1. Particle Physics: Status and Perspectives Part 6: Symmetries
Manfred Jeitler
SS 2015

2. SYMMETRIES

3. symmetries in physics
certain transformations do not change the laws of nature
translation in space
translation in time
for each continuous symmetry transformation, there is a conservation law (Noether theorem, 1918)
translation in space: conservation of momentum
translation in time: conservation of energy
Examples of conservation laws in neutron beta decay (n -> p + e + `
Energy conservation: mn > mp + me ( u > u u)
conservation of momentum and angular momentum (--> discovery of neutrino)
conservation of electric charge: 0 = +1 – 1 + 0
baryon number conservation: +1 =
lepton number conservation: 0 =

4. discrete symmetries p+ p-
Fundamental symmetry operations in particle physics:
parity transformation (spatial inversion P)
particle-antiparticle conjugation (charge conjugation C)
time inversion (T) p+ p-
parity: P ψ(r) = ψ(-r)
P2 = “unitary operator”
a wavefunction may or may not have a well-defined parity
parity can also be given for individual particles
may seem counterintuitive for macroscopic “balls”
but particles are not really “balls” but described by a wave function, which may be symmetric (eigenvalue P=1) or anti-symmetric (eigenvalue P=-1)
parity is multiplicative quantum number
C-eigenvalue exists only for particles which are their own anti-particle (neutral particles, without non-zero quantum numbers)
e.g. photon, neutral mesons
not for baryons (baryon quantum number!)
T-asymmetry: the obvious difference between forward and backward evolution (for example, when dropping and breaking a cup) is due to the different probability of the initial and final states, and not due to asymmetry in the interaction itself
According to the kind of interaction, the result of such a transformation may describe a physical state occurring with the same probability (“the symmetry is conserved”) or not (“the symmetry is broken” or “violated”).

5. Looking at particles more closely
Do particles behave just like balls in a game of billiards?
Is there room for any asymmetries? p e

6. intrinsic parity of the negative pion π-
consider pion capture: π- + d  n + n
total angular momentum J = 1
sd = 1, sπ = 0, capture occurs in S-state (L=0)
J = L + S = 1
two-neutron state:
L=0, S=1 or L=1, S=0 or L=1, S=1 or L=2, S=1
symmetry under neutron exchange: (-1)L+S+1
must be negative (neutrons are fermions  Fermi-Dirac statistics)
 L+S must be even !
orbital momentum L: even-numbered states are symmetric
spin: S=1 (“triplet”) is symmetric, S=0 (“singlet”) is antisymmetric
 L=1, S=1
parity = (-1)L = is negative
proton and neutron have intrinsic parity +1
 deuteron has positive parity
pion (π-) has negative parity !
for fermions, particle and antiparticle have opposite parities
for bosons, particle and antiparticle have the same parity

7. triplet and singlet spin states

10. parity violation
most physicists had thought that fundamental symmetries were never violated
this had not been proved, however, for Weak interactions
C.N.Yang and T.D.Lee conjectured that parity might not be conserved in Weak interactions
K+  2π (positive parity) and
K+  3π (negative parity): “τ-θ puzzle”
experiment made by C.S.Wu
β-decay of 60Cobalt
 the world’s mirror image differs from the world itself

11. parity violation Chien-Shiung Wu Chen Ning Yang and Tsung-Dao Lee
(Nobel prize 1957)

12. parity violation: Wu’s experiment
polarized matter
60Co at 0.01 Kelvin inside solenoid
high proportion of nuclei aligned
60Co (J=5)  60Ni* (J=4)
“Gamow-Teller transition”: lepton spin = 1
electron spin points in direction of 60Co spin J
conservation of angular momentum
degree of 60Co alignment determined from observation of 60Ni* γ-rays
observed electron intensity:
ϑ: angle between electron (p) and spin (J)
σ: unit vector in direction of electron spin
σ.p is pseudoscalar
mirror flips one but not the other

13. parity violation

14. helicity electrons have a “helicity”: H = -v/c
“left-handed”
the faster they are, the more the spin is aligned antiparallel to the momentum
positrons have opposite helicity: H = +v/c
“right-handed”
photons occur right-handed and left-handed with equal probability
 parity is conserved in electromagnetic interactions
what about the neutrino?
it is very light and therefore usually flies very fast (v/c ~ 1)
it is a lepton -- the “neutral sibling” of the electron
what should we expect?

15. the helicity of the neutrino
Sm* spin is parallel to electron spin and antiparallel to neutrino spin:
1 = 1/2 - (-1/2)
Sm* has the same polarization sense as the neutrino
whichever the neutrino helicity, the forward photons from Sm* deexcitation have the same helicity as the neutrino
only forward photons can be detected by resonance scattering
(conservation of momentum and energy)
detect polarization sense of these photons by passing them through magnetized iron
electron magnetic moment aligned anti-parallel to its spin:
μ ~ e.s exactly: μ = g (e`h/2mc.s

16. the helicity of the neutrino
experiment carried out by Goldhaber et al., 1958
result: helicity of neutrino is negative
H = -1 for massless neutrino
( ν has same lepton number (Le) as e- )

17. neutrinos P

18. neutrinos and antineutrinosP C CP ..

19. spinning neutrinos and antineutrinos
Parity
Charge CP
left-handed
neutrino
right-handed
neutrino X
Parity CP
Charge
particle-antiparticle
conjugation
In weak interactions P and C are “maximally violated” while the combined symmetry under CP is mostly conserved.
right-handed
antineutrino

20. pion decay dominant pion decay mode: could also decay as π+  e+ + νe
in both cases, lepton and neutrino have opposite helicity
they fly apart (opposite momentum)
pion has zero spin  conflict with angular momentum conservation
due to its higher mass, the muon is not so fast (not so “relativistic”) and “does not care” so much
suppression by factor 1 – v/c
therefore, π+  e+ + νe is strongly suppressed compared to π+  μ+ + νμ
suppressed by factor 104
pion mass: MeV/c^2
muon mass: MeV/c^2
electron mass: MeV/c^2

21. the neutral kaon system
kaons are mesons that contain one light quark (u, d,anti-u, anti-d) and one strange-type quark (i.e., strange or anti-strange)
quark model allows us to build two neutral kaons:
K0 = d anti-s anti-K0 = s anti-d
however, the physical particles we find are linear combinations of these two:
KL ~ K0 - anti-K KS ~ K0 + anti-K0
these are the eigenstates under Weak interaction, through which these particles decay (only Weak interactions can transform quarks of different generations into one another)
when a K0 or anti-K0 is formed by Strong interactions, these particles “oscillate” (transform into one another)

22. “oscillation” of neutral kaons due to transitions via virtual 2π and 3π states

23. box and penguin graphs

24. CP-eigenvalue K0L K0S particles can be attributed a “CP-eigenvalue”
like charge, mass, parity
this eigenvalue is multiplicative:
CP () = -1
CP () = +1
there are 2 kinds of “neutral K-mesons”
the (long-lived) K0L decays into -mesons
the (short-lived) K0S decays into -mesons
K0L and K0S differ by their CP-eigenvalue !
CP(K0L) = CP(K0S) = +1
K0L p
K0S
CP = -1
CP = +1

25. CP-violation K0L 1964: sometimes (0.3 percent) also K0L K0S p CP = -1

26. CP-violation: the first experiment
After parity violation had been found many physicists believed (hoped?) that symmetry might still hold under the product of parity (P) and charge (C) conjugation. In 1964, an experiment carried out with neutral K-mesons (“kaons”) showed, that also this combined “CP” operation showed symmetry breaking (violation of conservation of a certain quantity, the “CP eigenvalue”).
Christenson, Cronin, Fitch and Turlay: Brookhaven 1964
Nobel prize 1980

27. simple setup (by present standards):
spark chambers
scintillators
Cerenkov detectors
the first signal:
K0L  p+p-

28. CP and the decay of neutral kaons
CP eigenstates:
CP ( K1 ) = + K1
CP ( K2 ) = - K2
but the physical states are KL and KS:
KL  K2 + e  K1
KS  K1 + e  K2
CP-conserving:
KL (CP  -1)  ppp (CP = -1)
CP-violating:
KL (CP  -1)  pp (CP = +1)

29. Indirect and direct CP-violation
KL (CP  -1) = K2 (CP = -1) + e  K1 (CP = +1)  pp (CP = +1)
indirect CP-violation
(through mixing)
direct CP-violation
(in the decay)
indirect CP-violation: must be same for p0p0 and for p+p-
direct CP-violation: may be different for p0p0 and for p+p-

30. NA48: measuring the double ratio RKS p0 p+ p- KL p0 p+ p-
R  1:
direct CP-violation
exists
”frequent“
”rare“ /

31. Indirect and direct CP-violation
only indirect CP-violation: explanation by “Superweak model” possible (introducing a “5th interaction”)
direct CP-violation: explanation only through “Standard Model”
experimental results
R  1  Superweak model excluded

32. Layout of the NA48 experiment at CERN: the measurement of “direct CP-violation”
Ever more refined experiments allowed to finally established the so-called “direct” CP-violation.

33. The detector of the NA48 experiment at CERN
muon detector and anti-counters for background suppression
muon
ring anti
magnet
DCH
hadron calorimeter
electromagnetic liquid-krypton calorimeter for measuring p0p0-decays
hodoscope for exact timing
spectrometer (consisting of 4 drift chambers and a magnet) and hadron calorimeter for measuring p+p--decays

34. the tagging detector
distinguish KS- and KL-decays by “tagging” the protons directed at the “KS-target”
beam is “split up” onto several individual scintillators to reach good double-pulse resolution in spite of high rate (up to 30 MHz)
very fast flash-ADC (digitisation electronics) developed at HEPHY, Vienna

35. the Cabibbo-Kobayashi-Maskawa matrix
the coupling between “up-type” and “down-type” quarks is described by the Cabibbo-Kobayashi-Maskawa matrix:
so, the three generations of quarks are not completely separated, but “mix”, so that quarks of the heavier generations can decay into those of the lighter generations
in the Standard Model, CP-violation can be explained by a non-trivial complex phase in this matrix
this would not be possible with only two generations
so, third generation was actually predicted by CP-violation theory!

36.  Makoto Kobayashi
Toshihide Maskawa 
Nobel prize 2008
(together with Yoichiro Nambu)

37. Nicola Cabibbo

38. the Cabibbo-Kobayashi-Maskawa matrix and the”unitarity triangle“
this matrix desribes a mixture of states whose total number does not change; so the matrix has to be unitary:
V  V+ = V+  V = 1 or
this yields the relation
which can be graphically represented in the form of one of six so-called „unitarity triangles“

39. the unitarity triangleIm
i    1 Re
normalized to 1

40. CPT CPT (X) = X CP (X)  X  T (X)  X
There are powerful theoretical arguments and experimental evidence for CPT conservation, i.e.
CPT (X) = X
in this case we must conclude:
CP (X)  X  T (X)  X

41. Conservation of symmetry
C P CP T CPT
gravitation     
electromagnetism     
strong interaction     
weak interaction X X x x 

42. direct measurement of T-violation
As shown before, the physical states KL and KS are linear combinations:
KL  K2 + e  K1
KS  K1 + e  K2
However, K1 and K2 are also linear combinations :
K1 = K0 + 0
K2 = K0 - 0
What does a “linear combination of states” mean?

43. Comparing particles to coupled pendulumsKL KS K0 K0

44. Measure T-violation by comparing K0  K0 and K0 K0
Compare rates for neutral kaons which are created as K0 and decay as K0 with the inverse process: K0  K0 K0  K0 ? =

45. Identifying K0 and K0 at creation time
It is hard to separately create K0 and K0, usually both kinds of particles are produced.
Tagging due to “strangeness conservation in strong interactions”:
pp  K0 p+ K-
pp  K0 p- K+ S
K0, K+ +1
K0, K

46. Identifying K0 and K0 at decay time
Selecting “semileptonic decays” of neutral kaons, K  pen, we observe only
K0  p- e+ n
K0  p+ e- n
due to the so-called “DS = DQ rule”: d d K0 p- s u W+ e+ n

47. The CPLEAR experiment at CERN
Antiprotons p hit a
hydrogen target

48. Direct measurement of T-violation by CPLEAR at CERN

49. T-violation
T-violation has been seen via CP-violation in neutral kaons (CPT-conservation!).
T-violation has been observed directly by measuring the probabilities for K0  K0 and K0  K0.

50. The early universe
Soon after the time of the Big Bang, the energy density is very high: the energy per particle is of the order of the “Planck energy” (1019 GeV) where all four interactions are unified.
This is followed by expansion and cooling (decrease of energy density). At t = seconds, and an energy of 1016 GeV, Strong Interactions separate. At t = s and 102 GeV, electro-weak symmetry is broken, giving rise to the separate interactions of electromagnetism and Weak Interactions. Only much later, at 5 = 3 minutes, the universe cooles down to 0.1 MeV, and light nuclei are formed. Atoms are formed only after 105 years.
So, the development of the universe is characterized by particle physics, nuclear physics, and finally atomic physics. Astronomical observations and results from these disciplines in physics complement each other.

51. Matter in the universe
At the time of the Big Bang, particles and antiparticles were created in equal amounts.
Today, the universe consists mostly of matter, not antimatter.
Where has all the antimatter gone? How come we are still around?
CP-violation is an important condition for this “baryon asymmetry” (Sakharov, 1965).
arguments for the Big Bang:
redshift of distant galaxies
3K background radiation

53. Sakharov conditions
baryogenesis: to get from an originally symmetric state (matter = antimatter) to the present state (matter >> antimatter), the following three conditions must be met:
1) baryon number violation
obvious
2) no thermal equilibrium
else, for each process creating matter at the cost of antimatter, the inverse process would take place with equal probability
3) CP-violation
else, for each process creating matter at the cost of antimatter, the CP-mirrored, inverse process would take place with equal probability
 CP-violation of vital importance for us
or would you like to be one of zillions of photons, just playing with other photons?
but the known sources of CP-violation are not enough
your task: go and find the source of CP-violation that allows for baryogenesis!
proposed mechanism: leptogenesis
we do not know how many leptons (neutrinos) there are
mechanism that transforms leptons into baryons could create the observed baryon asymmetry
CP: e.g. same amount as left-handed baryons and right-handed antibaryons

55. proton decay?
if baryon number conservation is violated (Sakharov condition #1), the proton could decay
for example, p  e+ π0
no other conservation law forbids this
another argument for possible proton decay: there is no field that corresponds to baryon number conservation
in gauge theories, long-range fields give rise to absolutely conserved quantities
proton decay is definitely very slow
good for us: else, we would have problems with radiation damage!
experimental limits for decay rate very low
half-life > 1036 years (“Superkamiokande” detector, Japan)
spin-off: neutrino detectors
assume that radiation levels of 5 G (about 100 times the admissible dose) would prevent mammals from developing
estimate the minimum half-life of the proton under the assumption that all its rest energy (1 GeV) is absorbed into tissue
( 1 G = 1 J/kg ~ GeV/kg )