Dirac spinor’s transformation under Lorentz mappings

For a given Lorentz matrix, we deduce the Dirac spinor’s transformation in terms of four complex quantities.

Introduction

We have the Dirac equation for spin-1/2 particles [1-5] $\left[\left(x^{\mu }\right)=\left(t, x, y, z\right), \hslash =c=1\right]:$

$\left(i{\gamma }^{\mu }{\partial }_{\mu }-m_0\right)\psi =0, i=\sqrt{-1}, {\partial }_{\mu }=\frac{\partial }{\partial x^{\mu }} ,\text{ (1)}$

where $\psi$ is a 4-spinor with the ${\gamma }^{\mu }$ matrices verifying the anticommutator [6-8]:

$\left\{{\gamma }^{\mu },{\gamma }^{\nu }\right\}=2g^{\mu \nu } I_{4x4} , \left(g^{\mu \nu }\right)=Diag\left(1,-1,-1,-1\right).\text{ (2)}$

Here we shall use the Dirac-Pauli (or standard) representation [2,9]:

${\gamma }^0=\left(\begin{array}{cc}I& 0\\ 0& -I\end{array}\right), {\gamma }^j=\left(\begin{array}{cc}0& {\sigma }_j\\ -{\sigma }_j& 0\end{array}\right), j=1, 2, 3, \text{ (3)}$

with the Cayley [10]-Sylvester [11]-Pauli [12] matrices:

${\sigma }_1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right), {\sigma }_2=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right), {\sigma }_3=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),\text{ (4)}$

to analyze the transformation law of under the orthochronic and proper Lorentz group [13-19]:

${\tilde{x}}^{\mu }=L^{\mu }{}_{\nu } x^{\nu } ,\text{ (5)}$

which implies the existence [2, 7, 20, 21] of a non-singular matrix S such that:

$L^{\mu }{}_{\nu }S {\gamma }^{\nu }={\gamma }^{\mu }S ,\text{ (6)}$

and we deduce the relativistic invariance of (1) if the Dirac 4-spinor obeys the transformation rule:

$\tilde{\psi }=S \psi .\text{ (7)}$

Here we exhibit a method to findS for a given Lorentz matrix.

Construction of the matrix S for a given Lorentz mapping.

The arbitrary complex quantities α, β, γ, δ verifying the constraint αδ – βγ = 1, generate a Lorentz matrix $L=(L^{\mu }{}_{\nu })$ via the relations [13-15, 17, 22-26]:

$\begin{array}{l}{L}^0{}_0= \frac{1}{2}\left(\alpha \stackrel{-}{\alpha }+ \beta \stackrel{-}{\beta }+ \gamma \stackrel{-}{\gamma }+ \delta \stackrel{-}{\delta }\right), L^1{}_0= \frac{1}{2}\left(\stackrel{-}{\alpha } \gamma +\stackrel{-}{\beta }\delta \right)+ cc, L^2{}_0=- \frac{i}{2}\left(\alpha \stackrel{-}{\gamma }-\stackrel{-}{\beta }\delta \right)+ cc,\\ L^0{}_1= \frac{1}{2}\left(\stackrel{-}{\alpha }\beta +\stackrel{-}{\gamma }\delta \right)+ cc, L^1{}_1= \frac{1}{2}\left(\stackrel{-}{\alpha }\delta + \beta \stackrel{-}{\gamma }\right)+ cc, L^2{}_1=- \frac{i}{2}\left(\alpha \stackrel{-}{\delta }+ \beta \stackrel{-}{\gamma }\right)+ cc,\\ L^0{}_2=- \frac{i}{2}\left(\stackrel{-}{\alpha }\beta + \stackrel{-}{\gamma }\delta \right)+ cc, L^1{}_2=- \frac{i}{2}\left(\stackrel{-}{\alpha }\delta + \beta \stackrel{-}{\gamma }\right)+ cc, L^2{}_2= \frac{1}{2}\left(\stackrel{-}{\alpha }\delta -\stackrel{-}{\beta }\gamma \right)+ cc ,\\ L^0{}_3= \frac{1}{2}\left(\alpha \stackrel{-}{\alpha }-\beta \stackrel{-}{\beta }+\gamma \stackrel{-}{\gamma }-\delta \stackrel{-}{\delta }\right), L^1{}_3= \frac{1}{2}\left(\stackrel{-}{\alpha }\gamma -\stackrel{-}{\beta }\delta \right)+ cc, L^2{}_3=- \frac{i}{2}\left(\alpha \stackrel{-}{\gamma }+\stackrel{-}{\beta }\delta \right)+ cc ,\\ L^3{}_0= \frac{1}{2}\left(\alpha \stackrel{-}{\alpha }+\beta \stackrel{-}{\beta }-\gamma \stackrel{-}{\gamma }-\delta \stackrel{-}{\delta }\right), L^3{}_1= \frac{1}{2}\left(\stackrel{-}{\alpha }\beta -\stackrel{-}{\gamma }\delta \right)+ cc, L^3{}_2=- \frac{i}{2}\left(\stackrel{-}{\alpha }\beta -\stackrel{-}{\gamma }\delta \right)+ cc ,\\ L^3{}_3= \frac{1}{2}\left(\alpha \stackrel{-}{\alpha }-\beta \stackrel{-}{\beta }-\gamma \stackrel{-}{\gamma }+ \delta \stackrel{-}{\delta }\right), \alpha \delta -\beta \gamma =1,\text{ (8)}\end{array}$

where cc means the complex conjugate of all the previous terms.

The inverse problem is to obtain α, β, γ, δ if we know L, and the answer is [26-29]:

$\begin{array}{l}\alpha =\frac{1}{D} Q^1{}_1=\frac{1}{2D}\left[L^0{}_0+L^0{}_3+L^1{}_1+L^2{}_2+L^3{}_0+L^3{}_3-i \left(L^1{}_2-L^2{}_1\right)\right],\\ \beta =\frac{1}{D} Q^1{}_2=\frac{1}{2D}\left[L^0{}_1+L^1{}_0-L^1{}_3+L^3{}_1+i \left(L^0{}_2+L^2{}_0-L^2{}_3+L^3{}_2\right)\right],\\ \gamma =\frac{1}{D} Q^2{}_1=\frac{1}{2D}\left[L^0{}_1+L^1{}_0+L^1{}_3-L^3{}_1-i \left(L^0{}_2+L^2{}_0+L^2{}_3-L^3{}_2\right)\right],\\ \delta =\frac{1}{D} Q^2{}_2=\frac{1}{2D}\left[L^0{}_0-L^0{}_3+L^1{}_1+L^2{}_2-L^3{}_0+L^3{}_3+i \left(L^1{}_2-L^2{}_1\right)\right],\text{ (9)}\end{array}$

where $D^2=Q^1{}_1 Q^2{}_2-Q^1{}_2 Q^2{}_1$

From (6) are immediate the expressions [3, 30]:

$L^{\mu }{}_0=\frac{1}{4} tr \left({\gamma }^0 S^{-1 }{\gamma }^{\mu } S\right), L^{\mu }{}_k=-\frac{1}{4} tr \left({\gamma }^{k }{S}^{-1} {\gamma }^{\mu } S\right), \mu =0, \dots , 3, k=1, 2, 3,\text{ (10)}$

that is, if we know S then with (10) we can determine the Lorentz matrix; (10) generates the relations:

$\begin{array}{l}{L}^0{}_0=2\left(b_0^2-b_1^2-b_2^2-b_3^2\right)-1,L^0{}_1=2\left[ \left(b_2 d_3-b_3 d_2\right)+i \left(b_0 d_1-b_1 d_0\right) \right],\\ L^0{}_2=2\left[ \left(b_3 d_1-b_1 d_3\right)+i \left(b_0 d_2-b_2 d_0\right) \right],L^0{}_3=2\left[ \left(b_1 d_2-b_2 d_1\right)+i \left(b_0 d_3-b_3 d_0\right) \right],\\ L^1{}_0=2\left[- \left(b_2 d_3-b_3 d_2\right)+i \left(b_0 d_1-b_1 d_0\right) \right],L^1{}_1=2\left[ \left(b_0^2-b_1^2\right)+\left(d_2^2+d_3^2\right) \right]-1,\\ L^1{}_2=2\left[-\left(b_1 b_2+d_1 d_2\right)-i \left(b_0 b_3+d_0 d_3\right) \right],L^1{}_3=2\left[- \left(b_1 b_3+d_1 d_3\right)+i \left(b_0 b_2+d_0 d_2\right) \right],\\ L^2{}_0=2\left[-\left(b_3 d_1-b_1 d_3\right)+i \left(b_0 d_2-b_2 d_0\right) \right],L^2{}_1=2\left[- \left(b_1 b_2+d_1 d_2\right)+i \left(b_0 b_3+d_0 d_3\right) \right],\\ L^2{}_2=2\left[ \left(b_0^2-b_2^2\right)+\left(d_1^2+d_3^2\right) \right]-1,L^2{}_3=2\left[-\left(b_2 b_3+d_2 d_3\right)-i \left(b_0 b_1+d_0 d_1\right) \right],\\ L^3{}_0=2\left[-\left(b_1 d_2-b_2 d_1\right)+i \left(b_0 d_3-b_3 d_0\right) \right],L^3{}_1=2\left[- \left(b_1 b_3+d_1 d_3\right)-i \left(b_0 b_2+d_0 d_2\right) \right],\\ L^3{}_2=2\left[-\left(b_2 b_3+d_2 d_3\right)+i \left(b_0 b_1+d_0 d_1\right) \right],L^3{}_3=2\left[ \left(b_0^2-b_3^2\right)+\left(d_1^2+d_2^2\right) \right]-1,\text{ (11)}\end{array}$

which allow to obtain L if we have the expansion [31]:

$S=b_{0 }I+i{d}_{0 }{\gamma }^5+b_1 {\sigma }^{23}+b_2 {\sigma }^{31}+b_{3 }{\sigma }^{12}+ {\displaystyle \sum }_{j=1 }^3{d}_j{\sigma }^{oj}.\text{ (12)}$

However, here we have the inverse problem, that is, to obtain $b_{\mu } \& d_{\mu }, \mu =0, \dots , 3$ verifying (11) for a given Lorentz matrix. Our answer is the following:

$\begin{array}{l}{b}_0=\frac{1}{4}\left(\alpha +\stackrel{-}{\alpha }+\delta +\stackrel{-}{\delta }\right), b_1=\frac{1}{4}\left(\stackrel{-}{\beta }-\beta +\stackrel{-}{\gamma }-\gamma \right),b_2=\frac{i}{4} \left(\beta +\stackrel{-}{\beta }-\gamma -\stackrel{-}{\gamma }\right), b_3=\frac{1}{4}\left(\stackrel{-}{\alpha }-\alpha +\delta -\stackrel{-}{\delta }\right),\\ d_0=\frac{i}{4}\left(\alpha -\stackrel{-}{\alpha }+\delta -\stackrel{-}{\delta }\right), d_1=-\frac{i}{4}\left(\stackrel{-}{\beta }+\beta +\stackrel{-}{\gamma }+\gamma \right),d_2=\frac{1}{4} \left(\stackrel{-}{\beta }-\beta +\gamma -\stackrel{-}{\gamma }\right), d_3=\frac{i}{4}\left(\stackrel{-}{\delta }+\delta -\alpha -\stackrel{-}{\alpha }\right),\text{ (13)}\end{array}$

hence the expressions (8) are deduced if we apply (13) into (11). Besides, with (13) the matrix (12) acquires the structure:

$S=\left(\begin{array}{cc}A& E\\ E& A\end{array}\right), A=\frac{1}{2}\left(\begin{array}{cc}\stackrel{-}{\alpha }+\delta & \stackrel{-}{\beta }-\gamma \\ \stackrel{-}{\gamma }-\beta & \alpha +\stackrel{-}{\delta }\end{array}\right), E=\frac{1}{2}\left(\begin{array}{cc}\stackrel{-}{\alpha }-\delta & \stackrel{-}{\beta }+\gamma \\ \stackrel{-}{\gamma }+\beta & \stackrel{-}{\delta }-\alpha \end{array}\right) .\text{ (14)}$

Therefore, for a given Lorentz transformation first we employ (9) to determine α,β,γ,δ, then S is immediate via (14); this approach is an alternative to the process showed in [31] and to the explicit general formula obtained by Macfarlane [30]:

$S=\frac{1}{4\sqrt{G}} [G I+\frac{i}{2} {\epsilon }_{\mu \nu \alpha \beta } L^{\mu \nu } L^{\alpha \beta } {\gamma }^5+i \text{Γ}\left(L^2\right)-i \left(2+tr L\right)\text{ Γ}\left(L\right)]\text{ (15)}$

$\begin{array}{l}G=2 \left(1+tr L\right)+\frac{1}{2}\left[ {(tr L)}^2-tr L^{2 }\right], tr L= {\displaystyle \sum }_{\mu =0 }^3{L}^{\mu }{}_{\mu }, tr L^2={\displaystyle \sum }_{\nu ,\alpha =0 }^3{L}^{\nu }{}_{\alpha }{L}^{\alpha }{}_{\nu } ,\\ \Gamma \left(L\right)={\displaystyle \sum }_{\mu ,\nu =0 }^3{L}_{\mu \nu }{\sigma }^{\mu \nu },\Gamma \left(L^2\right)={\displaystyle \sum }_{\alpha ,\mu ,\nu =0 }^3{L}_{\mu \alpha }{L}^{\alpha }{}_{\nu }{\sigma }^{\mu \nu },\text{ (16)}\end{array}$

however, the possible physical applications are not evident in it. In our procedure, for example, the relations (9) are of great interest for the physicists working on supersymmetry [29], and the expressions (14) are very useful to study the relativistic motion of a classical point particle [28].

Conclusion

The Dirac equation is relativistic if the corresponding 4-spinor verifies the transformation (7) under Lorentz mappings, with the matrix S satisfying the condition (6). Here we showed a procedure to construct S for a given Lorentz matrix.

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