Lucious

The biggest crux of the idea of randomness comes from the uncertainty principle. Not only have scientists recently shown it's actually quite predictable, the location and trajectory, but it's all about how much information WE can extract from the system. That's a big ''WE'' because it becomes a subjective argument. It's about OUR limitations, NOT realities.

But then its not random, there are local variables which can dictate the spin during initial entanglement. I've proven this.

I don't believe in random systems, they are like or even akin to unicorns; Many think they exist but this is without reason. Just as you must have a reason, even on quantum scales. The inability to extract information is different to saying ''it does exist,'' to ''we cannot measure it in its entirity,'' is not evidence that quantum mechanics is naturally random. Let's admit it! I would not be here if not for some causal pre-existence.

To show how time emerged when matter is considered, we begin with an equation which is in the strong gravity range using the gravitational fine structure and a timeless action: [math]\alpha_G \cdot \hbar = \int p_{\gamma} \cdot q[/math] [math]\int \hbar\ dr_s = \lambda \int [m_{\gamma} v \cdot q][/math] We know this because of Maurapitus' principle [math]\int mv\ ds = \int p \cdot q[/math] [math]ds = cdt[/math] You can find an energy [math]\int E = \int p \cdot \dot{q}[/math] Our generalized coordinate [math]\dot{q}[/math] has absorbed the time term. You find the clock to matter by distributing a frequency [math]\int \hbar\ dr_s = \lambda \int [mv \cdot q][/math] The mass term requires a coefficient: [math](\frac{c^2}{\hbar})[/math] to become a frequency term itself. This would imply a relativistic change in the wavelength [math]\lambda_2 - \lambda_1 = \Delta \lambda[/math] since frequency and wavelength are related [math]\nu = \frac{c}{\lambda}[/math] [math]\lambda = \frac{c}{\nu}[/math] distributing the coefficient we have (including a factor of 2 as true zitter term) [math]\int d r_s\hbar\ (\frac{c^2}{\hbar}) = \lambda \int [2m(\frac{c^2}{\hbar})v \cdot q][/math] This gives time to matter! The equation simplifies to [math]\int c^2 \ dr_s = \lambda \int [2m(\frac{c^2}{\hbar})v \cdot q][/math] [math]\int c^2 r_s = \mu[/math] In the Planck sceme and using strong gravity [math]G_s[/math] we have [math]\int c^2 \ dr_s = \lambda \int [2m(\frac{c^2}{\hbar})v \cdot q][/math] Multiply by [math]\frac{G}{c^2}[/math] gives [math]G_s dr_s = \int c^2 \ dr_s \frac{G_s}{c^2}= \lambda \int [2m_P(\frac{c^2}{\hbar})v \cdot q]\frac{\ell}{m_P}[/math] multiply through by [math]m^2[/math] and cancel out the length [math]\hbar c = \lambda \int [2m_P(\frac{c^2}{\hbar})v \cdot q]G_s m = \lambda \int [2m_P(\frac{c^2}{\hbar})v \cdot q]\mu[/math] Knowing the Weyl quantization relationship [math]\hbar c =Gm^2[/math] means we have the charge. For a strongly bound photon in the high gravity range would move in a circulatory motion, this is of course zitter motion. Indeed, a photon trapped in a minimized proper change in time given by the equation [math]\lambda(\gamma_L|_{\delta [t_1,t_2]}) = \Delta t_0[/math] Where [math]\gamma_L[/math] be a Lipschitz-continuous function [math]\lambda(\gamma_L|_{[t_1,t_2]}) = |t_2 - t_1|[/math] In the proper frame, the phase of the orbital and rotation of the internal photon is [math]\phi = \omega_P (\lambda(\gamma_L|_{\delta [t_1,t_2]}))[/math] It's proper time must be described by the length of the confined photon [math]\omega_P \ell_P(\gamma|_{\delta [t_1,t_2]}) = \omega \gamma(\lambda(\gamma_L|_{\delta [t_1,t_2]}) - \frac{v \cdot \ell_P}{c^2})[/math] The phase is then related to the structure of gravity and the wavelength of the photon via [math]\sqrt{\alpha_{G_s}} = \sqrt{\frac{Gm_{P}^{2}}{\hbar c}}[/math] [math]= (\omega_P t_P) = 2 \pi f \gamma(\lambda(\gamma_L|_{\delta [t_1,t_2]}) - \frac{v \ell_P}{c^2})[/math] And we have a connection to time as well within this fine structure in the beginning as well from our action equation we first presented. ref 1. Hestenes, The zitter clock ref 2. https://books.google.co.uk/books?id=XUKXYiGs8KcC&pg=PA140&lpg=PA140&dq=mass+squared+term+importance&source=bl&ots=jrvxv3oGya&sig=HYWB3eEUCgW-KR7iRcmcVKnfpck&hl=en&sa=X&ei=KZDKVOD_PMz3UqjzgYAK&ved=0CFMQ6AEwCA#v=onepage&q=mass%20squared%20term%20importance&f=false

In a space with fields [math](\psi,(q) \hat{\psi}(q))[/math] where [math]q[/math] is a generalized coordinate second quantization leads to a description of the creation and annihilation operators [math](a^{-},a^{+})[/math] [math]\psi(q) = \sum_k a^{+}(k) e^{-ikq}[/math] [math]\hat{\psi}(q) = \sum_k a^{-}(k) e^{ikq}[/math] This is the quantization of the length. Written in the discrete form, for [math]k = \frac{2\pi n}{\ell}[/math] and a periodic space interval of [math]\int^{\frac{1}{2}}_{-\frac{1}{2}} e^{ikq} dq = 2\pi \sum_{k} a^{+}(k) \delta_{nm}[/math] with a value of zero at the centre (simple topological space example). Hitting them with a momentum operator can yield a complexified version of the momentum space in increment of unit length [math]n[/math] - for the case of [math]\delta_{nm}[/math] the operators work on both [math]-i \hbar \frac{\partial}{\partial t} \cdot 2 \pi \sum_k <n|a^{+}|m> \delta_{nm}[/math] Where [math]\delta_{nm} = <\phi_n|\phi_m>[/math] and the increments of [math]n[/math] are acting on the operator [math]a^{+}[/math] which yields the raising operation [math]<n|m + 1> \sqrt{n+1}[/math] ... the same can be done for lowering. In the Schwinger quantization of charge context, [math]\hbar[/math] is measured in terms of increment [math]<n>[/math] for the square of the charge with using the discrete basis picture, we have the relation [math]<\phi_n|\phi_m>= \delta_{nm}[/math]. This discrete version of the delta function [math]\delta(x−y)[/math] is best seen as the identity matrix [math]\mathcal{I}[/math], where [math]\delta_{nm} =\mathcal{1}[/math] if [math]m=n[/math] , [math]\delta_{nm}[/math] goes to zero if and only if [math]m \ne n[/math]. If [math]n[/math] acts on the creation operator, we can rewrite it in forms where [math]m=0[/math]. In this way, we can our quantization and second quantizated discrete fundamental space and rewrite an equivalence to the raising operator to a smooth periodic function with discplacement always at zero when they converge. To write this we can give it in a new set of forms, first of all with [math]m = 0[/math] and [math]k[/math] taking on the forms [math]\frac{2 \pi n}{\ell}[/math] would give us the simpler form of this picture we have adopted one might notice that we are dealing with [math]i \hbar \frac{\partial}{\partial q} \cdot 2 \pi(n \cdot n)\sum_k a^{+}(k) \delta_{nm}[/math] An unit vector dot product with another unit vector gives the value of the identity matrix and they effectively vanish. [math]-i \hbar \frac{\partial}{\partial q} \cdot 2 \pi \sum_k a^{+}(k)|n> \delta_{nm}[/math] Because the the ket acts on the creation operation space [math]a^{+}(k)|n>[/math] this can also be written as [math]<n|m + 1> \sqrt{n+1}[/math] with [math]m[/math] not acting on [math]m[/math] not acting on the topological space in this case for the creation operator gives us [math]-i \hbar \frac{\partial}{\partial q} \cdot 2 \pi \sum_k a^{+}(k) \delta(q) = \alpha \hbar n[/math] Notice the righthandside is the same form of the Schwinger quantization expression for his quantization of the charge and then used magnetic charge - we know today there are no magnetic monopoles or at least if any exist, inflation made them aloof as they were diluted during the verocious inflational stages of the universe, thanks to the work of Alan Guth. The equation presented before is possible because [math]\sum_k a^{+}(k) \int^{\frac{1}{2}}_{-\frac{1}{2}} e^{-ikq} \delta(q)[/math] and [math]\sum_k a^{-}(k) \int^{\frac{1}{2}}_{-\frac{1}{2}} e^{ikq} \delta(q)[/math] which are equal to the simple expression [math]2 \pi \delta(q)[/math]. The space topological charge which appears to indicate in previous Weyl quantization equation that there is also a presence of a topological charge-space through a relationship to working with the Weyl quantization equivalence to the generation of mass (aka. gravitational charge) via [math]\hbar c = Gm^2[/math] as mention before from a Motz paper on quantization of the mass [math]\hbar = \frac{Gm^2}{c}[/math] where the numerator plays a synonymous role of the quantization method of Schwinger [math]\frac{e^{2}_{1} - e^{2}_{2}}{c} = \pm \alpha \hbar n[/math] where this time it seen in the more accurate light of taking positive or negative values on the quantization of the angular momentum component. When dealing with the wave functions dictating the operations on the first and second quantized space, it must be kept in mind that they strictly work in the Banach space.

First read http://www.as.utexas.edu/astronomy/education/spring05/komatsu/lecture15.pdf The Gaussian curvature for a 3-sphere system with a proper density is actually identified with the square of the reciprocal of the length [math]\ell[/math] [math]8 \pi \rho_0 (\frac{G}{c^2})[/math] A three-dimensional hypersphere is [math]\frac{K}{6} = (\frac{\hbar}{mc})^{-2}[/math] Where [math]K[/math] is the Gaussian curvature [Appendix] which can be re-written as [math]K = \rho_0 (\frac{\ell}{M})[/math] Our factor of [math]6[/math] magically disappears when you plug in classical values. I’ll quickly show this proof: the density of a three dimensional hypersphere should be [math]\rho_0 = \frac{M}{(\frac{4}{3} \pi R^3)}[/math] The general relativistic relationship to the principle curvatures is found as [math]8 \pi \rho_0 (\frac{G}{c})[/math] sidenote *the term [math]8\pi[/math] arises often because it simplified several of the field equations describing gravity The expression above was noted by Motz on his paper of quantization, and by a reference I will have below this post. Let us plug in and solve the equation. Be careful to remember your algebra when solving, when dividing a whole number by a fraction. I calculate this as [math]\frac{M}{6 R^3}[/math] A three dimensional hypersphere is normally written as when equated with the Compton wavelength. Equating my expression with this term we have [math]\frac{M}{6R^3} (\frac{G}{c^2}) = \frac{K}{6}[/math] Cancelling out the 6's, we end up with a general relativistic interpretation of how the geometry of a 3-sphere relates to the proper The The proper density and the Gaussian Surface Curvature are related as [math]K = \rho_0 (\frac{G}{c^2}) = \rho_0 (\frac{L_p}{M_p})[/math] Where the last equality comes about from a knowing that it has dimensions of length over mass. What is the Gaussian curvature of a metric? Refer back now to the cited reference at the beginning: [math]ds^2 = c^2dt^2 - [\frac{dR^2}{1 - KR^2} + R^2(d\alpha^2 + sin \alpha d\phi][/math] Where [math]K[/math] is the curvature of the universe. [math](\frac{\dot{a}}{a})^2 = \frac{8 \pi}{3} \rho \frac{Rc^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}[/math] where [math]\sum_{i=(1,2..n)}^{N} m_i[/math] is the many body system and now allow [math]H^2 = (\frac{\dot{a}}{a})^2[/math] (the Hubble Constant) and multiply through by our Schwarzschild constant [math]H^2(\frac{G}{c^2}) = \frac{8 \pi K}{3} \frac{Rc^2}{(\sum_{i=(1,2..n)}^{N} m_i)} - \frac{k}{a^2}(\frac{G}{c^2})[/math] because [math]K = k_1k_2 = \rho_0 (\frac{G}{c^2})[/math] as we proved before. We also showed that this equation was actually a length over the mass, in exchange we may think of this as the radius of the universe and entire ensemble of statistical averages so that the equation doesn't hold two identical features on either side of the equal sign, it's just more aesthetic [math]H^2(\frac{G}{c^2}) = \frac{8 \pi K}{3} \frac{Rc^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}(\frac{R}{(\sum_{i=(1,2..n)}^{N} m_i)})[/math] There are of course limits we can look at for variety, the equation above can satisfy a cosmological model. If [math]R \rightarrow 0[/math] then the proper density goes to [math]\rho \rightarrow \infty[/math] which implies a breakdown in the presence of a singular region within this non-space. In other words, it can be argued for a universe, the disappearance of space itself is unthinkable. Making time disappear is another matter and one completely acceptable. The thing we must keep in mind however is that the equation [math](\frac{\dot{a}}{a})^2 = \frac{8 \pi}{3} \rho \frac{Rc^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}[/math] satisfies the matter-dominated region of the universe. If we really where to make the radius go to zero, with neglecting the range of density to infinite size, we still have the problem of an environment application, it wasn't until 100,000 give or take (according to Susskind) that our universe first appeared to have matter. In later work when investigating Pensrose, he concluded for me a finding I too independently came to, that if matter fields [math]\chi[/math] where to vanish in the universe, time ceases to exist. Because matter is in fact emergent, especially under the Wheeler model of Geometrogenesis, matter isn't and must not be fundamental, meaning the equation I derived [math]H^2(\frac{G}{c^2}) = \frac{8 \pi K}{3} \frac{Rc^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}(\frac{R}{(\sum_{i=(1,2..n)}^{N} m_i)})[/math] Satisfies the low energy range and emergent geometry with matter... and most of all, time itself. I can prove in a quantum approach how matter is ''given'' time, by finding a relationship between the zitter motion of particle clocks and with local gravity. To show how time emerged when matter is considered, we begin with an equation which is in the strong gravity range using the gravitational fine structure and a timeless action: [math]\alpha_G \cdot \hbar = \int p_{\gamma} \cdot q[/math] [math]\int \hbar\ dr_s = \lambda \int [m_{\gamma} v \cdot q][/math] We know this because of Maurapitus' principle [math]\int mv\ ds = \int p \cdot q[/math] [math]ds = cdt[/math] You can find an energy [math]\int E = \int p \cdot \dot{q}[/math] Our generalized coordinate [math]\dot{q}[/math] has absorbed the time term. You find the clock to matter by distributing a frequency [math]\int \hbar\ dr_s = \lambda \int [mv \cdot q][/math] The mass term requires a coefficient: [math](\frac{c^2}{\hbar})[/math] to become a frequency term itself. This would imply a relativistic change in the wavelength [math]\lambda_2 - \lambda_1 = \Delta \lambda[/math] since frequency and wavelength are related [math]\nu = \frac{c}{\lambda}[/math] [math]\lambda = \frac{c}{\nu}[/math] distributing the coefficient we have [math]\int d r_s\hbar\ (\frac{c^2}{\hbar}) = \lambda \int [m(\frac{c^2}{\hbar})v \cdot q][/math] This gives time to matter! The equation simplifies to [math]\int c^2 \ dr_s = \lambda \int [m(\frac{c^2}{\hbar})v \cdot q][/math] [math]\int c^2 r_s = \mu[/math] where [math]\mu[/math] is the gravitational parameter [math]GM[/math]. To finish off, we return to my equation of cosmology in the low energy range [math]H^2(\frac{G}{c^2}) = \frac{8 \pi K}{3} \frac{Rc^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}(\frac{R}{(\sum_{i=(1,2..n)}^{N} m_i)})[/math] The term of mass over length is in fact a reciprocal of the linear density expression [math]\bar{\rho} = \frac{M}{\ell}[/math] Before I go, there are some interesting things we can think upon. A dominated matter universe would have a relation allowed [math]8 \pi G \rho = 3 H^2[/math] Notice we can make [math]\frac{8 \pi G \rho}{3} = H^2[/math] Is the same LHS expression for the equation [math](\frac{\dot{a}}{a})^2 = \frac{8 \pi}{3} \rho \frac{Rc^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}[/math] except for a factor of [math]G[/math].

That's one interpretation I suppose. Just to make clear, this isn't any atomic bond, but some kind of information bond.

One of those differences is that time isn't even an observable. From a quantum point of view, there is no non-trivial operator for time either. This is the thing, there is no true directionality in space, the so-called ''arrow of time,'' arises from a psychological deception of there being past states and future states. There can only ever be a present state. If you made a journey into space, there is no concept of up, down, left or right, so conceptually there is no arrow in space to tell us where everything came from - not even one telling us where we are going, time is really a non-linear subject, it's not part of real space which belong to a set of observable positions. Even though everything moves relative to everything else, no one can absolutely be sure when events happen, which means that our universe is a lot stranger and doesn't satisfy the Newtonian linear view of time.

All I was doing was understanding the latex.

Abstract To look at the problem of entanglement is often one thought of a mystery because it was illogical how two systems could simultaneously know which state to be in when a collapse (or measurement) is made on the system. Using four new mathematical tools [math](\gamma_3,\gamma^3,A_i, A^i)[/math] you can find an internal symmetry between two superpositioned states directly related to their Chirality [math]\gamma^5[/math]. The relationship to the Chirality of a particle was simply [math]\gamma_3 \mathbf{A}^i [/math] or [math]\gamma^3 \mathbf{A}_i[/math] but they became the definition of the four common gamma matrices with the use of a unit psuedoscalar [math]i[/math]. To start off. ...all we needed was two new matrices [math]\gamma^3[/math] and [math]\gamma_3[/math] involving their own operators (that are also matrices) [math]A^i[/math] and [math]A_i[/math]. The author cannot specify as yet what these matrices are, but they could be a number of things we might tackle near the end. The work on describing hidden variables using new matrices has been an idea I had for a while now, but melding the two together, one can find my matrices answering for long equidistant signalling, not by any superluminal effects, but rather the spin is already determined using these new matrix coefficients on the equations. Let's start off. The eigenstates of a joint system [math]S_1 + S_2[/math] are products of the eignestates of their subsystsmes, which can be inolved with an interaction term. The two states are [math]U(S_0 \phi_{+}) = S_{+} \phi_{+}[/math] [math]U(S_0 \phi_{-}) = S_{-} \phi_{-}[/math] The initial state could be said to be in a superpositioning; [math]\alpha_{+} \phi_{+} + \alpha_{-} \phi_{-}[/math] Normally attempts of measurement yield measurable results of entanglement, but no specific way to ensure the eigenstates and their complex coefficients [math]\alpha_{+}\phi_{+}[/math] and [math]\alpha_{-}\phi_{-}[/math] enable any blueprint as to how. To do this, we must ensure a new principle, ''determinism at initial entanglement.'' To do this, we need some new tools, notably with the use of the gamma matrix [math]\gamma^3[/math] and it's sign change [math]\gamma_3[/math]- the answer will also be in a hidden set of ''Pilot Matrices,' which governs the spin at the initial state no matter the distance between the entangled states. The superpositioning can now be written [math]\mathbf{A}^i\alpha_{+} \phi_{+} + \mathbf{A}_i\alpha_{-} \phi_{-}[/math] If you hit this with the four common gamma matrices which give the Dirac basis using a unit pseudoscalar, [math]i \gamma^0 \gamma^1 \gamma^2 \gamma^3[/math] yields the appropriate spin designated from a pilot state. Chirality and the Dirac base show too much resemblance, the Dirac base is [math]i \gamma^0 \gamma^1 \gamma^2 \gamma^3 = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix}[/math] And there are only sign changes attributed to the matrices [math](\gamma^3, \gamma_3)[/math] through the use of the diagonal matrices [math]\mathbf{A}^i = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}[/math] and [math]\mathbf{A}_i = \begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}[/math] I can quickly show how these hidden variables play in harmony with other mathematical structures. You can not only derive Chirality, the thing which is the set of spins, but you can also derive fundamental rules: [math]\gamma_3 \mathbf{A}^i = i\gamma^0 \gamma^1 \gamma^2 \gamma^3 = \gamma^{5}[/math] [math]\gamma^3 \mathbf{A}_i = i\gamma_0 \gamma_1 \gamma_2 \gamma_3 = \gamma_{5}[/math] and the two corresponding anticommutator relationships are [math]<\ \gamma_3 \mathbf{A}^i, \gamma^{5}\ > = (\gamma^3 \mathbf{A}^i) \gamma^{5} + \gamma^{5}(\gamma^3\mathbf{A}^i)[/math] [math]<\ \gamma^3 \mathbf{A}_i, \gamma_{5}\ > = (\gamma_3 \mathbf{A}_i) \gamma_{5} + \gamma_{5}(\gamma_3\mathbf{A}_i)[/math] The matrices are also related to each other intrinsically it appears [math]\gamma^{3} \gamma^1 \gamma^0 = \begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & 1\\1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix} = \gamma_{3}[/math] And the product matrix of [math]\gamma^3\gamma_3[/math] turned out to be Hermitian even in using [math]\mathbf{A}^i\mathbf{A}_i[/math] which tended towards a diagonally dominant matrix. For those involving the Chirality formula re-written for our new hidden variable matrices [math]\psi_{R,L} = \frac{1 \pm \gamma^3\gamma_3}{2}\psi_{R,L}[/math] This only holds true naturally because it's eigenvalues depend on whether the diagonal entries are negative, again to do this we must use our operating matrices to dictate it whether it has any [math]\gamma^3\gamma_3 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i} \cdot i \gamma_0 \gamma_1 \gamma_2 \gamma_3 \mathbf{A}_{i} = \mathbf{I}_4[/math] And it turns out to be Hermitian. When you separate the left handedness from the right handedness in the equations above you find it's Eigenvalues satisfying [math]\pm 1[/math] because of [math](i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i})^2 = \mathbf{I}_4[/math] This is the same as saying [math](\gamma^5)^2 = \mathbf{I}_4[/math], The term [math]i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i}[/math] should also anticommute with the four gamma matrices [math]<\i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i}, \gamma^{\mu}> = (i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i}) \gamma^{\mu} + \gamma^{\mu} (i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i})[/math] Take the superpositioning state now by hitting it with the gamma matrices [math]\gamma_3\mathbf{A}^i\alpha_{+} \phi_{+} + \gamma^3\mathbf{A}_i\alpha_{-} \phi_{-}[/math] This is a superpositioning of a chirality using [math]\gamma_3 \mathbf{A}^i = i\gamma^0 \gamma^1 \gamma^2 \gamma^3 = \gamma^{5}[/math] [math]\gamma^3 \mathbf{A}_i = i\gamma_0 \gamma_1 \gamma_2 \gamma_3 = \gamma_{5}[/math] It shows that the chirality (spin) has been determined by both [math]\gamma_3\mathbf{A}^i[/math] and [math]\gamma^3\mathbf{A}_i[/math] Let's write [math]\gamma_3A^i \cdot \alpha_{+} \phi_{+} + \gamma^3\mathbf{A}_i\alpha_{-} \phi_{-}[/math] out in full matrix form for the new entries. [math]\begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix} \cdot \alpha_{+} \phi_{+} + \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\cdot \alpha_{-} \phi_{-}[/math] Keep in mind, these two matrices are really [math](\gamma_3\mathbf{A}^i, \gamma^3 \mathbf{A}_i)[/math], When you multiply them through you end up with [math]\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \alpha_{+} \phi_{+} + \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \alpha_{-} \phi_{-}[/math] Notice that we end up on both sides of the superposition is the Dirac basis - the only way we could find this connection is through the use of a second new Dirac gamma matrix, namely [math]\gamma_3[/math]. There are some interesting symmetries involving how we use the operating matrices [math](A^i,A_i)[/math], and instead make [math]A_i[/math] (for instance) act on [math]\gamma_3[/math] we get negative eigenvalues: For [math]\gamma_3A^i \cdot \alpha_{+} \phi_{+} + \gamma^3\mathbf{A}_i\alpha_{-} \phi_{-}[/math] written out in full matrix form for swapped operating matrices yields. [math]\begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix} \cdot \alpha_{+} \phi_{+} + \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix} \cdot \alpha_{-} \phi_{-}[/math] when it is calculated through you get [math]\begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix} \cdot \alpha_{+} \phi_{+} + \begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix} \cdot \alpha_{-} \phi_{-}[/math] Consider the states being composed of an initial state vector [math]\psi_0[/math]. The interaction [math]U[/math] will transform as [math]\psi = U \psi_0 = \alpha_{+} \phi_{+} + \alpha_{-} \phi_{-}[/math] [math]\psi = \alpha_{+} \chi_{+} \phi_{+} + \alpha_{-} \chi_{-} \phi_{-}[/math] The experiment will detect the joint state [math]\chi_{+} \phi_{+}[/math] with a probability of [math]|\alpha_{+}|^2[/math] and likewise one can make [math]|\alpha_{+}|^2 + |\alpha_{-}|^2 = \mathbf{1}[/math] Without some pilot matrices, the [math]\psi[/math] is said to be in a pure state of the joint system [math]S_1 + S_2[/math] insomuch it can be in either state [math]\chi_{+}\phi_{+}[/math] or [math]\chi_{-}\phi_{-}[/math] Solving the ambiguity of which state it really is in, may depend on subtle sign changes using the tools [math]\gamma^3\gamma_3[/math] and [math]A^iA_i[/math] So to finish off, the spin is already determined, probably during initial entanglement. The mix of states - [math]\gamma_3A^i \cdot \alpha_{+} \phi_{+} + \gamma^3\mathbf{A}_i\alpha_{-} \phi_{-}[/math] was just as example but doesn't say much. Say one such approach would be to find significance in hidden spin flips where one [math]\alpha_{+} \phi_{+}[/math] is denoted by positive matrix entries on the Dirac Basis and the spin down corresponds for negative entries... let's try this. Suggesting we want [math]\alpha_{+} \phi_{+}[/math] to be in an up state and for [math]\alpha_{-} \phi_{-}[/math] to be in a down state, we need to write the gamma matrices and the operating matrices correctly, we need some new representation of the superpositioning, instead we can write [math]\gamma^3A^i \cdot \alpha_{+} \phi_{+} + \gamma_3\mathbf{A}_i\alpha_{-} \phi_{-}[/math] [math]\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \cdot \alpha_{+}\phi_{-} + \begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix} \cdot \alpha_{-} \phi_{-}[/math] One can interpret that [math]\alpha_{+}\phi_{-}[/math] has for a particle, a spin up and the matrix [math]\begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix}[/math] would make the state [math]\alpha_{-} \phi_{-}[/math] take on a spin down. But this is only a theoretical example, there might be a better mathematical way of representing this I am yet to find. Though the symmetries are striking like jigsaw pieces filling in an entangled puzzle. index RANDOM THOUGHTS AND ADDITIVES To obtain say the matrix [math]\gamma^3[/math] we would introduce another matrix [math]i \gamma^0 \gamma^1 \gamma^2 \gamma^3 = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}[/math] [math]\gamma^3 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i} = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1\\-1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix}[/math] To find [math]\gamma_3[/math] you simply calculate [math]\gamma_3 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}_i = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}[/math] LuciousBass@mail.com

NOTE: It is very possible while making this I have somwhere a figure with the wrong index, just work out which one it should be just in case. I've taken considerable time writing this so I have tried to make it as accurate and clear as possible. mmm... latex... hold on.