x′ 1 = x 1 coshχ+ix 4 sinhχ
x′ = γx+βγi(ict)
x′ 4 = x 4 coshχ−ix 1 sinhχWe verify that a boost along theidirection is like a rotation in thei4 plane through an angleiχ.
We need tofind the transformation matricesS that satisfy the equationS−^1 γμS=γνaμν
for the Dirac equation to be covariant. Recalling that the 4 component equivalent ofσzis Σz=
[γ 1 ,γ 2 ]
2 i =
γ 1 γ 2
i , we will show that these matrices are (for a rotation in the xy plane and a boost in
the x direction).
Srot = cos
θ
2+γ 1 γ 2 sin
θ
2
Sboost = coshχ
2+iγ 1 γ 4 sinhχ
2Note that this is essentially the transformation that we derived forrotations of spin one-half states
extended to 4 components. For the case of the boost the angle is nowiχ.
Letsverify that this choice works for a boost.
(
cosh
χ
2+iγ 1 γ 4 sinh
χ
2)− 1
γμ(
cosh
χ
2+iγ 1 γ 4 sinh
χ
2)
= aμνγν
(
cosh
−χ
2+iγ 1 γ 4 sinh
−χ
2)
γμ(
cosh
χ
2+iγ 1 γ 4 sinh
χ
2)
= aμνγν
(
cosh
χ
2−iγ 1 γ 4 sinh
χ
2)
γμ(
cosh
χ
2+iγ 1 γ 4 sinh
χ
2)
= aμνγνcosh
χ
2
γμcosh
χ
2+ cosh
χ
2γμiγ 1 γ 4 sinh
χ
2−iγ 1 γ 4 sinh
χ
2γμcosh
χ
2−iγ 1 γ 4 sinh
χ
2γμiγ 1 γ 4 sinh
χ
2= aμνγνγμcosh^2χ
2+iγμγ 1 γ 4 coshχ
2sinhχ
2−iγ 1 γ 4 γμcoshχ
2sinhχ
2+γ 1 γ 4 γμγ 1 γ 4 sinh^2χ
2= aμνγνThe equation we must satisfy can be checked for eachγmatrix. First checkγ 1. The operations with
theγmatrices all come from the anticommutator,{γμ,γν}= 2δμν, which tells us that the square
of any gamma matrix is one and that commuting a pair of (unequal) matrices changes the sign.
γ 1 cosh^2χ
2+iγ 1 γ 1 γ 4 coshχ
2sinhχ
2−iγ 1 γ 4 γ 1 coshχ
2sinhχ
2+γ 1 γ 4 γ 1 γ 1 γ 4 sinh^2χ
2= a 1 νγνγ 1 cosh^2χ
2+iγ 4 coshχ
2sinhχ
2+iγ 4 coshχ
2sinhχ
2+γ 1 sinh^2χ
2= a 1 νγνγ 1 cosh^2χ
2
+ 2iγ 4 coshχ
2
sinhχ
2
+γ 1 sinh^2χ
2
= a 1 νγνcosh^2
χ
2+ sinh^2χ
2=
1
4
((eχ 2
+e− 2 χ
)^2 + (eχ 2
−e− 2 χ
)^2 ) =1
4
(eχ+ 2 +e−χ+eχ−2 +e−χ)=1
2
(eχ+e−χ) = coshχcoshχ
2sinhχ
2=
1
4
((eχ 2
+e− 2 χ
)(eχ 2
−e− 2 χ
)) =1
4
(eχ−e−χ) =1
2
sinhχ