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1 Neutral Kaon Mixing

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1 Neutral Kaon Mixing These lectures explore how the weak force gives rise to flavour oscillations and CP violation. Flavour oscillation means a reversible transmutation from one flavour to another. This phenomenon is observed within all neutral meson systems 0 0 0 0 0 0 (K K ; D D ; B B ; BsBs) and amongst the three neutrino species (νe; νµ; ντ). CP-violation is mathematical terminology for a difference between matter and antimatter. It is observed only in a handful of rare kaon and B-meson decays. The search for CP-violation in neutrinos is a major topic for the next generation of neutrino experiments. 1 Neutral kaon mixing Fig. 1 show early proof of neutral kaon mixing from a bubble chamber image. The strangeness of the neutral meson appears to change during the flight between creation and decay. This statement is under the [correct] assumption that the creation of the neutral kaon is via the strong force, which is flavour conserving. The transition, K+ + p+ ! K0 + p+ + π+ ; jK+i = jsu¯ i; jK0i = jsd¯ i; conserves antistrangeness. Conversely, the baryon-number-conserving transition, K0 + p+ ! Λ0 + π+ + π0 ; jK0i = jsd¯i; jΛ0i = jsudi; conserves strangeness. The neutral kaon has transmuted from jsd¯ i to jsd¯i in a few centimetres. For this to occur, the K0 and K0, cannot be the “mass eigenstates” of the Hamiltonian, those that propagate in time. The mass eigenstates (i.e. the 0 0 K-short, KS and K-long, KL as we know them today) are linear superpositions of the flavour eigenstates (states of definite flavour) K0 and K0 and hence have access to both in interactions. C B A Figure 1: a K+ beam enters the left edge of a bubble chamber and scatters off a proton atA. A pion and a neutral kaon is formed. The kaon flies to the right without leaving an trace and the recoiling proton and pion move to the bottom of the image. AtB, the neutral kaon scatters o ff another proton to form a Λ0 baryon with an associated π+, as well as an untraced π0. AtC, the Λ0 baryon decays into a proton and a π− in a characteristic “V” decay. 1 1.1 Formalism The time evolution of one neutral meson, in its rest frame, may be written as1, jK0(t)i = e−iMte−Γt=2jK0i i where the first exponential is a plane-wave solution, exp( ~ (p:x − Et)) for a state with energy E = M in its rest frame, jpj = 0, with ~ = 1. The second term describes the exponential decay for a state with proper lifetime τ (i.e. width Γ = ~/τ, again with ~ = 1) such that, jhK0jK0(t)ij2 / e−t/τ Generalise to a two state system with 2 × 2 matrices, M and Γ encoding the time evolution, 0 ! 0 ! jK (t)i jK i −i Mt−Γt=2 0 = Σ where Σ = e ; (1) jK (t)i jK0i where the presence (absence) of the bar above the K(t) shows the flavour of the state at t = 0: pure jK0i or pure jK0i . The states jK0i and jK0i are of well defined flavour so it is in this basis that it is appropriate to discuss their interaction by the weak force. Thus they are the weak eigenstates, as opposed to the mass eigenstates of the Hamiltonian. Apply Schrodinger’s¨ equation, i d =dt = H , identifies the Hamiltonian. 0 ! 0 ! d jK i jK i i 0 = H 0 with H = M − Γ: (2) dt jK i jK i 2 Any matrix, A can be decomposed in the form B + i C where B and C are hermitian because, A A Ay Ay A = + + − 2 2 2 2 1 1 = A + Ay + A − Ay 2 2 | {z } | {z } = B (= By) = iC (= −iCy) by inspection. ∗ ∗ So M and Γ are hermitian matrices: M21 = M 12 , Γ21 = Γ 12. This also implies observable quantities because the eigenvalues 0 0 of hermitian matrices are real. Last, we impose CPT-invariance (identical mass and lifetimes of the K and K ), M11 = M22 = M, Γ11 = Γ22 = Γ. So the hamiltonian of the meson-antimeson system becomes, ! ! MM12 i ΓΓ12 H = ∗ − ∗ (3) M12 M 2 Γ12 Γ The off-diagonal terms describe the transitions [mixing] between the meson and antimeson. Returning to the wavefunctions, the eigenstates of the K0 hamiltonian propagate in time with defined mass and life- times. We label the two mass eigenstates Heavy, Light and q K describe them as a combination of the flavour eigenstates, L 0 0 jKHi = p jK i − q jK i 0 0 jKLi = p jK i + q jK i ; (4) 0 where p2 + q2 = 1. The “p + q; p − q” form is required by p K orthogonality. Inverting this transformation, gives 0 1 jK i = 2p jKLi + jKHi (5) 0 1 jK i = 2q jKLi − jKHi : (6) KH 1This simplification of time-dependent perturbation theory is known as the Wigner-Weisskopf approximation. 2 Eq. 1 describes the time evolution of the flavour eigenstates and is related to the time evolution of the mass eigenstates by the similarity transformation, Σ = PDP−1, noting that the the time evolution in the mass basis is diagonal because mass eigenstates propagate in time with definite energy and lifetime. ! 0 1 1 1 ! jK(t)i B 2p 2p C jK (t)i = B C L jK t i @B 1 −1 AC jK (t)i ( ) 2q 2q H − − ! ! e i MLt ΓLt=2 0 jK i = 00 L −i MH t−ΓH t=2 0 e jKHi p q ! jK0i ! = 00 00 p −q jK0i In passing, it is noted that from linear algebra, the columns of P are eigenvectors of Σ. Multiplying out, 0 1 1 1 −i MLt−ΓLt=2 ! ! B 2p 2p C e 0 p q Σ = B C −i M t−Γ t=2 (7) @ 1 −1 A e H H p −q 2q 2q 0 0 g q g 1 B + p − C h − − − − i B C 1 i MLt ΓLt=2 ± i MH t ΓH t=2 = @B p AC where g± = 2 e e (8) q g− g+ Hence the time evolution of the decaying state is, q jK(t)i = g jK0i + g jK0i + p − p jK(t)i = g jK0i + g jK0i (9) q − + Or illustratively, K0 K0 g+(t) p q g−(t) K0 K0 q p g−(t) g+(t) K0 K0 From which we note, as a precursor to the discussion on CP violation, that a difference in the temporal evolution of neutral-meson mixing can arise if q and p are different, !2 q p q , ; , 1: p q p Applying Schrodinger’s¨ equation to the off diagonal part of Eq. 8, and equating to Eq. 3 noting that the time dependance is all contained within the g± factors, ! d q q i i i g (t) = i M − M − Γ − Γ = M − Γ dt p − p L H 2 L H 12 2 12 ! d p p i i i g (t) = i M − M − Γ − Γ = M∗ − Γ∗ dt q − q L H 2 L H 12 2 12 !2 i q M − Γ = 12 2 12 (10) p ∗ i ∗ M12 − 2 Γ12 0 Where we remind ourselves that M12 and Γ12 are the elements of the hamiltonian H that describe the action of jK i 0 appearing in the initially jK it=0 wavefunction. 3 1.2 Measuring K0-K0 oscillations For now, we concentrate on kaon oscillations and ignore CP violation, i.e. p = q. Neutral kaons can be produced in a state of definite antistrangeness by a strong interaction of negatively-charged pions on a proton target, see Fig. 2(a). By strangeness conservation (strong interaction) and baryon number conservation, it is impossible to make a Λ¯ 0. In the semileptonic decay, the charge of the muon must be that of the strange quark, see Fig. 2 (b) and (c). Hence by counting the number of µ+π− decays versus the number of µ−π+ decays as a function of decay time, this decay can pick out the proportion of K0 and K0 in the propagating neutral kaon wavefunction. d d µ+ µ− + p u u Λ0 νµ ν¯µ − u s W+ W u¯ s¯ s¯ u¯ s u π− K0 π− K0 π+ d d d d d¯ d¯ (a) (b) (c) Figure 2: Quark flow diagrams for (a) K0 production from pion-nucleon scattering (b) semileptonic decay of a K0 and (c) semleptonic decay of a K0. The charge of the muon unabiguously identifies the kaon flavour. As the system starts with a pure jK0i, Eq. 9 with p = q is appropriate. Writing out completely, 0 0 jK(t)i = g+ jK i + g− jK i − − − − − − − − = 1 ( e i MLt ΓLt=2 + e i MH t ΓH t=2 )jK0i + 1 ( e i MLt ΓLt=2 − e i MH t ΓH t=2 )jK0i (11) 2 | {z } | {z } 2 | {z } | {z } a b a b The K0 (K0) intensity is found from the probability of finding a K0 (K0) at time t from this combined wavefunction, 1 jhK0jK(t)ij2 = (a + b)(a + b)∗ 4 1 1 = (aa∗ + bb∗) + (ba∗ + ab∗) 4 4 1 jhK0jK(t)ij2 = (a − b)(a − b)∗ 4 1 1 = (aa∗ + bb∗) − (ba∗ + ab∗) 4 4 Taking first the direct “amplitude-squared” term in each line, Γ Γ Γ Γ ∗ ∗ (−iM − L )t (+iM − L )t (−iM − H )t (+iM − H )t aa + bb = e L 2 e L 2 + e H 2 e H 2 − − = e ΓLt + e ΓH t The interference term reveals a sinusoidal dependence, Γ Γ Γ Γ ∗ ∗ (−iM − H )t (+iM − L )t (−iM − L )t (+iM − H )t ba + ab = e H 2 e L 2 + e L 2 e H 2 Γ +Γ Γ +Γ − L H t −i(M −M )t − L H t i(M −M )t = e 2 e H L + e 2 e H L Γ +Γ − L H t = e 2 · 2 cos(∆Mt) So, 1 1 ΓL+ΓH 0 2 −ΓLt −ΓH t − t P 0 (t) = jhK jK(t)ij = e + e + e 2 cos(∆Mt) K 4 2 1 1 ΓL+ΓH 0 2 −ΓLt −ΓH t − t P 0 (t) = jhK jK(t)ij = e + e − e 2 cos(∆Mt) (12) K 4 2 4 2 which depends on the KL and KH lifetimes and their mass difference in the interference term.
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