Dirac electrons generate Dirac positrons ?

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http://en.wikipedia.org/wiki/Dirac_equation#Comparison_with_the_Pauli_theory

$\left( E - q \Phi \right) \Psi_+ - c \vec{\sigma}\circ\left(\hat{p}-q \vec{A}\right) \Psi_- = m c^2 \Psi_+$

$- \left( E - q \Phi \right) \Psi_- + c \vec{\sigma}\circ\left(\hat{p}-q \vec{A}\right) \Psi_+ = m c^2 \Psi_-$

If no anti-matter exists, i.e. $\Psi_- = 0$, then matter must exist in a "super-conducting state", to wit $\left( \vec{p} - q \vec{A} \right) = 0$.

So, oppositely, if matter does not propagate in such a "super-conducting BCS-similar state", then any residual $c\vec{\sigma}\circ\left(\hat{p}-q\vec{A}\right)\Psi_+ \ne 0$ will begin to generate anti-matter wave-functions $\Psi_- \ne 0$.

Can such solutions, seemingly representing matter-generated anti-matter wave-functions, be considered some sort of "friction", between the matter momentum $\vec{p}$, and the electro-magnetic momentum $q\vec{A}$ that the matter "should" have ? Could i make an analogy, to matter entrained in a flowing fluid, such that if the matter moves with some speed different from the bulk fluid flow, then "friction" and "vortices" and "eddy currents" and such are generated around it ? If so, then could some sort of "electro-magnetic drag", between moving matter, and the vector potential field in which said matter is immersed, conceivably create anti-matter ??

Note that the spin-axis $\left( \hat{z} \right)$ seems "special", since, if $D_{\mu} \equiv \hat{p}_{\mu} - q \vec{A}_{\mu}$

$\begin{bmatrix} c D_z & c \left( D_x - \imath D_y \right) \\ c \left( D_x + \imath D_y \right) & - c D_z \end{bmatrix} \Psi_+ = \begin{bmatrix} E + mc^2 - q \Phi & 0 \\ 0 & E + mc^2 - q \Phi \end{bmatrix} \Psi_-$

so that only the spin-axis derivative couples to the anti-matter (may the non-spin axes derivatives be non-zero?).

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$\left( E - q \Phi \right) \Psi_+ - c \vec{\sigma}\circ\left(\hat{p}-q \vec{A}\right) \Psi_- = m c^2 \Psi_+$

$- \left( E - q \Phi \right) \Psi_- + c \vec{\sigma}\circ\left(\hat{p}-q \vec{A}\right) \Psi_+ = m c^2 \Psi_-$

If, for each component, of the matter spinor, $\Psi_+ \rightarrow \Psi_+' e^{- \imath \frac{m c^2}{\hbar}t}$, then $\hat{E} \Psi_+ = e^{-\imath \frac{m c^2}{\hbar}t} \hat{E} \Psi_+' + m c^2 \Psi_+$. So, seemingly, the Dirac equation is (nearly) reducible to that of a ("pseudo-")massless particle.

Next, re-arranging terms, for the matter spinor, w/o any vector potential:

$\hat{E} \Psi_+ = c \vec{p} \circ \vec{\sigma} \Psi_- + m c^2 \Psi_+ + q \Phi \Psi_+$

comparison to the equivalent Schrodinger equation (including the oft-omitted rest-mass-energy term):

$\hat{E} \Psi_+ = \frac{1}{2m} \vec{p} \circ \vec{p} \Psi_+ + m c^2 \Psi_+ + q \Phi \Psi_+$

so seemingly suggests

$\vec{p} \Psi_+ = 2 m c \vec{\sigma} \Psi_-$

$\begin{bmatrix} \hat{p}_x \Psi_u \\ \hat{p}_x \Psi_d \end{bmatrix} = 2 m c \begin{bmatrix} -\imath \Psi_{-,d} \\ \imath \Psi_{-,u} \end{bmatrix}$

$\begin{bmatrix} \hat{p}_y \Psi_u \\ \hat{p}_y \Psi_d \end{bmatrix} = 2 m c \begin{bmatrix} \Psi_{-,d} \\ \Psi_{-,u} \end{bmatrix}$

$\begin{bmatrix} \hat{p}_z \Psi_u \\ \hat{p}_z \Psi_d \end{bmatrix} = 2 m c \begin{bmatrix} \Psi_{-,u} \\ - \Psi_{-,d} \end{bmatrix}$

linking the gradient of the matter wave-functions (in the Schrodinger equation) to the anti-matter wave-functions (from the Dirac equation). So, seemingly, anti-matter is implicitly present, in the Schrodinger equation (??). Order of magnitude:

$\hbar \vec{k} \approx 2 m c \Psi_-$

$\frac{k}{k_C} \approx \Psi_-$

Thus, the amount of anti-matter implicitly present, in the Schrodinger equation, scales as the ratio of the particle's propagation vector, to its Compton-k-vector (when the wavelength equals the Compton wavelength) -- where-ever the particle possesses momentum, there anti-particles are implicitly present, apparently (w/ some mixing between spin up vs. down components). Would that affect degenerate matter, within White Dwarves ?? Would the squished electron wave-functions begin generating copious quantities of anti-electrons ???

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