On the Second Dipole Moment of Dirac’s Particle

Abstract

An analysis is presented of the possible existence of the second anomalous dipole moment of Dirac’s particle next to the one associated with the angular momentum. It includes a discussion why, in spite of his own derivation, Dirac has doubted about its relevancy. It is shown why since then it has been overlooked and why it has vanished from leading textbooks. A critical survey is given on the reasons of its reject, including the failure of attempts to measure and the perceived violations of time reversal symmetry and charge–parity symmetry. It is emphasized that the anomalous electric dipole moment of the pointlike electron (AEDM) is fundamentally different from the quantum field type electric dipole moment of an electron (eEDM) as defined in the standard model of particle physics. The analysis has resulted into the identification of a third type Dirac particle, next to the electron type and the Majorana particle. It is shown that, unlike as in the case of the electron type, its second anomalous dipole moment is real valued and is therefore subject to polarization in a scalar potential field. Examples are given that it may have a possible impact in the nuclear domain and in the gravitational domain.

Appendix: Derivation of Dirac’s Anomalous Electric Dipole Moment (AEDM)

The aim in this “Appendix” is to give a refreshment of Dirac’s analysis that has resulted into the conclusion that a Dirac particle, in this “Appendix” not necessarily an electron, possesses two anomalous dipole moments, both purely quantum mechanical in nature. It is a generalization, though, with respect to two aspects. The first one is showing its relevancy beyond electromagnetism and the second one is showing a second option for Dirac’s decomposition of the squared Einsteinean energy expression, thereby revealing a fundamental difference in the properties of the second anomalous dipole moment. To symmetrise the analysis, exclusively within the scope of this “Appendix”, the Hawking metric $( + ,\; + ,\; + ,\; + )$ for (${\text{i}}ct,\;x,\,y,\;z$), ${\text{i}} = \sqrt { - 1}$ will be adopted and justified later by showing that the final result is the same as in the conventional metric $( - ,\; + ,\; + ,\; + )$ for ($ct,\;x,\;y,\;z$_).

It all starts from the Einsteinean energy expression of a generic free moving particle with rest mass $m_{0} .$ This reads as,

$$ E_{W} = \sqrt {(m_{0} c^{2} )^{2} + (c\left| {\mathbf{p}} \right|)^{2} } , $$

(41)

in which p is the three-vector momentum (${\text{d}}s/{\text{d}}t,$ not be confused with the four-vector momentum _p). Under adoption of the Hawking metric

$$ E_{W}^{2} = - p_{00}^{2} c^{2} = (m_{0} c^{2} )^{2} + c^{2} p_{1}^{2} + c^{2} p_{2}^{2} + c^{2} p_{3}^{2} , $$

(42)

which can be normalized as,

$$ p^{\prime 2}_{00} + p^{\prime 2}_{1} + p^{\prime 2}_{2} + p^{\prime 2}_{3} + 1 = 0;\quad p^{\prime}_{\upmu } = \frac{{p_{\upmu } }}{{m_{0} c}}. $$

(43)

As long as the temporal dimension is included, the bold italic notation for the vector p will be maintained.

Note: in the Hawking metric, time shows up as an imaginary quantity [29]. The merit of it is the full symmetry over the four dimensions as shown by (43). In most textbooks a preference is given to real time, hence a metric $( - ,\, + ,\, + ,\, + )$ for ($ct,\;x,\;y,\;z$). Perkins [30] prefers the Hawking metric. As will be shown, it simplifies Dirac’s analysis substantially. The main difference is the notation $p^{\prime}_{00}$ for the temporal momentum instead of $p^{\prime}_{0} ,$ in which $p^{\prime 2}_{00} = - p^{\prime 2}_{0} .$

Under particular number typing of a coefficient vector $\overline{\upalpha }(\upalpha_{0} ,\;\upalpha_{1} ,\;\upalpha_{2} ,\;\upalpha_{3} ),$, Eq. (43) can be factorized as,

$$ (\overline{\upalpha } \cdot {\user2{p}}^{\prime} + \upbeta ) \cdot (\overline{\upalpha } \cdot {\user2{p}}^{\prime} - \upbeta ) = (\overline{\upalpha } \cdot {\user2{p}}^{\prime}) \cdot (\overline{\upalpha } \cdot {\user2{p}}^{\prime}) - \upbeta^{2} = 0. $$

(44a)

Another possibility is full squaring as,

$$ (\overline{\upalpha } \cdot {\user2{p}}^{\prime} + \upbeta ) \cdot (\overline{\upalpha } \cdot {\user2{p}}^{\prime} + \upbeta ) = 0. $$

(44b)

These reflect the energy relationship (43) under particular conditions. For the first option,

$$ \upalpha_{\upmu } \upalpha_{\upnu } + \upalpha_{\upnu } \upalpha_{\upmu } = 0\quad {\text{if}}\quad \upmu \ne \upnu ;\quad {\text{and}}\quad \upalpha_{0}^{2} = 1\quad \upalpha_{i}^{2} = 1;\quad \upbeta^{2} = - 1, $$

(45a)

while for the second option,

$$ \upalpha_{\upmu } \upalpha_{\upnu } + \upalpha_{\upnu } \upalpha_{\upmu } = 0\quad {\text{if}}\quad \upmu \ne \upnu ;\quad \upalpha_{\upmu } \upbeta + \upbeta \upalpha_{\upmu } = 0\quad {\text{and}}\quad \upalpha_{0}^{2} = 1;\quad \upalpha_{i}^{2} = 1;\quad \upbeta^{2} = 1. $$

(45b)

The first option can be met if a coefficient vector $\overline{\upalpha }(\upalpha_{0} ,\;\upalpha_{1} ,\;\upalpha_{2} ,\;\upalpha_{3} )$ is constructed from the gamma matrices $\upgamma_{\upmu }$ shown before in Eq. (5) of the main text, supplemented by the identity matrix $I_{4} ,$ such that

$$ \upalpha_{1} = - {\text{i}}\upgamma_{1} ;\quad \upalpha_{2} = - {\text{i}}\upgamma_{2} ;\quad \upalpha_{3} = - {\text{i}}\upgamma_{3} ;\quad \upalpha_{0} = \upgamma_{0} ;\quad \upbeta = {\text{i}}I_{4} ; $$

$$ I_{4} = \left[ {\begin{array}{*{20}c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} I & 0 \\ 0 & I \\ \end{array} } \right]. $$

(46a)

The second option requires a different choice for the gamma matrices $\upgamma_{\upmu }$ and $I_{4} ,$ heuristically found as,

$$ \upgamma_{0} = \left[ {\begin{array}{*{20}c} 0 & { - I} \\ { - I} & 0 \\ \end{array} } \right];\quad \upgamma_{1} = \left[ {\begin{array}{*{20}c} {{\text{i}}\upsigma_{1} } & 0 \\ 0 & { - {\text{i}}\upsigma_{1} } \\ \end{array} } \right];\quad \upgamma_{2} = \left[ {\begin{array}{*{20}c} {{\text{i}}\upsigma_{2} } & 0 \\ 0 & { - {\text{i}}\upsigma {}_{2}} \\ \end{array} } \right];\quad \upgamma_{3} = \left[ {\begin{array}{*{20}c} {{\text{i}}\upsigma_{3} } & 0 \\ 0 & { - {\text{i}}\upsigma_{3} } \\ \end{array} } \right];\quad I^{\prime}_{4} = \left[ {\begin{array}{*{20}c} 0 & I \\ { - I} & 0 \\ \end{array} } \right]. $$

(46b)

The reader may verify that under this choice condition (45) is met. In both cases, we have,

$$ \upgamma_{\upmu } \upgamma_{\upnu } + \upgamma_{\upnu } \upgamma_{\upmu } = 0\quad {\text{if}}\quad \upmu \ne \upnu ;\quad {\text{and}}\quad \upgamma_{0}^{2} = 1\quad \upgamma_{i}^{2} = - 1. $$

(47)

Note the sign difference between (47) and (45).

Because of (45), for both cases,

$$ \begin{aligned} (\overline{\upalpha } \cdot {\user2{p}}^{\prime}) \cdot (\overline{\upalpha } \cdot {\user2{p}}^{\prime}) = (\upalpha_{0} p^{\prime}_{00} + \upalpha_{1} p^{\prime}_{1} + \upalpha_{2} p^{\prime}_{2} + \upalpha_{3} p^{\prime}) \cdot (\upalpha_{0} p^{\prime}_{00} + \upalpha_{1} p^{\prime}_{1} + \upalpha_{2} p^{\prime}_{2} + \upalpha_{3} p^{\prime}) \\ & = (\upalpha_{0} p^{\prime}_{00} )^{2} + (\upalpha_{0} p^{\prime}_{00} )(\upalpha_{1} p^{\prime}_{1} + \upalpha_{2} p^{\prime}_{2} + \upalpha_{3} p^{\prime}_{3} ) \\ & \quad + (\upalpha_{1} p^{\prime}_{1} )^{2} + (\upalpha_{1} p^{\prime}_{1} )(\upalpha_{0} p^{\prime}_{00} + \upalpha_{2} p^{\prime}_{2} + \upalpha_{3} p^{\prime}_{3} ) \\ & \quad + (\upalpha_{2} p^{\prime}_{2} )^{2} + (\upalpha_{2} p^{\prime}_{2} )(\upalpha_{0} p^{\prime}_{00} + \upalpha_{1} p^{\prime}_{1} + \upalpha_{3} p^{\prime}_{3} ) \\ & \quad + (\upalpha_{3} p^{\prime}_{3} )^{2} + (\upalpha_{3} p^{\prime}_{3} )(\upalpha_{0} p^{\prime}_{00} + \upalpha_{1} p^{\prime}_{1} + \upalpha_{2} p^{\prime}_{2} ) = U^{2} + \varepsilon , \\ \end{aligned} $$

(48)

in which

$$ U^{2} = (\upalpha_{0} p^{\prime}_{00} )^{2} + (\upalpha_{1} p^{\prime}_{1} )^{2} + (\upalpha_{2} p^{\prime}_{2} )^{2} + (\upalpha_{3} p^{\prime}_{3} )^{2} ,\;{\text{and}} $$

$$ \begin{aligned} \varepsilon & = \{ (a_{0} p^{\prime}_{00} )(a_{1} p^{\prime}_{1} ) + (a_{1} p^{\prime}_{1} )(a_{0} p^{\prime}_{00} )\} + \{ (a_{0} p^{\prime}_{00} )(a_{2} p^{\prime}_{2} ) + (a_{2} p^{\prime}_{2} )(a_{0} p^{\prime}_{00} )\} \\ & \quad + \{ (a_{0} p^{\prime}_{00} )(a_{3} p^{\prime}_{3} ) + (a_{3} p^{\prime}_{3} )(a_{0} p^{\prime}_{00} )\} + \{ (a_{1} p^{\prime}_{1} )(a_{2} p^{\prime}_{2} ) + (a_{2} p^{\prime}_{2} )(a_{1} p^{\prime}_{1} )\} \\ & \quad + \{ (a_{1} p^{\prime}_{1} )(a_{3} p^{\prime}_{3} ) + (a_{3} p^{\prime}_{3} )(a_{1} p^{\prime}_{1} )\} + \{ (a_{2} p^{\prime}_{2} )(a_{3} p^{\prime}_{3} ) + (a_{3} p^{\prime}_{3} )(a_{2} p^{\prime}_{2} )\} . \\ \end{aligned} $$

Hence, for the first option (44), we have

$$ (\overline{\upalpha } \cdot {\user2{p}}^{\prime}) \cdot (\overline{\upalpha } \cdot {\user2{p}}^{\prime}) - \upbeta^{2} = U^{2} + \varepsilon + 1, $$

(49a)

while for the second option,

$$ (\overline{\upalpha } \cdot {\user2{p}}^{\prime} + \upbeta ) \cdot (\overline{\upalpha } \cdot {\user2{p}}^{\prime} + \upbeta ) = U^{2} + \varepsilon + \delta + 1,\;{\text{in}}\;{\text{which}} $$

(49b)

$$ \delta = \upbeta (\overline{\upalpha } \cdot {\user2{p}}^{\prime}) + (\overline{\upalpha } \cdot {\user2{p}}^{\prime})\upbeta . $$

Obviously, $\varepsilon = 0$ as well as $\delta = 0,$ because of (45). This remains so for a particle moving under influence of a conservative field of forces with a (generic) field potential ${\user2{A}}^{\prime}(A^{\prime}_{0} ,\;A^{\prime}_{1} ,\;A^{\prime}_{2} ,\;A^{\prime}_{3} ).$ As before, ${\user2{A}}^{\prime}$ is signed for indicating the normalization by $m_{0} c.$ The field influence can be accounted for by,

$$ p^{\prime}_{\upmu } \to p^{\prime}_{\upmu } + A^{\prime}_{\upmu } . $$

(50)

The triviality $\varepsilon = 0$ disappears if the momenta are transformed into wave operators, like Dirac did by adopting the basic transform of quantum electrodynamics (QED),

$$ p^{\prime}_{\upmu } \to \hat{p}_{\upmu } \uppsi \quad {\text{with}}\quad \hat{p}^{\prime}_{\upmu } = \frac{1}{{m_{0} c}}\frac{\hbar }{{\text{i}}}\frac{\partial }{{\partial x_{\upmu } }}. $$

(51)

Note: $\hat{p}^{\prime}_{00} = \frac{1}{{m_{0} c}}\frac{\hbar }{{\text{i}}}\frac{\partial }{{\partial ({\text{i}}c\uptau )}},$ in which $\uptau$ is proper time, i.e., time in the center of frame.

As a consequence of the QED transform (51) and the minimum substitution rule (50), together known as the gauge covariant transform, the first term in $\varepsilon$ of (48) transforms as,

$$ a_{0} a_{1} (p^{\prime}_{00} + A^{\prime}_{0} )(\hat{p}^{\prime}_{1} + A^{\prime}_{1} ) + a_{1} a_{0} (\hat{p}^{\prime}_{1} + A_{1} )(\hat{p^{\prime}}_{00} + A^{\prime}_{0} ) = a_{0} a_{1} \hat{p^{\prime}}_{00} A^{\prime}_{1} + a_{1} a_{0} \hat{p}^{\prime}_{1} A^{\prime}_{0} . $$

(52)

Note that, quite some terms in $\varepsilon$ have disappeared because of $a_{0} a_{1} = - a_{1} a_{0} ,$ see (45) and, more importantly now, unlike in $\delta ,$ some remain because of the sequence sensitivity of the operator action.

Applying this on all terms of (48), the result is,

$$ \begin{aligned} \varepsilon & = \hat{p}^{\prime}_{00} (a_{0} a_{1} A^{\prime}_{1} + a_{0} a_{2} A^{\prime}_{2} + a_{0} a_{3} A^{\prime}_{3} ) + \{ a_{1} a_{0} \hat{p}^{\prime}_{1} A^{\prime}_{0} + a_{2} a_{0} \hat{p}^{\prime}_{2} A^{\prime}_{0} + a_{3} a_{0} \hat{p}^{\prime}_{3} A^{\prime}_{0} \} \\ & \quad + \{ \hat{p}^{\prime}_{1} (a{}_{1}a_{2} A^{\prime}_{2} ) + \hat{p}^{\prime}_{2} (a{}_{2}a_{1} A^{\prime}_{1} \} + \{ \hat{p}^{\prime}_{1} (a{}_{1}a_{3} A^{\prime}_{3} ) + \hat{p}^{\prime}_{3} (a_{3} a_{1} A^{\prime}_{1} )\} \\ & \quad + \{ \hat{p}^{\prime}_{2} (a{}_{2}a_{3} A^{\prime}_{3} ) + \hat{p}^{\prime}_{3} (a_{3} a_{2} A^{\prime}_{2} )\} \\ \end{aligned} $$

which can be rewritten as,

$$ \begin{aligned} \varepsilon & = (a_{0} a_{1} \hat{p}^{\prime}_{00} A^{\prime}_{1} + a_{1} a_{0} \hat{p}^{\prime}_{1} A^{\prime}_{0} ) + (a_{0} a_{2} \hat{p}^{\prime}_{00} A^{\prime}_{2} + a_{2} a_{0} \hat{p}^{\prime}_{00} A^{\prime}_{0} ) \\ & \quad + (a_{0} a_{3} \hat{p}^{\prime}_{00} A^{\prime}_{3} + a_{3} a_{0} \hat{p}^{\prime}_{00} A^{\prime}_{0} ) + (a_{1} a_{2} \hat{p}^{\prime}_{1} A^{\prime}_{2} + a_{2} a_{1} \hat{p}^{\prime}_{2} A^{\prime}_{1} ) \\ & \quad + (a_{1} a_{3} \hat{p}^{\prime}_{1} A^{\prime}_{3} + a_{3} a_{1} \hat{p}^{\prime}_{3} A^{\prime}_{1} ) + (a_{2} a_{3} \hat{p}^{\prime}_{2} A^{\prime}_{3} + a_{3} a_{2} \hat{p}^{\prime}_{3} A^{\prime}_{2} ). \\ \end{aligned} $$

(53)

Regrouping under consideration of (45) gives,

$$ \begin{aligned} \varepsilon & = a_{0} a_{1} (\hat{p}^{\prime}_{00} A^{\prime}_{1} - \hat{p}^{\prime}_{1} A^{\prime}_{0} ) + a_{0} a_{2} (\hat{p}^{\prime}_{00} A^{\prime}_{2} - \hat{p}^{\prime}_{2} A^{\prime}_{0} ) \\ & \quad + a_{0} a_{3} (\hat{p}^{\prime}_{00} A^{\prime}_{3} - \hat{p}^{\prime}_{00} A^{\prime}_{0} ) + a_{1} a_{2} (\hat{p}^{\prime}_{1} A^{\prime}_{2} - \hat{p}^{\prime}_{2} A^{\prime}_{1} ) \\ & \quad + a_{1} a_{3} (\hat{p}^{\prime}_{1} A^{\prime}_{3} - \hat{p}^{\prime}_{3} A^{\prime}_{1} ) + a_{2} a_{3} (\hat{p}^{\prime}_{2} A^{\prime}_{3} - \hat{p}^{\prime}_{3} A^{\prime}_{2} ). \\ \end{aligned} $$

(54)

Hence, from (54) and (51),

$$ \begin{aligned} & \varepsilon = \frac{\hbar }{{{\text{i}}m_{0} c}}\left\{ {\upsigma_{E1} \left( {\frac{{\partial A^{\prime}_{0} }}{\partial x} - \frac{{\partial A^{\prime}_{1} }}{{\partial ({\text{i}}c\uptau )}}} \right) + \upsigma_{E2} \left( {\frac{{\partial A^{\prime}_{0} }}{\partial y} - \frac{{\partial A^{\prime}_{2} }}{{\partial ({\text{i}}c\uptau )}}} \right) + \upsigma_{E3} \left( {\frac{{\partial A^{\prime}_{0} }}{\partial z} - \frac{{\partial A^{\prime}_{3} }}{{\partial ({\text{i}}c\uptau )}}} \right)} \right\} \\ & \quad + \frac{\hbar }{{{\text{i}}m_{0} c}}\left\{ {\upsigma_{B1} \left( {\frac{\partial }{\partial x}A^{\prime}_{y} - \frac{\partial }{\partial y}A^{\prime}_{x} } \right) - \upsigma_{B2} \left( {\frac{\partial }{\partial x}A^{\prime}_{z} - \frac{\partial }{\partial z}A^{\prime}_{x} } \right) + \upsigma_{B3} \left( {\frac{\partial }{\partial y}A^{\prime}_{z} - \frac{\partial }{\partial z}A^{\prime}_{y} } \right)} \right\}; \\ & \upsigma_{E1} = a_{0} a_{1} ;\quad \upsigma_{E2} = a_{0} a_{2} ;\quad \upsigma_{E3} = a_{0} a_{3} ; \\ & \upsigma_{B1} = a_{1} a_{2} ;\quad \upsigma_{B2} = a_{1} a_{3} ;\quad \upsigma_{B3} = a_{2} a_{3} . \\ \end{aligned} $$

(55)

It can be written in terms of the grad operator, the curl operator and Dirac’s–Pauli vector as

$$ \varepsilon = \frac{\hbar }{{{\text{i}}m_{0} c}}\left( {\overline{\upsigma }_{E} \cdot \nabla A^{\prime}_{0} - \overline{\upsigma }_{E} \cdot \frac{{\partial {\mathbf{A^{\prime}}}}}{{\partial ({\text{i}}c\uptau )}}} \right) + \frac{\hbar }{{{\text{i}}m_{0} c}}\overline{\upsigma }_{B} \cdot (\nabla \times {\mathbf{A^{\prime}}}). $$

(56)

Note that $\varepsilon$ still is a dimensionless quantity. It is an excess term to be included in the energy expression as a consequence of the particular characteristics of Dirac’s equation of motion. Hence,

$$ \frac{{E_{W}^{2} }}{{(m_{0} c^{2} )^{2} }} = 1 + \frac{{v^{2} }}{{c^{2} }} + \varepsilon , $$

(57)

in which $v$ is the velocity of the particle in motion. As long as $v/c \ll 1$ and $\varepsilon \ll 1,$, $E_{W}$ can be approximated as,

$$ E_{W} = m_{0} c^{2} \left( {1 + \frac{{v^{2} }}{{2c^{2} }} + \frac{\varepsilon }{2}} \right) = m_{0} c^{2} \left( {1 + \frac{{v^{2} }}{{2c^{2} }}} \right) + \Delta E;\quad \Delta E = \varepsilon \frac{{m_{0} c^{2} }}{2}, $$

(58)

in which $\varepsilon$ is given by (56).

We are almost done, but not quite. So far, the Dirac particle has been considered in general terms, i.e., without identifying it as an electron. To do so, a first step to do so is defining the four-vector potential as,

$$ {\user2{A}}^{\prime} = {\user2{A}}^{\prime}\left( {{\text{i}}\frac{\Phi /c}{{m_{0} c}},\;A^{\prime}_{x} ,\;A^{\prime}_{y} ,\;A^{\prime}_{z} } \right). $$

(59)

Note: the ${\text{i}}$ factor in the scalar component is due to the (Hawking) metric choice (+, +, +, +)/(ict, x, y, z). It can be easily seen from the Lorenz gauge

$$ \nabla \cdot {\mathbf{A}} + \frac{1}{{c^{2} }}\frac{\partial \Phi }{{\partial t}} = 0 \to \nabla \cdot {\mathbf{A}} + {\text{i}}\frac{\partial \Phi /c}{{\partial {\text{i}}ct}} = 0. $$

(60)

Note also that in (59) the dimension of $\Phi$ is energy. It is not the same as the electric potential $\Phi_{{\text{e}}} .$ The relationship between the two can be found from the force equity,

$$ F = e\frac{\partial }{\partial y}\Phi_{{\text{e}}} = \frac{\partial }{\partial y}\Phi \to \Phi_{{\text{e}}} = \frac{\Phi }{e}. $$

(61)

From the Lorenz gauge and (61) obviously,

$$ {\mathbf{A}}_{{\text{e}}} = \frac{{\mathbf{A}}}{e}. $$

(62)

Hence, from (58), (62) and (56),

$$ \begin{aligned} \Delta E & = \varepsilon \frac{{m_{0} c^{2} }}{2} = \frac{\hbar c}{{2{\text{i}}}}\left( {\overline{\upsigma }_{E} \cdot \nabla A^{\prime}_{0} - \overline{\upsigma }_{E} \cdot \frac{{\partial {\mathbf{A^{\prime}}}}}{{\partial ({\text{i}}c\uptau )}}} \right) + \frac{\hbar c}{{2{\text{i}}}}\overline{\upsigma }_{B} \cdot (\nabla \times {\mathbf{A^{\prime}}}) \\ & = \frac{\hbar c}{{2{\text{i}}}}\overline{\upsigma }_{E} \cdot \nabla \left( {e\frac{{{\text{i}}\Phi_{{\text{e}}} /c}}{{m_{0} c}} - \frac{{\partial {\mathbf{A^{\prime}}}}}{{\partial ({\text{i}}c\uptau )}}} \right) + \frac{\hbar c}{{2{\text{i}}}}\overline{\upsigma }_{B} \cdot (\nabla \times {\mathbf{A^{\prime}}}) = \frac{e\hbar }{{2m_{0} c}}( - \overline{\upsigma }_{E} \cdot {\mathbf{E}}) - \frac{e\hbar }{{2m_{0} }}{\text{i}}(\overline{\upsigma }_{B} \cdot {\mathbf{B}}). \\ \end{aligned} $$

(63)

From (55) and (46),

$$ \begin{aligned} & \upsigma_{E1} = a_{0} a_{1} ;\quad \upsigma_{E2} = a_{0} a_{2} ;\quad \upsigma_{E3} = a_{0} a_{3} ;\quad \to \upsigma_{E1} = - {\text{i}}\upgamma_{0} \upgamma_{1} ;\quad \upsigma_{E2} = - {\text{i}}\upgamma_{0} \upgamma_{2} ;\quad \upsigma_{E3} = - {\text{i}}\upgamma_{0} \upgamma_{3} , \\ & \upsigma_{B1} = a_{1} a_{2} ;\quad \upsigma_{B2} = - a_{1} a_{3} ;\quad \upsigma_{B3} = a_{2} a_{3} ;\quad \to \upsigma_{B1} = - \upgamma_{1} \upgamma_{2} ;\quad \upsigma_{B2} = \upgamma_{1} \upgamma_{3} ;\quad \upsigma_{B3} = - \upgamma_{2} \upgamma_{3} . \\ \end{aligned} $$

(64)

The different definitions of the gamma matrices for the first option as compared with the second option, such as expressed by (46), give a different result for the full expansion of this result in terms of 4 × 4 matrices. For the first option, we get,

$$ \begin{aligned} & - {\text{i}}\upsigma_{B1} = \left[ {\begin{array}{*{20}c} {\upsigma_{3} } & 0 \\ 0 & {\upsigma_{3} } \\ \end{array} } \right];\quad - {\text{i}}\upsigma_{B2} = \left[ {\begin{array}{*{20}c} { - \upsigma_{2} } & 0 \\ 0 & { - \upsigma_{2} } \\ \end{array} } \right];\quad - {\text{i}}\upsigma_{B3} = \left[ {\begin{array}{*{20}c} {\upsigma_{1} } & 0 \\ 0 & {\upsigma_{1} } \\ \end{array} } \right] \to - {\text{i}}\overline{\upsigma }_{B} = \overline{\upsigma }; \\ & \upsigma_{E1} = \left[ {\begin{array}{*{20}c} 0 & {{\mathbf{i}}\upsigma_{1} } \\ {{\mathbf{i}}\upsigma_{1} } & 0 \\ \end{array} } \right];\quad \upsigma_{E2} = \left[ {\begin{array}{*{20}c} 0 & {{\mathbf{i}}\upsigma_{2} } \\ {{\mathbf{i}}\upsigma_{2} } & 0 \\ \end{array} } \right];\quad \upsigma_{E3} = \left[ {\begin{array}{*{20}c} 0 & {{\mathbf{i}}\upsigma_{3} } \\ {{\mathbf{i}}\upsigma_{3} } & 0 \\ \end{array} } \right] \to \overline{\upsigma }_{E} = - {\text{i}}\overline{\upsigma }. \\ \end{aligned} $$

(65)

Hence, as in (6),

$$ \Delta E = \frac{e\hbar }{{2m_{0} }}\left[ {\begin{array}{*{20}c} {\overline{\upsigma } \cdot {\mathbf{B}}} & 0 \\ 0 & {\overline{\upsigma } \cdot {\mathbf{B}}} \\ \end{array} } \right] - \frac{e\hbar }{{2m_{0} c}}\left[ {\begin{array}{*{20}c} 0 & {{\text{i}}\overline{\upsigma } \cdot {\mathbf{E}}} \\ {{\text{i}}\overline{\upsigma } \cdot {\mathbf{E}}} & 0 \\ \end{array} } \right]. $$

(66)

This result is the same as obtained by Dirac. It confirms that, as could be expected, the choice of the metric has no influence on the result.

For the second option we get,

$$ \begin{aligned} & - {\text{i}}\upsigma_{B1} = \left[ {\begin{array}{*{20}c} {\upsigma_{3} } & 0 \\ 0 & {\upsigma_{3} } \\ \end{array} } \right];\quad - {\text{i}}\upsigma_{B2} = \left[ {\begin{array}{*{20}c} { - \upsigma_{2} } & 0 \\ 0 & { - \upsigma_{2} } \\ \end{array} } \right];\quad - {\text{i}}\upsigma_{B3} = \left[ {\begin{array}{*{20}c} {\upsigma_{1} } & 0 \\ 0 & {\upsigma_{1} } \\ \end{array} } \right] \to - {\text{i}}\overline{\upsigma }_{B} = \overline{\upsigma }, \\ & \upsigma_{E1} = \left[ {\begin{array}{*{20}c} 0 & { - \upsigma_{1} } \\ {\upsigma_{1} } & 0 \\ \end{array} } \right];\quad \upsigma_{E2} = \left[ {\begin{array}{*{20}c} 0 & { - \upsigma_{2} } \\ {\upsigma_{2} } & 0 \\ \end{array} } \right];\quad \upsigma_{E3} = \left[ {\begin{array}{*{20}c} 0 & { - \upsigma_{3} } \\ {\upsigma_{3} } & 0 \\ \end{array} } \right] \to \overline{\upsigma }_{E} = - \overline{\upsigma }. \\ \end{aligned} $$

(67)

Hence, as in (33)

$$ \Delta E = \frac{e\hbar }{{2m_{0} }}\left[ {\begin{array}{*{20}c} {\overline{\upsigma } \cdot {\mathbf{B}}} & 0 \\ 0 & {\overline{\upsigma } \cdot {\mathbf{B}}} \\ \end{array} } \right] + \frac{e\hbar }{{2m_{0} c}}\left[ {\begin{array}{*{20}c} 0 & {\overline{\upsigma } \cdot {\mathbf{E}}} \\ { - \overline{\upsigma } \cdot {\mathbf{E}}} & 0 \\ \end{array} } \right]. $$

(68)