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\begin{center}
\LARGE\bf Physics \course \\ Lecture \lect \\
\end{center}
%
\section{Introduction}
In the previous lecture, a relativistic wave equation was derived. This
equation yielded as free particle solution a scalar wave-function. Since
it has no degrees of freedom other than the spatial degrees, the equation
represents a spin zero particle. Further, the 4-current density had to
be reinterpreted from its interpretation in non-relativistic quantum, where
it is given a probabilistic interpretation. In the Klein-Gordon equation
it is interpreted as a charged current density to avoid the problem of
having a negative probability density. Historically, the negative energies
and probability density caused the Klein-Gordon equation to be ignored.
A second attempt at a relativistic equation was made, by making the
equation depend on a first order time derivative, to avoid the negative
probability densities. This equation, the Dirac equation will be discussed 
in this lecture.
%%%
\section{Dirac Equation}
A second attempt at finding an relativistic quantum equation was carried 
out by
Dirac. With the main problem being that 
the Klein-Gordon equation is second
order in the time derivative (recall the original interpretation of the
of the current density was in terms of probabilities, and negative
probabilities occurs), Dirac proposed an equation that was first 
order in all derivatives. In
this way he hoped to eliminate the problem of a 
negative probability density and maybe also
the negative energies.

To start with, he proposed an equation of the following form
\begin{equation} \label{eq:dirac1}
  \left[ \vec{\alpha} \cdot \vec{p} + \beta m \right] \psi = E \psi
  \Longrightarrow
  \left[-i \vec{\alpha} \cdot \vec{\nabla} + \beta m \right] \psi =
  i \frac{\partial \psi}{\partial t}.
\end{equation}
where the standard quantization conditions are imposed:
\begin{equation}
  \vec{p}=-i \vec{\nabla} \qquad E=i\frac{\partial}{\partial t}
\end{equation}
In order for this to be a valid relativistic equation, it must satisfy the
relativistic energy momentum relation for a free particle, that is it must
satisfy the Klein-Gordon equation. This will then impose a condition on the
$\vec{\alpha}$ and $\beta$ parameters in equation \ref{eq:dirac1}. Squaring
equation \ref{eq:dirac1} leads to:
\begin{align}
   E^2 \psi &= \left( \alpha_i p_i + \beta m\right)
   \left( \alpha_j p_j + \beta m\right) \psi \notag\\
   E^2 \psi &= \left[
   \alpha_i^2 p_i^2 + \left( \alpha_i \alpha_j +\alpha_j \alpha_i \right)
   p_i p_j + \left(\alpha_i \beta + \beta \alpha_i\right) p_i m +
   \beta^2 m^2 \right] \psi \label{eq:drsq}
\end{align}
where the condition $i>j$ is imposed on the second term in equation 
\ref{eq:drsq}.
Comparing equation \ref{eq:drsq} to the Klein-Gordon equation or the 
relativistic
energy-momentum relation imposes the following condition on the parameters
\begin{align}
  \alpha_i^2 = \beta^2 = 1 \label{eq:absqr}\\ 
  \alpha_i \alpha_j + \alpha_j \alpha_i = 2 \delta_{ij} \label{eq:ab}\\
  \alpha_i \beta + \beta \alpha_i = 0.
\end{align}
Obviously the only way for these relations to hold is for the
$\alpha_i$ and $\beta$ to all be matrices and the wave function cannot 
be a simple
function either.

To determine the form of the matrices, the following conditions need to be
imposed:
\begin{itemize}
  \item The wave function should be a column vector in order that the 
probability
         density be easily given as $\psi^\dagger \psi$. This imposes the 
condition
        that the matrices must be square.
  \item The Hamiltonian must be hermitian so that its eigenvalues are real. 
This
        forces the $\alpha_i$ and $\beta$ matrices to also be hermitian:
        $\alpha_i = \alpha_i^\dagger$ and $\beta = \beta^\dagger$.
\end{itemize}
Based on these conditions and equation \ref{eq:absqr} the matrices have
eigenvalues $\pm 1$. Further, using the trace theorem $\mathrm{Tr}[AB] =
\mathrm{Tr}[BA]$ and the relation $\alpha_i = -\beta \alpha_i \beta$, which
comes from equation \ref{eq:ab}, there are an equal number of eigenvalues with
values $+1$ and $-1$:
\begin{equation}
  \mathrm{Tr}[\alpha_i] = \mathrm{Tr}[\beta^2 \alpha_i]
  = \mathrm{Tr}[\beta \alpha_i \beta] = -\mathrm{Tr}[\alpha_i]
  \Longrightarrow \mathrm{Tr}[\alpha_i] = 0.
\end{equation}
Based on this result, there are four matrices with an even dimension. Since a
dimension of two only gives three anti-commuting matrices, the smallest 
dimension
that fulfills our requirement is four and therefore the wave function must be
a column vector of dimension four also. At this point it is not necessary to
introduce an explicit representation for the various matrices and column 
vectors.

%Given the relationships for the various matrices, it is time to find the 
%probability
%density and the probability current density. This will be approached in 
%the same
%manner as the Klein-Gordon case. Take the Dirac equation and multiply it 
%from the
%left with the hermitian conjugate wave function
%\begin{equation}
%   i \psi^\dagger \frac{\partial \psi}{\partial t} =
%   -i \psi^\dagger \vec{\alpha} \cdot \vec{\nabla} \psi + m \psi^\dagger
%   \beta \psi.
%\end{equation}
%Then take the hermitian conjugate of the Dirac equation and multiply it 
%from the
%right with the wave function
%\begin{equation}
%   -i \left(\frac{\partial \psi^\dagger}{\partial t}\right)  \psi=
%   i \left(\vec{\alpha} \cdot \vec{\nabla} \psi^\dagger\right)  \psi
%   + m \psi^\dagger \beta \psi.
%\end{equation}
%At this point the two equation are subtracted leaving
%\begin{equation}
%  \frac{\partial \psi^\dagger \psi}{\partial t} + \vec{\nabla} \cdot
%  \left(\psi^\dagger \vec{\alpha} \psi\right) =0
%\end{equation}
%giving a positive definite probability density $\rho = \psi^\dagger \psi
%=|\psi|^2$, 
%which is
%what was being searched for. One problem still exists, since each component 
%of the
%wave function still satisfies the Klein-Gordon equation, the energy eigenvalues
%can still be positive and negative.
\subsection{Covariant Form of Dirac Equation}
The form of the Dirac equation given above, is not in a form that easily
demonstrates its covariance. The main reason being that the time and
coordinates are not put on an equal footing. To transform the equation,
multiply both sides by $\beta$:
\begin{equation}
  \left[-i \vec{\alpha} \cdot \vec{\nabla} + \beta m \right] \psi =
  i \frac{\partial \psi}{\partial t} \quad\Rightarrow\quad
  \left[-i \beta\vec{\alpha} \cdot \vec{\nabla} +  m \right] \psi =
  i \beta\frac{\partial \psi}{\partial t}
\end{equation}
Next introduce the $\gamma$ matrices $\gamma^\mu=(\beta,\beta\vec{\alpha})$
and rewrite the Dirac equation as:
\begin{equation}
  \left[i\partial^\mu\gamma_\mu - m\right]\psi=0
\end{equation}
This equation puts both the time and position coordinates on an equal footing.

Before proceeding, a few properties of the $\gamma$ matrices are dervied.
First the anti-commutation relations are derived. These are derived 
from the anti-commutation relations of the $\vec{\alpha}$ and $\beta$
matrices:
\begin{alignat}{2}
  &\beta\alpha_i+\alpha_i\beta=0 &\quad &(\alpha_i)^2=(\beta)^2=1
  \label{eq:abacom}\\
  &\alpha_i\alpha_j+\alpha_j\alpha_i=2\delta_{ij} &\quad &\gamma^\mu=
  (\beta,\beta\vec{\alpha})
  \notag
\end{alignat}
Start with the first equation and multiply from the left by $\beta$:
\begin{equation}
  \beta(\beta\alpha_i)+(\beta\alpha_i)\beta=\gamma^0\gamma^i+
  \gamma^i\gamma^0=0
\end{equation}
Now take the second anti-commutation relation in equation \ref{eq:abacom}
and multiply from both the left and right by $\beta$:
\begin{equation}
 (\beta\alpha_i)(\alpha_j\beta)+(\beta\alpha_j)
 (\alpha_i\beta)=2\delta_{ij}\beta\beta \quad\Rightarrow\quad
 \gamma_i\gamma_j+\gamma_j\gamma_i =-2\delta_{ij}
\end{equation}
where the $\alpha$ $\beta$ anti-commutation relation was used. Putting the
previous two equations together yields:
\begin{equation}
  \gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu =2g^{\mu\nu}
\end{equation}

The hermiticity of the $\gamma^\mu$ matrices can be derived in a manner
similar to the commutation relations. Start with $\gamma^0$, since it
is equal $\beta$ and $\beta$ is hermitian $\gamma^{0\dagger}=\gamma^0$---it
is hermitian also. The other components are given by:
\begin{equation}
  \gamma^{i\dagger}=(\beta\alpha^i)^\dagger=(\alpha^i\beta)=-\gamma^i
\end{equation}
where the hermiticity of $\alpha$ and $\beta$ are used, and these components
are shown to be anti-hermitian. The hermitian conjugate can therefore be
written as:
\begin{equation}
  \gamma^{\mu\dagger}=\gamma^0\gamma^\mu\gamma^0=0
\end{equation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%
\subsection{The Conserved Current}
As in the case of the Klein-Gordon equation, a conserved current can be
derived. This current is derived in a manner analogous to that in the
Klein-Gordon equation. Start with the Dirac equation and take the 
hermitian conjugate:
\begin{equation}
  (i\gamma^\mu\partial_\mu\psi-m\psi=0)^\dagger \quad\Rightarrow\quad
  -i\partial_\mu\psi^\dagger\gamma^0\gamma^\mu\gamma^0-m\psi^\dagger=0
\end{equation}
Next multiply from the right by $\gamma^0$:
\begin{equation}
  (-i\partial_\mu\psi^\dagger\gamma^0\gamma^\mu\gamma^0-m\psi^\dagger
   =0)\gamma^0\quad\Rightarrow\quad i\partial_\mu\bar{\psi}\gamma^\mu+
   m\bar{\psi}=0
\end{equation}
To derive the conserved current, multipy the Dirac equation from the left
by $\bar{\psi}$ and the hermitian conjugate from the right by $\psi$. Then 
thenadd the two equation together:
\newcommand{\bpsi}{\bar{\psi}}
\newcommand{\partum}{\partial^\mu}
\newcommand{\partdm}{\partial_\mu}
\begin{equation}
  \left.
  \begin{array}{l}
    i\bpsi\gamma^\mu\partdm\psi-m\bpsi\psi=0\\
    i(\partdm\bpsi)\gamma^\mu\psi+m\bpsi\psi=0
  \end{array}\right\}\quad\Rightarrow\quad
  \partdm(\bpsi\gamma^\mu\psi)=\partdm j^\mu=0
\end{equation}
notice, that at this point $\bpsi\gamma^\mu\psi$ has not been shown to be
a 4-vector, this will be shown in a later lecture. In principle, $\rho(=j^0)$
in this case is a positive number, of course the normalization has yet
to be selected, and so it can again have both values.
where $\bar{\psi}=\psi^\dagger\gamma^0$.
%%%%%%
\subsection{The Gamma Matrices}
There are numerous ways of writing the gamma matrices explicitly. 
Starting with $\beta=\gamma^0$ and the relation that the square is 
the unit matrix, the eigenvalues of this matrix must be $\pm1$. 
This element is taken as diagonal:
 \begin{equation} \beta = \gamma^0=
   \begin{pmatrix}
       1&0\\0 & -1
   \end{pmatrix}
 \end{equation}
To satisfy the remaining relations, the alpha matrices are written 
in terms of the Pauli matrices:
\begin{equation}
  \vec{\alpha} =
  \begin{pmatrix}
  0 & \vec{\sigma}\\ \vec{\sigma}&0
  \end{pmatrix} \Rightarrow
  \gamma^i = \beta \vec{\alpha} =
  \begin{pmatrix}
    0&\vec{\sigma} \\ -\vec{\sigma}&0
  \end{pmatrix}
\end{equation}
{\em Note that each element of the $\gamma^\mu$ matrices is a 
$2\times 2$ matrix.}
As a reminder, the Pauli matrices are given by:
\begin{equation}
\sigma_x=
\begin{pmatrix}
  0&1\\1&0
\end{pmatrix}
\qquad
%%%
\sigma_y=
\begin{pmatrix}
  0&-i\\i&0
\end{pmatrix}
\qquad
%%%
\sigma_z=
\begin{pmatrix}
  1&0\\0&-1
\end{pmatrix}
\end{equation}
%%%%%%%%
%%%%%%%%
\section{The Free Particle Dirac Wave-Function}
Given that the Dirac equation is written in terms of $4\times4$ 
matrices and a 4 element column vector, it can be written as 4 
coupled differential equations. It has already been shown that each 
component of the Dirac equation must satisfy the Klein-Gordon equation; 
this was the condition that was imposed on the Dirac equation to get 
the form of the $\gamma$ matrices. Therefore the solution to the 
Dirac equation must be of the form:
\begin{equation}
  \psi(x) = u(p) e^{-i p_\mu x^\mu}
\end{equation}
where each of the variables are four vectors. This solution 
is substituted back into the Dirac equation giving:
\begin{equation}
  (i \gamma^\mu \partial_\mu -m) u(p) e^{-i p_\mu x^\mu} =
  ( \gamma^\mu p_\mu -m) u(p) e^{-i p_\mu x^\mu}=
  ( \gamma^\mu p_\mu -m) u(p)=0
\end{equation}
being a totally algebraic equation.

To arrive at the form of the free particle solutions, 
the matrix representation of the Dirac equation is used:
\begin{align}
  &\gamma^\mu p_\mu = E \begin{pmatrix} {\bf I} &0\\0 &-{\bf I}
  \end{pmatrix} - \vec{p} \cdot \begin{pmatrix}
  0&\vec{\sigma}\\ -\vec{\sigma}&0 \end{pmatrix} =
  \begin{pmatrix}
    E {\bf I} & -\vec{p}\cdot \vec{\sigma} \notag \\
    \vec{p} \cdot \vec{\sigma} & -E {\bf I}
  \end{pmatrix} \\
%
  &\Rightarrow \left(\gamma^\mu p_\mu -m\right) u(p) =
  \begin{pmatrix} E-m & -\vec{p} \cdot \vec{\sigma} \\
    \vec{p}\cdot \vec{\sigma} & -E -m \end{pmatrix}
  \begin{pmatrix} u_A\\u_B \end{pmatrix} =
  \begin{pmatrix} (E-m)u_A  -\vec{p} \cdot \vec{\sigma}u_B \\
    \vec{p}\cdot \vec{\sigma} u_A -(E+m) u_B
  \end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}
\end{align}
notice that the equation is broken into two pieces, 
even though this is a four component equation. Breaking up the two 
pieces gives:
\begin{equation}
  u_A=\frac{\vec{p} \cdot \vec{\sigma}}{E-m}u_B, \qquad
  u_B=\frac{\vec{p} \cdot \vec{\sigma}}{E+m}u_A
\end{equation}
which finally leads to:
\begin{equation}
  u_A=\frac{(\vec{p} \cdot \vec{\sigma})^2}{E^2-m^2}u_A
\end{equation}
This does not give a value for the wave-function, but it 
does impose a condition on the energy and momentum. This condition 
can be found by expanding out the numerator:
\begin{align}
  &\vec{p}\cdot \vec{\sigma} =p_x
  \begin{pmatrix} 0&1\\1&0 \end{pmatrix} +
  p_y \begin{pmatrix} 0&-i\\i&0 \end{pmatrix}+
  p_z \begin{pmatrix} 1&0\\0&-1 \end{pmatrix} =
  \begin{pmatrix} p_z & p_x-ip_y \\
     p_x+ip_y & -p_z \end{pmatrix} \\
%
  &\Rightarrow (\vec{p} \cdot \vec{\sigma})^2 = {\bf I}
  |\vec{p}|^2 \Rightarrow u_A = \frac{|\vec{p}|^2}{E^2-m^2}
  {\bf I} u_A
\end{align}
This equation imposes the condition that 
$E=\pm \sqrt{|\vec{p}|^2+m^2}$,
which is what would be expected. The positive energy, 
as before, is associated with the particle state while the
negative energy is associated with the anti-particle state.

Finally the wave-function is determined. Notice that there 
is some level of arbitrariness in the solution. The only
 requirement is that they be orthogonal and that they have 
energy eigenvalues corresponding to the positive and negative 
energy solutions. Going back to the relation 
between $u_A$ and $u_B$ and imposing that the solutions be 
orthogonal gives:
\begin{alignat}{4}
  &u_A=\begin{pmatrix}1\\0\end{pmatrix} &\quad
  &u_B=\frac{1}{E+m} \begin{pmatrix}
  p_z\\ p_x+ip_y \end{pmatrix} &\qquad
  &u_A =\begin{pmatrix} 0\\1\end{pmatrix} &\quad
  &u_B=\frac{1}{E+m} \begin{pmatrix}
  p_x-ip_y\\ -p_z \end{pmatrix} \notag \\
%
&u_B=\begin{pmatrix}1\\0\end{pmatrix} &\quad
  &u_A=\frac{1}{E-m} \begin{pmatrix}
  p_z\\ p_x+ip_y \end{pmatrix} &\qquad
  &u_B =\begin{pmatrix} 0\\1\end{pmatrix} &\quad
  &u_A=\frac{1}{E-m} \begin{pmatrix}
  p_x-ip_y\\ -p_z \end{pmatrix}
\end{alignat}
Notice that the first two equations must have $E>0$ 
otherwise the solution blows up when $\vec{p}=0$. 
For the second two equations the energy must be less than 
zero $(E<0)$ otherwise the solution blows up 
when $\vec{p}=0$. The particle solutions are therefore:
\begin{equation}
  u^1=N\begin{pmatrix} 1\\0\\ \frac{p_z}{E+m}\\
       \frac{p_x+ip_y}{E+m} \end{pmatrix}, \qquad
  u^2=N\begin{pmatrix} 0\\1\\ \frac{p_x-ip_y}{E+m}\\
     \frac{-p_z}{E+m}\\
        \end{pmatrix}
\end{equation}
while the anti-particle solutions are:
\begin{equation}
  u^3=N\begin{pmatrix} \frac{p_z}{E-m}\\
     \frac{p_x+ip_y}{E-m}\\
     1\\0
        \end{pmatrix}, \qquad
%
   u^4=N\begin{pmatrix}  
      \frac{p_x-ip_y}{E-m}\\
      \frac{-p_z}{E-m}\\0\\1
      \end{pmatrix}
\end{equation}
The prescription that has been used so far is to redefine 
the negative energy solutions as positive energy anti-particles. 
New states are then defined as:
\begin{equation}
  v^1=u^4(-p)=N\begin{pmatrix} \frac{p_x-ip_y}{E+m}\\
     \frac{-p_z}{E+m}\\0\\1
        \end{pmatrix}, \qquad
%
  v^2=-u^3(-p)=-N\begin{pmatrix}  \frac{p_z}{E+m}\\
       \frac{p_x+ip_y}{E+m} \\1\\0 \end{pmatrix}
\end{equation}
Notice that as the momentum approaches zero, 
the momentum dependent term approaches zero, which 
then gives the non-relativistic solution used in the Schr\"odinger 
equation. Finally the equations that govern the particle and anti-particle 
solutions are given as:
\begin{equation}
  \left(\gamma^\mu p_\mu -m\right) u(p)=0, \qquad
  \left(\gamma^\mu p_\mu +m\right) v(p)=0
\end{equation}

The wave-function normalization has yet to be selected. 
The normalization will be chosen the same as for bosons:
\begin{equation}
  \int \rho dV= \int\psi^\dagger \psi dV = u^\dagger u=2E
\end{equation}
where the number of particles per unit volume is 
given by the expression above. Applying this condition 
to the spinors gives  $N=\sqrt{|E|+m}$.
\end{document}