K.K. Gan L7: CP Violation 1

● CP is not conserved in neutral kaon decay. ☞ It makes more sense to use mass (or lifetime) eigenstates rather than |K1> and |K2>: ◆ “K-short”: short lifetime state with τS ≈ 9x10-11 sec.

◆ “K-long”: long lifetime state with τL ≈ 5x10-8 sec.

◆ ε is a (small) complex number that allows for CP violation through mixing. ● There are two types of CP violation in KL decay: ■ Indirect (“mixing”): KL → ππ because of its K1 component ■ Direct: KL → ππ because the amplitude for K2 allows K2 → ππ ■ Experimental observation: indirect >> direct! ● The strong interaction eigenstates with definite strangeness are:

■ Consider the strangeness operator:

◆ They are particle and anti-particle and by the CPT theorem have the same mass. ❍ Experimentally we find:

Direct and Indirect CP violation

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K 0 = s d and K 0 = sd

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KS =1

1+ |ε |2K1 +ε K2( )

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KL =1

1+ |ε |2K2 +ε K1( )

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S K 0 = K 0 and S K 0 = − K 0

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(mK 0 −mK 0 ) /m < 9×10−19

K.K. Gan L7: CP Violation 2

● The KL and KS states are not CP (or S) eigenstates:

■ In fact these states are not orthogonal:

■ If CP violation is due to mixing (“indirect”only): ☞ The amplitude for KL → ππ:

☞ The amplitude for KL → πππ:

◆ Experimental measurement: |ε| = 2.3x10-3. ◆ The standard model predicts a small amount of direct CP violation too!

Indirect CP violation

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KS KL =2Reε1+ |ε |2

≠ 0

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K1 KL =1

1+ |ε |2K1 K2 +ε K1 K1( ) =

ε

1+ |ε |2

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CP KS =1

1+ |ε |2CP K1 +εCP K2( ) =

11+ |ε |2

K1 −ε K2( ) ≠ KS

CP KL =1

1+ |ε |2CP K2 +εCP K1( ) =

11+ |ε |2

− K2 +ε K1( ) ≠ KL

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K2 KL =1

1+ |ε |2K2 K2 +ε K2 K1( ) =

11+ |ε |2

K.K. Gan L7: CP Violation 3

● Standard model predicts that quantities η+- and η00 should differ very slightly due to direct CP violation:

■ This is CP violation in the amplitude. ● CP violation is now described by two complex parameters, ε and ε′: ■ ε′ is related to direct CP violation. ■ The standard model estimates: Re(ε′/ε) ~ (4-30)x10-4! ■ Experimentally what is measured is the ratio of branching ratios:

■ There were multiple attempts to measure the ratio since 1970’s with some controversial results. ◆ A non-zero value has recently been measured by 2 different experiments: Re(ε′/ε) = (17.2 ± 1.8)x10-4

◆ Currently the measurement is more precise than the theoretical calculation! ◆ Calculating Re(ε′/ε) is presently one of the most challenging HEP theory projects.

Direct CP violation

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BR(KL →π +π− )BR(Ks →π +π− )BR(KL →π oπ o )BR(Ks →π oπ o )

=η+−

ηoo

2

=ε+ ′ ε 2

ε −2 ′ ε 2≈1+6Re ′ ε

ε

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η+− ≡Amp(KL →π +π− )Amp(Ks →π +π− )

ηoo ≡Amp(KL →π oπ o )Amp(Ks →π oπ o )

K.K. Gan L7: CP Violation 4

● We now consider how the neutral kaon state evolves with time: ☞ allow us to measure KL, KS mass difference and phases of the CP violation parameters, η+- and η00. ■ How a two-particle quantum system evolves in time is covered in many texts including: Particle Physics, Martin and Shaw, section 10.3 Introduction to High Energy Physics, Perkins Introduction to Nuclear and Particle Physics, Das and Ferbel Lectures on Quantum Mechanics, Baym, Ch 2. The Feynman Lectures, Vol III, section 11-5. ◆ Written in about 1963, before there was good experimental data on this topic! ★ CP violation was discovered in 1964. ■ The following derivation is for the neutral kaon system. ◆ It is also applicable to B-meson and neutrino oscillations.

Neutral Kaons and Strangeness Oscillations

K.K. Gan L7: CP Violation 5

● First, consider the case where CP is conserved. ■ CP eigenstates |K1> and |K2> are solutions to the time dependent Schrodinger equation:

◆ Heff is a phenomenological Hamiltonian that describes the system. ◆ Since our particles can decay, Heff is not a Hermitian operator! ◆ Since there are two states in this problem it is customary: ❍ Describe Heff by a 2x2 matrix (e.g. see Das and Ferbel Chapter XII). ❍ Describe |K1> and |K2> by column vectors. ◆ Heff matrix is written in terms of the masses (m1, m2) and lifetimes (τ1, τ2) of the two states:

❍ The eigenvalues and eigenvectors of Heff are:

❐ The states are orthogonal to each other: <K1(t)|K2(t)> = <K2(t)|K1(t)> = 0. ❍ The solutions to the Hamiltonian:

Kaon Oscillations

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i∂ ψ(t)∂t

= Heff ψ(t)

f(t) contains the time dependence.

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Heff K1(t) = (m1 −i

2τ1) K1(t) with K1(t) =

+ f1(t)− f1(t)

Heff K2 (t) = (m2 −i

2τ2) K2 (t) with K2 (t) =

+ f2 (t)− f2 (t)

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K1(t) = e−i

(m1−i

2τ1)tK1 K2 (t) = e

−i

(m2−i

2τ 2)tK2

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Heff =12 m1+m2( )− 14 i 1

τ1+ 1τ 2( ) 1

2 m2 −m1( )− 14 i 1τ 2− 1τ1( )

12 m2 −m1( )− 14 i 1

τ 2− 1τ1( ) 1

2 m1+m2( )− 14 i 1τ1

+ 1τ 2( )

K.K. Gan L7: CP Violation 6

● Consider an experiment which produces a beam of pure K0’s (at t = 0) using the strong interaction: π-p → ΛK0 ■ As in the previous lecture, we can express a K0 as of a mixture of |K1> and |K2> (and visa versa):

■ Using the time dependent solutions for K1 and K2 we can find the time dependent solution for K0:

■ The amplitude for finding a K0 in the beam at a later time (t) is given by:

Neutral Kaons and Strangeness Oscillations

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K1 =12

K 0 + K 0( ) K2 =12

K 0 − K 0( )K 0 =

12

K1 + K2( ) K 0 =12

K1 − K2( )

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K 0 (t) =12e−im1−

i2τ1

tK1 + e

−im2−

i2τ 2

tK2

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K 0 K 0 (t) =12

K1 + K2( ) e−im1−

i2τ1

tK1 + e

−im2−

i2τ 2

tK2

=12e−im1−

i2τ1

t

+ e−im2−

i2τ 2

t

K.K. Gan L7: CP Violation 7

● The probability to find a K0 in the beam at a later time is given by:

■ The third term in the above equation is an interference term. ☞ It causes an oscillation that depends on the mass difference between |K1> and |K2>. ☞ Observation of neutrino oscillation implies that neutrinos must have mass.

Neutral Kaons and Strangeness Oscillations

To make the units come out right in the cos term substitute (m2-m1)c2 for (m2-m1).

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K 0 K 0 (t)2

=14e

+im1+

i2τ1

t

+ e+im2+

i2τ 2

t

e−im1−

i2τ1

t

+ e−im2−

i2τ 2

t

=14e−tτ1 + e

−tτ 2 + e

−im1−

i2τ1

te

+im2+

i2τ 2

t

+ e+im1+

i2τ1

te−im2−

i2τ 2

t

=14e−tτ1 + e

−tτ 2 + e

im2−m1+

i2τ1

+i2τ 2

t

+ e+im1−m2+

i2τ1

+i2τ 2

t

=14e−tτ1 + e

−tτ 2 + e

−t21τ1

+1τ 2

eim2−m1( )t

+ e+im1−m2( )t

=14e−tτ1 + e

−tτ 2 +2e

−t21τ1

+1τ 2

cos t(m2 −m1)

K.K. Gan L7: CP Violation 8

Flavor Oscillations

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K 0 K 0 (t)2

=14

e−

tτ1 + e

−tτ 2 −2e

−t21τ1

+1τ 2

cos t(m2 −m1)

● Using the same procedure, ☞ we can calculate the probability that a beam initially consisting of K0’s contains K0’s at a later time:

◆ The sign of the interference term is now “-”. ◆ Since the strangeness of a K0 differs from a K0’s ☞ the strangeness content of the beam is changing as a function of time. ❍ This phenomena is called strangeness oscillations. ★ More generally we call this phenomena flavor oscillations: ❒ It also occurs with B-mesons (b quark oscillations) and neutrinos (e.g. νe ⇔ νµ). ■ We can measure the strangeness content of a beam as a function of time (distance): ◆ put some material in the beam ◆ counting the number of strong interactions with S = +1 in the final state vs. number with S = -1:

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K 0 p → K +n S = +1

K 0 p →Λp S = −1

Strangeness oscillations for an initially pure K0 beam. A value of (m2 - m1)τ1 = 0.5 is used.

K.K. Gan L7: CP Violation 9

● CP violation requires modifying Heff to include additional complex parameters. ■ The details of the derivation are given in many texts. ◆ e.g. Weak Interactions of Leptons and Quarks by Commins and Bucksbaum. ● CP violation can be measured using the following procedure: ■ produce a beam that is K0 initially ■ measure the yield of π+π- decays as a function of proper time, time measured in the rest frame of K0. ◆ This is a measure of the sum of the square of the amplitude |KL → π+π-> and |KS → π+π->. ☞ There will be an interference term in the number of π+π- decays per unit time (≡ I(t)). ❍ The yield of π+π- decays is given by:

☞ measuring this yield provides information on: ❒ the mass difference and the CP violation parameters η+- and φ+-.

CP Violation

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Iπ +π− (t) = Iπ +π− (0) e−tτ s + η+−

2e−tτL +2η+− e

−t21τ s

+1τL

cos t(m2 −m1)+φ+−

Event rate for π+π- decays as a function of proper time. The best fit requires interference between the KL and KS amplitudes: m2-m1 = (3.491 ± 0.009)x10-6 eV |η+-| = (2.29 ± 0.01)x10-3

φ+- = (43.7 ± 0.6)0

τS = 0.893x10-10 sec τL = 0.517x10-7 sec τ (10-10 sec)

τ (10-10 sec)