Anders Andreassen November 20, 2015
These notes are based on Prahar Mitra’s notes from 2012.
1 Introduction
Today we will be talking about the anomalous magnetic moment. In previous classes you may have seen that the Dirac equation implies in the non-relativistic limit a Hamiltonian of the form
~p2 e H = + V (~r) + B~ · L~ + gS~ (1) 2m 2m with g = 2 at tree level and L~ , S~ the orbital and spin angular momentum respectively. g is known as the “g-factor”. Before the Dirac equation, we didn’t know why there was a difference between the coefficients of the orbital and spin angular momentum.
All right, g = 2 was the tree-level result. Are there quantum corrections to this? Yes! Let’s look at them. We are interested in how the electron interacts with an electromagnetic field. The electric charge gives us a measure of how the electron interacts with a classical electric field and the magnetic dipole tells us how the electron precesses under a classical magnetic field. So we are interested in the following diagram
µ = −ieus(q2)Γ us0 (q1) (2)
where Γµ is to be determined. What form can it take? We introduce form factors to see what the most general form could be.
2 Form factors
Consider the vertex Insert diagram where the fermions are on-shell. The amplitude of the above vertex, will in general take the form
µ µ iM = −ieu¯s(q2)Γ us0 (q1) (3) where the parameter e above is the renormalized charge (actually eR, but I will suppress the subscript) and is defined to be the charge of the electron at p = 0. In the first part of this section, we will find the most general form that the above vertex can take. The process is done as follows
1 •Mµ must transform like a vector. Since the only objects with vector indices in our problem is µ µ µ µ µ q1 , q2 , p and γ , the most general form that Γ can take is
µ µ µ µ µ Γ = f1q1 + f2q2 + f3p + f4γ (4)
where fi are scalar functions of the remaining variables in our theory, namely f ≡ f p · q , p · q , q · q , p, q , q , p2, q2, q2 (5) i i 1 2 1 2 / /1 /2 1 2
µ µ µ • Momentum must be conserved at this vertex. We can write p = q1 − q2 and get
µ µ µ µ µ µ µ µ µ Γ = f1q1 + f2q2 + f3 (q1 − q2 ) + f4γ = (f1 + f3) q1 + (f2 − f3) q2 + f4γ (6)
• The fermions are on-shell. We can then replace 1 p · q → p2 1 2 1 p · q → − p2 2 2 2 2 q1 · q2 → 2m − p p/ → 0 (7) q → m /1 q → m /2 2 2 q1 = m 2 2 q2 = m Therefore, the form factors are only functions of p2 and m2.
• Either by gauge invariance (if the photon is internal) or by the Ward Identity (if the photon is µ external), we must have pµM = 0. Thus
µ µ pµM = −ieu¯s(q2)pµΓ us0 (q1) (8) 1 2 = − p ieu¯ (q ) [(f + f ) − (f − f )] u 0 (q ) 2 s 2 1 3 2 3 s 1
Thus, (f1 + f3) = (f2 − f3) = f0 (say). We then reduce the form factor to
µ µ µ µ Γ = f0 (q1 + q2 ) + f4γ (9)
The most general form of the amplitude is then
µ µ µ µ iM = −ieu¯s(q2)[f0 (q1 + q2 ) + f4γ ] us0 (q1) (10)
• Now we use the Gordon Identity, which is
µ µ µ µν u¯s(q2)(q1 + q2 ) u(q1) =u ¯s(q2) [2mγ + iσ pν] u(q1) (11)
The general amplitude is then
µ µ µν µ iM = −ieu¯s(q2)[f0 (2mγ + iσ pν) + f4γ ] us0 (q1) µ µν (12) = −ieu¯s(q2) [(2mf0 + f4) γ + iσ pνf0] us0 (q1)
2 Let us now write everything in standard convention and redefine 1 2mf + f → F (p2, m2), f → F (p2, m2) (13) 0 4 1 0 2m 2 The amplitude is then
µν µ µ 2 2 iσ pν 2 2 iM = −ieu¯ (q ) γ F (p , m ) + F (p , m ) u 0 (q ) (14) s 2 1 2m 2 s 1
µ 1 0 • Dimensional Analysis: We must have [M ] = 1. Since [us(p)] = [¯us (p)] = 2 , the form factors 2 2 F1 and F2 are dimensionless. This must imply that they are only functions of p /m . Thus, in full generality,
2 µν 2 µ µ p iσ pν p iM = −ieu¯ (q ) γ F + F u 0 (q ) (15) s 2 1 m2 2m 2 m2 s 1
The form factors F1 and F2 above contain complete information about how electrons interact with an electromagnetic field. In particular, it contains information about the electric charge and magnetic moment of the electrons.
Important note For the rest of this section, we will assume the electron is placed in a background bg electromagnetic field Aµ (x). In our analysis, we will not quantize this field as it is assumed to a background. Think of this as the Dirac Lagrangian coupled to some classical potential in the background. The interaction Hamiltonian Z 3 ¯ µ HI = e d xψΓ ψAµ (16) where Γµ is the vertex evaluated above and therefore contains all the quantum corrections.
3 Electric Charge and Magnetic Moment of the Electron
In this section, I will explain how one extracts the charge and magnetic moment of the electron from the amplitude (15). How does one measure these quantities? For the electric charge, we simply put an electron in a background electrostatic potential φbg(~x) and look at how the electron scatters (in the non-relativistic limit). Using the Born approximation, we can then extract the non-relativistic potential that gives rise to such scattering. The potential due to electrostatic interactions is simply given by
bg V (~x) = qeφ (~x) (17)
The coefficient qe is read off as the charge of the electron. To measure the magnetic moment, we put the electron in a magnetic field and measure the precession of the electron. Again by the Born approximation, one can extract the potential which we expect to be of the form e V (~x) = −~µ · B~ = −g S~ · B~ (18) 2m We can then read of the g-factor from this expression. Let us extract these quantities from the form factors F1 and F2.
3 3.1 P&S Problem 4.4 I will be using results from P&S problem 4.4 in the rest of the section. I am listing the problem and the results here. The solution to the problem is given in the Appendix.
Prob 4.4 The cross-section for scattering of an electron by the electromagnetic field can be com- puted without quantizing the electromagnetic field. Instead, we treat the field as a given classical bg potential Aµ (x). The interaction Hamiltonian is Z 3 ¯ µ bg HI = e d xψΓ ψAµ (19)
The vertex factor here is Γµ which includes all quantum corrections to the vertex shown above.
(a) The T -matrix element is given by
0 µ ˜ hq2, s| iT |q1, s i = −ieu¯s(q2)Γ us0 (q1)Aµ(q2 − q1) (20) ˜ where Aµ is the four-dimensional Fourier transform of Aµ(x).
(b) Define the quantity Mss0 as follows
0 hq2, s| iT |q1, s i = iMss0 (2π)δ(E2 − E1) (21)
The cross-section for scattering of an electron is given by 1 1 d3p 1 2 2 (22) dσ = 3 |M| (2π)δ(E2 − E1) vi 2Ei (2π) 2E2 We can integrate out this expression to give dσ |M|2 = (23) dΩ 16π2
3.2 The Born Approximation
The Born approximation forms the crux of scattering calculations in non-relativistic quantum mechanics. I will not derive this here as you must have seen it before. The important result is
dσ m2 2 = V˜ (q − q ) (24) dΩ 4π2 2 1
3.3 Relation between the amplitude and the potential
Given the two results, we can extract |M| V˜ (q − q ) = (25) 2 1 2m A more careful analysis actually tells us that
M V˜ (q − q ) = − (26) 2 1 2m This is the important relation that we will be using in this section.
4 4 Electric Charge of the Electron
As explained above, we place the electron in a background electrostatic field Z bg ~ bg 4 bg ip·x 0 ˜ ~ Aµ = (φ(~x), 0) =⇒ Aµ (p) = d xAµ (x)e = 2πδ(p )φ(~p), 0 (27) where Z φ˜(~p) = d3xφ(~x)e−i~p·~x (28)
The amplitude of an electron scattering over this background field is then
2 0i 2 0 p iσ pi p iM 0 = −ieφ˜(~p)¯u (q ) γ F + F u 0 (q ) (29) ss s 2 1 m2 2m 2 m2 s 1
We now assume that the electrostatic field is slowly varying over a large region.1 This implies that ˜ ~ µ µ φ(~p) has support only in the neighbourhood of ~p = 0. This also implies that q1 ∼ q2 . The amplitude is then
˜ 0 iMss0 ∼ −ieφ(~p)F1(0)¯us(q1)γ us0 (q1) (30)
Now
0 † 0 † u¯s(q1)γ us0 (q1) ∼ us(q1)us0 (q1) = 2q1ξsξs0 ∼ 2mδss0 (31)
Thus
h ˜ i iMss0 = −i(2m) eφ(~p)F1(0) δss0 (32)
This immediately tells us that the interaction of the electron with the electric field is independent of its spin. Further, an electric field does not change the spin of the electron. For fixed given initial spin of the electron
h ˜ i iM = −i(2m) eφ(~p)F1(0) (33)
The potential is then
˜ ˜ V (p) = eφ(~p)F1(0) =⇒ V (~x) = eF1(0)φ(~x) (34)
Thus the charge of the electron (at zero momentum) is qe = eF1(0). However, since e is the charge of the electron exactly, we must have F1(0) = 1 (to all orders in perturbation theory!). Thus
F1(0) = 1 (exact) (35)
5 Magnetic Moment of the Electron
Consider an electron scattering of a static background vector potential
bg ~ ˜bg 0 ~˜ (36) Aµ (x) = (0, A(~x)) =⇒ Aµ (x) = (0, 2πδ(p )A(~p))
1This assumptions keeps the electric field small and hence does not disturb the electron much.
5 The scattering amplitude is then
2 iν 2 i i p iσ pν p iM 0 = ieA˜ (~p)¯u (q ) γ F + F u 0 (q ) (37) ss s 2 1 m2 2m 2 m2 s 1
Again, we assume that the magnetic field is slowly varying over a large region. The above expression therefore is only supported in the neighbourhood of ~p = 0. However, we must now be careful. The magnetic field interacts with the electron via the Lorenz force and a spin-coupling. We are only interested in extracting the term proportional to the spin coupling. As we will see below, when the µ (37) is simplified there are terms proportional to ~q1 + ~q2 and terms proportional to ~q1 − ~q2 = p . We know that the Lorenz force does not change the momentum of the electron. Therefore, terms proportional to pµ must be the ones coming from the spin coupling. Therefore, we can’t just set ~p = 0 in the term above. We should expand in small ~p and keep only leading order term.
√ √ ~p·~σ 1 − ξs p · σξs m − ~p · ~σξs √ 2m 2 us(p) = √ = √ = m + O(~p ) (38) p · σξ¯ s m + ~p · ~σξs ~p·~σ 1 + 2m ξs
Thus ~q1·~σ i 1 − ξs0 i † ~q2·~σ † ~q2·~σ 0 1 0 σ 2m u¯s(q2)γ us0 (q1) = m ξs 1 − ξs 1 + i 2m 2m ~q1·~σ 1 0 −σ 0 0 1 + 2m ξs † ~q2 · ~σ i ~q1 · ~σ † ~q2 · ~σ i ~q1 · ~σ = m −ξ 1 − σ 1 − ξ 0 + ξ 1 + σ 1 + ξ 0 s 2m 2m s s 2m 2m s † j j i j i j = ξs q2σ σ + q1σ σ ξs0 † j ji jik k j ij ijk k = ξs q2 δ + i σ + q1 δ + i σ ξs0 † i i j ijk k = ξs q1 + q2 − ip σ ξs0 (39)
Also, ipν iν pj † † 0 1 i j ξs0 u¯s(q2)σ us0 (q1) = − ξs ξs [γ , γ ] 2m 4 1 0 ξs0 k ipj ijk † † σ 0 ξs0 (40) = ξs ξs k 2 0 σ ξs0 † j ijk k = −ξs ip σ ξs0 Plugging all this into the amplitude, we get
˜i † ijk j k iM = ieA (~p)ξs −i p σ [F1(0) + F2(0)] ξs0 (41)
Given that the electron is at rest before and after the interaction, we get
˜i ijk j † k iM = ie (F1(0) + F2(0)) A (~p) −i p ξsσ ξs0 (42)
i ijk k ˜i ijk j ˜k Now since B (~x) = ∂jA (~x) =⇒ B (~p) = −i p A (~p). Thus
† k ˜k iM = ie (F1(0) + F2(0)) ξsσ ξs0 B (~p) k (43) e † σ k = −i(2m) − [F (0) + F (0)] ξ ξ 0 B˜ (~p) m 1 2 s 2 s
6 σk Recall that for spinors 2 is simply the spin matrix. Thus h e D E i iM = −i(2m) − [F (0) + F (0)] S~ · B~˜(~p) (44) m 1 2 The potential can then be read off as e D E V (~x) = − [F (0) + F (0)] S~ · B~ (45) m 1 2 The electron g-factor is then
g = 2F1(0) + 2F2(0) = 2 + 2F2(0), (exact) (46)
6 Calculating F2(0)
Now that we have a better understanding of where g = 2 + 2F2(0) comes from, we can follow the rest of the calculation in chapter 17 of Matt Schwartz’s book for calculating F2(0).
A P & S Problem 4.4 Solution http://www-personal.umich.edu/~jbourj/peskin/4-4.pdf
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