11.3: Lorentz Invariance of Dirac Equation

Consider two inertial frames, \( S'\) . Let the \( x^{\,\mu'}\) 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

\( a^{\,\mu}_{~\nu}\) are real numerical coefficients that are independent of the \( x_{\mu'} = a_{\mu}^{~\nu}\,x_\nu.\) \ref{1146}

Now, since [see Equation \ref{1102}]

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

Moreover, it is easily shown that

\( = a_\nu^{~\mu}\,x^{\nu'},\) \ref{1149} \( = a^\nu_{~\mu}\,x_{\nu'}.\) \ref{1150}

By definition, a 4-vector \( x^{\,\mu}\) . Thus,

\( = a^{\,\mu}_{~\nu}\,p^\nu,\) \ref{1151} \( = a_\nu^{~\mu}\,p^{\nu'},\) \ref{1152}

etc.

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

Let \( S'\) . Suppose that

\( A\) is a \( x^{\,\mu}\) . (Hence, \( A\) commutes with the \( {\mit\Phi}_\mu\) .) Multiplying \ref{1153} by \( A\) , we obtain

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

Suppose that

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

Here, we have assumed that the \( A\) and the \( \left[\gamma^{\,\mu}\left(p_{\mu'}- \frac{e}{c}\,{\mit\Phi}_{\mu'}\right)-m_e\,c\right]\psi' = 0.\) \ref{1159}

A comparison of this equation with \ref{1153} reveals that the Dirac equation takes the same form in frames \( S'\) . In other words, the Dirac equation is Lorentz invariant. Incidentally, it is clear from \ref{1153} and \ref{1159} that the \( A\) that satisfies \ref{1158}. Consider an infinitesimal Lorentz transformation, for which

\( {\mit\Delta}\omega_\mu^{~\nu}\) are real numerical coefficients that are independent of the \( {\mit\Delta}\omega^{\,\mu\,\nu} + {\mit\Delta}\omega^{\nu\,\mu} = 0.\) \ref{1161}

Let us write

\( \sigma_{\mu\,\nu}\) are \( 4\times 4\) matrices. To first order in small quantities,

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

To first order in small quantities, Equations \ref{1158}, \ref{1160}, \ref{1162}, and \ref{1163} yield

\ref{1165}

Hence, making use of the symmetry property \ref{1161}, we obtain

\ref{1166}

where \( {\mit\Delta}\omega^{\alpha\,\beta}\) , we deduce that

\( \sigma_{\mu\,\nu} = \frac

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{2}\,[\gamma_\mu,\gamma_\nu].\) \ref{1168}

Hence,

\( = 1 + \frac{1}{8}\,[\gamma_\mu,\gamma_\nu]\,{\mit\Delta}\omega^{\,\mu\,\nu},\) \ref{1169} \( = 1 - \frac{1}{8}\,[\gamma_\mu,\gamma_\nu]\,{\mit\Delta}\omega^{\,\mu\,\nu}.\) \ref{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 \ref{1127}, as well as \( A^\dag = 1-\frac{1}{8}\,\gamma^0\,[\gamma_\mu,\gamma_\nu]\,\gamma^0\,{\mit\Delta}\omega^{\,\mu\,\nu} = \gamma^0\,A^{-1}\,\gamma^0.\) \ref{1171}

Hence, \ref{1158} yields

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

or

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

where the \( j^{\,\mu}\) transform as the contravariant components of a 4-vector.

Contributors

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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