Lorentz Invariance of Dirac Equation

Consider two inertial frames, $S$ and $S'$ . Let the $x^{\,\mu}$ and $x^{\,\mu'}$ be the space-time coordinates of a given event in each frame, respectively. These coordinates are related via a Lorentz transformation, which takes the general form

$$x^{\,\mu'} = a^{\,\mu}_{~\nu}\,x^{\,\nu},$$ (11.52)

where the $a^{\,\mu}_{~\nu}$ are real numerical coefficients that are independent of the $x^{\,\mu}$ . We also have

$$x_{\mu'} = a_{\mu}^{~\nu}\,x_{\,\nu}.$$ (11.53)

Now, because [see Equation (11.5)]

$$x^{\,\mu'}\,x_{\mu'} = x^{\,\mu}\,x_\mu,$$ (11.54)

it follows that

$$a^{\,\mu}_{~\nu}\,a_\mu^{~\lambda} = g_\nu^{~\lambda}.$$ (11.55)

Moreover, it is easily shown that

$$x^{\,\mu} = a_\nu^{~\mu}\,x^{\nu'},$$ (11.56)

$$x_\mu = a^\nu_{~\mu}\,x_{\nu'}.$$ (11.57)

By definition, a 4-vector, $p^{\,\mu}$ , has analogous transformation properties to the $x^{\,\mu}$ . Thus,

$$p^{\,\mu'} = a^{\,\mu}_{~\nu}\,p^\nu,$$ (11.58)

$$p^{\,\mu} = a_\nu^{~\mu}\,p^{\nu'},$$ (11.59)

et cetera.

In frame $S$ , the Dirac equation is written

$$\left[\gamma^{\,\mu}\left(p_\mu- \frac{e}{c}\,{\mit\Phi}_\mu\right)-m_e\,c\right]\psi = 0.$$ (11.60)

Let $\psi'$ be the wavefunction in frame $S'$ . Suppose that

$$\psi' = A\,\psi,$$ (11.61)

where $A$ is a $4\times 4$ transformation matrix that is independent of the $x^{\,\mu}$ . (Hence, $A$ commutes with the $p_\mu$ and the ${\mit\Phi}_\mu$ .) Multiplying Equation (11.60) by $A$ , we obtain

$$\left[A\,\gamma^{\,\mu}\,A^{-1}\left(p_\mu- \frac{e}{c}\,{\mit\Phi}_\mu\right)-m_e\,c\right]\psi' = 0.$$ (11.62)

Hence, given that the $p_\mu$ and ${\mit\Phi}_\mu$ are the covariant components of 4-vectors, we obtain

$$\left[A\,\gamma^{\,\mu}\,A^{-1}\,a^{\nu}_{~\mu}\left(p_{\nu'}- \frac{e}{c}\,{\mit\Phi}_{\nu'}\right)-m_e\,c\right]\psi' = 0.$$ (11.63)

Suppose that

$$A\,\gamma^{\,\mu}\,A^{-1}\,a^{\nu}_{~\mu} = \gamma^{\,\nu},$$ (11.64)

which is equivalent to

$$A^{-1}\,\gamma^{\,\nu}\,A = a^{\nu}_{~\mu}\,\gamma^{\,\mu}.$$ (11.65)

Here, we have assumed that the $a^{\nu}_{~\mu}$ commute with $A$ and the $\gamma^{\,\mu}$ (because they are just numbers). If Equation (11.64) holds then Equation (11.63) becomes

$$\left[\gamma^{\,\mu}\left(p_{\mu'}- \frac{e}{c}\,{\mit\Phi}_{\mu'}\right)-m_e\,c\right]\psi' = 0.$$ (11.66)

A comparison of this equation with Equation (11.60) reveals that the Dirac equation takes the same form in frames $S$ and $S'$ . In other words, the Dirac equation is Lorentz invariant. Incidentally, it is clear from Equations (11.60) and (11.66) that the $\gamma^{\,\mu}$ matrices are the same in all inertial frames.

It remains to find a transformation matrix $A$ that satisfies Equation (11.65). Consider an infinitesimal Lorentz transformation, for which

$$a_{~\mu}^{\nu} = g_{~\mu}^{\nu} + \delta\omega_{~\mu}^{\nu},$$ (11.67)

where the $\delta\omega_{~\mu}^{\nu}$ are real numerical coefficients that are independent of the $x^{\,\mu}$ , and are also small compared to unity. To first order in small quantities, Equation (11.55) yields

$$\delta\omega^{\,\mu\,\nu} + \delta\omega^{\,\nu\,\mu} = 0.$$ (11.68)

Each of the six independent non-vanishing $\delta\omega^{\,\mu\,\nu}$ generates a particular infinitesimal Lorentz transformation. For instance,

$$\delta\omega^{\,01} =-\delta\omega^{\,10}= \delta \beta$$ (11.69)

for a transformation to a coordinate system moving with a velocity $c\,\delta\beta$ along the $x^{\,1}$ -direction. Furthermore,

$$\delta\omega^{\,21} = -\delta\omega^{\,12} = \delta\varphi$$ (11.70)

for a rotation through an angle $\delta\varphi$ about the $x^{\,3}$ -axis, and so on.

Let us write

$$A = 1 - \frac{{\rm i}}{4}\,\sigma_{\mu\,\nu}\,\delta\omega^{\,\mu\,\nu},$$ (11.71)

where the $\sigma_{\mu\,\nu}$ are ${\cal O}(1)$ $4\times 4$ matrices. To first order in small quantities,

$$A^{-1} = 1 + \frac{{\rm i}}{4}\,\sigma_{\mu\,\nu}\,\delta\omega^{\,\mu\,\nu}.$$ (11.72)

Moreover, it follows from Equation (11.68) that

$$\sigma_{\mu\,\nu} = -\sigma_{\nu\,\mu}.$$ (11.73)

To first order in small quantities, Equations (11.65), (11.67), (11.71), and (11.72) yield

$$\delta\omega^{\nu}_{~\beta}\,\gamma^{\,\beta} = -\frac{\rm i}{4}\... ...{\,\nu}\,\sigma_{\alpha\,\beta}- \sigma_{\alpha\,\beta}\,\gamma^{\,\nu}\right).$$ (11.74)

Hence, making use of the symmetry property (11.68), we obtain

$$\delta\omega^{\,\alpha\,\beta}\,(g^\nu_{~\alpha}\,\gamma_\beta -g... ...\gamma^{\,\nu}\,\sigma_{\alpha\,\beta}-\sigma_{\alpha\,\beta}\,\gamma^{\,\nu}),$$ (11.75)

where $\gamma_\mu = g_{\mu\,\nu}\,\gamma^{\,\nu}$ . Because this equation must hold for arbitrary $\delta\omega^{\,\alpha\,\beta}$ , we deduce that

$$2\,{\rm i}\,(g^\nu_{~\alpha}\,\gamma_\beta -g^\nu_{~\beta}\,\gamma_\alpha) = [\gamma^{\,\nu}, \sigma_{\alpha\,\beta}].$$ (11.76)

Making use of the anti-commutation relations (11.29), it can be shown that a suitable solution of the previous equation is

$$\sigma_{\mu\,\nu} = \frac{{\rm i}}{2}\,[\gamma_\mu,\gamma_\nu].$$ (11.77)

(See Exercise 7.) Hence,

$$A = 1 + \frac{1}{8}\,[\gamma_\mu,\gamma_\nu]\,\delta\omega^{\,\mu\,\nu},$$ (11.78)

$$A^{-1} = 1 - \frac{1}{8}\,[\gamma_\mu,\gamma_\nu]\,\delta\omega^{\,\mu\,\nu}.$$ (11.79)

Now that we have found the correct transformation rules for an infinitesimal Lorentz transformation, we can easily find those for a finite transformation by building it up from a large number of successive infinitesimal transforms [9]. (See Exercises 8 and 9.)

Making use of Equation (11.34), as well as $\gamma^{\,0}\,\gamma^{\,0}=1$ , the Hermitian conjugate of Equation (11.78) can be shown to take the form

$$A^\dag = 1-\frac{1}{8}\,\gamma^{\,0}\,[\gamma_\mu,\gamma_\nu]\,\gamma^{\,0}\,\delta\omega^{\,\mu\,\nu} = \gamma^{\,0}\,A^{-1}\,\gamma^{\,0}.$$ (11.80)

Hence, Equation (11.65) yields

$$A^\dag\,\gamma^{\,0}\,\gamma^{\,\mu}\,A = a^{\,\mu}_{~\nu}\,\gamma^{\,0}\,\gamma^{\,\nu}.$$ (11.81)

It follows that

$$\psi^\dag\,A^\dag\,\gamma^{\,0}\,\gamma^{\,\mu}\,A\,\psi= a^{\,\mu}_{~\nu}\,\psi^\dag\,\gamma^{\,0}\,\gamma^{\,\nu}\,\psi,$$ (11.82)

or

$$\psi^{\dag '}\,\gamma^{\,0}\,\gamma^{\,\mu}\,\psi'= a^{\,\mu}_{~\nu}\,\psi^\dag\,\gamma^{\,0}\,\gamma^{\,\nu}\,\psi,$$ (11.83)

which implies that

$$j^{\,\mu'} = a^{\,\mu}_{~\nu}\,j^{\,\nu},$$ (11.84)

where the $j^{\,\mu}$ are defined in Equation (11.46). This proves that the $j^{\,\mu}$ transform as the contravariant components of a 4-vector.