Particle Physics II -The higgs mechanism Lecture 5 review fermion by nml23533

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```									Particle Physics II – The higgs mechanism

Lecture 5: review fermion masses & quark masses in some detail

Slides based on (stolen from) the course from
Marcel Merk, Wouter Verkerke and Niels Tuning
The Standard Model Lagrangian

Introduce the massless fermion fields
Require local gauge invariance     existence of gauge bosons

Introduce Higgs potential with <φ> ≠ 0   GSM = SU (3)C × SU (2)L ×U (1)Y → SU (3)C ×U (1)Q

Spontaneous symmetry breaking            W+, W-,Z0 bosons acquire a mass

Ad hoc interactions between Higgs field & fermions

Ivo van Vulpen (2)
(2)

Notation of the fields
Fields: Notation
⎛ 1− γ 5 ⎞           ⎛ 1+ γ 5 ⎞
Fermions:              ψL = ⎜        ⎟ψ   ; ψR = ⎜        ⎟ψ         with ψ = QL, uR, dR, LL, lR, νR
⎝ 2 ⎠                ⎝ 2 ⎠
Interaction rep.
I t    ti
Quarks:                                                                                               bord

Under SU2:
Left handed doublets
•
Right hander singlets
Li
SU(3)C SU(2)L Hypercharge Y
Left-handed
generation index

•                                       •

Leptons:
•
Li

•                                       •
Scalar field:                                                         Interaction representation:
•                                             standard model Ivo van Vulpen (4)
Fields: explicitly

Explicitly:
• The left handed quark doublet                  :
⎛ urI , u g , ubI ⎞ ⎛ crI , cg , cbI ⎞ ⎛ trI , t g , tbI ⎞
I                  I                   I
T3 = + 1   2
QLi (3, 2,1 6) = ⎜ I I I ⎟ , ⎜ I I I ⎟ , ⎜ I I I ⎟
I
(Y = 1
⎜ d , d , d ⎟ ⎜ s , s , s ⎟ ⎜b ,b ,b ⎟                                         T3 = − 1   2
⎝ r g b ⎠L ⎝ r g b ⎠L ⎝ r g b ⎠L

• Similarly for the quark singlets:

u Ri (3,1, 2 3) =
I
(u , u , u ) , (c , c , c ) , (t , t , t )
I
r
I
r
I
r    R
I
r
I
r
I
r   R
I
r
I
r
I
r       R
( Y = 2 3)
d Ri (3,1, −1 3) =
I
(         )         ( d , d , d ) , ( s , s , s ) , (b , b , b )
I
r
I
r
I
r   R
I
r
I
r
I
r   R
I
r
I
r
I
r       R
(Y   = − 1 3)

⎛ν eI ⎞ ⎛ν μ ⎞ ⎛ν τI ⎞
I
T3 = + 1 2
• The left handed leptons: LLi (1, 2, − 1 2) = ⎜ I ⎟ , ⎜ I ⎟ , ⎜ I ⎟
I
⎜ ⎟ ⎜ ⎟ ⎜ ⎟                                              T3 = − 1 2
⎝ e ⎠ L ⎝ μ ⎠ L ⎝τ ⎠ L

• And similarly the (charged) singlets:                          lRi (1 1 −1) = eR , μ R ,τ R
I
(1,1,       I     I    I
(Y
Ivo van Vulpen (5)
(2)

Lagrangian:   kinetic part
L_kin 1/2

:The Kinetic Part

: Fermions + gauge bosons + interactions

Procedure: Introduce the fermion fields and demand that the theory is local gauge invariant
under SU(3)CxSU(2)LxU(1)Y transformations.

Replace:

Fields:        Gaμ : 8 gluons
Wbμ : weak bosons: W1, W2, W3
Bμ : hypercharge boson

Generators:       La : Gell-Mann matrices:          ½ λa      (3x3)     SU(3)C
σb : Pauli Matrices:              ½ τb      (2x2)     SU(2)L
Y : Hypercharge:
yp        g                                      ( )
U(1)Y

Ivo x Vulpen Y
For the remainder we only consider Electroweak: SU(2)L van U(1)(7)
L_kin 2/2

: The Kinetic Part

L kinetic : iψ (∂ μ γ μ )ψ → iψ ( D μ γ μ )ψ
with ψ = QLi , uRi , d Ri , LILi , lRi
I     I      I            I

Example: the term with QLiI becomes:

L kinetic (QLi ) = iQLiγ μ D μ QLi
I        I          I

i             i         i
= iQLiγ μ (∂ μ +
I
g s Gaμ λa + gWbμτ b + g ′B μ ) QLi
W                  I
⎛0 1⎞
2             2         6                              τ1 = ⎜    ⎟
⎝1 0⎠
⎛ 0 −i ⎞
g        y          p            q
Writing out only the weak part for the quarks:                                                            τ2 = ⎜      ⎟
⎝i 0 ⎠
I
⎛ μ i                              ⎞⎛ u ⎞
= i( u , d ) Lγ μ ⎜ ∂ + g (W1 τ 1 + W2 τ 2 + W3 τ 3 ) ⎟ ⎜ ⎟
I                 μ        μ         μ                                 ⎛1 0 ⎞
L    Weak
(u, d )   I
bord       τ3 = ⎜     ⎟
kinetic             L                                                                                     ⎝ 0 −1⎠
⎝   2                              ⎠ ⎝ d ⎠L
g I                     g I
= iuLγ μ ∂ μ uL + id L γ μ ∂ μ d L −
I           I       I       I
u L γ μW − μ d L −
I
d L γ μW + μ u L
I
−
2                       2
uLI

W+μ
g
dLI                                            L=JμWμ                                                 Ivo van Vulpen (8)
(1)

Lagrangian:   Higgs part
Higgs 1/1

: The Higgs Potential

V(φ)                                                    V(φ)
Symmetry                                           Broken symmetry

0
v
φ                                                          φ

Spontaneous Symmetry Breaking: The Higgs field adopts a non-zero vacuum expectation value

P    d
Procedure:    =       =                             Substitute:

And rewrite the Lagrangian (tedious):   1 GSM : ( SU (3)C × SU (2) L × U (1)Y ) → ( SU (3)C × U (1) EM )
1.        .
2. The W+,W-,Z0 bosons acquire mass
3. The Higgs boson H appears                  Ivo van Vulpen (10)
(4)

Lagrangian:   Yukawa part
Yukawa 1/4

: The Yukawa Part

Since we have a Higgs field we can add (ad-hoc) interactions
between Higgs field and the fermions in a gauge invariant way
doublets
The result is:                               singlet

− L Yukawa =         Yij ψ(      Li   φ     )ψ   Rj            +      h .c .

= Y       d
ij   (   Q φ d +Y
I
Li     )    I
Rj
u
ij     (    I ~
Q φ uRj + Yijl LILi φ lRj + h.c.
Li
I
)              I
(           )
i, j : indices for the 3 generations!
% = iσ φ * = ⎛ 0 1 ⎞ φ * = ⎛ φ ⎞
0

With:   φ             ⎜      ⎟      ⎜ −⎟
⎝ −1 0 ⎠      ⎝ −φ ⎠
2

(The CP conjugate of φ)

y     p                   p
are arbitrary complex matrices which operate
Yijd Yiju Yijl                   in family space (3x3)   flavour physics
Ivo van Vulpen (12)
Yukawa 2/4

: The Yukawa Part

Writing the first term explicitly:

⎛ϕ + ⎞ I
d       I   I
Y (u , d )i ⎜ 0 ⎟ d Rj
ij       L⎜ϕ ⎟
L                   =
⎝ ⎠
⎛ d I I                 ⎛ϕ + ⎞                    ⎛ϕ + ⎞              ⎛ϕ + ⎞ ⎞
(
⎜ Y11 uL , d L
⎜
)   ⎜ϕ ⎟
⎝ ⎠
d
(I
⎜ 0 ⎟ Y12 uL , d L
I
)   ⎜ 0 ⎟ Y13 uL , d L ⎜ 0 ⎟ ⎟
⎜ϕ ⎟
⎝ ⎠
d
(I     I
)⎜ϕ ⎟
⎝ ⎠⎟
⎟ ⎛ dR ⎞
I
⎜
⎛ϕ + ⎞                    ⎛ϕ + ⎞              ⎛ϕ + ⎞ ⎟ ⎜ I ⎟
⎜ 21 L L(
⎜ Y d cI , sI
)   ⎜ϕ ⎟
⎝ ⎠
d
(
I
⎜ 0 ⎟ Y22 cL , sL
I
)   ⎜ 0 ⎟ Y13 cL , sL ⎜ 0 ⎟ ⎟ • ⎜ sR ⎟
⎜ϕ ⎟
⎝ ⎠
d
(
I    I
)⎜ϕ ⎟
⎝ ⎠⎟ ⎜ I ⎟
⎜
⎜                                                                              ⎝ bR ⎠
⎛ϕ + ⎞                    ⎛ϕ ⎞                ⎛ϕ ⎞ ⎟
(           )               (         )               (     )
+                   +
⎜ Y31 t L , bL
d  I    I                   d  I    I
⎜ 0 ⎟ Y32 t L , bL        ⎜ 0 ⎟ Y33 t L , bL ⎜ 0 ⎟ ⎟
d  I    I
⎜                       ⎜ϕ ⎟                      ⎜ϕ ⎟                ⎜ϕ ⎟ ⎟
⎝                       ⎝ ⎠                       ⎝ ⎠                 ⎝ ⎠⎠

Ivo van Vulpen (13)
(3)

Fermion masses
M_fermion 1/3
SSB
: The Fermion Masses

⎛ϕ + ⎞ I
− LYuk    = Yijd (u L , d L )i ⎜ 0 ⎟ d Rj
I     I
⎜ϕ ⎟         + Yiju (...) + Yijl (...)
⎝ ⎠
Spontaneous symmetry breaking

After which the following mass term emerges:

, with
bord

Ivo van Vulpen (15)
M_fermion 2/3
SSB
: The Fermion Masses
Writing in an explicit form:
⎛dI ⎞                                      ⎛uI ⎞                                 ⎛ eI ⎞
⎛   ⎞                                                                           ⎛
LMass = ( d                     ) Md     ⎜   ⎟
⎜ I⎟
(u , c , t ) M
⎛   u⎞   ⎜ I⎟
( e , μ ,τ ) M      ⎜   l⎞
⎟
⎜ I⎟
−                                                    ⎜s ⎟ +                                     ⎜c ⎟ +                                ⎜μ ⎟      + h.c.
I
, s I , bI                                I   I   I
⎜    ⎟
I      I   I

⎜   ⎟                                     ⎜    ⎟                                ⎜    ⎟
L
⎝   ⎠   ⎜ bI ⎟
L
⎝    ⎠   ⎜ tI ⎟
L
⎝    ⎠   ⎜τ I ⎟
⎝ ⎠R                                       ⎝ ⎠ R                                 ⎝ ⎠R

The matrices M can always be diagonalised by unitary matrices
y        g          y       y                                                             VLf and VRf                   such that:
⎡                                            ⎛dI ⎞         ⎤
V M V =M
L
f           f
R
f†               f
diagonal                                  (
⎢ I I I
⎢ d , s ,b     )                             ⎜ ⎟
VLf † VLf M f VRf † VRf ⎜ s I ⎟
⎥
⎥
L
⎢                                            ⎜ bI ⎟        ⎥
⎣                                            ⎝ ⎠          R⎦

Then the real fermion mass eigenstates are given by:                                                                                                   bord

dLi = (VLd ) ⋅ dLj
I
dRi = (VRd ) ⋅ dRj
I
ij                                            ij

uLi = (VLu ) ⋅uLj
I
uRi = (VR ) ⋅uRj
u     I
ij                                       ij

lLi = (VLl ) ⋅lLj
I
lRi = (VR ) ⋅lRj
l     I
ij                                           ij

dLI , uLI , lLI        are the weak interaction eigenstates
dL , uL , lL           are the mass eigenstates (“physical particles”)
Ivo van Vulpen (16)
M_fermion 3/3
SSB
: The Fermion Masses
In terms of the mass eigenstates:
⎛ md         ⎞ ⎛d ⎞                       ⎛ mu           ⎞   ⎛u ⎞
− L Mass =    ( d , s, b )   L
⎜
⎜    ms
⎟
⎟
⎜ ⎟
(
⎜ s ⎟ + u , c, t   )   L
⎜
⎜      mc
⎟
⎟
⎜ ⎟
⎜c⎟
⎜            ⎟
mb ⎠ ⎜b⎟                        ⎜              ⎟
mt ⎠   ⎜t ⎟
⎝              ⎝ ⎠R                       ⎝                  ⎝ ⎠R
⎛ me         ⎞ ⎛e⎞
+    ( e, μ ,τ )    L
⎜
⎜    mμ
⎟⎜ ⎟
⎟ ⎜ μ ⎟ + h.c.
⎜                  ⎟
mτ ⎟ ⎜ τ ⎠ R
⎝            ⎠⎝
− L Mass =         mu uu              +     m c cc      +       mt tt
+       m d dd             +     m s ss      +       mb bb
+       m e ee            +      m μ μμ       +      mτ ττ
In flavour space one can choose:

Weak basis: The gauge currents are diagonal in flavour space, but the flavour
mass matrices are non-diagonal

Mass basis: The fermion masses are diagonal, but some gauge currents (charged
k interactions) are not di
weak i        i   )                l in flavour space
diagonal i fl

What happened to the charged current interactions (in LKinetic) ?                                Ivo van Vulpen (17)
(3)

Charged current
ChargedC 1/2

: The Charged Current
The charged current interaction for quarks in the interaction basis is:
g
− LW +       =                   I
u Li       γμ     d Li Wμ+
I

2
The charged current interaction for quarks in the mass basis is:
g
− LW +   =                  uLi VLu       γ μ VLd † d Li Wμ+
2

VCKM = (VLu ⋅ VLd † )                     VCKM ⋅ VCKM = 1
†
The unitary matrix:                                            with:
is the Cabibbo Kobayashi Maskawa mixing matrix:
⎛d ⎞
g
−LW +    =                 ( u , c , t ) L (VCKM ) ⎜ s ⎟
⎜ ⎟        γ μ Wμ+          Note: alleen down-
2                                ⎜b⎟                         type roteert
yp
⎝ ⎠L

Lepton sector: similarly                  VMNS = (VL ⋅ VLl † )
ν

However, for massless neutrino’s: VLν = arbitrary.
Choose it such that VMNS = 1  no mixing in the lepton sector
Ivo van Vulpen (19)
ChargedC 2/2

Charged Currents

The charged current term reads:
g I μ − I                  g I μ + I                 μ−         μ+
LCC =        uLiγ Wμ d Li       +       d Liγ Wμ uLi       = J CC Wμ− + J CC Wμ+
2                          2
g ⎛ 1− γ 5 ⎞ μ − ⎛ 1− γ 5 ⎞                          g    ⎛ 1− γ 5 ⎞ μ + † ⎛ 1− γ 5 ⎞
=   ui ⎜     ⎟ γ Wμ Vij ⎜   ⎟dj                  +       dj ⎜        ⎟ γ Wμ V ji ⎜    ⎟ ui
2 ⎝ 2 ⎠              ⎝ 2 ⎠                           2 ⎝ 2 ⎠                  ⎝ 2 ⎠
g                                   g
=      uiγ μWμ−Vij (1 − γ 5 ) d j   +      d j γ μWμ+Vij* (1 − γ 5 ) ui
2                                   2

Under the CP operator this gives:
(together with (x,t)   (-x,t))

g                                               g
LCC    ⎯⎯→CP
d j γ μWμ+Vij (1 − γ 5 ) ui           +         uiγ μWμiVij* (1 − γ 5 ) d j
2                                               2
CP is conserved only if        Vij = Vij*

Ivo van Vulpen (20)
ComplexPhase 1/1

Why complex phases matter
• CP conjugation of a W boson vertex involves complex
conjugation of coupling constant

W−                            W+
b                             b
gVub
b                          gV*ub
u                             u

Above process violates CP if Vub ≠ Vub*

• With 2 generations Vij is always real and Vij≡Vij*
• With 3 generations Vij can be complex
CP violation built into weak decay mechanism!
Ivo van Vulpen (21)
(3)

Standard Model Lagrangian
recap
L_recap 1/1

The Standard Model Lagrangian (recap)

• LKinetic : IIntroduce the massless ffermion fi ld
t d     th      l         i fields
Require local gauge invariance      gives rise to existence of gauge bosons

• LHiggs :
gg      Introduce Higgs potential with <φ> ≠ 0    GSM = SU (3)C × SU (2) L × U (1)Y → SU (3)C × U (1)Q

Spontaneous symmetry breaking               The W+, W-,Z0 bosons acquire a mass

Ad hoc interactions between Higgs field & fermions
CP violating with a single phase
(3)

CKM matrix in more detail

Parameters:     2
Measurements:   2
Reenking        2
Vckm_param 1/2

Ok…. we’ve got the CKM matrix, now what?

• It’s unitary
– “probabilities add up to 1”:
– d’=0.97 d + 0.22 s + 0.003 b   (0.972+0.222+0.0032=1)

• How many free parameters?
– How many real/complex?

• How do we normally visualize these parameters?

Ivo van Vulpen (25)
Vckm_meas 1/2

How do you measure those numbers?

• Magnitudes are typically determined from ratio of decay
rates
• Example 1 – Measurement of Vud
– Compare decay rates of neutron
decay and muon decay
– Ratio proportional to Vud2
|
– |Vud| = 0.9735 ± 0.0008
– Vud of order 1

Ivo van Vulpen (26)
Vckm_meas 2/2

What do we know about the CKM matrix?

• Magnitudes of elements have been measured over time
– Result of a large number of measurements and calculations

⎛ d ' ⎞ ⎛ Vud     Vus Vub ⎞ ⎛ d ⎞
⎜ ⎟ ⎜                     ⎟⎜ ⎟
⎜ s ' ⎟ = ⎜ Vcd   Vcs Vcb ⎟ ⎜ s ⎟
⎜ b'⎟ ⎜V          Vts Vtb ⎟ ⎜ b ⎟
⎝ ⎠ ⎝ td                  ⎠⎝ ⎠
t
4 parameters
•3 real
•1 phase

⎛ Vud     Vus   Vub ⎞ ⎛ 0.9738 ± 0.0002 0.227 ± 0.001 0.00396 ± 0.00009 ⎞
⎜                   ⎟ ⎜                                                   ⎟
⎜ Vcd     Vcs   Vcb ⎟ = ⎜ 0.227 ± 0.001 0.9730 ± 0.0002 0.0422 ± 0.0005 ⎟
⎜V
⎝ td      Vts   Vtb ⎟ ⎜ 0.0081 ± 0.0005 0.0416 ± 0.0005 0.99910 ± 0.00004 ⎟
⎠ ⎝                                                   ⎠
only,
Magnitude of elements shown only no information of phase

Ivo van Vulpen (27)
Vckm_ranking 1/2

Exploit apparent ranking for a convenient parameterization
• Given current experimental precision on CKM element values,
we usually drop λ4 and λ5 terms as well
– Effect of order 0.2%...

⎛        λ2                               ⎞
⎜    1−              λ      Aλ ( ρ − iη ) ⎟
3

⎜         2                               ⎟⎛ d ⎞
⎛ d′⎞
⎜ ′⎟      ⎜                     λ2                  ⎟⎜ ⎟
⎜s ⎟     =⎜      −λ          1−          Aλ  2
⎟⎜ s ⎟
⎜ b′ ⎟    ⎜ 3                    2                  ⎟⎜ b ⎟
⎝ ⎠L      ⎜ Aλ (1 − ρ − iη ) − Aλ 2        1        ⎟ ⎝ ⎠L
⎜                                         ⎟
⎝                                         ⎠

f    k     f t     d d              (λ
• Deviation of ranking of 1st and 2nd generation ( vs λ2)
parameterized in A parameter
• Deviation of ranking between 1st and 3rd generation,
parameterized through |ρ-iη|
i
• Complex phase parameterized in arg(ρ-iη)
Ivo van Vulpen (28)
Vckm_ranking 2/2

Approximately diagonal form

• Values are strongly ranked:
– Transition within generation favored
– Transition from 1st to 2nd generation suppressed by cos(θc)
– Transition from 2nd to 3rd generation suppressed bu cos2(θc)
– Transition from 1st to 3rd generation suppressed by cos3(θc)

g
CKM magnitudes
d     s    b              Why the ranking?
We don’t know (yet)!
u           λ         λ3

If you figure this out,
c     λ               λ2         you will win the nobel
prize
t
λ3    λ2

λ=cos(θc)=0.23
Ivo van Vulpen (29)

```
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