Brian Meadows, U. Cincinnati Discrete Symmetries Noether’s theorem – (para-phrased) “A...

Brian Meadows, U. Cincinnati

Discrete Symmetries

Noether’s theorem – (para-phrased)

“A symmetry in an interaction Lagrangian corresponds to a conserved quantity.”


Conserved Quantities

Strong E/M Weak

4-momentum Yes Yes Yes

Charge Yes Yes Yes

Baryon # Yes Yes Yes

Spin Yes Yes Yes

Lepton # (e,,) -- Yes Yes

Flavour (S,C,B,T)

Yes YesNo (CKM)

(or Q = F)

Iso-spin Yes No No

P Yes Yes No

CPT, CP, C Yes, Yes, Yes Yes, Yes, Yes Yes, No, No


Parity P

Particles have “intrinsic parity” =± 1 P |> = - |> ; P |q> = +1 (q is a quark); etc..

We define parity of quarks (ie the proton) to be positive. (ie P=+1)

It is usually possible to devise an experiment to measure the “relative parity” of other particles.

Parity of 2-body system is thereforeP = (-1)l 1 2

Example: parity of Fermion anti-Fermion pair (e.g. e+e-):Whatever intrinsic parity the e- has, the e+ is opposite(actually a requirement of the Dirac theory)So, P = (-1)(l+1)


Parity Violation

Parity is strictly conserved in strong and electromagnetic interactions

Helicity can be +1 or -1 for almost any particle. It can flip if you view particle from a different coordinate system BUT not if the particle travels at c!

Real photons have both +1 and -1 helicities (not zero) Consequence of conservation of parity in e/m interactions

Not so for neutrinos In + + + helicity of + is ALWAYS = -1 (“left-handed”)

The neutrino is LEFT-HANDED (always!) Parity is “maximally violated” in weak interactions.


Charge Conjugation C

Operator C turns particle into anti-particle. C |+> = |-> ; C |K+> = |K-> ; C |q> = |q> ; etc.

C2 has eigenvalue 1 Therefore C=± 1

Since C reverses charges, E- and B-fields reverse under C. Therefore, the has C=-1

C is conserved in strong and E/M interactions. Since 0 2, then C|0> = +|0> Since 0 2, then C|0> = +|0> AND 0 cannot decay to 3

(experimentally, 0 3 / 0 2 < 3 10-8)


Time Reversal T

This, in effect, reverses the direction of time It does not reverse x, y or z.


CPT and Time-Reversal

There is compelling reason to believe that CPT is strictly conserved in all interactions It is difficult to define a Lagrangian that is not invariant under CPT

T is an operator that reverses the time No states have obviously good quantum numbers for this,

but you can define CP quantum number e.g. CP |+-> = (-1)L (why?)

Even CP is not conserved e.g. K0 observed to decay into +- (CP=+1) as well as into -+0

(CP=-1) B0 decays to J/psi Ks, J/psi KL and +-


CP Conservation

Recall that P is not conserved in weak interactions since ’s are left-handed (and anti-’s are right-handed).

Therefore, C is not conserved in weak interactions either:+ + +

Makes a left-handed + (because is spin 0)

C(+ + + ) (- - + )

makes a left-handed - (C only converts particle to anti-particle).

BUT – the - has to be right-handed because the anti- is right-handed.

However, the combined operation CP restores the situationCP(+ + + ) (- - + )

Because P reverses momenta AND helicities


CP and the K0 Particle

The K0 is a pseudo-scalar particle (P=-1), thereforeP |K0> = - |K0> and P |K0> = - |K0>

The C operator just turns K0 into K0 and vice-versaC |K0> = + |K0> and C |K0> = + |K0>

Therefore, the combined operator CP isCP |K0> = - |K0> and CP |K0> = - |K0>

Neither |K0> nor |K0> are CP eigen-states

We can define odd- and even-CP eigen-states K1 and K2:

|K1> = (|K0> - |K0>) / \/2 CP |K1> = (+1) |K1>

|K2> = (|K0> + |K0>) / \/2 CP |K2> = (-1) |K2>


CP and K0-K0 Mixing Experimentally, it is observed that there are two K0 decay modes labeled as KL and

Ks:

Ks +- (s = 0.893 x 10-10 s)

KL +-0 (L = 0.517 x 10-7 s)

The decay products of the Ks have P = (-1)L = (-1)0 = +1

For the KL the products have P = -1

It is tempting to assign KL to K1 and Ks to K2

However, this is not exactly correct:

V. Fitch and J. Cronin observed, in an experiment at Brookhaven, that about 1 in 500 times, either

Ks 3 or

KL 2So one defines

KL=1/sqrt(1+2) (K2+ K1) where is the deviation from CP conservation


CP and K0-K0 Mixing

It is possible for a K0 to become a K0 ! The main diagram contributing to mixing in the K0 system:

This contributes to the observation of CP violation in the K0K0 system. It generates a difference in mass between K1 and K2

It is described by a phase in the CKM matrice.

d

d

s

s

u, c, t u, c, t

W

W

K0 K0


Strangeness Oscillations

• Graph shows I(K0) and I(K0) as function of t

for ms/ ~ = 0.5

• Experimentally, measure hyperon production in matter (due to K0, not K0) as function of distance from source of K0)

m s/ ~ = 0.498.

• This corresponds to m/m ~ 5 x 10-15 !

Observation of K0-K0 Oscillations

K0->3 is only 34%, 39% of the decays are leptonic Observe the asymmetry in the leptonic sector

Use the sign of lepton in decays K0+e-e K0-e+ e


Gjesdal et al, Phys.Lett.B52:113,1974

World Average:


Other examples of “Mixing”

Evidence now also exists for mixing in other neutral meson systems: K0 - K0 (ds) - observed in ~1960 B0 - B0 (bd) - observed in ~1992 Bs - Bs (bs) - observed in 2005

D0 - D0 (cu) - observed in April 2007 !

by BaBar and (almost simultaneously) by Belle

Similar mass oscillation versus “flavor observations” areObserved with neutrinos, revealing that neutrino have mass.

Brian Meadows, U. Cincinnati Discrete Symmetries Noether’s theorem – (para-phrased) “A...

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