Solutions to Problems 2 and 31

4447, The Standard Model of Elementary ParticlePhysics, 2011/2012

Lecturer: Stefan Sint

Problem 2, Noether currents and conservation laws

Suppose the Lagrangian L(ϕ, ∂µϕ) has a global symmetry under a group Gwith (hermitian) generators T a, (a = 1.2, . . . , nG), satisfying the Lie algebra

[T a, T b] = ifabcT c (1)

This means that the infinitesimal transformation ϕk → ϕk + iεaT aklϕl leaves

L invariant. Using the equations of motion and the Leibniz rule one obtains

0 = δaL (2)

=∂L

∂ϕk

iT aklϕl +

∂L

∂(∂µϕk)iT a

kl∂µϕl (3)

= ∂µJaµ (4)

where

Jaµ(x) =

∂L

∂(∂µϕk)iT a

klϕl. (5)

The conservation of the current implies that there is a time-independent(i.e. conserved) charge,

Qa =

∫

d3x Ja0 (x), ∂0Q

a = 0. (6)

In canonically quantized QFT the charges become operators in Hilbert spaceand form a representation of the Lie algebra in the Hilbert space, i.e.

[Qa, Qb] = ifabcQc. (7)

Important examples for such charges are the baryon number, lepton numberor lepton number for a given lepton family (e, µ, τ). These cases are obtainedfrom global U(1) symmetries.

1Solution to problem 1 has been discussed in the tutorial in March 2012.

1. The Lagrangian density of a massless fermion doublet has a globalSU(2)×SU(2) symmetry. Work out the corresponding Noether cur-rents.

2. The full symmetry of this Lagrangian density includes additional U(1)×U(1)phase symmetries, which is tantamount of saying that the global sym-metry really is U(2)×U(2). Work out the corresponding left/right-handed Noether currents of the additional U(1) symmetries.

Note: in QFT only the sum of these 2 currents (the vector current) isconserved, the axial vector current is anomalous, i.e. the correspondingsymmetry does not survive the quantization procedure in QFT.

3. Show that the SM Lagrangian density is invariant under the globalphase transformations

ψL → exp(iα)ψL, ψL → ψL exp(−iα), (8)

ψR → exp(iα)ψR, ψR → ψR exp(−iα), (9)

(10)

where ψ stands for any of the leptons in a given lepton family. Workout the Noether current for such a transformation. The correspondingconserved charge is the Lepton number for the given lepton family. Isthere a similar symmetry for a family of quarks?

Solution

1. The Lagrangian density for a massless fermion doublet ψ can be writtenas

L(x) =4

∑

α,β=1

2∑

f=1

ψα,f (x)i(γµ)αβ∂µψβ,f (x) = ψ(x)iγµ∂µψ(x) (11)

where the last notation hides the flavour (f, g, . . .) and spin indices(α, β, . . .). Massless fermions can be further decomposed into left andright handed components by means of the projections:

ψL =1

2(1− γ5)ψ, ψR =

1

2(1 + γ5)ψ (12)

and

ψL = ψ1

2(1 + γ5), ψR = ψ

1

2(1− γ5). (13)

Using this notation one gets

L(x) = ψL(x)iγµ∂µψL(x) + ψR(x)iγ

µ∂µψR(x) (14)

and therefore a decoupling of left and right handed components. Onemay therefore transform left and right handed components using inde-pendent SU(2) transformations UL and UR:

ψL,R → UL,RψL,R (15)

ψL,R → ψL,RU†L,R (16)

Expanding the SU(2) elements around the identity

UL,R = exp

[

i

3∑

a=1

ωaL,R

τa

2

]

≈ 1 + i

3∑

a=1

ωaL,R

τa

2+ . . . (17)

one obtains the infinitesimal left handed variations of the fields as

δaLψL = i

3∑

a=1

ωaL

τa

2ψL, δaLψL = ψL(−i)

3∑

a=1

ωaL

τa

2, (18)

whereasδaLψR = 0 = δaLψR. (19)

The infinitesimal right handed variations are obtained by exchangingL ↔ R. Applying the Noether procedure one then obtains the corre-sponding symmetry currents2

Laµ = −ψLγµ

τa

2ψL, Ra

µ = −ψRγµτa

2ψR. (20)

2. The additional U(1) symmetries are simply obtained by a transforma-tion where the Pauli matrices (τa)fg are replaced by the unit matrix inflavour space, hence

ψL,R → exp(iωL,R)ψL,R, ψL,R → ψL,R exp(−iωL,R) (21)

2The annoying minus signs disappear if the spinors are assumed to be anticommuting

Grassmann fields. Since we did not assume this in class I here treated the spinor compo-

nents the same way as a scalar field. Note that after quantization the normalization of

the currents is fixed by the Lie algebra between the corresponding charges.

The corresponding currents are

Lµ = −ψLγµψL, Rµ = −ψRγµψR. (22)

The vector and axial currents are given by

Vµ = Rµ + Lµ = −ψγµψ, Aµ = Rµ − Lµ = −ψγµγ5ψ. (23)

In gauge theories axial U(1) charges are afflicted by anomalies: thismeans that the classical symmetry of the lagrangian density cannot bemaintained in the quantized theory. The U(1) vector charge does notsuffer from this problem and corresponds to fermion number (in QFTit is shown that it counts the number of fermionic modes).

3. Assuming neutrinos to be strictly massless (this is a good approxima-tion but not exactly true!) the Noether current for the electron leptonfamily (other families are analogous) is obtained as

Jµ = −ψνe,Lγµψνe,L − ψeγµψe (24)

The conserved charge in this case is the lepton number of the electronfamily. Its conservation implies that the electron lepton number onboth sides of any reaction must be the same, for instance in β-decayn→ p+ e− + νe this is the case as the electron lepton number of e− is1 and of νe is −1, adding to zero on both sides of the reaction. Thusthe neutrino must be an electron anti-neutrino, the conservation lawwould be violated if it was a muon or tau (anti-) neutrino,

In the quark sector there is mixing between the families and no suchquark number conservation per family exists. The electron (muon, tau)lepton numbers are only approximately conserved, as there is also somemixing between the families. This in turn is evidence for neutrinos notbeing exactly massless.

One may also define a total lepton number, counting all leptons regard-less of family. Similarly one may define a total quark number, whichis usually referred to as baryon number, since baryons are the boundstates of quarks and observed in experiments. However, total baryonnumber B and lepton number L are not conserved separately in theStandard Model but their difference B − L is.

Problem 3, Yukawa potential

Given the classical potential

V (r) =

∫

d3q

(2π)3V (q2)e−i~q·~r, V (q2) =

−4πα

q2 +m2(25)

with r = |~r|, q = |~q|. Compute the integral for constant α and show that

V (r) = −αe−mr

r(26)

Here m is the mass of the exchanged bosonic particle and the function V (q2)arises from the boson propagator.

Hint: introduce spherical coordinates, integrate over the angles and sym-metrically extend the integration for the radial coordinate to the real axis.Then apply the residue theorem.

Solution:

Using spherical coordinates and extending the integration to the real axisone arrives at

V (r) = −α

πr

∫ ∞

−∞

dqq sin(qr)

q2 +m2(27)

Decomposing the denominator, q2 + m2 = (q + im)(q − im) the poles areseen to be located on the imaginary axis, either in the upper or the lowerhalf plane. Rewriting sin(qr) in terms of exponentials one obtains

V (r) = −α

2πir

∫ ∞

−∞

dq

[

f+(q)eiqr

q − im− f−(q)

e−iqr

q + im

]

, (28)

with f±(q) = q/(q ± im). Now, the 2 integrals can be completed to contourintegrals by integrating over a half circle of infinite radius either clockwise(lower half plane) or anti-clockwise (upper half plane). The section of thecontour integral along the semicircle can be seen to vanish as the radius issent to infinity due to the exponentials in the integrand. So the integralsover the real axis are equal to the contour integrals and the residue theoremcan thus be applied:

V (r) = −α

re−mr [f+(im) + f−(−im)] = −

α

re−mr. (29)

Hence the boson mass m determines the range of the interactions and theheavy mass of the weak gauge bosons explains the weakness of the weakinteractions.