Advanced Particle Physics 04/05

Dr Gavin Davies - Problem Sheet 6 Answers

1. The free Lagrangian density for the two massless fermions is

L = iψ ν γ µ (∂µ ψν ) + iψ e γ µ (∂µ ψe )
à !
³ ´
µ ψν
= i ψν ψe γ ∂µ
ψe
= iΨγ µ (∂µ Ψ)

For a special unitary transformation

Ψ0 = U Ψ

where
U = e−iαi σi /2
then, since U is unitary, meaning U † U = I, and U 6= U (xµ )
0
L0 = iΨ γ µ (∂µ Ψ0 )
= iΨU † γ µ (∂µ U Ψ)
= iΨγ µ (∂µ Ψ) = L

Hence, the Lagrangian density is invariant to tranformations under U . Using

∂Ψ i ∂Ψ i
= − σi Ψ, = Ψσi
∂αi 2 ∂αi 2
and
∂L ∂L
= iΨγ µ , =0
∂(∂µ Ψ) ∂(∂µ Ψ)
then the Nöther currents associated with these transformations are
i i 1
JVµ i = (iΨγ µ )(− σi Ψ) + ( Ψσi )(0) = Ψγ µ σi Ψ
2 2 2
In terms of ψν and ψe , these are therefore
1³ ´
JVµ 1 = ψ ν γ µ ψe + ψ e γ µ ψν
2
i³ ´
JVµ 2 = ψ e γ µ ψν − ψ ν γ µ ψe
2
1³ ´
JVµ 3 = ψ ν γ µ ψν − ψ e γ µ ψe
2

The corresponding axial currents are

µ 1³ ´
JA1 = ψ ν γ µ γ 5 ψe + ψ e γ µ γ 5 ψν
2
µ i³ ´
JA2 = ψ e γ µ γ 5 ψν − ψ ν γ µ γ 5 ψe
2
µ 1³ ´
JA3 = ψ ν γ µ γ 5 ψν − ψ e γ µ γ 5 ψe
2

1
and hence we can form the combinations
1 µ
Jiµ = µ
(J − JAi )
2 Vi
which are also conserved. Under local gauge invariance, these currents can couple to fields
Wiµ . The charged interaction term for the currents J1µ and J2µ is

LC = gW W1µ J1µ + gW W2µ J2µ

gW 1 1
· ¸
= W1µ ψ ν γ µ (1 − γ 5 )ψe + ψ e γ µ (1 − γ 5 )ψν +
2 2 2
gW 1 µ1
· ¸
µ 5 5
W2µ i ψ e γ (1 − γ )ψν − ψ ν γ (1 − γ )ψe
2 2 2
gW 1 µ1
· ¸
µ 5 5
= (W1µ − iW2µ ) ψ ν γ (1 − γ )ψe + (W1µ + iW2µ ) ψ e γ (1 − γ )ψν
2 2 2
µ√ √
gW 1 1
¶
= 2Wµ− ψ ν γ µ (1 − γ 5 )ψe + 2Wµ+ ψ e γ µ (1 − γ 5 )ψν
2 2 2
gW 1 1
· ¸
− µ 5 + µ 5
= √ Wµ ψ ν γ (1 − γ )ψe + Wµ ψ e γ (1 − γ )ψν
2 2 2
gW h − µ + µ
i
= √ Wµ ψ νL γ ψeL + Wµ ψ eL γ ψνL
2
as is observed.
The current which interacts with the Y µ field is generally

JYµ = aψ νL γ µ ψνL + bψ νR γ µ ψνR + cψ eL γ µ ψeL + dψ eR γ µ ψeR

The total neutral interaction term is then

gY
LN = gW W3µ J3µ + Yµ JYµ
2
In terms of the left- and right-handed terms, this is
gW gW gY
LN = W3µ ψ νL γ µ ψνL − W3µ ψ eL γ µ ψeL + a Yµ ψ νL γ µ ψνL
2 2 2
gY gY gY
+ b Yµ ψ νR γ ψνR + c Yµ ψ eL γ ψeL + d Yµ ψ eR γ µ ψeR
µ µ
2 2 2
Rotating the W3µ and Y µ fields into each other, then

Z µ = W3µ cos θW − Y µ sin θW , Aµ = W3µ sin θW + Y µ cos θW

which can be inverted to give

W3µ = Z µ cos θW + Aµ sin θW , Y µ = −Z µ sin θW + Aµ cos θW

Therefore
gW gW
LN = (Zµ cos θW + Aµ sin θW ) ψ νL γ µ ψνL − (Zµ cos θW + Aµ sin θW ) ψ eL γ µ ψeL
2 2
gY gY
+ a (−Zµ sin θW + Aµ cos θW ) ψ νL γ µ ψνL + b (−Zµ sin θW + Aµ cos θW ) ψ νR γ µ ψνR
2 2
gY gY
+ c (−Zµ sin θW + Aµ cos θW ) ψ eL γ µ ψeL + d (−Zµ sin θW + Aµ cos θW ) ψ eR γ µ ψeR
2 2

2
which can be expressed as
1 1
LN = Aµ (gW sin θW + agY cos θW ) ψ νL γ µ ψνL + bAµ gY cos θW ψ νR γ µ ψνR
2 2
1 1
+ Aµ (−gW sin θW + cgY cos θW ) ψ eL γ ψeL + dAµ gY cos θW ψ eR γ µ ψeR
µ
2 2
1 1
+ Zµ (gW cos θW − agY sin θW ) ψ νL γ µ ψνL − bZµ gY sin θW ψ νR γ µ ψνR
2 2
1 1
+ Zµ (−gW cos θW − cgY sin θW ) ψ eL γ ψeL − dZµ gY sin θW ψ eR γ µ ψeR
µ
2 2
The neutrino couplings to the photon have to be zero for both the left- and right-handed
parts, so
gW sin θW + agY cos θW = 0, b=0
The electron terms must be equal to the QED coupling
1 ³ ´
−eAµ ψ e γ µ ψe = −eAµ ψ e γ µ (1 − γ 5 + 1 + γ 5 )ψe = −eAµ ψ eL γ µ ψeL + ψ eR γ µ ψeR
2
so
1 1
(−gW sin θW + cgY cos θW ) = −e, dgY cos θW = −e
2 2
All the above are satisfied by

gW sin θW = gY cos θW , e = gW sin θW

and
a = −1, b = 0, c = −1, d = −2
as can be verified by direct substitution.
The general term for the Z coupling to neutrinos is
gZ h i
Zµ cνL ψ νL γ µ ψνL + cνR ψ νR γ µ ψνR
2
Since b = 0, the Z coupling to the neutrino is purely left-handed and the term is
gZ 1
Zµ cL ψ νL γ µ ψνL = Zµ (gW cos θW + gY sin θW ) ψ νL γ µ ψνL
2 2
gW sin θW 1
µ ¶
= Zµ gW cos θW + sin θW ψ γ µ ψνL
cos θW 2 νL
gW ´1
µ ¶³
= Zµ cos2 θW + sin2 θW ψ γ µ ψνL
cos θW 2 νL
1 gW
µ ¶
= Zµ ψ νL γ µ ψνL
2 cos θW

and so has a strength gZ = gW / cos θW and couplings

cνL = 1, cνR = 0

Using cV = (cL + cR )/2 and cA = (cL − cR )/2, these are

1 1
cνV = , cνA =
2 2

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 The Z coupling terms to the electron are both left- and right-handed
1
Zµ (−gW cos θW + gY sin θW ) ψ eL γ µ ψeL + Zµ gY sin θW ψ eR γ µ ψeR
2
gW sin θW 1 gW sin θW
µ ¶
= Zµ −gW cos θW + sin θW ψ eL γ µ ψeL + Zµ sin θW ψ eR γ µ ψeR
cos θW 2 cos θW
1 gW
µ ¶ h³ ´ i
= Zµ − cos2 θW + sin2 θW ψ eL γ µ ψeL + 2 sin2 θW ψ eR γ µ ψeR
2 cos θW
1 gW
µ ¶ h³ ´ i
= Zµ −1 + 2 sin2 θW ψ eL γ µ ψeL + 2 sin2 θW ψ eR γ µ ψeR
2 cos θW
Hence, the overall strength gZ is the same and the couplings are

ceL = −1 + 2 sin2 θW , ceR = 2 sin2 θW

so that
1 1
ceV = − + 2 sin2 θW , ceA = −
2 2
2. The general formula for the partial width is
|M |2 ρ
Γ=
2MZ
with the phase space for the two-body decay, which is independent of solid angle, being
1
ρ=
8π
Since the matrix element for the Z to decay to f f is given as
gW2 M2 ³ ´
h|M |2 i = Z
c 2
f V + c 2
f A
3 cos2 θW
the partial width is
gW2 M2 ³ ´ 1 1 2 M
gW ³ ´
Z 2 2 Z 2 2
Γ(Z → f f ) = c + c = c + c
3 cos2 θW f V fA
8π 2MZ 48π cos2 θW f V fA

Using
MW
cos θW =
MZ
then the partial width becomes
2 M3 ³
gW ´ 2
gW MZ3 ³ 2 ´ GF MZ3 ³ 2 ´
Z 2 2 2 2
Γ(Z → f f ) = 2 c fV + c fA = 2 6π c fV + c fA = √ c fV + c fA
48πMW 8MW 2 6π

For any neutrino, cνV = 0.5 and cνA = 0.5 so the partial width is

Γ(Z → νe ν e ) = 0.166 GeV

This is the same for the muon and tau neutrinos also.
The mass of the Z is less than twice the top mass so the allowed decays of the Z are to
all the charged leptons and neutrinos, Z → l + l− and Z → νl ν l , and all the quarks except
the top, Z → qq. The charged leptons have ceV = −0.037 and ceA = −0.5 so the partial
width is
Γ(Z → e+ e− ) = 0.083 GeV

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 and again is the same for the muon and tau also.
The u and c quarks have cuV = 0.192 and cuA = 0.5 so for one colour

Γ(Z → uu) = 0.095 GeV

For d, s and b, cdV = −0.346 and cdA = −0.5, so

Γ(Z → dd) = 0.123 GeV

The Z width to hadrons for all three colours is therefore

Γ(Z → hadrons) = 6Γ(Z → uu) + 9Γ(Z → dd) = 1.677 GeV

The visible partial width is thus

Γvisible = 3Γ(Z → e+ e− ) + Γ(Z → hadrons) = 1.926 GeV

The invisible width can be determined from the measured visible and total widths i.e.

Γinvisible = ΓZ − Γvisible

Above technique was used to constrain the number of light neutrinos and hence the number
of generations.
The total width is

ΓZ = 3Γ(Z → νe ν e ) + 3Γ(Z → e+ e− ) + Γ(Z → hadrons) = 2.424 GeV

which corresponds to a lifetime of 2.7 × 10−25 s. The leptonic branching fractions are each

Γ(Z → νe ν e ) Γ(Z → e+ e− )
B(Z → νe ν e ) = = 6.8%, B(Z → e+ e− ) = = 3.4%
ΓZ ΓZ
and the branching fraction to hadrons is

Γ(Z → hadrons)
B(Z → hadrons) = = 69.1%
ΓZ
The fractions of hadronic events containing c and b quarks are

3Γ(Z → cc) 3Γ(Z → bb)

Rc = = 17.0%, Rb = = 22.0%,
Γ(Z → hadrons) Γ(Z → hadrons)

The total width and branching fraction to hadrons are very similar to those for W ± decay
as might be expected from SU(2) symmetry, although the detailed quark production rates,
particularly for b quarks, are very different.

3. (i) The coupling of the Higgs to a fermion is proportional to the fermion mass. Hence,
the amplitude for a process involving this will go as the mass, while the rate goes as
the square of the mass.
(ii) Principal production mode for the Higgs at LEP2 is e+ e− → Z 0∗ → Z 0 H and so
Feynman diagram is as overleaf.
For a given centre of mass energy, Ecm , then the Higgs mass must be mH < Ecm −MZ ,
which for E = 209 GeV means mH < 118 GeV.

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 e- Z

Z*

e+ H0

From energy and momentum conservation in the centre of mass

Ecm = EH + EZ pH = p Z

Squaring the second of these

2
EH − m2H = EZ2 − m2Z

so
m2H − m2Z = EH
2
− EZ2 = (EH + EZ )(EH − EZ ) = Ecm (EH − EZ )
Hence
m2H − m2Z
EH − E Z =
Ecm
Adding this to the energy conservation equation gives

m2H − m2Z
2EH = Ecm +
Ecm
so
2 + m2 − m2
Ecm H Z
EH =
2Ecm
For mH = 115 GeV, EH = 116.3 GeV.
At the kinematic limit, the Higgs and Z 0 would be produced at rest and so have zero
phase space and so zero cross-section. Even at 115 GeV, the energy is only slightly
greater than the mass and so the phase space is small. The highest limit possible at
LEP2 would be somewhat below the kinematic value, at about 115GeV.
(iii) A Higgs of 115 GeV can decay into any quark-antiquark pair, except for tt, or any
lepton-antilepton pair. It is too light to decay to Z 0 Z 0 or W + W − . Hence, since the
rate to any pair is proportional to m2 and neglecting any differences in the phase
space since mH À 2mf , then the branching ratio to a fermion pair f f is

m2f
B(f f ) = P 2
f mf

where each of the quarks must be included three times in the sum to account for
colour. The main decay mode is clearly the one to the heaviest particle, which is the
b quark, so the dominant decay is H → bb. With

m2f = 62.1 GeV2

X

6
 then
3m2
B(bb) = P b 2 = 0.852
f mf

Answer for experimental identification should include

– Higgs is predominantly decaying to b-quarks, so need to identify these to separate
from backgrounds involving light quarks.
– B-quarks have non negligible lifetime, so can travel far enough to be identified
via tracks with a large impact parameter or a secondary vertex.
– Reconstruct the Higgs mass from the jet masses.
– Can use the Z decay products to help reject backgrounds, in particular two
electrons / muons or missing energy totalling the Z mass.
(iv) A Higgs of 250GeV will predominantly decay to W or Z pairs. Thus easiest way to
identify these will be through their leptonic decay modes, particularly at the hadron
colliders. ZZ decay to four muons is the gold-plated discovery channel for a higgs of
this mass at LHC.

4. (i) The Feynman diagram for e− µ− → e− µ− scattering is

e- e-
e

e
µ- µ-

There are two vertices in the diagram, each with a power of e, so the amplitude is
proportional to e2 and hence the cross section to e4 or α2 , as given.
(ii) Since the reaction is elastic, the electron energy E does not change. By scattering
through an angle θ, then the momentum changes by p(1 − cos θ) along the initial
electron direction and p sin θ perpendicular to it. Hence, the change in the four-
momentum is

q 2 = −p2 sin2 θ − p2 (1 − cos θ)2

= −p2 sin2 θ − p2 − p2 cos2 θ + 2p2 cos θ = −2p2 (1 − cos θ)

In the centre-of-mass, the muon momentum is also p and, neglecting masses, then
the square of the centre-of-mass energy s = 4p2 , so
1
q 2 = − s(1 − cos θ)
2
Hence
d 2 d
2
=
d(q ) s d(cos θ)
Also, using cos θ = 1 − 2 sin2 (θ/2), then

q 2 = −s sin2 (θ/2)

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 or
q2
sin2 (θ/2) = −
s
so that
q2
cos2 (θ/2) = 1 − sin2 (θ/2) = 1 +
s
Therefore the cross section becomes
¤2
2 πα2 1 + 1 + (q 2 /s)
£
dσ 2 dσ
= =
d(q 2 ) s d(cos θ) s s (q 4 /s2 )
" #
2πα2 q2 q4
= 4
1+1+2 + 2
q s s
" Ã !#
2πα2 q 4 q2
= + 2 1 +
q4 s2 s

(iii) For a quark with momentum fraction x, then in the ep centre-of-mass, the total eq
energy and momentum are

Eeq = p + xp = (1 + x)p, Peq = p − xp = (1 − x)p

so the eq centre-of-mass energy is

2 2
ŝ = Eeq − Peq = (1 + x)2 p2 − (1 − x)2 p2
= (1 + 2x + x2 − 1 + 2x − x2 )p2 = 4xp2 = xs

(iv) The cross section for scattering from quark type i with fractional momentum x is
¶2 " Ã !# ¶2 " Ã !#
dσ 2πα2 ei q4 q2 2πα2 ei q4 q2
µ µ
2
= +2 1+ = +2 1+
d(q ) q4 e ŝ 2 ŝ q4 e 2
x s 2 xs

so the total cross section is

" Ã !#
dσ 2πα2 X ei 2 q4 q2
µ ¶
= p i (x) + 2 1 + dx
d(q 2 ) q4 i e x2 s2 xs

or " Ã !#
dσ 2πα2 X ei 2 q4 q2
µ ¶
= p i (x) + 2 1 +
dx d(q 2 ) q4 i e x2 s2 xs
Comparing with the conventional expression for the cross section, then
X µ e i ¶2 F2 (x, q 2 ) X ei µ ¶2
2
F1 (x, q ) = pi (x), =2 pi (x)
i
e x i
e

Hence, the quark model predicts the structure functions are related by

F2 (x, q 2 ) = 2xF1 (x, q 2 )

5. (a) The oscillations are due to a “beat” effect of having different frequencies. With all
neutrinos having zero mass, all the phases would remain equal at all times.

8
(b) Solving the two equations for νµ and ντ , then

νµ cos θ = ν1 cos2 θ − ν2 sin θ cos θ, ντ sin θ = ν1 sin2 θ + ν2 sin θ cos θ

so
ν1 = νµ cos θ + ντ sin θ
Similarly

νµ sin θ = ν1 sin θ cos θ − ν2 sin2 θ, ντ cos θ = ν1 sin θ cos θ + ν2 cos2 θ

so
ν2 = −νµ sin θ + ντ cos θ
An initially pure muon neutrino beam at time t = 0 is in a state

ψ(0) = νµ = ν1 cos θ − ν2 sin θ

Each of the states νi change with time according to the standard quantum mechanical
time dependence e−Ei t , so at a later time t, the state is

ψ(t) = ν1 e−E1 t cos θ − ν2 e−E2 t sin θ

Substituting for ν1 and ν2 , then this is

ψ(t) = (νµ cos θ + ντ sin θ)e−E1 t cos θ − (−νµ sin θ + ντ cos θ)e−E2 t sin θ
= νµ (e−E1 t cos2 θ + e−E2 t sin2 θ) + ντ (e−E1 t sin θ cos θ − e−E2 t sin θ cos θ)

so the amplitude for muon neutrinos is

Aµ = e−iE1 t cos2 θ + e−iE2 t sin2 θ

and for tau neutrinos is

³ ´
Aτ = cos θ sin θ e−iE2 t − e−iE1 t

(c) Writing
E2 + E 1 E2 − E 1 E2 + E 1 E2 − E 1
E2 = + , E1 = −
2 2 2 2
then the amplitude for having a tau neutrino in the beam is
h i
Aτ = cos θ sin θ e−i(E2 +E1 )t/2 e−i(E2 −E1 )t/2 − e−i(E2 +E1 )t/2 ei(E2 −E1 )t/2
1 h i
= sin 2θe−i(E2 +E1 )t/2 e−i(E2 −E1 )t/2 − ei(E2 −E1 )t/2
2
1
= −i sin 2θe−i(E2 +E1 )t/2 sin [(E2 − E1 )t/2]
2
Hence, the probability of having a tau neutrino is

(E2 − E1 )t
· ¸
2 2 2
Pτ = |Aτ | = sin 2θ sin
2
The general equation for the energy and momentum is
q ³ ´1/2
E= p2 + m 2 = p2 + m 2

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 If E is much greater than m, then so is p, so approximating the square root by a
binomial expansion gives
à !1/2 à !
m2 1 m2 m2
E =p 1+ 2 ≈p 1+ =p+
p 2 p2 2p
Hence E ≈ p and
m22 − m21 ∆(m2 ) ∆(m2 )
E2 − E 1 ≈ = ≈
2p 2p 2E
The distance gone is l = βt, but as the energy is much greater than the mass, then
the velocity is effectively β = 1, so
" #2
2 2 ∆(m2 )l
Pτ ≈ sin 2θ sin
4E
(d) The cosmic rays interact strongly in the atmosphere and so produce many pions. Of
these, the charged pions then mostly decay to a muon and a muon neutrino
π − → µ− ν µ
and its charge conjugate. The muons themselves subsequently decay to an electron,
an electron neutrino and another muon neutrino
µ− → e − ν e ν µ
and its charge conjugate. Hence, every charged pion results in two muon neutrinos
and one electron neutrino.
Under the hypothesis of maximal mixing, the muon neutrinos have mixed to (unde-
tected) tau neutrinos and so the ratio of muon to electron neutrinos can be less than
two. The probability of a muon neutrino remaining in the beam is Pµ = 1−Pτ = R/2,
where R is the muon to electron neutrino ratio. For neutrinos coming vertically down-
wards, the limit on Pτ is therefore given by
Pτ < 1 − 0.5R = 0.1
For maximal mixing, sin2 2θ = 1, so
" #
∆(m2 )l
sin2 < 0.1
4E
or
∆(m2 )l
< 0.32 rad
4E
Hence, with E = 0.5 GeV and l the thickness of the atmosphere = 20 km or 1.01×10 20
GeV−1 , then
∆(m2 ) < 6 × 10−21 GeV2 = 6 × 10−3 eV2
For neutrinos coming vertically upwards, then the sin2 term must be averaging to 0.5
as the ratio is independent of the angle (and hence path length) and energy. Hence,
∆(m2 )l
À 2π rad
4E
so with l the diameter of the Earth = 12800 km or 6.5 × 1022 GeV−1 , then
∆(m2 ) À 2 × 10−22 GeV2 = 2 × 10−4 eV2
These set the limits on ∆(m2 ).

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