92
4. Lorentz Transformations and Discrete Symmetries
transforms under rotations. Of course, it should behave as a 3-vector, and
this is checked in problem 4.2(a).
We now turn to the behaviour of the spinors φ and χ under boosts, which
mix x and t, or equivalently p and E. For example, consider a Lorentz
velocity transformation (boost) from a frame S to a frame S
′ which is moving
with speed u with respect to S along the common x-axis. Then the energy E
′
and momentum p x of a particle in S are transformed to E
′ and p in S
′ where
x
(cf (D.1))
E
′
= coshϑ E − sinh ϑ p x
(4.36)
′
p
= coshϑ p x − sinh ϑ E,
(4.37)
x
2 )
−1/2
where cosh ϑ = (1 − u
≡ γ(u), and sinh ϑ = γ(u)u. As before, we
start with an infinitesimal transformation, where ϑ is replaced by η x such
that cosh η x ≈ 1 and sinh η x ≈ η x . Then (4.36) and (4.37) become E
′ =
′
E − η x p x , p = p x − η x E. For the general infinitesimal boost parametrized
x
by η = (η x , η y , η z ), the transformation law for (E, p) is
E
′ = E − η · p
(4.38)
′
p = p − ηE.
(4.39)
Once again, we have to determine φ
′ and χ
′ such that the transformed versions
of (4.14) and (4.15) are
(E
′
− σ · p
′ )φ
′
= mχ
′
(4.40)
′
(E
′ + σ · p
′ )χ
′ = mφ .
(4.41)
Note that this time E does transform, according to (4.38).
The required φ
′ and χ
′ are
φ
′ = (1 − σ · η/2)φ, χ
′ = (1 + σ · η/2)χ.
(4.42)
The spinors φ and χ behaved the same under rotations, but they transform
differently under boosts. There are two kinds of 2-component spinors, φ-type
and χ-type, in the representation (3.40), which are distinguished by their
behaviour under boosts. The group theory behind this will be explained in
appendix M of volume 2.
To verify the rule (4.42), take equation (4.14) in the form (4.40) and multiply from the left by the matrix (1 + σ · η/2), to obtain
(1 + σ · η/2)(E − σ · p)φ = mχ
′ ,
(4.43)
or equivalently
(1 + σ · η/2)(E − σ · p)(1 + σ · η/2)φ
′ = mχ
′ ,
(4.44)
where we have used (1− σ · η/2)
−1
≈ (1 + σ · η/2). For (4.44) to be consistent
with (4.40) we require
′
(1 + σ · η/2)(E − σ · p)(1 + σ · η/2) = E
′
− σ · p .
(4.45)
4. Lorentz Transformations and Discrete Symmetries
transforms under rotations. Of course, it should behave as a 3-vector, and
this is checked in problem 4.2(a).
We now turn to the behaviour of the spinors φ and χ under boosts, which
mix x and t, or equivalently p and E. For example, consider a Lorentz
velocity transformation (boost) from a frame S to a frame S
′ which is moving
with speed u with respect to S along the common x-axis. Then the energy E
′
and momentum p x of a particle in S are transformed to E
′ and p in S
′ where
x
(cf (D.1))
E
′
= coshϑ E − sinh ϑ p x
(4.36)
′
p
= coshϑ p x − sinh ϑ E,
(4.37)
x
2 )
−1/2
where cosh ϑ = (1 − u
≡ γ(u), and sinh ϑ = γ(u)u. As before, we
start with an infinitesimal transformation, where ϑ is replaced by η x such
that cosh η x ≈ 1 and sinh η x ≈ η x . Then (4.36) and (4.37) become E
′ =
′
E − η x p x , p = p x − η x E. For the general infinitesimal boost parametrized
x
by η = (η x , η y , η z ), the transformation law for (E, p) is
E
′ = E − η · p
(4.38)
′
p = p − ηE.
(4.39)
Once again, we have to determine φ
′ and χ
′ such that the transformed versions
of (4.14) and (4.15) are
(E
′
− σ · p
′ )φ
′
= mχ
′
(4.40)
′
(E
′ + σ · p
′ )χ
′ = mφ .
(4.41)
Note that this time E does transform, according to (4.38).
The required φ
′ and χ
′ are
φ
′ = (1 − σ · η/2)φ, χ
′ = (1 + σ · η/2)χ.
(4.42)
The spinors φ and χ behaved the same under rotations, but they transform
differently under boosts. There are two kinds of 2-component spinors, φ-type
and χ-type, in the representation (3.40), which are distinguished by their
behaviour under boosts. The group theory behind this will be explained in
appendix M of volume 2.
To verify the rule (4.42), take equation (4.14) in the form (4.40) and multiply from the left by the matrix (1 + σ · η/2), to obtain
(1 + σ · η/2)(E − σ · p)φ = mχ
′ ,
(4.43)
or equivalently
(1 + σ · η/2)(E − σ · p)(1 + σ · η/2)φ
′ = mχ
′ ,
(4.44)
where we have used (1− σ · η/2)
−1
≈ (1 + σ · η/2). For (4.44) to be consistent
with (4.40) we require
′
(1 + σ · η/2)(E − σ · p)(1 + σ · η/2) = E
′
− σ · p .
(4.45)