71
3.2. The Dirac equation
so
4
∑
ρ =
|ψ a |
2 > 0
(3.53)
a=1
and we see that ρ is a scalar density which is explicitly positive-definite. This
is one property we require of a probability density: in addition, we require
a conservation law, coming from the Dirac equation, and a corresponding
probability current density. In fact (see problem 3.5) we can demonstrate,
using the Dirac equation,
i∂ψ/∂t = (−iα · ∇ + βm)ψ
(3.54)
and its Hermitian conjugate
← −
−i∂ψ
† = ψ
† (+iα · ∇ + βm)
(3.55)
that there is a conservation law of the required form
∂ρ/∂t + ∇ · j = 0.
(3.56)
−
The notation ψ
† ←
∇ requires some comment: it is shorthand for three row
matrices
−
ψ
† ←
∇ x ≡ ∂ψ
† /∂x
etc.
(recall that ψ
† is a row matrix).
In equation (3.56), with ρ being given by (3.51), the probability current
density j is
j(x) = ψ
† (x)αψ(x)
(3.57)
representing a 3-vector with components
(ψ
† α 1 ψ, ψ
† α 2 ψ, ψ
† α 3 ψ).
(3.58)
We therefore have a positive-definite ρ and an associated j satisfying the
required conservation law (3.56), which, as usual, we can write in invariant
form as ∂ μ j
μ = 0, where
j
μ = (ρ, j).
(3.59)
Thus j
μ is an acceptable probability current, unlike the current for the KG
equation – as we might have anticipated.
The form of equation (3.56) implies that j
μ of (3.59) is a contravariant
4-vector (cf equation (D.4)), as we verified explicitly in the KG case. The
corresponding verification is more difficult in the Dirac case, since the Dirac
spinor ψ transforms non-trivially under Lorentz transformations, unlike the
KG wavefunction φ. We shall come back to this problem in chapter 4.
We now turn to further discussion of the spin degree of freedom, postponing
consideration of the negative-energy solutions until section 3.4.
3.2. The Dirac equation
so
4
∑
ρ =
|ψ a |
2 > 0
(3.53)
a=1
and we see that ρ is a scalar density which is explicitly positive-definite. This
is one property we require of a probability density: in addition, we require
a conservation law, coming from the Dirac equation, and a corresponding
probability current density. In fact (see problem 3.5) we can demonstrate,
using the Dirac equation,
i∂ψ/∂t = (−iα · ∇ + βm)ψ
(3.54)
and its Hermitian conjugate
← −
−i∂ψ
† = ψ
† (+iα · ∇ + βm)
(3.55)
that there is a conservation law of the required form
∂ρ/∂t + ∇ · j = 0.
(3.56)
−
The notation ψ
† ←
∇ requires some comment: it is shorthand for three row
matrices
−
ψ
† ←
∇ x ≡ ∂ψ
† /∂x
etc.
(recall that ψ
† is a row matrix).
In equation (3.56), with ρ being given by (3.51), the probability current
density j is
j(x) = ψ
† (x)αψ(x)
(3.57)
representing a 3-vector with components
(ψ
† α 1 ψ, ψ
† α 2 ψ, ψ
† α 3 ψ).
(3.58)
We therefore have a positive-definite ρ and an associated j satisfying the
required conservation law (3.56), which, as usual, we can write in invariant
form as ∂ μ j
μ = 0, where
j
μ = (ρ, j).
(3.59)
Thus j
μ is an acceptable probability current, unlike the current for the KG
equation – as we might have anticipated.
The form of equation (3.56) implies that j
μ of (3.59) is a contravariant
4-vector (cf equation (D.4)), as we verified explicitly in the KG case. The
corresponding verification is more difficult in the Dirac case, since the Dirac
spinor ψ transforms non-trivially under Lorentz transformations, unlike the
KG wavefunction φ. We shall come back to this problem in chapter 4.
We now turn to further discussion of the spin degree of freedom, postponing
consideration of the negative-energy solutions until section 3.4.