Problems
83
Hence, in (3.109),
1
ˆ
H 1 Ψ ≈
{σ · (−i∇ − qA)}
2 Ψ + qA
0 Ψ.
(3.114)
2m
The right-hand side of (3.114) should therefore be the non-relativistic energy
operator for a spin1 particle of charge q and mass m in a field A
μ .
2
Consider then the case A
0 = 0 which is sufficient for the discussion of g.
We need to evaluate
{σ · (−i∇ − qA)}
2 Ψ.
(3.115)
2
This requires care, because although it is true that (for example) (σ · p)
2 = p
if p = (p x , p y , p z ) are ordinary numbers which commute with each other,
the components of ‘−i∇ − qA’ do not commute due to the presence of the
differential operator ∇, and the fact that A depends on r. In problem 3.10
it is shown that
{σ · (−i∇ − qA)}
2 Ψ = (−i∇ − qA)
2 Ψ − qσ · BΨ.
(3.116)
The first term on the right-hand side of (3.116) when inserted into (3.114),
gives precisely the spin-0 non-relativistic Hamiltonian appearing on the lefthand side of (3.103) (see appendix A), while the second term in (3.116) yields
1
exactly (3.105) with g = 2, recalling that S = σ. Thus the non-relativistic
2
reduction of the Dirac equation leads to the prediction g = 2 for a spin1
2
particle.
In actual fact, the measured g-factor of the electron (and muon) is slightly
greater than this value: g exp = 2(1 + a). The ‘anomaly’ a, which is of order
10
−3 in size, is measured with quite extraordinary precision (see section 11.7)
for both the e
− and e
+ . This small correction can also be computed with
equally extraordinary accuracy, using the full theory of QED, as we shall
briefly explain in chapter 11. The agreement between theory and experiment is
phenomenal and is one example of such agreement exhibited by our ‘paradigm
theory’.
It may be worth noting that spin1 hadrons, such as the proton, have g2
factors very different from the Dirac prediction. This is because they are, as
we know, composite objects and are thus (in this respect) more like atoms in
nuclei than ‘elementary particles’.
Problems
(a) In natural units ħ = c = 1 and with 2m = 1, the Schr¨ odinger
equation may be written as
−∇
2 ψ + V ψ − i∂ψ/∂t = 0.
3.1
83
Hence, in (3.109),
1
ˆ
H 1 Ψ ≈
{σ · (−i∇ − qA)}
2 Ψ + qA
0 Ψ.
(3.114)
2m
The right-hand side of (3.114) should therefore be the non-relativistic energy
operator for a spin1 particle of charge q and mass m in a field A
μ .
2
Consider then the case A
0 = 0 which is sufficient for the discussion of g.
We need to evaluate
{σ · (−i∇ − qA)}
2 Ψ.
(3.115)
2
This requires care, because although it is true that (for example) (σ · p)
2 = p
if p = (p x , p y , p z ) are ordinary numbers which commute with each other,
the components of ‘−i∇ − qA’ do not commute due to the presence of the
differential operator ∇, and the fact that A depends on r. In problem 3.10
it is shown that
{σ · (−i∇ − qA)}
2 Ψ = (−i∇ − qA)
2 Ψ − qσ · BΨ.
(3.116)
The first term on the right-hand side of (3.116) when inserted into (3.114),
gives precisely the spin-0 non-relativistic Hamiltonian appearing on the lefthand side of (3.103) (see appendix A), while the second term in (3.116) yields
1
exactly (3.105) with g = 2, recalling that S = σ. Thus the non-relativistic
2
reduction of the Dirac equation leads to the prediction g = 2 for a spin1
2
particle.
In actual fact, the measured g-factor of the electron (and muon) is slightly
greater than this value: g exp = 2(1 + a). The ‘anomaly’ a, which is of order
10
−3 in size, is measured with quite extraordinary precision (see section 11.7)
for both the e
− and e
+ . This small correction can also be computed with
equally extraordinary accuracy, using the full theory of QED, as we shall
briefly explain in chapter 11. The agreement between theory and experiment is
phenomenal and is one example of such agreement exhibited by our ‘paradigm
theory’.
It may be worth noting that spin1 hadrons, such as the proton, have g2
factors very different from the Dirac prediction. This is because they are, as
we know, composite objects and are thus (in this respect) more like atoms in
nuclei than ‘elementary particles’.
Problems
(a) In natural units ħ = c = 1 and with 2m = 1, the Schr¨ odinger
equation may be written as
−∇
2 ψ + V ψ − i∂ψ/∂t = 0.
3.1