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    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 (\gamma_3,\gamma^3,A_i, A^i) you can find an internal symmetry between two superpositioned states directly related to their Chirality \gamma^5. The relationship to the Chirality of a particle was simply \gamma_3 \mathbf{A}^i or
    \gamma^3 \mathbf{A}_i but they became the definition of the four common gamma matrices with the use of a unit psuedoscalar i.


    To start off.

    ...all we needed was two new matrices \gamma^3 and \gamma_3 involving their own operators (that are also matrices) A^i and A_i. 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 S_1 + S_2 are products of the eignestates of their subsystsmes, which can be inolved with an interaction term.

    The two states are

    U(S_0 \phi_{+}) = S_{+} \phi_{+}

    U(S_0 \phi_{-}) = S_{-} \phi_{-}

    The initial state could be said to be in a superpositioning;

    \alpha_{+} \phi_{+} + \alpha_{-} \phi_{-}

    Normally attempts of measurement yield measurable results of entanglement, but no specific way to ensure the eigenstates and their complex coefficients \alpha_{+}\phi_{+} and \alpha_{-}\phi_{-} 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 \gamma^3 and it's sign change \gamma_3- 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


    \mathbf{A}^i\alpha_{+} \phi_{+} + \mathbf{A}_i\alpha_{-} \phi_{-}


    If you hit this with the four common gamma matrices which give the Dirac basis using a unit pseudoscalar, i \gamma^0 \gamma^1 \gamma^2 \gamma^3 yields the appropriate spin designated from a pilot state. Chirality and the Dirac base show too much resemblance, the Dirac base is

    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}

    And there are only sign changes attributed to the matrices (\gamma^3, \gamma_3) through the use of the diagonal matrices

    \mathbf{A}^i = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}


    and


    \mathbf{A}_i = \begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}

    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:


    \gamma_3 \mathbf{A}^i = i\gamma^0 \gamma^1 \gamma^2 \gamma^3 = \gamma^{5}

    \gamma^3 \mathbf{A}_i = i\gamma_0 \gamma_1 \gamma_2 \gamma_3 = \gamma_{5}


    and the two corresponding anticommutator relationships are


    <\ \gamma_3 \mathbf{A}^i, \gamma^{5}\ > = (\gamma^3 \mathbf{A}^i) \gamma^{5} + \gamma^{5}(\gamma^3\mathbf{A}^i)

    <\ \gamma^3 \mathbf{A}_i, \gamma_{5}\ > = (\gamma_3 \mathbf{A}_i) \gamma_{5} + \gamma_{5}(\gamma_3\mathbf{A}_i)


    The matrices are also related to each other intrinsically it appears

    \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}


    And the product matrix of \gamma^3\gamma_3 turned out to be Hermitian even in using \mathbf{A}^i\mathbf{A}_i which tended towards a diagonally dominant matrix. For those involving the Chirality formula re-written for our new hidden variable matrices


    \psi_{R,L} = \frac{1 \pm \gamma^3\gamma_3}{2}\psi_{R,L}


    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

    \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

    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 \pm 1 because of (i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i})^2 = \mathbf{I}_4 This is the same as saying (\gamma^5)^2 = \mathbf{I}_4, The term i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \mathbf{A}^{i} should also anticommute with the four gamma matrices

    <\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})


    Take the superpositioning state now by hitting it with the gamma matrices

    \gamma_3\mathbf{A}^i\alpha_{+} \phi_{+} + \gamma^3\mathbf{A}_i\alpha_{-} \phi_{-}

    This is a superpositioning of a chirality using

    \gamma_3 \mathbf{A}^i = i\gamma^0 \gamma^1 \gamma^2 \gamma^3 = \gamma^{5}

    \gamma^3 \mathbf{A}_i = i\gamma_0 \gamma_1 \gamma_2 \gamma_3 = \gamma_{5}

    It shows that the chirality (spin) has been determined by both \gamma_3\mathbf{A}^i and \gamma^3\mathbf{A}_i

    Let's write

    \gamma_3A^i \cdot \alpha_{+} \phi_{+} + \gamma^3\mathbf{A}_i\alpha_{-} \phi_{-}

    out in full matrix form for the new entries.

    \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}\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}\alpha_{-} \phi_{-}


    Keep in mind, these two matrices are really (\gamma_3\mathbf{A}^i, \gamma^3 \mathbf{A}_i), When you multiply them through you end up with


    \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_{-}


    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 \gamma_3.

    There are some interesting symmetries involving how we use the operating matrices (A^i,A_i), and instead make A_i (for instance) act on \gamma_3 we get negative eigenvalues: For


    \gamma_3A^i \cdot \alpha_{+} \phi_{+} + \gamma^3\mathbf{A}_i\alpha_{-} \phi_{-}

    written out in full matrix form for swapped operating matrices yields.

    \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}\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}\alpha_{-} \phi_{-}


    when it is calculated through you get


    \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_{-}


    Consider the states being composed of an initial state vector \psi_0.

    The interaction U will transform as

    \psi = U \psi_0 = \alpha_{+} \phi_{+} + \alpha_{-} \phi_{-}

    \psi = \alpha_{+} \chi_{+} \phi_{+} + \alpha_{-} \chi_{-} \phi_{-}

    The experiment will detect the joint state \chi_{+} \phi_{+} with a probability of |\alpha_{+}|^2 and likewise one can make

    |\alpha_{+}|^2 + |\alpha_{-}|^2 = \mathbf{1}

    Without some pilot matrices, the \psi is said to be in a pure state of the joint system S_1 + S_2 insomuch it can be in either state \chi_{+}\phi_{+} or \chi_{-}\phi_{-}

    Solving the ambiguity of which state it really is in, may depend on subtle sign changes using the tools \gamma^3\gamma_3 and A^iA_i


    So to finish off, the spin is already determined, probably during initial entanglement. The mix of states -

    \gamma_3A^i \cdot \alpha_{+} \phi_{+} + \gamma^3\mathbf{A}_i\alpha_{-} \phi_{-}

    was just as example but doesn't say much. Say one such approach would be to find significance in hidden spin flips where one \alpha_{+} \phi_{+} 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 \alpha_{+} \phi_{+} to be in an up state and for \alpha_{-} \phi_{-} 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

    \gamma^3A^i \cdot \alpha_{+} \phi_{+} + \gamma_3\mathbf{A}_i\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_{-}  + \begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix} \cdot \alpha_{-} \phi_{-}


    One can interpret that \alpha_{+}\phi_{-} has for a particle, a spin up and the matrix \begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix} would make the state \alpha_{-} \phi_{-} 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 \gamma^3 we would introduce another matrix

    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}



    \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}


    To find \gamma_3 you simply calculate




    \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}
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    This is a study help forum.
    What exactly is your question?

    It is not a place to post abstracts or papers.

    Maybe you could remove this yourself and post it in a more appropriate forum. (Physics Forums, maybe.)
 
 
 
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