Quantization of the Dirac Field

The classical free field Lagrangian density for the Dirac electron field is.

\begin{displaymath}\bgroup\color{black} {\cal L}=-c\hbar\bar{\psi}\gamma_\mu{\partial\over\partial x_\mu}\psi-mc^2\bar{\psi}\psi \egroup\end{displaymath}

The independent fields are considered to be the 4 components of \bgroup\color{black}$\psi$\egroup and the four components of \bgroup\color{black}$\bar{\psi}$\egroup. This Lagrange density is a Lorentz scalar that depends only on the fields. The Euler-Lagrange equation using the \bgroup\color{black}$\bar{\psi}$\egroup independent fields is simple since there is no derivative of \bgroup\color{black}$\bar{\psi}$\egroup in the Lagrangian.

{\partial\over\partial x_\mu}\left({\partial{\cal L}\over\part...
...\mu{\partial\over\partial x_\mu}+{mc\over\hbar}\right)\psi=0 \\

This gives us the Dirac equation indicating that this Lagrangian is the right one. The Euler-Lagrange equation derived using the fields \bgroup\color{black}$\psi$\egroup is the Dirac adjoint equation,

{\partial\over\partial x_\mu}\left({\partial{\cal L}\over\part...
...tial x_\mu}\bar{\psi}\gamma_\mu+{mc\over\hbar}\bar{\psi}&=&0 \\

again indicating that this is the correct Lagrangian if the Dirac equation is assumed to be correct.

To compute the Hamiltonian density, we start by finding the momenta conjugate to the fields \bgroup\color{black}$\psi$\egroup.

\begin{displaymath}\bgroup\color{black} \Pi={\partial{\cal L}\over\partial\left(...
...}=i\hbar\psi^\dagger\gamma_4\gamma_4=i\hbar\psi^\dagger \egroup\end{displaymath}

There is no time derivative of \bgroup\color{black}$\bar{\psi}$\egroup so those momenta are zero. The Hamiltonian can then be computed.

{\cal H}&=&{\partial\psi\over\partial t}\Pi-{\cal L} \\
...ma_k{\partial\over\partial x_k}+mc^2\gamma_4\right)\psi d^3x \\

We may expand the field \bgroup\color{black}$\psi$\egroup in the complete set of plane waves either using the four spinors \bgroup\color{black}$u^{(r)}_{\vec{p}}$\egroup for \bgroup\color{black}$r=1,2,3,4$\egroup or using the electron and positron spinors \bgroup\color{black}$u^{(r)}_{\vec{p}}$\egroup and \bgroup\color{black}$v^{(r)}_{\vec{p}}$\egroup for \bgroup\color{black}$r=1,2$\egroup. For economy of notation, we choose the former with a plan to change to the later once the quantization is completed.

\begin{displaymath}\bgroup\color{black} \psi(\vec{x},t)=\sum\limits_{\vec{p}}\su...
... u^{(r)}_{\vec{p}} e^{i(\vec{p}\cdot\vec{x}-Et)/\hbar} \egroup\end{displaymath}

The conjugate can also be written out.

\begin{displaymath}\bgroup\color{black} \psi^\dagger(\vec{x},t)=\sum\limits_{\ve...
...dagger}_{\vec{p}} e^{-i(\vec{p}\cdot\vec{x}-Et)/\hbar} \egroup\end{displaymath}

Writing the Hamiltonian in terms of these fields, the formula can be simplified as follows

H&=&\int\psi^\dagger\left(\hbar c\gamma_4\gamma_k{\partial\ove...
... u^{(r)}_{\vec{p}} e^{i(\vec{p}\cdot\vec{x}-Et)/\hbar} d^3x \\

...{\vec{p}}\sum\limits_{r=1}^4 E c_{\vec{p},r}^*c_{\vec{p},r} \\

where previous results from the Hamiltonian form of the Dirac equation and the normalization of the Dirac spinors have been used to simplify the formula greatly.

Compare this Hamiltonian to the one used to quantize the Electromagnetic field

\begin{displaymath}\bgroup\color{black} H=\sum\limits_{k,\alpha}\left({\omega\ov...
...\alpha}c_{k,\alpha}^*+c_{k,\alpha}^*c_{k,\alpha}\right] \egroup\end{displaymath}

for which the Fourier coefficients were replaced by operators as follows.

c_{k,\alpha}&=&\sqrt{\hbar c^2\over 2\omega}a_{k,\alpha} \\
...alpha}^*&=&\sqrt{\hbar c^2\over 2\omega}a_{k,\alpha}^\dagger \\

The Hamiltonian written in terms of the creation and annihilation operators is.

H&=&{1\over 2}\sum\limits_{k,\alpha}\hbar\omega\left[a_{k,\alp...
..._{k,\alpha}^\dagger+a_{k,\alpha}^\dagger a_{k,\alpha}\right] \\

By analogy, we can skip the steps of making coordinates and momenta for the individual oscillators, and just replace the Fourier coefficients for the Dirac plane waves by operators.

H&=&\sum\limits_{\vec{p}}\sum\limits_{r=1}^4 E b^{(r)^\dagger...
...{(r)\dagger}_{\vec{p}} e^{-i(\vec{p}\cdot\vec{x}-Et)/\hbar} \\

(Since the Fermi-Dirac operators will anti-commute, the analogy is imperfect.)

The creation an annihilation operators \bgroup\color{black}$b^{(r)^\dagger}_{\vec{p}}$\egroup and \bgroup\color{black}$b^{(r)}_{\vec{p}}$\egroup satisfy anticommutation relations.

...{(r)}_{\vec{p}}&=&b^{(r)^\dagger}_{\vec{p}}b^{(r)}_{\vec{p}} \\

\bgroup\color{black}$N^{(r)}_{\vec{p}}$\egroup is the occupation number operator. The anti-commutation relations constrain the occupation number to be 1 or 0.

A state of the electrons in a system can be described by the occupation numbers (0 or 1 for each plane wave). The state can be generated by operation on the vacuum state with the appropriate set of creation operators.

Jim Branson 2013-04-22