Question 2.10: Features of Dirac Matrices Start with the anticommutator {γ^......
Features of Dirac Matrices
Start with the anticommutator
\left\{\gamma^{\mu}, \gamma^{\nu}\right\}=2 g^{\mu \nu} \mathbb{1} (1)
and show that the following relations hold:
(a) \not a\not b=-\not b \not a+2 a \cdot b. (2)
(b) \left(\not k-m_{0}\right)\left(\not k+m_{0}\right)=\left(\not k+m_{0}\right)\left(\not k-m_{0}\right)
=0, \quad \text { for } k^{2}=m_{0}^{2}. (3)
(c) \hat{\Lambda}_{+}(k)=\left(\not k+m_{0}\right), (4)
which eliminates the negative energy parts of an arbitrary spinor, and
\hat{\Lambda}_{-}(k)=\left(\not k-m_{0}\right) (5)
which eliminates the positive energy parts of an arbitrary spinor.
(d) Employ the explicit forms
\begin{aligned} u(k, s) & =\sqrt{\omega+m_{0}}\left(\begin{array}{c} \phi^{s} \\ \frac{\hat{\sigma} \cdot {k}}{\omega+m_{0}} \phi^{s} \end{array}\right), \\ \phi^{1} & =\left(\begin{array}{l} 1 \\ 0 \end{array}\right), \quad \phi^{2}=\left(\begin{array}{l} 0 \\ 1 \end{array}\right), & (6)\\ v(k, s) & =\sqrt{\omega+m_{0}}\left(\begin{array}{c} \frac{\hat{\sigma} \cdot {k}}{\omega+m_{0}} \chi^{s} \\ \chi^{s} \end{array}\right), \\ \chi^{1} & =\left(\begin{array}{l} 0 \\ 1 \end{array}\right), \quad \chi^{2}=\left(\begin{array}{l} 1 \\ 0 \end{array}\right) & (7) \end{aligned}
and show how the 4 \times 4 matrices \sum_{s} u(k, s) \bar{u}(k, s) and \sum_{s} v(k, s) \bar{v}(k, s) depend on \hat{\Lambda}_{+}(k) and \hat{\Lambda}_{-}(k), respectively.
(a) \not a \not b+\not b \not a =a_{\mu} \gamma^{\mu} b_{\nu} \gamma^{\nu}+b_{\nu} \gamma^{\nu} a_{\mu} \gamma^{\mu}=a_{\mu} b_{\nu}\left\{\gamma^{\mu}, \gamma^{\nu}\right\}
=2 a_{\mu} b_{\nu} g^{\mu \nu} \mathbb{1}=2 a \cdot b \mathbb{1} ; (8)
from this equation follows in particular that \not a^{2}=a^{2} \mathbb{1}.
(b) \left(\not k-m_{0} \mathbb{1}\right)\left(\not k+m_{0} \mathbb{1}\right)=\left(\not k^{2}-m_{0}^{2}\right) \mathbb{1}
=\left(k^{2}-m_{0}^{2}\right) \mathbb{1}=0 (9)
(c) The wave function of a particle with positive energy and four-momentum k_{\mu} can be written as
\Psi^{(+)}(x)=\sum\limits_{s} b_{s} u(k, s) \mathrm{e}^{-\mathrm{i} k \cdot x} (10)
Correspondingly a solution with negative energy and four-momentum k_{-\mu} is
\Psi^{(-)}(x)=\sum\limits_{s} d_{s} v(k, s) \mathrm{e}^{\mathrm{i} k \cdot x} (11)
Both wave functions must obey the Dirac equation, i.e.,
\begin{aligned} \left(\mathrm{i} \not \nabla-m_{0}\right) \Psi^{(+)}(x) & =\left(\not k-m_{0}\right) \Psi^{(+)}(x) \\ & =\hat{\Lambda}_{-(k)} \Psi^{(+)}(x)=0 & (12) \end{aligned}
and
\begin{aligned} \left(\mathrm{i} \not \nabla-m_{0}\right) \Psi^{(-)}(x) & =-\left(\not k+m_{0}\right) \Psi^{(-)}(x) \\ & =-\hat{\Lambda}_{+(k)} \Psi^{(-)}(x)=0. & (13) \end{aligned}
Equations (12) and (13) can also be put into the form
\begin{aligned} & \not k \Psi^{(+)}(x)=m_{0} \Psi^{(+)}(x), \\ & \not k \Psi^{(-)}(x)=-m_{0} \Psi^{(-)}(x) . & (14) \end{aligned}
From (14) we immediately obtain
\begin{aligned} & \hat{\Lambda}_{+} \Psi=2 m_{0} \Psi^{(+)}, \\ & \hat{\Lambda}_{-} \Psi=-2 m_{0} \Psi^{(-)}. & (14) \end{aligned}
Now we employ the Hermiticity of the Pauli matrices, \hat{\sigma}^{\dagger}=\hat{\sigma}, and make use of the identity \sum_{s} \phi^{s} \phi^{s \dagger}=\mathbb{1} (cf. (2.165)).
\begin{aligned} \sum\limits_s \phi^s \phi^{s \dagger} & =\phi^1 \phi^{1 \dagger}+\phi^2 \phi^{2 \dagger} \\ & =\left(\begin{array}{l} 1 \\ 0 \end{array}\right)(1,0)+\left(\begin{array}{l} 0 \\ 1 \end{array}\right)(0,1) \\ & =\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right)+\left(\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right)=\mathbb{1} & (2.165) \end{aligned}
Equation (16) then assumes the form
\left(\omega+m_{0}\right)\left(\begin{array}{cc} \mathbb{1} & -\frac{\hat{{\sigma}} \cdot {k}}{\omega+m_{0}} \\ \frac{\hat{{\sigma}} \cdot {k}}{\omega+m_{0}} & -\frac{(\hat{{\sigma}} \cdot {k})^{2}}{\left(\omega+m_{0}\right)^{2}} \end{array}\right). (17)
But since we have
\begin{aligned} (\hat{{\sigma}} \cdot {k})^{2} & =\frac{1}{2} k_{i} k_{j}\left(\hat{\sigma}_{i} \hat{\sigma}_{j}+\hat{\sigma}_{j} \hat{\sigma}_{i}\right)=k_{i} k_{j} \delta_{i j} \\ & ={k}^{2}=\omega^{2}-m_{0}^{2}, \end{aligned}
the result becomes
\left(\begin{array}{cc} \left(\omega+m_{0}\right) \mathbb{1} & -\hat{{\sigma}} \cdot {k} \\ \hat{{\sigma}} \cdot {k} & \left(-\omega+m_{0}\right) \mathbb{1} \end{array}\right). (18)
Employing the \gamma-matrix representation
\gamma^{0}=\left(\begin{array}{cc} \mathbb{1} & 0 \\ 0 & -\mathbb{1} \end{array}\right), \quad {\gamma}=\left(\begin{array}{cc} 0 & \sigma \\ -\sigma & 0 \end{array}\right) , (19)
we obtain the result
\begin{aligned} \sum_{s} u(k, s) \bar{u}(k, s) & =\omega \gamma^{0}-{k} \cdot {\gamma}+m_{0} \\ & =\left(\not k+m_{0}\right)=\hat{\Lambda}_{+}(k) . \end{aligned}
An analogous calculation for negative solutions yields
\begin{array}{rl} \sum\limits_{s} & v(k, s) \bar{v}(k, s) \\ & =\left(\omega+m_{0}\right) \sum\limits_{s}\left(\begin{array}{c} \frac{\hat{{\sigma}} \cdot {k}}{\omega+m_{0}} \chi^{s} \\ \chi^{s} \end{array}\right)\left(\chi^{s \dagger} \frac{\hat{{\sigma}} \cdot {k}}{\omega+m_{0}},-\chi^{s}\right) \\ & =\left(\omega+m_{0}\right)\left(\begin{array}{cc} \frac{\omega^{2}-m_{0}^{2}}{\left(\omega+m_{0}\right)^{2}} & -\frac{\hat{{\sigma}} \cdot {k}}{\omega+m_{0}} \\ \frac{\hat{{\sigma}} \cdot {k}}{\omega+m_{0}} & -\mathbb{1} \end{array}\right) \\ & =\left(\begin{array}{cc} \left(\omega-m_{0}\right) \mathbb{1} & -\hat{{\sigma}} \cdot {k} \\ \hat{{\sigma}} \cdot {k} & \left(-\omega-m_{0}\right) \mathbb{1} \end{array}\right) \\ & =\omega \gamma^{0}-{k} \cdot {\gamma}-m_{0} \\ & =\left(\not k-m_{0}\right)=-\hat{\Lambda}_{-}(k) . & (20) \end{array}