Editing Dirac spinor (section)

==Derivation from Dirac equation==
The Dirac equation has the form
<math display="block">\left(-i \vec{\alpha} \cdot \vec{\nabla} + \beta m \right) \psi = i \frac{\partial \psi}{\partial t} </math>

In order to derive an expression for the four-spinor {{mvar|ω}}, the matrices {{mvar|α}} and {{mvar|β}} must be given in concrete form. The precise form that they take is representation-dependent. For the entirety of this article, the Dirac representation is used. In this representation, the matrices are
<math display="block">
  \vec\alpha = \begin{bmatrix}
    \mathbf{0} & \vec{\sigma} \\
    \vec{\sigma} & \mathbf{0}
  \end{bmatrix} \quad \quad
  \beta = \begin{bmatrix}
    \mathbf{I} & \mathbf{0} \\
    \mathbf{0} & -\mathbf{I}
  \end{bmatrix}
</math>

These two 4×4 matrices are related to the [[gamma matrices|Dirac gamma matrices]]. Note that {{math|'''0'''}} and {{math|'''I'''}} are 2×2 matrices here.

The next step is to look for solutions of the form
<math display="block">\psi = \omega e^{-i p \cdot x} = \omega e^{ -i \left(E t - \vec{p} \cdot \vec{x}\right) },</math>
while at the same time splitting {{mvar|ω}} into two two-spinors:
<math display="block">\omega = \begin{bmatrix} \phi \\ \chi \end{bmatrix} \,.</math>

===Results===
Using all of the above information to plug into the Dirac equation results in
<math display="block">
  E \begin{bmatrix}
    \phi \\
    \chi
  \end{bmatrix} = 
  \begin{bmatrix}
    m \mathbf{I} & \vec{\sigma}\cdot\vec{p} \\
    \vec{\sigma}\cdot\vec{p} & -m \mathbf{I}
  \end{bmatrix}\begin{bmatrix}
    \phi \\
    \chi
  \end{bmatrix}.
</math>
This matrix equation is really two coupled equations:
<math display="block">\begin{align}
  \left(E - m \right) \phi &= \left(\vec{\sigma} \cdot \vec{p} \right) \chi \\
  \left(E + m \right) \chi &= \left(\vec{\sigma} \cdot \vec{p} \right) \phi
\end{align}</math>

Solve the 2nd equation for {{mvar|&chi;}} and one obtains 
<math display="block">\omega = \begin{bmatrix}
    \phi \\
    \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \phi
\end{bmatrix} .</math>

Note that this solution needs to have <math display="inline">E = +\sqrt{\vec p^2 + m^2}</math> in order for the solution to be valid in a frame where the particle has <math>\vec p = \vec 0</math>.

Derivation of the sign of the energy in this case. We consider the potentially problematic term <math display="inline">\frac{\vec\sigma\cdot \vec{p}}{E + m} \phi</math>.

* If <math display="inline">E = +\sqrt{p^2 + m^2}</math>, clearly <math display="inline">\frac{\vec\sigma\cdot\vec p}{E + m} \rightarrow 0</math> as <math>\vec p \rightarrow \vec 0</math>.
* On the other hand, let <math display="inline">E = -\sqrt{p^2 + m^2}</math>, <math>\vec p = p\hat{n}</math> with <math>\hat n</math> a unit vector, and let <math>p \rightarrow 0</math>.

<math display="block">\begin{align}
     E = -m\sqrt{1 + \frac{p^2}{m^2}} &\rightarrow -m\left(1 + \frac{1}{2}\frac{p^2}{m^2}\right) \\
  \frac{\vec\sigma\cdot\vec p}{E + m} &\rightarrow p\frac{\vec\sigma\cdot\hat n}{-m - \frac{p^2}{2m} + m} \propto \frac{1}{p} \rightarrow \infty
\end{align}</math>

Hence the negative solution clearly has to be omitted, and <math display="inline">E = +\sqrt{p^2 + m^2}</math>. End derivation.

Assembling these pieces, the full '''positive energy solution''' is conventionally written as
<math display="block">\psi^{(+)} = u^{(\phi)}(\vec p)e^{-i p \cdot x} = \textstyle \sqrt{\frac{E + m}{2m}} \begin{bmatrix}
    \phi \\
    \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \phi
\end{bmatrix} e^{-i p \cdot x}</math>
The above introduces a normalization factor <math display="inline"> \sqrt{\frac{E+m}{2m}},</math> derived in the next section.

Solving instead the 1st equation for <math>\phi </math> a different set of solutions are found:
<math display="block">\omega = \begin{bmatrix} -\frac{\vec{\sigma} \cdot \vec{p}}{-E + m} \chi \\ \chi \end{bmatrix} \,.</math>

In this case, one needs to enforce that <math display="inline">E = -\sqrt{\vec p^2 + m^2}</math> for this solution to be valid in a frame where the particle has <math>\vec p = \vec 0</math>. The proof follows analogously to the previous case. This is the so-called '''negative energy solution'''. It can sometimes become confusing to carry around an explicitly negative energy, and so it is conventional to flip the sign on both the energy and the momentum, and to write this as
<math display="block">\psi^{(-)} = v^{(\chi)}(\vec p) e^{i p \cdot x} = \textstyle\sqrt{\frac{E + m}{2m}} \begin{bmatrix}
    \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \chi \\ \chi 
\end{bmatrix} e^{i p \cdot x}</math>

In further development, the <math>\psi^{(+)}</math>-type solutions are referred to as the [[particle]] solutions, describing a positive-mass spin-1/2 particle carrying positive energy, and the <math>\psi^{(-)}</math>-type solutions are referred to as the [[antiparticle]] solutions, again describing a positive-mass spin-1/2 particle, again carrying positive energy. In the laboratory frame, both are considered to have positive mass and positive energy, although they are still very much dual to each other, with the flipped sign on the antiparticle plane-wave suggesting that it is "travelling backwards in time". The interpretation of "backwards-time" is a bit subjective and imprecise, amounting to hand-waving when one's only evidence are these solutions. It does gain stronger evidence when considering the quantized Dirac field. A more precise meaning for these two sets of solutions being "opposite to each other" is given in the section on [[charge conjugation]], below.