93
4.1. Lorentz transformations
Keeping only first order terms in η, the left hand side of (4.45) is
1
E − σ · p + Eσ · η − (σ · p σ · η + σ · η σ · p)
(4.46)
2
= E − η · p − σ · (p − ηE)
(4.47)
′
E
′
=
− σ · p
(4.48)
as required for the right hand side of (4.45).
For a finite boost φ and χ transform by the ‘exponentiation’ of (4.42),
namely
φ
′ = exp(−σ · ϑ/2) φ,
χ
′ = exp(σ · ϑ/2) χ
(4.49)
where the three real parameters ϑ = (ϑ x , ϑ y , ϑ z ) specify the direction and
magnitude of the boost. In contrast to (4.28), the transformations (4.49) are
†
not unitary. If we denote the matrix exp(−σ · ϑ/2) by B, we have B = B
−1
†
rather than B = B . So B does not leave φ
† φ and χ
† χ invariant. Actually
this is no surprise. We already know from section 4.1.2 that the density
φ
† φ + χ
† χ ought to transform as the fourth component ρ of the 4-vector
j
μ = (ρ, j). Let us check this for our infinitesimal boost:
ρ
′
φ
′† φ
′ + χ
′† χ
′
=
= φ
† (1 − σ · η/2)(1 − σ · η/2)φ + χ
† (1 + σ · η/2)(1 + σ · η/2) χ
= φ
† φ + χ
† χ − φ
†
σφ · η + χ
†
σχ · η
= ρ − η · j
(4.50)
as required by (4.38). Similarly, it may be verified (problem 4.2(b)) that j
transforms as the 3-vector part of the 4-vector j
μ , under this infinitesimal
boost.
On the other hand, the products φ
† χ and χ
† φ are clearly invariant under
the transformation (4.49), since the exponential factors cancel. This means
that the quantity ω
† βω is a Lorentz invariant.
At this point it is beginning to be clear that a more ‘covariant-looking’
notation would be very desirable. In the case of the KG probability current,
the 4-vector index μ was clearly visible in the expression on the right-hand side
of (3.20), but there is nothing similar in the Dirac case so far. In problem 4.3
the four ‘γ matrices’ are introduced, defined by γ
μ = (γ
0 , γ) with γ
0 = β and
γ = βα, together with the quantity ψ ¯ ≡ ψ
† γ
0 , in terms of which the Dirac
¯
¯
ρ of (3.51) and j of (3.57) can be written as ψ(x)γ
0 ψ(x) and ψ(x)γψ(x)
respectively. The complete Dirac 4-current is then
j
μ = ψ ¯ (x)γ
μ ψ(x).
(4.51)
For free particle solutions, we (and problem 4.2) have established that j
μ
of (4.51) indeed transforms as a 4-vector under infinitesimal rotations and
boosts. We have also just seen that the quantity ¯
ψψ is an invariant.
We end this section by illustrating the use of the finite boost transformations (4.49). Consider two frames S and S
′ , such that in S a particle is at rest
4.1. Lorentz transformations
Keeping only first order terms in η, the left hand side of (4.45) is
1
E − σ · p + Eσ · η − (σ · p σ · η + σ · η σ · p)
(4.46)
2
= E − η · p − σ · (p − ηE)
(4.47)
′
E
′
=
− σ · p
(4.48)
as required for the right hand side of (4.45).
For a finite boost φ and χ transform by the ‘exponentiation’ of (4.42),
namely
φ
′ = exp(−σ · ϑ/2) φ,
χ
′ = exp(σ · ϑ/2) χ
(4.49)
where the three real parameters ϑ = (ϑ x , ϑ y , ϑ z ) specify the direction and
magnitude of the boost. In contrast to (4.28), the transformations (4.49) are
†
not unitary. If we denote the matrix exp(−σ · ϑ/2) by B, we have B = B
−1
†
rather than B = B . So B does not leave φ
† φ and χ
† χ invariant. Actually
this is no surprise. We already know from section 4.1.2 that the density
φ
† φ + χ
† χ ought to transform as the fourth component ρ of the 4-vector
j
μ = (ρ, j). Let us check this for our infinitesimal boost:
ρ
′
φ
′† φ
′ + χ
′† χ
′
=
= φ
† (1 − σ · η/2)(1 − σ · η/2)φ + χ
† (1 + σ · η/2)(1 + σ · η/2) χ
= φ
† φ + χ
† χ − φ
†
σφ · η + χ
†
σχ · η
= ρ − η · j
(4.50)
as required by (4.38). Similarly, it may be verified (problem 4.2(b)) that j
transforms as the 3-vector part of the 4-vector j
μ , under this infinitesimal
boost.
On the other hand, the products φ
† χ and χ
† φ are clearly invariant under
the transformation (4.49), since the exponential factors cancel. This means
that the quantity ω
† βω is a Lorentz invariant.
At this point it is beginning to be clear that a more ‘covariant-looking’
notation would be very desirable. In the case of the KG probability current,
the 4-vector index μ was clearly visible in the expression on the right-hand side
of (3.20), but there is nothing similar in the Dirac case so far. In problem 4.3
the four ‘γ matrices’ are introduced, defined by γ
μ = (γ
0 , γ) with γ
0 = β and
γ = βα, together with the quantity ψ ¯ ≡ ψ
† γ
0 , in terms of which the Dirac
¯
¯
ρ of (3.51) and j of (3.57) can be written as ψ(x)γ
0 ψ(x) and ψ(x)γψ(x)
respectively. The complete Dirac 4-current is then
j
μ = ψ ¯ (x)γ
μ ψ(x).
(4.51)
For free particle solutions, we (and problem 4.2) have established that j
μ
of (4.51) indeed transforms as a 4-vector under infinitesimal rotations and
boosts. We have also just seen that the quantity ¯
ψψ is an invariant.
We end this section by illustrating the use of the finite boost transformations (4.49). Consider two frames S and S
′ , such that in S a particle is at rest