The Dirac equation is given by:
[iγμ∂μ−m]ψ(x)=0 .
We can prove that it's Lorentz invariant when:
ψ(x)→S−1ψ′(x′) and ∂μ→Λνμ∂′ν, where
S(Λ)=1−i4σμνϵμν
which applying that for Dirac equation covariance that ΛνμSγμS−1=γν
Now the Weyl spinors equations are given by:
iˉσ.∂ψL=0, (1)
iσ.∂ψR=0, (2)
where σμ≡(1,σ_) and ˉσμ≡(1,−σ_)
Look for instance Peskin's book "An Introduction To Quantum Field Theory"
where the Lorentz transformation of ψL and ψR are given by:
ψL→(1−iθ.σ2−β.σ2)ψL,
ψR→(1−iθ.σ2+β.σ2)ψR,
How one can show that (1) and (2) are Lorentz invariant ?
Answer
The key thing is you are contracting ∂μ into either σμ,ˉσμ. Consider contracting a general vector pμ σμpμ=(p0−p3−p1+ip2−p1−ip2p0+p3) Note that the determinant is just the norm of the vector: (p0)2−(p1)2−(p2)2−(p3)2.
If you sandwich this matrix between some U and U−1 the determinant is unchanged det so it corresponds to some Lorentz transformation.
So we can write \sigma^\mu (\Lambda^{\mu'}_\mu p_{\mu'})=U(\Lambda)\sigma^\mu p_\mu U(\Lambda)^{-1} for some 2 by 2 matrix U(\Lambda).
It turns out that U(\Lambda) is the very same matrix you are using to transform your right spinor, so that is why the Weyl equations are invariant.
For instance as an exercise, take a finite rotation about the z-axis: I-i\frac{\theta}{2}\sigma^3+\dots=\exp(-i\frac{\theta}{2}\sigma^3)=\left(\begin{array}{cc} \exp(-i\frac{\theta}{2})& 0\\ 0 &\exp(+i\frac{\theta}{2}) \end{array}\right) You can show that \exp(-i\frac{\theta}{2}\sigma^3)\,\sigma^\mu p_\mu\,\exp(+i\frac{\theta}{2}\sigma^3) does transform the p^1,p^2 components as a rotation by angle \theta.
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