Solving Spooky Entanglement with a Luggage Matrix

[Lucious]

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

Edited January 11, 2015 by Lucious

[Lucious]

No thoughts?

[John Cuthber]

What's a luggage matrix?

[Phi for All]

What's a luggage matrix?

 

A rectangular mathematical array that helps you carry on?

[imatfaal]

What's a luggage matrix?

 

[andrewcellini]

What's a luggage matrix?

where else would you put socks in the equation?

[imatfaal]

and as any fule kno socks are fermions and so you cannot pack them in your luggage too closely together - thus a need for a luggage matrix

[Lucious]

What's a luggage matrix?

 

 

Well the new operating matrices [math]A_iA^i[/math] are ''like a new luggage'' we would have to carry on in the equation because when they act with the gamma matrices [math]\gamma^3\gamma_3[/math] they produce the chirality [math]\gamma^5\gamma_5[/math]. The Dirac Basis has non-trivial sign changes to describe up and down spin states, and just attaching this object which is a matrix [math]\gamma^3A_i[/math] onto just one of the mixed states would allow it to tell the system which state it would be in. It's almost like a mathematical Akashic record, some sort of information written into the system during initial entanglement.

Edited January 15, 2015 by Lucious

[imatfaal]

 

 

Well the new operating matrices [math]A_iA^i[/math] are ''like a new luggage'' we would have to carry on in the equation because when they act with the gamma matrices [math]\gamma^3\gamma_3[/math] they produce the chirality [math]\gamma^5\gamma_5[/math]. The Dirac Basis has non-trivial sign changes to describe up and down spin states, and just attaching this object which is a matrix [math]\gamma^3A_i[/math] onto just one of the mixed states would allow it to tell the system which state it would be in. It's almost like a mathematical Akashic record, some sort of information written into the system during initial entanglement.

 

Ah come on - this is getting needlessly theosophic.

 

Sorry Lucious. You will have to be more explicit in your working and declare your variables more simply. I get a few lines in and think - that does not look like a superposition to me. There are people who can help you here - but you need to be absolutely clear in what you are attempting and assume that your audience does not already have a working knowledge of your ideas.

 

The Dirac Basis has non-trivial sign changes to describe up and down spin states, and just attaching this object which is a matrix onto just one of the mixed states would allow it to tell the system which state it would be in.

 

The dirac basis

 

is the way of making gamma matrices - a most usual way. For instance

 

[latex] 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} [/latex]

is just wrong. In the dirac basis

[latex] 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} [/latex]

unless these are not gamma matrices or are not in the dirac basis - in which case you need to be more specific and stop having matrices and calling them gamma-zero to gamma-three.

[John Cuthber]

 

 

Well the new operating matrices [math]A_iA^i[/math] are ''like a new luggage'' we would have to carry on in the equation because when they act with the gamma matrices [math]\gamma^3\gamma_3[/math] they produce the chirality [math]\gamma^5\gamma_5[/math]. The Dirac Basis has non-trivial sign changes to describe up and down spin states, and just attaching this object which is a matrix [math]\gamma^3A_i[/math] onto just one of the mixed states would allow it to tell the system which state it would be in. It's almost like a mathematical Akashic record, some sort of information written into the system during initial entanglement.

perhaps I should rephrase the question.

What's a luggage matrix?

 

BTW,

http://en.wikipedia.org/wiki/Chirality_(disambiguation)

What do you mean by chirality?

[Lucious]

 

 

Ah come on - this is getting needlessly theosophic.

 

Sorry Lucious. You will have to be more explicit in your working and declare your variables more simply. I get a few lines in and think - that does not look like a superposition to me. There are people who can help you here - but you need to be absolutely clear in what you are attempting and assume that your audience does not already have a working knowledge of your ideas.

 

 

The dirac basis

 

is the way of making gamma matrices - a most usual way. For instance

 

[latex] 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} [/latex]

is just wrong. In the dirac basis

[latex] 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} [/latex]

unless these are not gamma matrices or are not in the dirac basis - in which case you need to be more specific and stop having matrices and calling them gamma-zero to gamma-three.

 

 

 

I apologize, this equation

 

[latex] 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} [/latex]

expressed the appearance of a new matrix which wasn't accounted for initially on the RSHS, it really has the appearance;

[latex] 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} [/latex]

 

Where

 

[latex]\mathbf{A}_i = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix} [/latex]

 

perhaps I should rephrase the question.

What's a luggage matrix?

 

BTW,

http://en.wikipedia.org/wiki/Chirality_(disambiguation)

What do you mean by chirality?

 

 

 

Chirality is a product of the four common gamma matrices including the pseudoscalar, it is given as [latex]\gamma^5[/latex] represents chirality and chiraity explains the ''handedness'' or spin of a system.

Another beautiful but still mysterious part of the equations is that the product of the operating matrices create a diagonally-dominent matrix with negative eigenvalues [latex]A_i \cdot A^i[/latex]

 

 

Ah come on - this is getting needlessly theosophic.

 

Sorry Lucious. You will have to be more explicit in your working and declare your variables more simply. I get a few lines in and think - that does not look like a superposition to me. There are people who can help you here - but you need to be absolutely clear in what you are attempting and assume that your audience does not already have a working knowledge of your ideas.

 

 

The dirac basis

 

is the way of making gamma matrices - a most usual way.

 

 

 

 

Indeed, but mine is represented as a dirac basis in the definition of the fifth gamma matrix, chirality in other words. There is a reference to it in wiki

 

http://en.wikipedia.org/wiki/Gamma_matrices#Dirac_basis

[latex]\gamma_3 \mathbf{A}^i = i\gamma^0 \gamma^1 \gamma^2 \gamma^3 = \gamma^{5}[/latex]

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

We must keep in mind, that even though the above tranformation into two Chiraities hold true using the new definitions

[math](\gamma_3 \mathbf{A}^i,\gamma^3 \mathbf{A}_i)[/math]

a useful known identity for chirality is [latex]\gamma^5 \gamma^5 = \mathbf{1}[/latex] - the only logical question is whether the notation we have used [math]\gamma^5\gamma_5[/math] holds the same as the product [math]\gamma_3 \mathbf{A}^i \cdot \gamma^3 \mathbf{A}_i[/math]

Let's work it out

[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]

So [math]\gamma^3A_i = [/math]

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

[math]\gamma_3 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}

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

Now to calculate the final matrix product [latex]\gamma_3 A^i \gamma^3A^i[/latex]

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

The only way to retain a symmetry of some kind then it seems that we can still retrieve a diagonally-dominant

matrix if the order or multipication (for some unknown reason) did not matter. In this case

[math]\gamma^3\gamma_3A^iA_i[/math]

[math]= \mathbf{I}_4 \cdot \begin{pmatrix} -1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}[/math]

The fact we can still get the diagonal negative entries by assuming the order of multiplication doesn't matter needs further justification.

when they compute they give a Diagonally Dominant Matrix, if all its diagonal elements are negative, then the real parts of its eigenvalues are negative. These results can be shown from the Gershgorin circle theorem. This further indicates that there are two solutions in which the possible eigenvalues [math]\pm[/math] can take this as a property of Chirality, which should be thought of as a ''handedness'' to particle systems.

note:

[math]Det[A_i] = Det[\begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}] = \mathbf{1}[/math]

[math]Det[A^i] = Det[\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}] = \mathbf{0}[/math]

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

Edited January 18, 2015 by Lucious

[Bignose]

note:

 

[math]Det[A_i] = Det[\begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}] = \mathbf{1}[/math]

 

[math]Det[A^i] = Det[\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}] = \mathbf{0}[/math]

Little worried about the middle stuff because there is an elementary matrix algebra mistake here...

 

[math]\det(-\mathbf{A}) = (-1)^n \det(\mathbf{A})[/math] where n is the dimension of the matrix, in this case n=4 so the determinant of the negative of a matrix is the determinant of that matrix. In your nomeclature [math]Det[A_i] = Det[A^i][/math]. Not sure where you use the stuff in the quote, but if you did, there is a mistake.

[imatfaal]

...So [math]\gamma^3A_i = [/math]

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

 

The above is incorrect

 

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

---

Merged post follows:

[/mp]

[math]\gamma_3 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}= \begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \\-1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix}[/math]

 

The above is also wrong

 

[math]\gamma_3 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}= \begin{pmatrix} 0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \\-1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \end{pmatrix}[/math]

 

[mp]

 

Now to calculate the final matrix product [latex]\gamma_3 A^i \gamma^3A^i[/latex]

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

 

 

 

This above calculation is also incorrect and mathmatically should read as

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

 

 

But as detailed above both the first terms in this have already been calculated incorrectly

[latex]\gamma_3 A^i \gamma^3A^i[/latex] should be calculated ACCORDING TO YOUR LOGIC as

 

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

[Lucious]

Thank you for the corrections. I wrote this up yesterday and it appears my signs got muddled up nastily.

 

 

Again, thank you to you both, you have written it correctly.