1 History. The origin of Nuclear Physics The emergence of particle physics: the Standard Model and Hadrons CERN LHC Highlights 7 Units: [Length, Mass and Energy] (+ Thomson) S-­‐Relativity – Relativistic kinematics (App. B ) (+ Thomson) 2 Relativity and Antiparticles (Schrödinger, Klein-­‐Gordon, Dirac) Angular momentum and Spin 3 Space-­‐Time Symmetries and Conservation Laws. Parity
Charge Conjugation
Time Reversal 4 Interactions and Feynman Diagrams. 5 Particle Exchange: Forces and Potentials. Range of forces, The Yukawa potential 6 Observable Quantities Amplitudes, Cross-­‐sections, Decay rates of unstable particles 7 Units: Length, Mass and Energy. 28/01/14
F. Ould-Saada
1
¡
S.I. units §  [kg, m, s] everyday objects, macroscopic phenomena, but ▪  Mass: me=9.1x10-­‐31kg ▪  Distance: 1 fm=10-­‐15 m ▪  Cross-­‐section (area): 1 barn=10-­‐28 m2 , 1pb=10-­‐12b, 1e=10-­‐15b ¡
Natural units §  [h,c,GeV] ▪  Unit of action in QM: !=h/2π=1.055x10–34 J.s = 6.58x10–16 eV.s ▪  Speed of light (relativity): c = 2.998 x 108 m/s=2.998 x 1023 fm/s ▪  Proton rest mass energy: mp~1GeV=109 eV=1.602x10-­‐10 J ▪  !c = 0.197 GeV.fm §  !=c=1 ▪  E2 =p2c2 +m2c4 à E2 =p2+m2 ▪  Momenta: eV/c; MeV/c; GeV/c; TeV/c ▪  Masses: eV/c2 : Me=0.51 MeV/c2 ; Mp=0.94 GeV/c2 ▪  !=c=1 à [M]=[E]=[P]=[L–1]=[T–1] à GeV 28/01/14
F. Ould-Saada
2
Quantity
[kg, m, s] [!, c, GeV ]
! = c =1
Energy
kg m 2 s −2
GeV
GeV
Momentum kg m s −1
GeV / c
GeV
Mass
kg
GeV / c 2
GeV
Time
s
(GeV / !)
Length
m
(GeV / !c)
−1
GeV −1
Area
m2
(GeV / !c)
−2
GeV −2
¡
−1
GeV −1
1 MeV = 1.53 × 1021 s–1 1 MeV–1 = 197 fm 1 s = 3 × 1023 fm 1 s–1= 6.5 × 10-­‐16 eV 1 ps–1= 0.65 meV 1 m = 5.07 × 106 eV–1 1 m–1 = 1.97 × 10–7 eV–1 Example §  Root-­‐mean-­‐square charge radius of proton §  NUàSI (Dimensions!) 28/01/14
F. Ould-Saada
r2
1/2
= 4.1GeV −1 → ×!c [GeV fm ]
= 4.1GeV −1 ( 0.197GeV fm ) = 0.8 fm
3
§  Distances: 1 fm=10-­‐15 m ▪  Proton Radius: ~0.8 fm ▪  Range of Strong nuclear fore: ~1-­‐2 fm ▪  Range of weak force: ~10-­‐3 fm ▪  Radius of atom: ~ 105 fm §  Cross sections – unit for area. Barn: 1 b=10-­‐28 m2 ▪  ππ scattering (strong process): few 10 mb, as pp scattering (previous figure) ▪  νp scattering (weak process): few 10 e (1e= 10-­‐15 b) §  Energies: 1 eV=1.6 X 10-­‐19 joules ▪  1keV= 103 eV; 1MeV= 106 eV; 1GeV= 109 eV; 1TeV= 1012 eV; ▪  Atomic ionisation: few eV ▪  Binding of nucleons in heavy nuclei: 7-­‐8 MeV per particle §  Momenta: eV/c; MeV/c; GeV/c; TeV/c §  Masses: eV/c2 : Me=0.51 MeV/c2 ; Mp=0.94 GeV/c2 ; ▪  MW=80.4 GeV/c2 ; MZ=91.2 GeV/c2 ▪  SI units: 1 MeV/c2 = 1.78 X 10-­‐30 kg ▪  Atomic mass unit: 1 u=mC/12 = 1.661 X 10-­‐27 kg = 931.5 MeV/c2 28/01/14
F. Ould-Saada
4
¡
Heaviside-­‐Lorentz (HL) units + Natural units §  Classical electrodynamics e2
F=
4πε 0 r 2
ε 0 : permitivity of free space
e2
ε0 = 1 ⇒ F =
4π r 2
1 / ε0µ0 = c 2
µ 0 : permeability of free space
HL + NU ⇒ ! = c = ε 0 = µ 0 = 1
¡
Strength of Quantum Electrodynamics §  fine structure constant α §  α dimensionless: same numerical value regardless of unit system used 28/01/14
F. Ould-Saada
!
#
e2
1
α
=
≈
"
4π 137
#
! = c = ε 0 = µ 0 = 1$
e2
α=
4πε 0 !c
5
28/01/14
F. Ould-Saada
6
¡  v=constant "t ' = t
$
$x ' = x
Galilei #
$ y' = y
$% z' = z − vt
"v(light) = c = constant
$ 2 2 2
2 2
$x + y + z = c t
Einstein # 2
2
2
2 2
$ x ' + y' + z' = c t '
$c 2 t 2 − x 2 − y 2 − z 2 = c 2 t '2 − x '2 − y'2 − z'2
%
"ct ' = γ ( ct − β z )
$
$x ' = x
Lorentz #
$ y' = y
$ z' = γ ( p − β ct )
%
z
28/01/14
F. Ould-Saada
β=
γ=
v
c
1
(1− β )
2
v << c → β << 1 → γ ≈ 1
7
¡
¡
Time dilatation Distance contraction 28/01/14
F. Ould-Saada
è
Time and space coordinate
s make up a 4-­‐vector è
Space-­‐
time unification 8
¡  4-­‐vectors, matrix notation §  Inverse LT: Σ’àΣ (và-­‐v) §  (Remember c can be omitted in NU with h=c=1) !
X = (ct, x)
X ' = ΛX
X = Λ −1 X '
(ΛΛ
# ct '& # γ
% ( %
%x' ( = % 0
% y' ( % 0
% ( %%
$ z' ' $ −γβ
0 0 −γβ &# ct &
(% (
1 0 0 (% x (
0 1 0 (% y (
((% (
0 0 γ '$ z '
# ct & # γ
% ( %
%x ( = % 0
%y ( % 0
% ( %%
$ z ' $ +γβ
0 0 +γβ &# ct '&
(% (
1 0 0 (% x ' (
0 1 0 (% y' (
((% (
0 0 γ '$ z' '
28/01/14
F. Ould-Saada
−1
= I)
9
¡
Goal §  Express physical predictions – interaction cross sections, decay rates – in explicitly Lorentz Invariant (LI) form valid in all inertial frames ¡
Contravariant 4-­‐vectors §  Set of quantities measured in 2 inertial frames and related by LT x µ = (ct, x, y, z)
µ = {0,1, 2, 3}; (Einstein summation for repeated indices)
x 'µ = Λ µ ν x ν
Λ µ ν : represents elements of Λ
§  (magnitude of normal 3-­‐vector x2 is invariant under rotations) ¡
xµ = (ct, −x, −y, −z)
Covariant 4-­‐vector §  LI space-­‐time interval: c2t2-­‐x2-­‐y2-­‐z2 28/01/14
µ = {0,1, 2, 3}
x µ xµ = x 0 x0 + x1 x1 + x 2 x2 + x 3 x3 = c 2 t 2 − x 2 − y 2 − z 2
x 'µ = Λ µ ν xν
F. Ould-Saada
Λ µ ν : represents elements of Λ −1
10
x 'µ = Λ µ ν xν
¡
¡
# ct ' & # γ
% ( %
% −x '( = % 0
% −y' ( % 0
% ( %%
$ −z' ' $ +γβ
0 0 +γβ &# ct &
(% (
1 0 0 (% −x (
0 1 0 (% −y (
((% (
0 0 γ '$ −z '
Covariant – contravariant xµ = gµν xν
§  Metric tensor gµν #
%
gµν ≡ %
%
%
$
Scalar product §  Guaranteed to be LI gµν : metric tensor
1 0 0 0 &
(
0 −1 0 0 (
0 0 −1 0 (
0 0 0 −1 ('
a µ bµ = aµ b µ = gµν a µ b µ
28/01/14
F. Ould-Saada
11
!
!
!
E = γ mc ,
p = γ mv = γ mcβ,
!
§  Contravariant pµ = (E/c, p) !
" = c = 1 ⇒ E = γ m,
p = γ mβ ;
transforms (as xµ) p µ = (E / c, px , py , pz ) = (E, px , py , pz )
according to Λ ¡
Energy/momentum ¡
Scalar product §  E/p/m relation 2
! 2!
v =c p/E
! !
β = p/E
2
§  (Lorentz)-­‐Invariant mass ! E $ !2
!2
µ
2
p pµ = # & − p = E − p
"c%
µ
particle at rest: p = (mc, 0, 0, 0) = (m, 0, 0, 0)
→ p µ pµ = m 2 c 2 = m 2 (LI!)
2
! E $ !2
!2 2
2 2
2
⇒ # & − p = m c ⇒ E − p c = m2c 4
"c%
!
c = 1: E 2 − p 2 = m 2
28/01/14
F. Ould-Saada
12
a µ = (a 0 , a1, a 2 , a 3 )
Scalar product ¡
a ⋅ b ≡ a µ bµ ≡ gµν a µ bν = a 0 b0 − a1b1 − a 2 b2 − a 3b3
!2
2
2
2
p = p⋅ p = m = E − p
§  a.b N
¡
System of particles N particles:
E = ∑ Ei
i=1
! N !
p = ∑ pi
i=1
2
"N % "N !%
µ
p pµ = $ ∑ Ei ' − $ ∑ pi '
# i=1 & # i=1 &
¡
Particle decay aà1+2 a →1+ 2
( p1 + p2 )
µ
2
( p1 + p2 )µ = paµ paµ = ma2
! ! 2
2
ma2 = ( E1 + E2 ) − ( p1 + p2 )
28/01/14
F. Ould-Saada
13
¡
ΣàΣ’ (LT in z-­‐direction) §  t’=γ(t-­‐βz); x’=x; y’=y; z’=γ(z-­‐βt); Inverse: t=γ(t’+βz’); z=γ(z’+βt’) ∂ " ∂z % ∂ " ∂t % ∂
∂
∂
= $ ' + $ ' = γ + βγ
∂z' # ∂z' & ∂z # ∂z' & ∂t
∂z
∂t
∂ " ∂z % ∂ " ∂t % ∂
∂
∂
= $ ' + $ ' = βγ + γ
∂t ' # ∂t ' & ∂z # ∂t ' & ∂t
∂z
∂t
"∂ / ∂t ' % " γ
0 0 +γβ %"∂ / ∂t %
'$
$
' $
'
$∂ / ∂x '' = $ 0 1 0 0 '$∂ / ∂x '
$∂ / ∂y' ' $ 0 0 1 0 '$∂ / ∂y '
''$
$
' $$
'
#∂ / ∂z' & # +γβ 0 0 γ &#∂ / ∂z &
28/01/14
F. Ould-Saada
14
¡
4-­‐derivative §  transforms as covariant (+ sign in front of βγ!) "∂
∂
∂
∂%
∂
,
+
,
+
,
+
=
∂
=
$
' µ
µ
∂t
∂x
∂y
∂z
∂x
#
&
"∂
∂
∂
∂% µ
∂
,
−
,
−
,
−
=
∂
=
$
'
∂xµ
# ∂t ∂x ∂y ∂z &
!2
Laplacian: ∇
covariant!
contravariant
∂2
∂2
∂2
= 2+ 2+ 2
∂x ∂y ∂y
∂2 ∂2
∂2
∂2
d'Alembertian: = ∂ ∂µ = 2 − 2 − 2 − 2
∂t ∂x ∂y ∂y
µ
28/01/14
F. Ould-Saada
15
¡
¡
Reaction 1+2à3+4 4-­‐vectors scalar products s = ( p1 + p2 ) = ( p3 + p4 )
§  Lorentz invariant, can be evaluated in any frame t = ( p1 − p3 ) = ( p2 − p4 )
§  s-­‐channel, t-­‐channel, u-­‐channel 2
2
2
2
▪  u-­‐ relevant when identical particles in final state ¡
2
u = ( p1 − p4 ) = ( p2 − p3 )
2
Centre of mass system §  s: center of mass energy squared §  Problem appendix B-­‐1 !
!
p1 = ( E1*, p* ) ; p2 = ( E2*, − p* )
!* !* 2
2
*
* 2
s = ( p1 + p2 ) = ( E1 + E2 ) − ( p − p )
= ( E + E
*
1
28/01/14
* 2
2
)
s + t + u = m12 + m22 + m32 + m42
⇒ s = E1* + E2*
F. Ould-Saada
16
¡
¡
Protons @ LHC §  Speed=? Cosmic muons §  paradox 28/01/14
F. Ould-Saada
17
¡  Read appendix ¡  Solve problems p358-­‐359 §  B.1 – B.10 ¡  Exercise session 28/01/14
F. Ould-Saada
18
Elementary PP = High Energy Physics (HEP) ¡  In collisions of 2 particles, enough energy needed to produce new particles è E=mc2 ¡  to explore structure of particle, wavelength λ = h/p of probe < structure r è p and T~p²/2m must be large ¡  To explore structure of proton λp~10-­‐15 m è pe ~1GeV >> me §  Ultra-­‐relativistic electrons h 2π !c
2π !c
=
⇔ pc =
p
pc
λ
6.28 × 0.197GeV ⋅ fm
λ = 1 fm ⇒ pc =
~ 1.2GeV
1 fm
6.28 × 0.197GeV ⋅ fm
pc = 1TeV ⇒ λ =
~1.2 × 10 −18 m ~ am
1000GeV
F. Ould-Saada
λ=
28/01/14
19
¡
Classical physics: §  E,p, … of particle dynamical variables represented by time-­‐dependent quantities ¡
QM (Schrödinger picture) §  Wavefunction (WF) postulated to contain all info about state, including E,p, … §  Time-­‐independent operators act on time-­‐dependent WF §  Observable quantities A correspond to action of operator on WF §  Result of measurement of A are eigenvalues of operator ψ (x, t) = Ne−i(Et−px )/!
Âψ = aψ
a : eigenvalue, real
*
 : Hermitian ⇒ ∫ ψ Âψ 2 dt = ∫ $% Âψ1 &' ψ 2 dt
*
1
28/01/14
F. Ould-Saada
20
¡
What is the wave equation for matter-­‐waves? ω = 2πν ; v = λν
§  Wave formula ψ (x, t) = Ne−iω (t−x/v) = Ne−2 π i(vt−x/λ )
E = hν = 2π !ν "
$
§  Wave – particle duality −i(Et−px )/!
h 2π ! # ⇒ ψ (x, t) = Ne
λ= =
p
p $%
2
2
§  Wave equation – (equivalent of ∂
φ 1 ∂ φ
) ∂x
2
=
v 2 ∂t 2 ! 2 ∂2ψ
∂ψ
$
−
+V ψ = i!
∂ψ
p
∂ψ
E
2
= − 2 ψ ; = −i ψ
2m ∂x
∂t
&&
2
∂x
!
∂t
!
%⇒
∂ψ
2
2
Ĥ
ψ
=
i!
= Eψ
p
pψ
&
∂t
E = H = T +V =
+V → Eψ =
+V ψ &
'
"
" −iEt/!
2m
2m
ψ ( x, t) = φ ( x)e
2
2
28/01/14
F. Ould-Saada
21
¡
Expectation value of operator §  If operator commutes with Hamiltonian, corresponding observable A is a conserved quantity §  For an eigenstate of H, <A> is constant ▪  Conserved quantity 28/01/14
F. Ould-Saada
!
〈 Â〉 = ∫ ψ Âψ d x
*
3
d〈 Â〉 i
= 2 〈[ Ĥ, Â]〉
dt
"
[ Ĥ, Â] = ĤÂ − ÂĤ
Ĥ ψ 0 = Eψ 0 ⇒ d〈 Â0 〉
= 0 dt
22
¡
!
!
* !
ρ dV: probability of finding particle density:ρ ( x, t) = ψ ( x, t)ψ ( x, t) !
3!
in volume element dV probability: ρ ( x, t)d x
! ! 1 ∂ρ
∇⋅ j +
= 0 ⇒ ∂µ j µ = 0
c ∂t
!2 2
∂ψ $
−
∇ ψ = i!
&& !
! *
i! * !
2m
∂t
ψ ∇ψ − ψ∇ψ
%⇒ j =−
2
*
2m
!
∂ψ &
2 *
−
∇ ψ = −i!
2m
∂t &'
(
§  Plane wave describes free particle (E,p) with ρ=|N|2 and j=p|N|2 /m 28/01/14
F. Ould-Saada
ψ (x, t) = Ne
)
!!
−i(Et− px )/!
!
"
!
2 p
⇒j=N
≡ nv
m
23
¡
HEP requires: §  Quantum theory §  Theory of special relativity §  Particle – antiparticle symmetry Quantum Field Theory 28/01/14
F. Ould-Saada
24
Equations of motion ¡
Quantum mechanics §  Schrödinger equation (linear in time, not in space) è +
p2
E = T +V =
+V
-2m
,
!
#
∂ & #!
∂ &% E →ih ( ; % p →− ih∇ = −ih ! (
$
∂t ' $
∂x ' -.
! !
3! !
∂ψ ( x, t) 0 h 2 ∂2
⇒ i"
= 2−
+V 5ψ ( x, t)
2
∂t
1 2m ∂x
4
§  Plane wave describes free particle of energy-­‐momentum (E,p) with probability density ρ=|N|2 and current j=p|N|2 /m 28/01/14
F. Ould-Saada
25
Equations of motion ¡
Relativistic version §  Klein-­‐Gordon equation (quadratic in time and space derivatives) §  Expressed in LI form §  Spin-­‐0 fields 2
E =± p +m
¡
!
%
E 2 = m2c 4 + p2c2
' * 1 ∂2 " 2 - m 2 c 2
" & ⇒ , 2 2 − ∇ /φ + 2 φ = 0
∂ "
!
E →ih ; p →− ih∇ ' + c ∂t
.
(
∂t
2
*
,
,
,,
∂
p µ →ih ∂µ = ih
⇒+
∂xµ
,
,
,
,-
Inconsistency §  Negative solutions lead to negative probability densities 28/01/14
F. Ould-Saada
$ ∂2
m2c2 '
&&
+ 2 ))φ = 0
µ
! (
% ∂xµ∂x
(∂ ∂
µ
2
+
m
)φ = 0
µ
m2c2
φ+ 2 φ =0
!
! !
ρ ∝ E, j ∝ p ⇒ j µ ∝ p µ
26
¡
Solutions §  m=0: propagation of EM wave §  φ interpreted as potential U in coordinate space, or as wave amplitude of associated photons §  Static potential U(r) ¡
!
m2c2
1 ∂ % 2 ∂U (
2
φ (r, t) → U(r) ⇒ ∇ U = 2 U = 2 ' r
*
!
r ∂r & ∂r )
g 1 −r/R
Solution: ⇒ U(r) =
e
4π r
!
Range: R =
; g: "intensity strength"
mc
Analogy EM – “Nuclear force” §  QED infinite range, mγ=0 §  NF: short range, massive EM : R = ∞, m = 0, ∇ 2U = 0 ⇒ U =
exchange particle §  Yukawa: pion … R ≈ 1.2fm ⇒ mc
28/01/14
F. Ould-Saada
2
=
Q
r
!c
≈ 150MeV
R
27
¡
Dirac’s goal: §  Linear equation in both space and time derivatives §  Reproducing KG equation – satisfying E2=p2+m2 §  Relativistic Quantum Mechanics + Spin & Magnetic moment of spin ½ fermions 28/01/14
F. Ould-Saada
28
¡
Start with Schrödinger equation ¡
Choose parameters αi
and β such that
§  Equation is linear §  And solutions are also solutions to Klein-­‐
Gordon equation ∂ψ
i!
= Hψ
∂t
! !
! !
2
H = c α. p + β mc = −ihc α.∇ + β mc 2
!
α = (α 1, α 2, α 3) and β ⇒ to be fixed
&
∂ψ #
∂
∂
∂
i
= % −iα x − iα y − iα z + β m (ψ
∂t $
∂x
∂y
∂z
'
square + compare to KG ⇒
{αi, α j} =2δij ; {αi, β } = 0 ; β 2 = 1
{αi, α j} = αiα j + α jαi
⇒ α i , β ⇒ 4 × 4 matrices
28/01/14
F. Ould-Saada
29
¡
¡
γ i ≡ βα i ; γ 0 ≡ β "$
µ
⇒ (i
γ
∂µ −m)ψ =0
#
µ
0 1 2
3
γ = (γ , γ , γ , γ ) $%
Introduce Dirac γ 4x4 matrices and their properties (γ 0 )2 = I
invariance γ 5 ≡i γ 0γ 1γ 2γ 3 with {γ µ , γ 5 } = 0; γ 5γ 5 = I
#
%
§  Anti-­‐commutation i 2
(γ ) = −I
$ ⇒ {γ µ , γ ν } ≡ γ µγ ν + γ ν γ µ = 2g µν
and hermiticity %
µ ν
ν µ
relations γ γ = −γ γ for µ ≠ υ &
Ensure Lorentz γ i↑ =−γ i ; γ 0↑ = γ 0
"I 0%
" 0 σ i% 5 "0 I %
i
γ =$
' ; γ =$ i
' ; γ =$
'
#0 − I &
# −σ 0 &
# I 0&
" 0 1%
"0 − i%
"1 0 %
"1
2
1
3
σ =$ ' ;σ =$
' ;σ =$
' ; 1= $
#1 0 &
#i 0 &
# 0 −1&
#0
F. Ould-Saada
0
28/01/14
0%
'
1&
30
4-­‐component wave-­‐functions – Spinors ¡
§  2 spins (up and down) of e-­‐ §  2 spins (up and down) of e+ §  (“e-­‐ with negative energy”) "
$
γ0 =$
$
$
#
"
$
2
γ =$
$
$
#
1
0
0
0
0
0
0
−i
"
0 0 0 %
'
$
1 0 0 '
; γ 1 = $
$
0 −1 0 '
$
0 0 −1 '&
#
"
0 0 −i %
'
$
0 i 0 '
3
; γ = $
$
i 0 0 '
$
0 0 0 '&
#
28/01/14
0 0 0
0 0 1
0 −1 0
−1 0 0
0
0
−1
0
0
0
0
1
1
0
0
0
1 0
0 −1
0 0
0 0
F. Ould-Saada
%
'
'
'
'
&
%
'
'
'
'
&
(iγ µ∂µ −m)ψ =0
"∂ / ∂t %
$
'
∂
/
∂x
'
∂µ = $
$∂ / ∂y '
$
'
#∂ / ∂z &
iγ 0
!ψ1 $
# &
ψ2 &
#
ψ
#ψ 3 &
# &
"ψ 4 %
∂ψ
∂ψ
∂ψ
∂ψ
+ iγ 1
+ iγ 2
+ iγ 3
− mψ = 0
∂t
∂x
∂y
∂z
31
¡
Wavefunctions 4-­‐component spinors §  Complex conjugate à Hermitian conjugates
ψ † = (ψ1*, ψ 2*, ψ3*, ψ 4* )
†
2
2
2
2
ψ = ψ1 + ψ 2 + ψ3 + ψ 4 !
†!
Current: j = ψ γψ !
µ
† 0 µ
Covariant current: j = ( ρ, j ) = ψ γ γ ψ
Density: ρ = ψ
ψ ≡ ψ †γ 0 ⇒ j µ = ψγ µψ ∂ρ ! !
+ ∇ ⋅ j = 0 ⇒ ∂µ j µ = 0
∂t
28/01/14
F. Ould-Saada
!ψ1 $
# &
ψ2 &
#
ψ
#ψ 3 &
# &
"ψ 4 %
ψ ≡ ψ †γ 0 = (ψ * )T γ 0 = (ψ1*, ψ 2*, −ψ3*, −ψ 4* )
32
¡
Dirac interpretation negative energy solutions as holes in the vacuum that correspond to antiparticle states §  Vacuum corresponds to the state where all negative energy states are occupied ¡
In Dirac “sea” picture, Pauli exclusion principle prevents e-­‐ (E>0) from falling into states (E<0) §  Holes correspond to positive energy antiparticles (e+) §  Limitation of interpretation: bosons (Pauli principle does not hold) have also antiparticles 28/01/14
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33
¡
E<0 solutions interpreted as negative energy particles propagating backwards in time §  Physical E>0 antiparticles propagating forwards in time §  Time dependence of WF: exp(-­‐iEt) unchanged under simultaneous transformation (Eà-­‐E & tà-­‐t) §  The 2 pictures mathematically equivalent è e
[−iEt ]
≡e
[−i(−E )(−t )]
¡
Electron-­‐positron annihilation ¡
e+ and e-­‐ can be created and destroyed just like γ’s, the quanta of the EM field 28/01/14
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34
From Braibant et al.
28/01/14
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35
¡
Positron discovered in 1933 by Carl Anderson in a cloud chamber photograph of cosmic rays ¡
Antiproton discovered in 1955 by Chamberlain et al., required high energy accelerators Emilio Segrè and Owen Chamberlain
were awarded the
Nobel Prize in Physics 1959 for the
discovery of the antiproton
Show that
p+ p→ p+ p+ p+ p
.
E thr
p = 7 m p = 6.6 GeV
C.D. Anderson,
Physical Review 43,
491 (1933).
Carl Anderson won
the 1936 Nobel Prize
for Physics for this
discovery.
Charged particle with spin necessarily has a magnetic moment ¡
!
! q !
q !
s
e!
µ=
s=g
s = gµ 0 with magneton µ 0 =
mc
2mc
"
2mc
e!
Bohr magneton (electron): µ B =
= 5.7884 ×10 −15 MeV / G
2mec
Nuclear magneton (proton): µ N =
¡
¡
e!
= 3.1525 ×10 −18 MeV / G
2m p c
Information about the structure of a particle is contained in the g factor – gyromagnetic factor Dirac equation predicts for point-­‐like, spin-­‐1/2 particle of mass m (e,µ) 28/01/14
F. Ould-Saada
g = 2 ⇒ µD =
e!
mc
!
!
e !
e !
µ p = 2.79 s and µ n = −1.91 s
mp
mn
37
¡
QED: relativistic quantum field theory describing the interaction of electrically charged particles via photons. §  Higher order radiative corrections à g or a=(g-­‐2)/2 2
3
4
"α %
"α %
"α %
µe
1α
= 1+
− 0.32848 $ ' +1.1765$ ' − 0.8 $ ' = 1.001159652307(11)
#π &
#π &
#π &
µB
2π
Experimental value:
= 1.001159652193(10)
⇒
α −1 = 137.035 999 084 ± 0.000 000 051
§  Anomalous magnetic moment leads to the most accurate determination of the fine structure constant α 28/01/14
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¡  Go through and study the slides ¡  Work out intermediate steps if needed as well as examples and proposed exercises ¡  Read ¡  Appendix A.1, A.2, A.3 ¡  Appendix B ¡  Solve problems ¡  B.1-­‐B.10 28/01/14
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39