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

$$x^{\,\mu'} = a^{\,\mu}_{~\nu}\,x^\nu,$$ (1145)

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

$$x_{\mu'} = a_{\mu}^{~\nu}\,x_\nu.$$ (1146)

Now, since [see Equation (1102)]

$$x^{\,\mu'}\,x_{\mu'} = x^{\,\mu}\,x_\mu,$$ (1147)

it follows that

$$a^{\,\mu}_{~\nu}\,a_\mu^{~\lambda} = g_\nu^{~\lambda}.$$ (1148)

Moreover, it is easily shown that

$$x^{\,\mu} = a_\nu^{~\mu}\,x^{\nu'},$$ (1149)

$$x_\mu = a^\nu_{~\mu}\,x_{\nu'}.$$ (1150)

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

$$p^{\,\mu'} = a^{\,\mu}_{~\nu}\,p^\nu,$$ (1151)

$$p^{\,\mu} = a_\nu^{~\mu}\,p^{\nu'},$$ (1152)

etc.

In frame $S$ , the Dirac equation is written

$$\left[\gamma^{\,\mu}\left(p_\mu- \frac{e}{c}\,{\mit\Phi}_\mu\right)-m_e\,c\right]\psi = 0.$$ (1153)

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

$$\psi' = A\,\psi,$$ (1154)

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 (1153) by $A$ , we obtain

$$\left[A\,\gamma^{\,\mu}\,A^{-1}\left(p_\mu- \frac{e}{c}\,{\mit\Phi}_\mu\right)-m_e\,c\right]\psi' = 0.$$ (1155)

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

$$\left[A\,\gamma^{\,\mu}\,A^{-1}\,a^\nu_{~\mu}\left(p_{\nu'}- \frac{e}{c}\,{\mit\Phi}_{\nu'}\right)-m_e\,c\right]\psi' = 0.$$ (1156)

Suppose that

$$A\,\gamma^{\,\mu}\,A^{-1}\,a^\nu_{~\mu} = \gamma^\nu,$$ (1157)

which is equivalent to

$$A^{-1}\,\gamma^\nu\,A = a^\nu_{~\mu}\,\gamma^{\,\mu}.$$ (1158)

Here, we have assumed that the $a^\nu_{~\mu}$ commute with $A$ and the $\gamma^{\,\mu}$ (since they are just numbers). If (1157) holds then (1156) becomes

$$\left[\gamma^{\,\mu}\left(p_{\mu'}- \frac{e}{c}\,{\mit\Phi}_{\mu'}\right)-m_e\,c\right]\psi' = 0.$$ (1159)

A comparison of this equation with (1153) 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 (1153) and (1159) that the $\gamma^{\,\mu}$ matrices are the same in all inertial frames.

It remains to find a transformation matrix $A$ that satisfies (1158). Consider an infinitesimal Lorentz transformation, for which

$$a_\mu^{~\nu} = g_\mu^{~\nu} + {\mit\Delta}\omega_\mu^{~\nu},$$ (1160)

where the ${\mit\Delta}\omega_\mu^{~\nu}$ are real numerical coefficients that are independent of the $x^{\,\mu}$ , and are also small compared to unity. To first order in small quantities, (1148) yields

$${\mit\Delta}\omega^{\,\mu\,\nu} + {\mit\Delta}\omega^{\nu\,\mu} = 0.$$ (1161)

Let us write

$$A = 1 - \frac{{\rm i}}{4}\,\sigma_{\mu\,\nu}\,{\mit\Delta}\omega^{\,\mu\,\nu},$$ (1162)

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

$$A^{-1} = 1 + \frac{{\rm i}}{4}\,\sigma_{\mu\,\nu}\,{\mit\Delta}\omega^{\,\mu\,\nu}.$$ (1163)

Moreover, it follows from (1161) that

$$\sigma_{\mu\,\nu} = -\sigma_{\nu\,\mu}.$$ (1164)

To first order in small quantities, Equations (1158), (1160), (1162), and (1163) yield

$${\mit\Delta}\omega^\nu_{~\beta}\,\gamma^{\,\beta} = -\frac{\rm i}... ...(\gamma^\nu\,\sigma_{\alpha\,\beta}- \sigma_{\alpha\,\beta}\,\gamma^\nu\right).$$ (1165)

Hence, making use of the symmetry property (1161), we obtain

$${\mit\Delta}\omega^{\alpha\,\beta}\,(g^\nu_{~\alpha}\,\gamma_\bet... ...beta}\,(\gamma^\nu\,\sigma_{\alpha\,\beta}-\sigma_{\alpha\,\beta}\,\gamma^\nu),$$ (1166)

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

$$2\,{\rm i}\,(g^\nu_{~\alpha}\,\gamma_\beta -g^\nu_{~\beta}\,\gamma_\alpha) = [\gamma^\nu, \sigma_{\alpha\,\beta}].$$ (1167)

Making use of the anti-commutation relations (1122), it can be shown that a suitable solution of the above equation is

$$\sigma_{\mu\,\nu} = \frac{{\rm i}}{2}\,[\gamma_\mu,\gamma_\nu].$$ (1168)

Hence,

$$A = 1 + \frac{1}{8}\,[\gamma_\mu,\gamma_\nu]\,{\mit\Delta}\omega^{\,\mu\,\nu},$$ (1169)

$$A^{-1} = 1 - \frac{1}{8}\,[\gamma_\mu,\gamma_\nu]\,{\mit\Delta}\omega^{\,\mu\,\nu}.$$ (1170)

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.

Making use of (1127), as well as $\gamma^0\,\gamma^0=1$ , the Hermitian conjugate of (1169) can be shown to take the form

$$A^\dag = 1-\frac{1}{8}\,\gamma^0\,[\gamma_\mu,\gamma_\nu]\,\gamma^0\,{\mit\Delta}\omega^{\,\mu\,\nu} = \gamma^0\,A^{-1}\,\gamma^0.$$ (1171)

Hence, (1158) yields

$$A^\dag\,\gamma^0\,\gamma^{\,\mu}\,A = a^\mu_{~\nu}\,\gamma^0\,\gamma^\nu.$$ (1172)

It follows that

$$\psi^\dag\,A^\dag\,\gamma^0\,\gamma^{\,\mu}\,A\,\psi= a^{\,\mu}_{~\nu}\,\psi^\dag\,\gamma^0\,\gamma^\nu\,\psi,$$ (1173)

or

$$\psi'^{\dag }\,\gamma^0\,\gamma^{\,\mu}\,\psi'= a^{\,\mu}_{~\nu}\,\psi^\dag\,\gamma^0\,\gamma^\nu\,\psi,$$ (1174)

which implies that

$$j^{\,\mu'} = a^{\,\mu}_{~\nu}\,j^{\,\nu},$$ (1175)

where the $j^{\,\mu}$ are defined in Equation (1139). This proves that the $j^{\,\mu}$ transform as the contravariant components of a 4-vector.