User:UniversalExplanation/appendix3

Appendix 3: The Equivalence of the Two Representations
Given that $$\vec\Psi(\vec x,t)$$ is the solution to the fundamental equation

$$\left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j}\beta_{ij}\delta(\vec x_i - \vec x_j)\right\}\vec{\Psi} = K\frac{\partial}{\partial t}\vec{\Psi} = iKm\vec{\Psi}$$

and the anticommutation relations given for the defined alpha and beta matrices, we know that $$\alpha_{kx}\alpha_{ix} = \delta_{ik} - \alpha_{ix}\alpha_{kx}$$. It follows that if we left multiply the fundamental equation by $$\alpha_{kx}$$ we will obtain

$$\alpha_{kx}\left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j}\beta_{ij}\delta(\vec x_i - \vec x_j)\right\}\vec{\Psi} = \frac{\partial}{\partial x_k}\vec\Psi - \left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j}\beta_{ij}\delta(\vec x_i - \vec x_j)\right\}\alpha_{kx}\vec{\Psi}$$ ,

which, when summed over k, becomes:

$$\sum_k\frac{\partial}{\partial x_k}\vec\Psi - \left\{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j}\beta_{ij}\delta(\vec x_i - \vec x_j)\right\}\sum_k\alpha_{kx}\vec{\Psi} = K\frac{\partial}{\partial t}\sum_k\alpha_{kx}\vec{\Psi}$$.

If we require the constraint $$\sum_i\vec\alpha_i\vec\Psi = \sum_{i \neq j}\beta_{ij}\vec\Psi = 0$$, every component of the vector sum must vanish exactly. This result implies $$\sum_i\frac{\partial}{\partial x_i}\vec\Psi = 0$$ (note that k is nothing but an index being summed over and is thus equivalent to the i given here).

Exactly the same arguments made with respect to the $$\tau$$ component of alpha suffice to show that the fundamental equation will also yield $$\sum_i\frac{\partial}{\partial \tau_i}\vec\Psi = 0$$ and, finally, application of the anticommutation relations together with a sum over beta matrices will yield $$\sum_{i \neq j}\delta(\vec x_i - \vec x_j)\vec\Psi = 0$$.

These results are not exactly the expressions deduced in the original presentation; however, as shown in appendix 1, the original proof of the differential constraints constitute exactly the standard proof of conservation of momentum in classical quantum mechanics. It follows directly from an analogy to classical quantum mechanics that the requirement $$\kappa_x = \kappa_\tau = 0$$ implies that the fundamental equation is valid only in the "center of mass system" of our model (points in the x,$$\tau$$ plane). Since mass has not been defined in this model, let alone center of mass, I will define the "center of mass system" to be that system where $$\kappa_x = \kappa_\tau = 0$$.

In actual fact, it is a trivial matter to convert a solution to $$\sum_i\frac{\partial}{\partial x_i}\vec\Psi = 0$$ into a solution of $$\sum_i\frac{\partial}{\partial x_i}\vec\Psi = i\kappa_x\vec\Psi$$. Simple substitution will confirm that, if $$\vec\Psi_0$$ is a solution to the first equation then

$$\vec\Psi_1 = e^{\sum_{j=1}^n ik_x x_j}\vec\Psi_0\;\;\text{ where }\;\;nk_x = \kappa_x$$

is a solution to the second. The transformation is clearly reversible. Exactly the same arguments may be made with regard to both the $$\tau$$ and time components of the fundamental equation. Since the exponential term commutes with the Dirac delta function $$\sum_{i \neq j}\delta(\vec x_i - \vec x_j)\vec\Psi = 0$$ still holds for either solution and we can conclude that any solution to the fundamental equation is, for all practical purposes, a solution to the to the original constraints deduced.

As an aside, this result should be viewed as analogous to Newton's "inertial frame". Just as Newtonian mechanics simplify to F=ma when one uses an "inertial frame", the constraints on a self consistent explanation simplify to the given fundamental equation when one works in the "center of mass system". It should be seen as nothing more than a mathematical convenience.

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