Dirac equation

How to construct

Starting from the relativistic relation between energy and momentum:

$E^{2}={\vec {p}}\;^{2}c^{2}+m^{2}c^{4}$

or $E=c{\sqrt {p^{2}+m^{2}c^{2}}}$

From this equation we can not directly replace $E,{\vec {p}}$ by the corresponding operators since we don't have the definition for the square root of an operator. Therefore, first we need to linearize this equation as follows:

$E=c{\sqrt {p^{2}+m^{2}c^{2}}}=c{\sqrt {(p_{x}^{2}+p_{y}^{2}+p_{z}^{2})+m^{2}c^{2}}}=c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^{2}$

where ${\mathbf {\alpha }}_{x},\alpha _{y},\alpha _{z}$ and ${\mathbf {\beta }}$ are some operators independent of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec p} .

From this it follows that:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c^2(p_{x}^2+p_{y}^2+p_{z}^2+m^2c^2)=[c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^2] . [c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^2]}

Expanding the right hand side and comparing it with the left hand side, we obtain the following conditions for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \alpha _{x},\alpha _{y},\alpha _{z}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \beta}  :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha _{i}^2=\beta ^2=1}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \alpha_ {i}\alpha_ {j}+\alpha_ {j}\alpha_ {i}=\{\alpha_ {i},\alpha_ {j}\}=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \alpha_ {i} \beta+\alpha_ {j} \beta=\{\alpha_ {i},\beta\}=0}

where $i=1,2,3$ corresponds to $x,y,z$

In order to describe both particle (positive energy state) and antiparticle (negative energy state); spin-up state and spin-down state, the wave function must have 4 components and all operators acting on such states correspond to 4 by 4 matrices. Therefore, ${\mathbf {\alpha }}_{x},\alpha _{y},\alpha _{z}$ and ${\mathbf {\beta }}$are 4 by 4 matrices. It is convention that these matrices are given as follows (in the form of block matrices for short):

$\alpha _{x}=\left({\begin{array}{cc}0&\sigma _{x}\\\sigma _{x}&0\end{array}}\right)$; $\qquad \alpha _{y}=\left({\begin{array}{cc}0&\sigma _{y}\\\sigma _{y}&0\end{array}}\right)$; $\qquad \alpha _{z}=\left({\begin{array}{cc}0&\sigma _{z}\\\sigma _{z}&0\end{array}}\right)$; $\qquad \beta =\left({\begin{array}{cc}1&0\\0&-1\end{array}}\right)$

where $\sigma _{x},\;\sigma _{y},\;\sigma _{z}$ are 2 by 2 Pauli matrices.

Let's define:

${\vec {\alpha }}=\alpha _{x}{\hat {x}}+\alpha _{y}{\hat {y}}+\alpha _{z}{\hat {z}}$

Then we can write:

$E=c{\vec {\alpha }}{\vec {p}}+\beta mc^{2}$

Substituting all quantities by there corresponding operators, we obtain Dirac equation:

$i\hbar {\frac {\partial \psi }{\partial t}}=H\psi$

where $H=c{\vec {\alpha }}{\vec {p}}+\beta mc^{2}$

Dirac equation can also be written explicitly as follows:

$i\hbar {\frac {\partial \psi _{1}}{\partial t}}=c(p_{x}-ip_{y})\psi _{4}+cp_{z}\psi _{3}+mc^{2}\psi _{1}$

$i\hbar {\frac {\partial \psi _{2}}{\partial t}}=c(p_{x}+ip_{y})\psi _{3}-cp_{z}\psi _{4}+mc^{2}\psi _{2}$

$i\hbar {\frac {\partial \psi _{3}}{\partial t}}=c(p_{x}-ip_{y})\psi _{2}+cp_{z}\psi _{1}-mc^{2}\psi _{3}$

$i\hbar {\frac {\partial \psi _{4}}{\partial t}}=c(p_{x}+ip_{y})\psi _{1}-cp_{z}\psi _{2}-mc^{2}\psi _{4}$