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 ${\vec {p}}$.

From this it follows that:

$c^{2}(p_{x}^{2}+p_{y}^{2}+p_{z}^{2}+m^{2}c^{2})=[c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^{2}].[c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^{2}]$

Expanding the right hand side and comparing it with the left hand side, we obtain the following conditions for ${\mathbf {\alpha }}_{x},\alpha _{y},\alpha _{z}$ and ${\mathbf {\beta }}$ :

$\alpha _{i}^{2}=\beta ^{2}=1$

${\mathbf {\alpha }}_{i}\alpha _{j}+\alpha _{j}\alpha _{i}=\{\alpha _{i},\alpha _{j}\}=0$

${\mathbf {\alpha }}_{i}\beta +\alpha _{j}\beta =\{\alpha _{i},\beta \}=0$

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 their corresponding operators, we obtain Dirac equation:

$i\hbar {\frac {\partial \psi }{\partial t}}=H_{D}\psi$

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

Dirac equation can also be written explicitly as follows:

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

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

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

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

In the present of electromagnetic field, Dirac equation becomes:

$(i\hbar {\frac {\partial }{\partial t}}-e\phi )\psi =\left[c{\vec {\alpha }}({\frac {\hbar }{i}}{\vec {\nabla }}-{\frac {e}{c}}{\mathbf {A}})+\beta mc^{2}\right]\psi$

Continuity equation

Dirac equation and its adjoint equation:

$i\hbar {\frac {\partial \psi }{\partial t}}=(-i\hbar c{\vec {\alpha }}{\vec {\nabla }}+mc^{2}\beta )\psi$

$-i\hbar {\frac {\partial \psi ^{\dagger }}{\partial t}}=(i\hbar c{\vec {\nabla }}\psi ^{\dagger }{\vec {\alpha }}+mc^{2}\psi ^{\dagger }\beta )$

Multiplying Dirac equation by $\psi ^{\dagger }$from the left and the adjoint equation by ${\mathbf {\psi }}$ from the right, we get:

$i\hbar \psi ^{\dagger }{\frac {\partial \psi }{\partial t}}=-i\hbar c\psi ^{\dagger }{\vec {\alpha }}{\vec {\nabla }}\psi +mc^{2}\psi ^{\dagger }\beta \psi$

$-i\hbar {\frac {\partial \psi ^{\dagger }}{\partial t}}\psi =i\hbar c{\vec {\nabla }}\psi ^{\dagger }{\vec {\alpha }}\psi +mc^{2}\psi ^{\dagger }\beta \psi$

Subtracting one from the other, we get:

$i\hbar \left(\psi ^{\dagger }{\frac {\partial \psi }{\partial t}}+{\frac {\partial \psi }{\partial t}}\psi ^{\dagger }\right)=-i\hbar c\left[\psi ^{\dagger }{\vec {\alpha }}{\vec {\nabla }}\psi +{\vec {\nabla }}\psi ^{\dagger }{\vec {\alpha }}\psi \right]=-i\hbar c{\vec {\nabla }}\left(\psi ^{\dagger }{\vec {\alpha }}\psi \right)$

$\Rightarrow {\frac {\partial }{\partial t}}\left(\psi ^{\dagger }\psi \right)+{\vec {\nabla }}\left(c\psi ^{\dagger }{\vec {\alpha }}\psi \right)=0$

Therefore, we can define:

$\rho =\psi ^{\dagger }\psi$ as probability density

${\vec {j}}=c\left(\psi ^{\dagger }{\vec {\alpha }}\psi \right)$ as probability current density