Download F34TPP Particle Physics 1 Lecture one

F34TPP Particle Physics
1
Lecture one
1.
γ µ ∂µ ψ +
mc
ψ = 0,
~
(1.1)
2.
∂µ ∂ µ φ −
mc 2
φ = 0
(1.2)
~
3. As the spin s already has units of angular momentum (i.e. units of ~) this equation needs
no extra factors.
4. take m and replace it with
mc
.
~
5. V µ Vµ = V 0 V0 + V 1 V1 + V 2 V2 + V 3 V3 = V0 V 0 + V1 V 1 + V2 V 2 + V3 V 3 = Vν V ν .
6. Mµ ρ = ηµν η νρ is a matrix with rows labelled by µ and columns labelled by ρ, e.g. M0 0 =
η0ν η ν0 = η00 η 00 + η01 η 10 + η02 η 20 + η03 η 30 = η00 η 00 = 1. If we continue to look at all
components we will see that Mµ ρ = diag(1, 1, 1, 1), which is just the Kronecker delta.
7.
Vµ = ηµν V ν
⇒ η ρµ Vµ = η ρµ ηµν V ν
= δνρ V ν
= Vρ
(1.3)
(1.4)
(1.5)
(1.6)
η µν ηµν = η 00 η00 + η 01 η01 + ...
= η 00 η00 + η 11 η11 + η 22 η22 + η 33 η33
= 4
(1.7)
(1.8)
(1.9)
8.
2
Lecture two
1. For example,
γ5γ1 =
=
=
=
iγ 0 γ 1 γ 2 γ 3 γ 1
−iγ 0 γ 1 γ 2 γ 1 γ 3
iγ 0 γ 1 γ 1 γ 2 γ 3
−iγ 1 γ 0 γ 1 γ 2 γ 3
= −γ 1 γ 5
1
(2.10)
(2.11)
(2.12)
(2.13)
(2.14)
(2.15)
the rest follow a similar pattern.
2.
(γ 5 )2 =
=
=
=
=
=
=
=
−γ 0 γ 1 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3
γ0γ1γ2γ3γ3γ0γ1γ2
γ0γ1γ2γ0γ1γ2
γ0γ1γ2γ2γ0γ1
γ0γ1γ0γ1
−γ 0 γ 1 γ 1 γ 0
−γ 0 γ 0
I
(2.16)
(2.17)
(2.18)
(2.19)
(2.20)
(2.21)
(2.22)
(2.23)
3.
T r(γ µ ) =
=
=
=
=
=
T r(Iγ µ )
T r((γ 5 )2 γ µ )
T r(γ 5 γ 5 γ µ )
T r(γ 5 γ µ γ 5 ) (by cyclic property)
−T r(γ 5 γ 5 γ µ ) (as γ 5 anticommutes with γ µ )
−T r(γ µ )
(2.24)
(2.25)
(2.26)
(2.27)
(2.28)
(2.29)
(2.30)
so the result follows
4.
1
1
T r(γ µ γ ν ) = T r(γ µ γ ν ) + T r(γ µ γ ν )
2
2
(2.31)
1
1
T r(γ µ γ ν ) + T r(γ ν γ µ )
2
2
1
µ ν
=
T r(γ γ + γ ν γ µ )
2
1
=
T r(2η µν I)
2
= 4η µν
=
(2.32)
(2.33)
(2.34)
(2.35)
5.
T r(γ 5 ) =
=
=
=
and the result follows
2
−T r(γ 0 γ 0 γ 5 )
−T r(γ 0 γ 5 γ 0 )
T r(γ 0 γ 0 γ 5 )
−T r(γ 5 )
(2.36)
(2.37)
(2.38)
(2.39)
(2.40)
6.
1 µ
(γ γµ + γ µ γµ )
2
1 µ
=
(2η µ I)
2
= 4I
γ µ γµ =
3
(2.41)
(2.42)
(2.43)
Lecture three
1. just slog away with the matrix multiplication
2.
1 2
1
1
S ,S
= S 1S 2 − S 2S 1 = − γ2γ3γ3γ1 + γ3γ1γ2γ3
4
4
1 2 1 1 3 3 1 2
= − γ γ + γ γ γ γ
4
4
1 2 1 1 1 2
= − γ γ + γ γ
4
4
1 1 2
=
γ γ
2
i 1 2
= i − γ γ = iS 3
2
etc.
3.
(σ 3 )2 = I, (σ 3 )3 = σ 3 , (σ 3 )4 = I,
e−iθ/2 σ
3
1
1
(−iθ/2 σ 3 )2 + (−iθ/2 σ 3 )3 + ...
2!
3!
1
1
= I − iθ/2 σ 3 − (θ/2)2 I + i (θ/2)3 σ 3 ...
2!
3!
1
1
= I(1 − (θ/2)2 + ...) − i(θ/2 − (θ/2)3 ...)σ 3
2!
3!
3
= I cos(θ/2) − i sin(θ/2)σ
= I + (−iθ/2 σ 3 ) +
4. just work through the algebra using (γ 5 )2 = I.
4
Lecture four
1.
(Dµ Dν ψ)0 = [(∂µ − iqAµ )(∂ν ψ − iqAν ψ)]0
= (∂µ − iqA0µ )(∂ν ψ 0 − iqA0ν ψ 0 )
3
1
1
−iΛ
−iΛ
=
∂µ − iq(Aµ − ∂µ Λ)
∂ν (e ψ) − iq(Aν − ∂ν Λ)(e ψ)
q
q
= ...
= e−iΛ ψDµ Dν ψ
2.
[Dµ , Dν ]ψ =
=
=
=
Dµ Dν ψ − (µ ↔ ν)
(∂µ − iqAµ )(∂ν ψ − iqAν ψ) − (µ ↔ ν)
...
−iq(∂µ Aν − ∂ν Aµ )
so
Fµν = ∂µ Aν − ∂ν Aµ
(4.44)
3. for example
F01 = ∂0 A1 − ∂1 A0
= ∂t A1 + ∂x φ
= −E x
and
F12 = ∂1 A2 − ∂2 A1
= ∂x Ay + ∂y Ax
= Bz
(4.45)
4. Dµ Dµ φ − m2 φ = 0
5
Lecture five
1.
−∂t2 φ + (∇ − iqA)2 φ − m2 φ = 0
∂t2 φ − (p − qA)2 φ − m2 φ = 0
then use φ ∼ e−iEt , with E ∼ m for non-relativistic systems, to lead us to the ansatz
φ = e−imt ϕ
(5.46)
we then find
∂t2 φ = (ϕ̈ − 2imϕ̇ − m2 ϕ)
(5.47)
leading to the Klein-Gordon equation
(p − qA)2
ϕ = i∂t ϕ
(5.48)
2m
i.e. just the Schrödinger equation, with no B.S term, which is good, because scalar fields
have no spin.
4
6
Lecture six
1.
x =
=
=
=
1 + bx
1 + b(1 + bx)
1 + b(1 + b(1 + bx))
1 + b + b2 + b3 + ...
(6.49)
(6.50)
(6.51)
(6.52)
but we also know from te original equation that
x = 1/(1 − b)
(6.53)
2. First of all, define τ = −E + p2 /2m, for fixed p (i.e. do the E integral first) dτ = −dE.
We may then perform the τ integral to find
Z
dp
i
xf − x
2
G0 (tf , xf , t, x) = Θ(tf − t)
exp −
(tf − t) p − 2m
p
(6.54)
2πi
2m
tf − t
We now complete the square of the exponent by defining
p−m
xf − x
= p̃
tf − t
(6.55)
so that
Z
dp̃
i 2
i (xf − x)2
exp −
p̃ (tf − t)(6.56)
G0 (tf , xf , t, x) = −iΘ(tf − t) exp
2m tf − t
2πi
2m
r
i (xf − x)2
m
= −iΘ(tf − t) exp
(6.57)
2m tf − t
2πi(tf − t)
3. just do the substitution. Now note that using one of the representations of the Dirac delta
function
r
α
δ(x) = lim
exp iαx2 /
(6.58)
→0
πi
we see that the propogator solution to the Schrödinger equation, given in the previous
equation, just corresponds to a particle localized at x = 0 at t = 0, which is as it should
be.
4. We start with
(γ µ ∂µ + m)ψ = 0
(6.59)
(γ µ ∂µ + m)S0 = δ
(6.60)
which leads to
5
where S is the standard notation for the Dirac propogator. Now, this is linear, so we look
for the Fourier solution
Z
d4 p
S0 (x) =
S0 (p)eip.x
(6.61)
(2π)4
giving
Z
d4 p
S0 (p)(iγ µ pµ + m)e−p.x = δ
(2π)4
⇒ S0 (p)(iγ µ pµ + m) = 1
(6.62)
(6.63)
(6.64)
Now, remember that the γ µ are matrices, so we need to be a bit careful how we proceed.
It is simplest to multiply by (iγ µ pµ + m) and use the Dirac algebra, to find
S0 (p) = −
7
Lecture seven
1.
•
=
=
=
=
0
0
0,
0,
∂µ ∂ µ φ + λψ̄ψ − gφ3 = 0
∂µ ∂ µ ϕ − M 2 ϕ + λφ? φ = 0
6
(7.66)
(7.67)
(7.68)
(7.69)
1
+ m2
(7.70)
λ
(7.71)
1
p2 + m2
(7.72)
p2
•
(6.65)
∂µ ∂ µ φ − m2 φ2 − λφ3
∂µ ∂ µ φ − m2 φ − (φφ? )φ
γ µ ∂µ ψ + λφψ
∂µ ∂ µ φ − m2 φ + λφϕ
2.
iγ µ pµ − m
p2 + m2
•
λ
(7.73)
γ µ pµ
p2
(7.74)
1
+ m2
(7.75)
λ
(7.76)
g
(7.77)
p2
1
+ M2
(7.78)
p2
1
+ m2
(7.79)
p2
•
7
λ
3.
8
Lecture eight
(7.80)
(7.81)
1. Consider first GL(n). if A, B ∈GL, then by definition det(A) 6= 0, det(B) 6= 0, thus for
C = AB we have det(C) = det(AB) = det(A)det(B) 6= 0. As C is also n × n then C ∈GL,
proving closure. As det(A) 6= 0 then A has an inverse. The identity is just the standard
matrix identity I, and associativity follows from standard linear algebra.
As for the rest of the groups, I’ll leave that to you.
2. V 0T V 0 = (OV )T (OV ) = (V T OT )(OV ) = (V T OT OV ) = (V T V )
3.


V 1 · V 1 V 1 · V 2 ...


OT O =  V 2 · V 1 V 2 · V 2 ... 
..
.
(8.82)
so we find that the set of vectors are orthogonal, hence the name orthogonal group.
4. Hamilton’s equations are
∂H
∂p
∂H
ṗ = −
∂q
q̇ =
(8.83)
(8.84)
which translate as
∂H
∂H
= Ω12
∂η2
∂η2
∂H
∂H
= −
= Ω21
∂η1
∂η1
η̇1 =
(8.85)
η̇2
(8.86)
and may be combined as a single symplectic equation
η̇a = Ωab
8
∂H
∂ηb
(8.87)
9
Lecture nine
1. This follows a similar, if a bit more tedious, route as the question in lecture 3. We then
see that the matrix U is of the form
a
b
(9.88)
U =
−b? as tar
with |a|2 + |b|2 = 1, which is just the general form of an SU(2) matrix, as can be seen by
direct calculation.
2. As a U(1) element is given by gα = eiα and so gα = gα+2π we see that the periodic
identification just defines a circle. Another way to see it is to write g = x + iy subject to
x2 + y 2 = 1, but this is just a circle.
3. We have already seen that as SU(2) matrix is of the form
a
b
U =
−b? as tar
(9.89)
with |a|2 + |b|2 = 1, so by writing a = x + iy, b = z + iw we have x2 + y 2 + z 2 + w2 = 1,
which is just a three-sphere.
4. the field strength for a non-Abelian gauge field is defined analogously to the Abelian one,
[Dµ , Dν ]φ = −iqFµν φ
(9.90)
• As (Dµ φ)0 = U Dµ φ, then ([Dµ , Dν ]φ)0 = U [Dµ , Dν ]φ so
([Dµ , Dν ]φ)0 = −iq(Fµν φ)0
0
⇒ U [Dµ , Dν ]φ = −iqFµν
Uφ
0
⇒ −iqU Fµν φ = −iqFµν U φ
(9.91)
(9.92)
(9.93)
0
⇒ U Fµν U † U φ = Fµν
Uφ
(9.94)
0
⇒ U Fµν U † = Fµν
(9.95)
•
[Dµ , Dν ]φ = Dµ Dν φ − µ ↔ ν
⇒ [Dµ , Dν ]φ = (∂µ − iqAµ )(∂ν φ − iqAν φ) − µ ↔ ν
..
.
[Dµ , Dν ]φ = −iq(∂µ Aν − ∂ν Aµ − iq[Aµ Aν − Aν Aµ ])
which then reduces to the Abelian case when the gauge fields commute.
9
(9.96)
10
1.
Lecture ten
•
j
+
=
0 1
0 0
,
−
j =
0 0
1 0
,
3
j =
1 0
0 −1
(10.97)
•
1 1
| , i =
2 2
1
0
,
1 1
| ,− i =
2 2
0
1
(10.98)
•
1 1
j +| , i
2 2
1
1
j +| , − i
2 2
1 1
j −| , i
2 2
1
1
j −| , − i
2 2
= 0
1 1
= | , i
2 2
1 1
= | ,− i
2 2
= 0
(10.99)
(10.100)
(10.101)
(10.102)
2. For example, use j 1 = 21 (j + + j − ) then
1
1
hj, m|j + |j, m0 i + hj, m|j − |j, m0 i
2
2
1
h 12 , 12 |j + | 21 , 21 i
h 21 , 21 |j + | 12 , − 21 i
=
2 h 12 , − 12 |j + | 21 , 12 i h 21 , − 12 |j + | 12 , − 12 i
1 1 − 1 1
1
h 2 , 2 |j | 2 , 2 i
h 12 , 21 |j − | 12 , − 21 i
+
2 h 12 , − 21 |j − | 12 , 21 i h 21 , − 12 |j − | 12 , − 12 i
1 0 0
1 0 1
+
=
2 0 0
2 1 0
(j 1 )mm0 =
= σ 1 /2
(10.103)
(10.104)
(10.105)
(10.106)
the rest follow a similar pattern. What this means is that we have a way of constructing an
explicit representation, out of the general algebra. If you’re feeling brave, try confirming
the j = 3/2 rep given in the lectures.
11
Lecture eleven
1. If isospin is conserved then it commutes with the Hamiltonian, so
1
h0, 0|Ĥint |1, 0i = h0, 0|Ĥint I 2 |1, 0i
2
1
h0, 0|I 2 Ĥint |1, 0i
=
2
1
=
h0, 0|I 2 Ĥint |1, 0i
2
= 0
10
(11.107)
(11.108)
(11.109)
(11.110)
2. For this I have put a hat on the isospin operator to make it explicitly clear
operator, and which is eigenvalue; it’s not usually necessary to do this as it is
clear from the context.
1
hI, I 3 |Ĥint |I, I 03 i = hI, I 3 |Ĥint 03 Iˆ3 |I, I 03 i
I
1
3 ˆ3
= 03 hI, I |I Ĥint |I, I 03 i
I
1 3 ˆ3
= 03 hI, I |I Ĥint |I, I 03 i
I
I3
= 03 hI, I 3 |Ĥint |I, I 03 i
I
which is
normaly
(11.111)
(11.112)
(11.113)
(11.114)
(11.115)
and hence that it vanishes unless I 3 = I 03 .
3.
hI 0 , I 3 |Ĥint |I, I 3 i = hI 0 , I 3 |Ĥint
2
1
Î |I, I 3 i
I(I + 1)
2
1
hI 0 , I 3 |Î Ĥint |I, I 3 i
I(I + 1)
1
0 3 2
=
hI , I |Î Ĥint |I, I 3 i
I(I + 1)
I 0 (I 0 + 1) 0 3
=
hI , I |Ĥint |I, I 3 i
I(I + 1)
=
(11.116)
(11.117)
(11.118)
(11.119)
and hence that it vanishes unless I(I +1) = I 0 (I 0 +1), meaning that I = I 0 , or I = −I 0 −1,
the latter is spurious as both I and I 0 are positive.
4. given that
|π + i|pi = |3/2, 3/2i
r
r
2
1
|3/2, 1/2i −
|1/2, 1/2i
|π − i|pi =
3
3
M1 := h1/2, 1/2|Ĥint |1/2, 1/2i
M3 := h3/2, 3/2|Ĥint |3/2, 3/2i
(11.120)
hπ + p|Ĥint |π + pi = M3
2
1
hπ − p|Ĥint |π − pi =
M1 + M3
3
3
(11.124)
(11.121)
(11.122)
(11.123)
show that
(11.125)
The first is just a restatement of the question, as |π + pi is just notation for |π + i|pi. To do
the second we have
!
!
r
r
r
r
2
1
2
1
hπ − p|Ĥint |π − pi =
h3/2, 1/2| −
h1/2, 1/2| Ĥint
|3/2, 1/2i −
|1/2, 1/2i
3
3
3
3
11
2
h3/2, 1/2|Ĥint |3/2, 1/2i +
3
2
=
h3/2, 3/2|Ĥint |3/2, 3/2i +
3
=
1
Ĥint |1/2, 1/2i
3
1
Ĥint |1/2, 1/2i
3
where the last step use the raising operator to convert the |3/2, 1/2i to a |3/2, 3/2i.
5. ↑↓, The full list of possibilities is a list like ↑↑↑, ↑↑↓, ↑↓↑,... But the objects in this list
typically have no symmetry under the interchange of pairs, so we form them into the
following
totally anti − symmetric
anti − symmetric on f irst pair
anti − symmetric on second pair
totally symmetric
: none
: (↑↓ − ↓↑) ↑, (↑↓ − ↓↑) ↓
: ↑ (↑↓ − ↓↑), ↓ (↑↓ − ↓↑)
: ↑↑↑, ↓↑↑ + ↑↓↑ + ↑↑↓, ↓↓↑ + ↓↑↓ + ↑↓↓, ↓↓↓
The first two sets have two states in each, and the last has four, showing that
2⊗2⊗2 = 2⊕2⊕4
(11.126)
which is at least consistent with the usual arithmetic of 2 × 2 × 2 = 2 + 2 + 4. This will be
used when we look at baryons, because each quark forms a spin 1/2 doublet, and there are
three quarks in a baryon. The possible spin states are thus either j = 21 (the doublets),
or j = 32 (the quadruplet).
6. To do this, start with the prodtuct-state as |1/2, 1/2i|1/2, 1/2i, this is the highest possible
isospin state, and so
|1, 1i = |1/2, 1/2i|1/2, 1/2i
(11.127)
which we can check explicitly by using the operators
I 2total = I 2 ⊗ I + I ⊗ I 2
3
Itotal
= I 3 ⊗ I + I ⊗ I 3,
(11.128)
now we lower the I 3 isospin eigenvalue with
−
Itotal
= I− ⊗ I + I ⊗ I−
(11.129)
to construct the |1, 0i and |1, −1i states. The |0, 0i state is the one that is anihilated by
−
+
both Itotal
and Itotal
, which turns out to be uniquely given, once you start with a general
combination of all four possible product states.
12
Lecture twelve
1. we have that
u
d
→U
12
u
d
(12.130)
so
d → −b? u + a? d
u → au + bd,
(12.131)
¯ → ((ū, d)U
¯ † so
But we also have (ū, d)
¯ + bū,
¯ + a? ū
−d¯ → a(−d)
ū → −b? (−d)
u
−d¯
so
and
transform in the same way.
d
ū
(12.132)
2. The triplet is formed out of
¯
|1/2, 1/2i|1/2, 1/2i = −|ui|di
(12.133)
1
1
¯ (12.134)
√ (|1/2, 1/2i|1/2, −1/2i + |1/2, −1/2i|1/2, 1/2i) = √ (|ui|ūi − |di|di)
2
2
|1/2, −1/2i|1/2, −1/2i = |di|ūi
(12.135)
and the singlet is
1
¯ (12.136)
√ |1/2, 1/2i|1/2, −1/2i − |1/2, −1/2i|1/2, 1/2i) = |ui|ūi + |di|di
2
3.
• the lowest energy mesonsare s-states with no angular momentum, l = 0.
• We may combine the spins in four ways ↑↑, ↑↓, ↓↑ and ↓↓. These are, as usual,
broken up into states that are symmetric
↑↑,
1
√ (↑↓ + ↓↑),
2
↓↓
(12.137)
which form a spin-one triplet, and a state that is anti-symmetric
1
√ (↑↓ − ↓↑)
2
(12.138)
to form a spin-zero singlet. The spin-one mesons are called vector-mesons.
• there are 3×3 possibilities.
• √13 (uū + dd¯ + ss̄) is an SU(3) singlet.
•
ds̄ (K 0 )
S=1
S=0
us̄ (K + )
dū (π − )
√1
6
S = −1
¯ (π 0 )
− dd)
(uū + dd¯ − 2ss̄) (η8 )
−ud¯ (π + )
√1 (uū
2
−sd¯ (K − )
↑
↑
I 3 = −1 I 3 = −1/2
↑
I3 = 0
sū (K̄ 0 )
↑
I 3 = 1/2
↑
I 3 = 3/2
and the singet state √13 (uū + dd¯ + ss̄) is the η0 . Note the minus signs in states that
¯ that is because −d,
¯ ū) transforms in the same way as (u, d)
have a d,
13
13
Lecture thirteen
1. The magnetic dipole moment is a pseudo vector, so particles are allowed to have one, as
neither it, nor the spin would change under parity.
2. One way to do it would be to set up a capcitor, with negative charge on the top plate,
and the bottom plate earthed. To specify negative charge, we just say the lighter particle
of the hydrogen atom - the lightest of the elements - is negatively charged. Then we place
our particle that has an edm in the electric field of the capacitor, which will tend to align
its dipole moment with the electric field to minimize its energy. Then, if the edm and
spin are aligned the spin is up, in which case the nearest face of the particle, as we look
at it, will move from left to right as it spins. What analogous experiment would you use
for magnetic fields and magnetic dipole moments, and why would it not work?
14
Lecture fourteen
1. The way to do this would be to align the cobalt nuclei, using a magnetic field, in such a
way that the electrons are emmitted mostly downwards. This means that the spin vector
is pointing up, so as we look at the nuclei, the front facing us would rotate from left to
right.
2. are the following quantities even under T (T : Φ → Φ), or odd (T : Φ → −Φ).
• x, even
• t, odd
• p, odd
• the electric field E, even
• the magnetic field B, odd. Think about how a circular current creates a magnetic
field, and how T reverse the current direction.
• electric charge, even
• angular momentum, odd
• spin, odd
• electromagnetic vector potential A, odd, as B = ∇ × A
• magnetic dipole moment, odd
• electric dipole moment, even
3. No.
15
Lecture fifteen
1. If it was me, i’d run away, because they are made of anti-matter. To see this, think
about how an anti-cobalt nuclei would decay. We know that the anti-neutrino that is
14
emitted in the decay is spinning in the same direction as the nucleus, so for the anticobalt the neutrino would also spin in the same direction as the nucleus, but we know
that all neutrinos are right-handed, so if the neutrino is emmitted upwards, then the spin
vector of the nucleus (and neutrino) point downwards. This means that as we look at the
nuclei, the nearest face of it moves from right to left.
2. As previously, assume there exists an electrically charged, neutral particle with an electric
dipole moment. Use the same capacitor experiment to tell our alien friends the difference
between left and right. Now suppose they live in an anti-matter galaxy, would they agree
with you about what is left and right when you meet with them?
What we told our alien friends to do was to make a capacitor, where the top plate is
charged with the lightest particle of the lightest element, in the case of anti-matter aliens
this would be the positrons orbiting in the anti-hydrogen atoms. The bottom plate is
Earth; this means that the electric field in the capacitor is now pointing down. So, we
put in our hypothetical partical, and it aligns its electric dipole moment with the electric
field, meaning that its spin is now pointing down, where in our galaxy it would have been
pointing up. So, our definitions of left and right don’t match.
16
1.
Lecture sixteen
νµ p → µ+ n,
F orbidden as lepton number is not conserved
− +
νe p → e π n, F orbidden as electric charge is not conserved
K + → π 0 µ+ νµ allowed
2.
(16.139)
(16.140)
(16.141)
µ
s
W
u
νµ
(16.142)
W
µ
d
this has an amplitude ∼ gw4 sin θc cos θc . If we add the charm then there is also
µ
s
W
c
νµ
W
d
and this has an amplitude ∼ −gw4 sin θc cos θc .
15
µ
(16.143)
3. Draw Feynman diagrams for the following decays of the D0 (= cū)
d
d
Z
Z
u
c
W
d
d
u
u
u
s
c
d
W
u
d
(16.144)
u
d
D0 → K − π + π 0 ,
D0 → π − π + π 0
(16.145)
¯ These two processes
In fact, you could also have a uū at the top, as π 0 = √12 (uū − dd).
simply differ by the replacement of a csw vertex with a cdw vertex, so the ratio of the
rates is
2
2
gcs
/gcd
= (tan2 θc )−1 > 1
(16.146)
the first process dominates
4. In these decays the final particle content is the same, whereas the initial state differs by
the replacement of s → d. So, there is a us vertex in the first decay, and a ud vertex in
2
2
the second, the ratio of rates is therefore tan2 θc = gus
/gud
.
5. estimate the Cabbibo angle from the decays of a n(= udd) and Λ0 (= uds)
n → pe− ν̄e , with time constant = 900s, branching ratio = 1
energy scale = 1.3MeV
0
Λ → pe− ν̄e , with time constant = 2.63 × 1 −−10 s, branching ratio = 8.4 × 10−4
energy scale = 177.3MeV
where you may assume that the phase-space factor varies with energy as E 5 .
The Feynman diagrams for these processes are
u
u
d
d
W
u
u
d
d
d
u
s
u (16.147)
νe
W
νe
e
e
so, given that the phase space factor varies as E 5 , then the ratio of the rates is
5
5
2 2
gw En
gw4 cos2 θc
1.3
gud
= 4
(16.148)
2 g2
gus
EΛ
gw sin2 θc 177.3
w
which we equate to the observed ratio of rates. The calculation of the observed rate
involves two pieces of information: how long the process takes (the time constant), and the
16
probability of that process happening as opposed some other allowed decay (the branching
ratio). The net rate is then
rate = (branching ratio)/(time constant)
(16.149)
so
1
tan2 θc
1.3
177.3
5
=
1 2.63 × 10−10
900 8.3 × 10−4
(16.150)
giving
θc ∼ 13.8◦
17
(16.151)
Lecture eighteen
1.
m
1.6 × 10−27 kg
∼ 1.4 × 1016 m
= −47 −2
σρlead
10 m 11400kg m−3
∼ 1.5 light years.
l =
(17.152)
(17.153)
so, a neutrino can travel a long way without hitting anything, even in lead.
2.

c12 c13
s12 c23
s13
=  −s12 c23 − c12 s23 s13 c12 c23 − s12 s23 s13 s23 c13 
s12 s23 − c12 c23 s13 −c12 s23 − s12 c23 s13 c23 c13

UM N S
(17.154)
3.


νe
 νµ  = UM N S 
ντ



ν1
−1

⇒  ν2  = UM
NS
ν3


ν1
ν2 
ν3

ν1
ν2 
ν3
(17.155)
(17.156)
†
−1
but we also have that UM N S is unitary so UM
N S = UM N S , and as we have set the complex
−1
T
phase to zero, UM
N S = UM N S , giving,
ν1 = c12 νe + (−s12 c23 − c12 s23 s13 )νµ + (s12 s23 − c12 c23 s13 )νµ
ν2 = s12 c23 νe + (c12 c23 − s12 s23 s13 )νµ + (−c12 s23 − s12 c23 s13 )νµ
ν3 = s13 νe + s23 c13 νµ + c23 c13 νµ
17
(17.157)
(17.158)
(17.159)
which become, once the data is substituted (using θ12 = 34.5◦ , θ23 = 34◦ , θ13 = 10◦ .).
ν1 = 0.82νe − 0.51νµ + 0.29νµ
ν2 = 0.42νe + 0.53νµ − 0.63νµ
ν3 = 0.17νe + 0.67νµ + 0.72νµ
(17.160)
(17.161)
(17.162)
and we find that a2i + b2i + c2i = 1 for each vaue of i, this is just a restatement that UM N S
is unitary.
18
1.
Lecture nineteen
• just follow the route laid out in the lectures, write down the action, then vary either
φ or φ? - not both together as they are treated as independent.
λ
∂µ ∂ µ φ = − (φφ? − η 2 /2)φ
2
(18.163)
√
• The potential is λ4 (φφ? − η 2 /2)2 , which is minimized at |φ| = η/ 2.
• yes
•
1
λ
1
L ' − ∂µ χ∂ µ χ − η 2 ∂µ ξ∂ µ ξ − η 2 χ2 + ...
2
2
4
(18.164)
• we see from the above that there is no quadratic mass term for ξ, so it is massless these are called Goldstone bosons. The χ field hass mass given by m2χ = 21 λη 2 .
This calculation shows a generic result, namely Goldstone’s theorem, that symmetry
breaking is accompanied by massless fields - in this case the ξ - unless there is a gauge
boson to eat them. It is this eating of a massless Goldstone boson that gives the gauge
bosons their mass!
In the standard model we do not have any massless scalars, so if there is symmetry breaking
we had better have gauge bosons to soak them up; we do.
18