90
4. Lorentz Transformations and Discrete Symmetries
As before, we start with the infinitesimal rotation (4.9). Since p is a vector,
it transforms in the same way as x, so that under an infinitesimal rotation p
′
becomes p where
′
p = p − ∈ × p.
(4.16)
The question for us now is: how do the spinors φ and χ transform under this
same rotation of the coordinate system?
The essential point is that in the new coordinate system the defining equations (4.14) and (4.15) should take exactly the same form, namely
′ φ
′
Eφ
′
= σ · p + mχ
′
(4.17)
′ χ
′
Eχ
′ = −σ · p + mφ
′
(4.18)
where φ
′ and χ
′ are the spinors in the new coordinate system, and we have
used the fact that both E and m do not change under rotations. Our task is
to find φ
′ and χ
′ in terms of φ and χ.
Since both φ and χ are 2-component spinors, we might guess from (4.11)
that the answer is
φ
′
χ
′
= (1 + iσ · ∈/2)φ,
= (1 + iσ · ∈/2)χ,
(4.19)
since the σ/2 are the spin-1/2 matrices, taking the place of L ˆ . To check that
this is, in fact, the correct transformation law, we proceed as follows.
1 First,
multiply (4.14) from the left by the matrix (1 + iσ · ∈/2): then, since E and
m commute with all matrices, the result is
Eφ
′
= (1 + iσ · ∈/2)σ · pφ + mχ
′
(4.20)
= (1 + iσ · ∈/2)σ · p(1 − iσ · ∈/2)φ
′ + mχ
′
(4.21)
where we have used
(1 + iσ · ∈/2)
−1
≈ (1 − iσ · ∈/2)
(4.22)
to first order in ∈. Keeping only first order terms in ∈, the first term on the
right hand side of (4.21) is
(σ · p +
1 iσ · ∈ σ · p −
1 iσ · p σ · ∈)φ
′ .
(4.23)
2
2
This can be simplified using the result from problem 3.4(b):
σ · a σ · b = a · b + iσ · a × b,
(4.24)
provided all the components of a and b commute. Applying (4.24), (4.23)
becomes
[σ · p +
i (∈ · p + iσ · ∈ × p) −
i (∈ · p + iσ · p × ∈)]φ
′
(4.25)
2
2
′ φ
′
= (σ · p − σ · ∈ × p)φ
′ = σ · p .
(4.26)
1 We shall derive (4.19), and the corresponding rule for velocity transformations, equation
(4.42) below, in appendix M of volume 2 using group theory.
4. Lorentz Transformations and Discrete Symmetries
As before, we start with the infinitesimal rotation (4.9). Since p is a vector,
it transforms in the same way as x, so that under an infinitesimal rotation p
′
becomes p where
′
p = p − ∈ × p.
(4.16)
The question for us now is: how do the spinors φ and χ transform under this
same rotation of the coordinate system?
The essential point is that in the new coordinate system the defining equations (4.14) and (4.15) should take exactly the same form, namely
′ φ
′
Eφ
′
= σ · p + mχ
′
(4.17)
′ χ
′
Eχ
′ = −σ · p + mφ
′
(4.18)
where φ
′ and χ
′ are the spinors in the new coordinate system, and we have
used the fact that both E and m do not change under rotations. Our task is
to find φ
′ and χ
′ in terms of φ and χ.
Since both φ and χ are 2-component spinors, we might guess from (4.11)
that the answer is
φ
′
χ
′
= (1 + iσ · ∈/2)φ,
= (1 + iσ · ∈/2)χ,
(4.19)
since the σ/2 are the spin-1/2 matrices, taking the place of L ˆ . To check that
this is, in fact, the correct transformation law, we proceed as follows.
1 First,
multiply (4.14) from the left by the matrix (1 + iσ · ∈/2): then, since E and
m commute with all matrices, the result is
Eφ
′
= (1 + iσ · ∈/2)σ · pφ + mχ
′
(4.20)
= (1 + iσ · ∈/2)σ · p(1 − iσ · ∈/2)φ
′ + mχ
′
(4.21)
where we have used
(1 + iσ · ∈/2)
−1
≈ (1 − iσ · ∈/2)
(4.22)
to first order in ∈. Keeping only first order terms in ∈, the first term on the
right hand side of (4.21) is
(σ · p +
1 iσ · ∈ σ · p −
1 iσ · p σ · ∈)φ
′ .
(4.23)
2
2
This can be simplified using the result from problem 3.4(b):
σ · a σ · b = a · b + iσ · a × b,
(4.24)
provided all the components of a and b commute. Applying (4.24), (4.23)
becomes
[σ · p +
i (∈ · p + iσ · ∈ × p) −
i (∈ · p + iσ · p × ∈)]φ
′
(4.25)
2
2
′ φ
′
= (σ · p − σ · ∈ × p)φ
′ = σ · p .
(4.26)
1 We shall derive (4.19), and the corresponding rule for velocity transformations, equation
(4.42) below, in appendix M of volume 2 using group theory.