Relativistic Quantum Mechanics
3.5 Inclusion of electromagnetic interactions via the gauge principle: the Dirac prediction of g = 2gaugeprinciple:theDiracpredictionofg= 2
for the electron
Having set up the relativistic spin-0 and spin-12 free-particle wave equations, we are now in a position to use the machinery developed in chapter 2, in order to include electromagnetic interactions. All we have to do is make the replacement
∂μ → Dμ ≡ ∂μ + iqAμ (3.99) for a particle of charge q. For the spin-0 KG equation (3.10) we obtain, after some rearrangement (problem 3.9),
(❗ +m 2)φ = −iq(∂μAμ +Aμ∂μ)φ +q 2A2φ (3.100)
= −Vˆ
KGφ. (3.101)
81 3.5. Inclusion ofelectromagnetic interactionsviathegauge principle
Note that the potential VˆKG contains the differential operator ∂μ; the sign of VˆKG is a convention chosen so as to maintain the same relative sign between
∇2 and Vˆ as in the Schr¨odinger equation – for example that in (A.5).
For the Dirac equation the replacement (3.99) leads to i ∂ψ
= [α ·(−i∇ − qA) + βm+ qA0]ψ (3.102)
∂t
where Aμ = (A0 ,A). The potential due to Aμ is therefore VˆD= qA01−qα·A, which is a 4 ×4 matrix acting on the Dirac spinor.
The non-relativistic limit of (3.102) is of great importance, both physically and historically. It was, of course, first obtained by Dirac; and it provided, in 1928, a sensational explanation of why the g-factor of the electron had the value g = 2, which was then the empirical value, without any theoretical basis.
By way of background, recall from appendix A that the Schr¨odinger equa
tion for a non-relativistic spinless particle of charge q in a magnetic field B described by a vector potential Asuch that B= ∇ ×Ais states of different magnetic quantum number) of the (2l+ 1)-fold degeneracy associated with a state of definite l. In particular, though, there should be no splitting of the hydrogen ground state which has l = 0. But experimentally splitting into two levels is observed, indicating a two-fold degeneracy and thus (see earlier) a j = -like degree of freedom. 12
Uhlenbeck and Goudsmit (1925) suggested that the doubling of the hy
drogen ground state could be explained if the electron were given an addi
tional quantum number corresponding to an angular-momentum-like observ
1 1
able, having magnitude j = . The operators 2 S = 2σwhich we have already met serve to represent such a spin angular momentum. If the contribution to the energy operator of the particle due to its spin S enters into the effective Schr¨odinger equation in exactly the same way as that due to its orbital an
gular momentum, then we would expect an additional term on the left-hand side of (3.103) of the form
− q B · S. (3.104)
2m
The corresponding wavefunction must now have two (spinor) components, acted on by the 2×2 matrices in S.
The energy difference between the two levels with eigenvalues Sz = ±1 would then be qB/2m in magnitude. Experimentally the splitting was found to be just twice this value. Thus empirically the term (3.104) was modified to
−g q B · S (3.105)
2m
where g is the ‘gyromagnetic ratio’ of the particle, with g ≈ 2. Let us now see
2
how Dirac deduced the term (3.105), with the precise value g = 2, from his equation.
To achieve a non-relativistic limit, we expect that we have somehow to reduce the four-component Dirac equation to one involving just two compo
nents, since the desired term (3.105) is only a 2 × 2 matrix. Looking at the explicit form (3.72) for the free-particle positive-energy solutions, we see that the lower two components are of order v (i.e. v/c with c = 1) times the upper two. This suggests that, to get a non-relativistic limit, we should regard the lower two components of ψ as being small (at least in the specific representa
tion we are using for αand β). However, since (3.102) includes the Aμ-field, this will have to be demonstrated (see (3.112)). Also, if we write the total energy operator as m + Hˆ1, we expect Hˆ1 to be the non-relativistic energy operator.
We let ( )
ψ = Ψ (3.106)
Φ
where Ψ and Φ are not free-particle solutions, and they carry the space–time dependence as well as the spinor character (each has two components). We set
Hˆ1= α · (−i∇ − qA) + βm+qA0 −m (3.107) where a 4×4 unit matrix multiplying the last two terms is understood. Then
( ) ( ) ( )
Ψ 0 σ · (−i∇ − qA) Ψ
Hˆ1 =
Φ σ ·(−i∇ − qA) 0 Φ
( ) ( )
− 2m 0
+ qA0 Ψ
. (3.108)
Φ Φ
Multiplying out (3.108), we obtain
Hˆ1Ψ = σ · (−i∇ − qA)Φ + qA0Ψ (3.109) Hˆ1Φ = σ · (−i∇ − qA)Ψ + qA0Φ−2mΦ. (3.110) From (3.110), we obtain
(Hˆ1−qA0 + 2m)Φ = σ ·(−i∇ −qA)ψ. (3.111) So, if Hˆ1 (or rather any matrix element of it) is ≪ m and if A0is positive or, if negative, much less in magnitude than m/e, we can deduce
Φ ∼ (velocity)×Ψ (3.112)
as in the free case, provided that the magnetic energy ∼ σ ·Ais not of order m. Further, if Hˆ1≪ m and the conditions on the fields are met, we can drop Hˆ1 and qA0on the left-hand side of (3.111), as a first approximation, so that
σ · (−i∇ − qA)
Φ ≈ Ψ. (3.113)
2m
Problems 83 Hence, in (3.109),
ˆ 1
H1Ψ ≈ {σ · (−i∇ − qA)}2Ψ +qA0Ψ. (3.114) 2m
The right-hand side of (3.114) should therefore be the non-relativistic energy operator for a spin-12 particle of charge q and mass min a field Aμ .
Consider then the case A0 = 0 which is sufficient for the discussion of g.
We need to evaluate
{σ · (−i∇ − qA)}2Ψ. (3.115) This requires care, because although it is true that (for example) (σ ·p)2= p2
if p = (px, py, pz) 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 Adepends on r. In problem 3.10 it is shown that
{σ · (−i∇ − qA)}2Ψ = (−i∇ − qA)2Ψ−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 left-hand side of (3.103) (see appendix A), while the second term in (3.116) yields exactly (3.105) with g= 2, recalling that S = 12 σ. Thus the non-relativistic reduction of the Dirac equation leads to the prediction g = 2 for a spin-1 2 particle.
In actual fact, the measured g-factor of the electron (and muon) is slightly greater than this value: gexp = 2(1 + a). The ‘anomaly’ a, which is of order 10−3in 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 spin-12 hadrons, such as the proton, have g
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
−∇2ψ+V ψ−i∂ψ/∂t= 0.
3.1
3.2
Multiply this equation from the left by ψ∗ and multiply the complex conjugate of this equation by ψ (assume V is real). Subtract the two equations and show that your answer may be written in the form of a continuity equation
∂ρ/∂t+∇ · j= 0 where ρ= ψ∗ψ and j= i−1[ψ∗(∇ψ) −(∇ψ∗)ψ].
(b) Perform the same operations for the Klein–Gordon equation and derive the corresponding ‘probability’ density current. Show also that for a free-particle solution
−ip·x
φ = Ne
with pμ = (E,p), the probability current jμ = (ρ,j) is proportional to pμ .
(a) Prove the following properties of the matrices αi and β:
(i) αi and β (i = 1,2,3) are all Hermitian [Hint: what is the Hamiltonian?].
(ii) Trαi = Trβ = 0 where ‘Tr’ means the trace, i.e. the sum of the diagonal elements [Hint: use Tr(AB) = Tr(BA) for any matrices Aand B – and prove this too!].
(iii) The eigenvalues of αi and β are ±1 [Hint: square αi and β].
(iv) The dimensionality of αi and β is even [Hint: the trace of a matrix is equal to the sum of its eigenvalues].
(b) Verify explicitly that the matrices αand β of (3.31), and of (3.40), satisfy the Dirac conditions (3.34) – (3.36).
3.3For free-particle solutions of the Dirac equation
−ip·x
ψ = ωe
the four-component spinor ω may be written in terms of the two-component
spinors ( )
ω = φ . χ From the Dirac equation for ψ
i∂ψ/∂t= (−iα · ∇ +βm)ψ using the explicit forms for the Dirac matrices
( ) ( )
0 σ 1 0
α= σ 0 β = 0 − 1
3.4
Problems 85
show that φ and χ satisfy the coupled equations (E −m)φ = σ · pχ (E + m)χ = σ · pφ where pμ = (E,p).
(a) Using the explicit forms for the 2 × 2 Pauli matrices, verify the commutation (square brackets) and anticommutation (braces) rela
tion [note the summation convention for repeated indices: ∈ijk σk ≡
∑3
k=1 ∈ijk σk]:
[σi, σj ] = 2i∈ijk σk {σi, σj } = 2δij 1 where ∈ijk is the usual antisymmetric tensor
( +1 for an even permutation of 1, 2, 3
∈ijk = −1 for an odd permutation of 1, 2, 3 0 if two or more indices are the same,
δij is the usual Kronecker delta, and 1is the 2 × 2 matrix. Hence show that
σiσj = δij 1+ i∈ijk σk . (b) Use this last identity to prove the result
(σ ·a)(σ ·b) = a · b1+ iσ ·a ×b.
Using the explicit 2 ×2 form for
( )
pz px −ipy σ · p=
px +ipy − pz show that
(σ · p)2= p21.
3.5Verify the conservation equation (3.56).
3.6Check that h(p) as given by (3.66) does commute with α · p+ βm, the momentum–space free Dirac Hamiltonian.
3.7Let φ be an arbitrary two-component spinor, and let uˆ be a unit vector.
(a) Show that 1 2 (1 + σ · uˆ )φ is an eigenstate of σ · uˆ with eigenvalue +1. The operator 1 2 (1 + σ · uˆ ) is called a projector operator for the σ · uˆ = +1 eigenstate since when acting on any φ this is what it ‘projects out’. Write down a similar operator which projects out the σ ·uˆ = −1 eigenstate.
(b) Construct two two-component spinors φ+ and φ− which are eigen
states ofσ·uˆ belonging to eigenvalues±1, and normalized toφ† rφs = δrs for (r, s) = (+,−), for the case ( )1 uˆ = (sinθcosφ,sinθsinφ,cosθ) [Hint: take the arbitrary φ = 0 ].
3.8Positive-energy spinors u(p, s) are defined by
( )
φs
u(p, s) = (E +m)1/2( σ ·p ) s = 1,2 φs
E +m with φs†φs = 1. Verify that these satisfy u u † = 2E.
In a similar way, negative-energy spinors v(p, s) are defined by ( σ ·p )
χs m)1/2( E +m )
v(p, s) = (E + s = 1,2
χs with χs†χs = 1. Verify that v v † = 2E.
3.9Using the KG equation together with the replacement ∂μ →∂μ + iqAμ , find the form of the potential VˆKG in the corresponding equation
(❗ +m 2)φ = −Vˆ
KGφ in terms of Aμ .
3.10Evaluate
{σ ·(−i∇ −qA)}2ψ
by following the subsequent steps (or doing it your own way):
(a) Multiply the operator by itself to get
{(σ · −i∇)2 + iq(σ ·∇)(σ ·A) + iq(σ ·A)(σ ·∇) + q 2(σ ·A)2}ψ.
2A2
The first and last terms are, respectively, −∇2and q where the 2×2 unit matrix 1is understood. The second and third terms are iq(σ ·∇)(σ ·Aψ) and iq(σ ·A)(σ ·∇ψ). These may be simplified using the identity of problem 4.4(b), but we must be careful to treat
∇ correctly as a differential operator.
(b) Show that (σ ·∇)(σ ·A)ψ = ∇ ·(Aψ) + iσ · {∇ ×(Aψ)}. Now use
∇ ×(Aψ) = (∇ ×A)ψ −A ×∇ψ to simplify the last term.
(c) Similarly, show that (σ ·A)(σ ·∇)ψ = A ·∇ψ + iσ ·(A ×∇ψ).
(d) Hence verify (3.116).