like

 

[math]|\psi> = \psi_A \otimes \psi_B[/math]

 

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

 

[math]|\psi_{AB}>[/math]

 

 

When there is no statistical mixing (pure state) then you write the system as outer product

[math]\rho_{AB} = |\psi><\psi|[/math]

 

Edited November 10, 2017 by Dubbelosix

[Dubbelosix]

The projector in 2 dimensions can be written in the form

[math]\mathbf{P} = \frac{(\mathbf{I} + n \cdot \sigma )}{2} = |\psi_n><\psi_n|[/math]

This does introduce the traceless Pauli spin matrices where [math]n \in \mathbf{R}^3[/math] which is known as the Bloch sphere. 

So [math]n[/math] has to be the real unit three vector [math](n_x,n_y,n_z)[/math]. We have been speculating in the Hilbert space which deals with the projective space so here is one avenue to incorporate spin into the model.

I'm familiar with some of this algebra so I can understand this. 

I'll get to work see what I can do with it.

This approach though seems to be good for pure states only?

Seems right. 

But I have already wrote down the relationships that ties the inner and outer products as you know, so will write it up later Mordred. 

[math]\mathbf{P} = \frac{(\mathbf{I} + n \cdot \sigma )}{2} = |\psi_n><\psi_n|[/math]

This does introduce the traceless Pauli spin matrices where [math]n \in \mathbf{R}^3[/math] which is known as the Bloch sphere. 

So [math]n[/math] has to be the real unit three vector [math](n_x,n_y,n_z)[/math]. We have been speculating in the Hilbert space which deals with the projective space so here is one avenue to incorporate spin into the model.

[math]\mathbf{P} = \frac{(\mathbf{I} + n \cdot \sigma )}{2} = |\psi_n><\psi_n|[/math]

 

https://physics.stackexchange.com/questions/70436/differences-between-pure-mixed-entangled-separable-superposed-states

 

[math]\sigma_3 = \begin{pmatrix} 1 & 0 \\0 & -1 \end{pmatrix}[/math]

(it has eignevalues of +1 or -1)

[math]\sigma_1 = \begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}[/math]

A property of these two matrices is that if you square them, you just get the identity back. The last one is

[math]\sigma_2 = \begin{pmatrix} 0 & -i \\i & 0 \end{pmatrix}[/math]

[math]| n \cdot \sigma>\ = |1>[/math]

[math]|n \cdot \sigma>\ = 1[/math]

[math]n \cdot n\sigma|n \cdot \sigma>\ = 1| n \cdot \sigma>\ = |1>[/math]

These are eigenstates of [math]|\sigma \cdot n>[/math] and I use the notation of 1 to denote identity. Same for another definition of the vector for 

[math]|m \cdot \sigma>\ = 1[/math]

and so a probability amplitude is

[math]|n \cdot \sigma>\ = 1[/math]

The probability is

[math]|<m \cdot \sigma |n \cdot \sigma>|^2[/math]

 and 

[math]n \cdot \sigma = \begin{pmatrix} n_3 & n_{-} \\ n_{+} & -n_3 \end{pmatrix}[/math]

 

Which explains one component of this projective space.

 

Edited November 10, 2017 by Dubbelosix

[Dubbelosix]

Sorry, that equation should be [math]n \cdot \sigma |n \cdot \sigma>[/math] just posted this and cannot edit.

Edited November 10, 2017 by Dubbelosix

[Dubbelosix]

[math]Tr(R_{ij} \rho) = Tr(R_{ij} \sum_i p_i|\psi_i><\psi_i|)[/math]
[math]= \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>[/math]

[math]\mathbf{P} = \frac{(\mathbf{I} + n \cdot \sigma )}{2} = |\psi_n><\psi_n|[/math]

Now, we have traceless Pauli spin matrices in second equation, this posits investigation. I know there are traces for the products of Paul matrices. 

Edited November 10, 2017 by Dubbelosix

[Dubbelosix]

The probability amplitude wrote down before is incomplete:

 

[math] < m \cdot \sigma|n\cdot \sigma> [/math] 

 

you square for probability obviously. 

[Dubbelosix]

[points]

 

Just a bit on traces, the trace of a commutator is zero. The product of a symmetric and an antisymmetric matrix has zero trace, A trace is a linear algebra example. 

 
Ricci tensor can be symmetric, but is antisymmetric in its last two indices. (seems true if we have metric compatiility)

 If the covariant derivative of the metric is non-zero it doesn't have to be true. Ricci tensor is not necessarily symmetric. It has been known that while an antisymmetric matrix has no trace, its square can have a trace - right? I am trying to find cases of this with the Pauli matrices, but I am guessing the relationship holds [math]n^2 \cdot \sigma^2 = \mathbf{I}[/math]. See I need to be extra careful about how the trace enters the theory and how it enters this projective space. 

 

[math]P = \frac{\mathbf{I} + n \cdot \sigma}{2} = |\psi><\psi|[/math]

Of course, we are not just dealing with the antisymmetric parts, we are dealing with the full Ricci curvature which tends to be a symmetric object. But finding enough reading material on this is proving hard. 

 

The Ricci tensor is antisymmeric in the last two indices such that it has the property

 

[math]R^{a}_{bcd} = -R^{a}_{bdc}[/math]

 

Common notation to denote the antisymmetry is use of brackets

 

[math]R^{a}_{b[cd]}[/math]

 

So from this, is a very basic question, is the Ricci tensor symmetric or skew? There are two existing answers but the far most common is that it is symmetric. 

 

(a question more directed towards Mordred)

The full Riemann equation (in usual standard form) with LHS  showing commutation in indices, we have

[math]R^{\sigma}_{\rho\ [i,j]} = \partial_i \Gamma^{\sigma}_{j\rho} - \partial_i \Gamma^{\sigma}_{j\rho} + \Gamma^{\sigma}_{ie} \Gamma^{e}_{j\rho} - \Gamma^{\sigma}_{je} \Gamma^{e}_{i\rho}[/math]

The curvature and torsion is given by

[math]T^{\sigma}_{[i,j]} = \Gamma^{\sigma}_{ij} - \Gamma^{\sigma}_{ji}[/math]

 

So it is interesting to note, while the curvature and torsion part explains it is terms of the antisymmetric indices, it is often set to zero in general relativity (which is actually the most boring case).

You can obtain the skew symmetric part from the Bianchi identities as well, which this link explains 

https://mathoverflow.net/questions/69374/geometrical-meaning-of-the-ricci-tensor-and-its-symmetry

 

It seems we are ok, I finally found something which kind of answers the question

 

https://math.stackexchange.com/questions/349817/proving-the-symmetry-of-the-ricci-tensor

 

The tensor is skew symmetric in the last two indices and the tensor is symmetric from the Bianchi identities (something I suspected to look into to answer this)

 

 

Reading further though, he has set the antisymmetric part zero. We want to avoid this. 

I wonder how those antisymmetric parts influence the first Bianchi identities if it is non-zero? Equally with an antisymmetric property which is non-zero could be beneficial to create quantum Bianchi identities. 

Edited November 13, 2017 by Dubbelosix

[Dubbelosix]

So when taking the trace relationships for the curvature it is important to realize how limited those relationships really are since those traces will not be satisfied with the Ricci tensor since it is not completely symmetric when the torsion + curvature is non-zero. 

I went to see if anyone has constructed a theory for quantum Bianchi identities based exsctly on the issue of a non-vanishing antisymmetric part (as I have linked this in the past) to the non-zero phase space of non-commutivity (quantum domain) and it seems at least one author I have found seems to liken the issues correctly as I have and speaks about the role of the commutator in regards to the symmetries. And behold, he uses the key word I searched for, ''quantum Bianchi identities.''

''I ?;r the classical domain the existence of any gauge group in a field theory gives rise to a set of identities which the field variables satisfy. A convenient method of obtaining these so-called Bianchi identities' for any gauge theory is to invoke the invariance of the action integral under the given gauge transformations of the theory. This procedure was used by Schrodinger3 to obtain the contracted Bianchi identities for the classical theory of general relativity. Even if the action integral of a gauge is invariant under a group of c-number gauge transformations, the above procedure does not in general hold in the quantum domain because of the need to maintain operator orderings. Essentially, the basic problem resides in the dual role the commutator plays in the quantum theory.''

 

http://sci-hub.bz/10.1103/PhysRevD.3.2325

[Mordred]

I really like that last link extremely informative gonna study it a bit myself.

This lemma will help

the difference of a square matrix and its conjugate transpose is skew Hermitian

[math]A=-A^\dagger[/math] being anti Hermitean

Edited November 13, 2017 by Mordred

[Dubbelosix]

Symmetry under interchange of the first pair of indices with second pair

[math]R_{abcd} = R_{cdab}[/math]

Antisymmetry

[math]R_{abcd} = -R_{bacd} = -R_{abdc} = R_{bacd}[/math]

Cylicity is the sym of the permtutations of the last three indices which vanish,

[math]R_{abcd} + R_{acdb} + R_{adbc} = 0[/math]

A contraction can give the Ricci tensor

[math]R_{cd}[/math]

It should be antisymmetric in its last two indices, but as a consequence of the Bianchi identities the Ricci tensor is symmetric for a Riemannian manifold. Thus it is the cyclicity which implies the curvature part vanishes for the totally antisymmetric part of the Riemann tensor. 

The antisymmetric part is actually just

[math]T^{e}_{cd} = \Gamma^e_{cd} - \Gamma^e_{dc} = 2\Gamma^e_{[cd]}[/math]

There does exist a gauge symmetry between curvature and torsion

''Gauge fixing torsion to 0, we end up with general relativity, whereas fixing curvature to 0, we end up with its teleparallel equivalent.''

https://physics.stackexchange.com/questions/103576/why-can-we-assume-torsion-is-zero-in-gr
 

Edited November 13, 2017 by Dubbelosix

[Mordred]

yep basically correct, we had a small cross post I added to my last reply for skew symmetric lemma

I like the reply on the Levi-Cevita given in the last link. (It is why I wanted to see you looking into the Levi-Cevita applications glad to see you are.

Edited November 13, 2017 by Mordred

[Dubbelosix]

What I am trying to figure out, is whether the classical domain has (this antisymmetric part zero) whereas it may be non-zero in the quantum domain, just as non-commutation gives non-zero results.  It's funny, because literature is not very clear sometimes. I have been confused from several different papers. 

This paper will be beneficial for me when I get back to the pure spin space, it has loads of useful identities in it

 

http://peeterjoot.com/archives/math/pauli_matrix.pdf

[Mordred]

yeah you have to be careful particularly if your jumping from canonical to conformal states etc that and not all papers get various aspects of group theory entirely correct depending on which lemmas they are following. 

 I myself have oft gotten confused when looking at different treatnents. Its often very frustrating but challenges are fun lol