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11.3: 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

 

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

 

 

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

 

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

 

 

Now, since [see Equation (1102)]

 

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

 

 

it follows that

 

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

 

 

Moreover, it is easily shown that

 

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

 

 

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

 

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

 

 

etc.

In frame $ S$ , the Dirac equation is written

 

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

 

 

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

 

$\displaystyle \psi' = A\,\psi,$ (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

 

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

 

 

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

 

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

 

 

Suppose that

 

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

 

 

which is equivalent to

 

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

 

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

 

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

 

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

 

 

Let us write

 

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

 

 

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

 

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

 

 

Moreover, it follows from (1161) that

 

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

 

 

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

 

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

 

 

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

 

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

 

 

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

 

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

 

 

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

 

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

 

 

Hence,

 

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

 

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

 

 

Hence, (1158) yields

 

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

 

 

It follows that

 

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

 

 

or

 

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

 

which implies that

 

$\displaystyle j^{\,\mu'} = a^{\,\mu}_{~\nu}\,j^{\,\nu},$ (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.

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