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129A Lecture Notes
Weak Interactions I
1 Nuclear β-decay
The nuclear β-decay caused a great deal of anxiety among physicists. Both α- and γ-rays are emitted with discrete spectra, simply because of the energy conservation. The energy of the emitted particle is the same as the difference between the initial and final state of the nucleus. It was much more difficult to see what was going on with the β-decay, the emission of electrons from nuclei. Chadwick once reported that the energy spectrum of β electrons is con- tinuous. The energy could take any value between 0 and a certain maximum value. This observation was so bizzarre that many more experiments fol- lowed up. In fact, Otto Han and Lise Meitner, credited for their discovery of nuclear fission, studied the β spectrum and claimed that it was discrete. They argued that the spectrum may appear continuous because the electrons can easily lose energy by breamsstrahlung in material. The maximum energy observed is the correct discrete spectrum, and we see lower energies because of the energy loss. The controversy went on over a decade. In the end a definitive experiment was done by Ellis and Wooseley using a very simple idea. Put the β-emitter in a calorimeter. This way, you can measure the total energy deposit. They have demonstrated that the total energy was about a half of the maximum energy on average. The spectrum is indeed continuous. The fact that the β-spectrum is continuous was su puzzling to people, and let Niels Bohr to say even this:
At the present stage of atomic theory, however, we may say that we have no argument, either empirical or theoretical, for uphold- ing the energy principle in the case of β-ray disintegrations.
He was ready to give up the energy conservation! This quote shows how desperate people were. Pauli, also desperate about this problem, wrote a letter to people who attended a meeting in T¨ubingen in December 1930. He himself could not go to the meeting because he had to attend a ball. Nonetheless, he could not
resist to propose a speculation to solve this problem even though somewhat reluctantly. Here is a quote from his letter:
4th December 1930
Dear Radioactive Ladies and Gentlemen,
As the bearer of these lines, to whom I graciously ask you to listen, will explain to you in more detail, how because of the "wrong" statistics of the N and Li6 nuclei and the continuous beta spectrum, I have hit upon a desperate remedy to save the "exchange theorem" of statistics and the law of conservation of energy. Namely, the possibility that there could exist in the nuclei electrically neutral particles, that I wish to call neutrons, which have spin 1/2 and obey the exclusion principle and which further differ from light quanta in that they do not travel with the velocity of light. The mass of the neutrons should be of the same order of magnitude as the electron mass and in any event not larger than 0.01 proton masses. The continuous beta spectrum would then become understandable by the assumption that in beta decay a neutron is emitted in addition to the electron such that the sum of the energies of the neutron and the electron is constant...
Remember this letter was written before the discovery of the neutron by Chadwick in 1932. The picture of 14 N was that it composed of 14 protons and 7 electrons. On the other hand, the molecular spectrum showed that 14 N was a boson. He attempted to hit two birds with one stone: this statistics puzzle and the apparent energy non-conservation. He proposed that the 14 N is made of 14 protons, 7 electrons, and 7 neutrinos. The β-decay emits one electron and one neutrino at the same time. Because you don’t detect the neutrino, by assumption, it appears that the energy is not conserved. On the other hand, this composition makes the 14 N a boson. When Chadwick discovered the neutron in 1932, the picture of the nuclei was established fairly quickly. 14 N is made of 7 protons and 7 neutrons, and there is no statistics puzzle. But what about the β-decay? It was clear that Chadwick’s neutron is not Pauli’s “neutron” which was supposed to be at least as light as the electron. It was Fermi who formalized the Pauli’s idea into the first theory of weak interaction. First of all, he renamed
Fermi’s interaction is not complete. It has to be supplemented by something else. Considering all possible 16 Dirac matrices, ¯upun (S) and ¯upγμun (V) reduce to the Fermi transition in the non-relativistic limit, while ¯upσμν^ un (T) and ¯upγμγ 5 un (A) to the Gamov–Teller transition u† p~σun. The last possibil- ity ¯upγ 5 un (P) does not cause any transitions in the non-relativistic limit. Therefore, it is clear that you need either S or V, and T or A. The true answer turned out to be V and A as we will see below.
2 Parity Violation
As strange the strange particles were, one very puzzling feature emerged. One particle, called τ +, decays into three pions π+π+π−^ or π+π^0 π^0. Another one, called θ+, decays into two pions π+π^0. Both are spin zero particles of strangeness one. The analysis of the final state showed that the τ +^ decays into a parity odd state, while the θ+^ into a parity even state. The mystery was that both particles appeared to have the same mass and the same lifetime. Could this just be a coincidence? T.D. Lee and C.N. Yang pointed out in 1956 that maybe these two parti- cles could be the same particle.^1 Of course it is possible only if the parity is not preserved in these decays. They examined carefully the available evidence for parity conservation, and concluded that there were many good evidence for parity conservation in the strong and the electromagnetic interactions, while there was none in the weak interaction. They further proposed various ways the parity (non)conservation can be tested experimentally in the weak interaction. C.S. Wu did such an experimental test quickly. Read Chapter 6 of Cahn– Goldhaber about the experiment. She has observed a correlation between the spin of 60 Co nucleus and the emitted β-electrons. If the parity were conserved, a final state and its parity conjugate must have the same probability. Under parity, the spin does not change, while the direction of the electron flips. Therefore there should not be any correlation between the direction of the spin and the direction of the electron. She observed that there was a clear correlation in a beautiful experiment. (This paper is a must-read!) The parity, which was long believed to be the true symmetry of nature, fell in
(^1) See anecdote that involves Martin Bloch, Richard Feynman and Lee–Yang in Cahn– Goldhaber, Chapter 6.
The words spread quickly that the parity is violated, and the violation is large. Hearing this, R. L. Garwin, L. M. Lederman, and M. Weinrich, decided to test parity violation in muon decay. According to Lederman’s book “The God Particle,” his student was mounting his thesis experiment at a cyclotron. Garwin, Lederman, and Weinrich, all excited, stromed the lab, dismounted poor student’s experiment, and set up the muon decay experi- ment quickly. There, they found a strong correlation between the muon spin and the electron direction. Their papers are publisehd next to each other in Physical Review. Around the same time, there was a big confusion if the nuclear beta decay was due to the S, T combination or V, A combination. It was once “established” that it was due to the S, T combination, that turned out to be false. In the end it was decided that V and A are the correct form of the weak interaction. What combination of V and A, then? An extremely clever experiment was performed by M. Goldhaber, L. Grodzins, and A. W. Sunyar, to determine the helicity of the neutrino emitted in the nuclear beta decay. It sounds like an impossible task because you don’t see neutrinos. How can you measure the helicity of the particle you don’t see? They made a very clever choice of the parent and daughter nuclei so that the daughter nucleus undergoes γ-decay and the helicity of the photon basically tells you what the helicity of the neutrino is. They have found the result that was consistent with 100% left-handed neutrino. Because the helicity and the chirality are in one- to-one correspondence in the ultra-relativistic limit, it follows that the left- handed helicity of neutrinos means the chirality γ 5 = −1. In other words, the projection operator (1 − γ 5 )/2 is need. (This experiment used the capture of an orbital electron by the nucleus, and hence the net reaction is e−p → νen.) This point determines that the combination of V (¯uν γμue) and A (¯uν γμγ 5 ue) to be of the V − A form ¯uν γμ(1 − γ 5 )ue. Because both V and A appear with 50:50 mixture, this form of the interaction is said to be “maximal violation of parity.” If V dominates over A, you can treat A as a small perturbation that violates parity, and vice versa. But they come with 50:50 mixture, and neither of them is “small,” hence the maximal violation. Parity can’t be violated more! Another important point is this. It is only the left-handed chirality par- ticles that are involved in the weak interaction. This point is very important in the later construction of the SU (2) × U (1) gauge theory. If neutrinos are indeed all left-handed, it implies that they are exactly
Figure 1: Test of V − A nature in the muon decay. 90% CL experimental limit on possible interactions. Taken from A. Pich, http://www.arxiv.org/ abs/hep-ph/9701263.
decay probabilities is well-understood as
BR(π−^ → e−^ ν¯e) BR(π−^ → μ−^ ν¯μ)
( me mμ
) 2 (1 − m^2 e/m^2 π)^2 (1 − m^2 μ/m^2 π)^2
This observation demonstrates both the V − A nature of the interaction and also the universal strength between the electron and muon coupling. In fact, the universal strength of the weak interaction became more and more apparent. The nuclear β-decay, pion decay (both into electron and muon), muon decay appear all described by the same Fermi constant. This property is very similar to that of the electromagnetism and the quark-gluon interaction. In either case, once you specify the “charge,” the electric charge and the color, respectively, the coupling constants α, and αs describe the strength of the interaction no matter what the particle is. The force does not care the nature of the particle except for its charge. This point strongly suggests that the weak interaction is of the same kind as the other forces, namely the gauge force mediated by the spin one boson similar to the photon and the gluon. In the modern picture, what causes the nuclear beta decay is the process d → ue−^ ν¯e. For example, one of the down quarks in the neutron (udd) un- dergoes this decay, converting the neutron to a proton (udd). The amplitude for this process is exactly the same as the muon decay with the V − A form.^2 Let us crudely test the universality based on dimensional analysis. First take the muon decay. All particles in the final state can be approximated massless relative to the muon mass. The decay amplitude is proportional to GF which has the dimension of energy inverse squared. The decay with is proportional to G^2 F. To obtain the decay width that has the dimension of energy, we need to provide a factor with the dimension of energy to the fifth. The only quantity you can use is the muon mass, and hence Γ ∼ G^2 F m^5 μ. In fact, much more careful calculation gives
Γμ =
192 π^3
G^2 F m^5 μ. (4)
The factor 1/ 192 π^3 is typical of three-body phase space. Using the measured muon lifetime 2.19703(4) μs (with higher order corrections), we obtain the Fermi constant GF = 1.16639(1) × 10 −^5 GeV^2.
(^2) Note that the strength is slightly weaker because of the Cabibbo mixing, as we will discuss later.
two neutrinos have distinct electron and muon numbers and hence must be different particles. The accelerator-based neutrino beam was first produced for this experiment. They brought in a ship wreck as a steel shield to avoid any muons sneaking into the detector from the beam!
3 Cabibbo Angle
OK, the Fermi interaction together with the V − A theory seems to describe many weak interactions well. But what about the decay of the strange parti- cles? After all, the strangeness was introduced to explain the pair production of new types of particles via the strong interaction, but their longevity (com- pared to 10−^23 sec, a characteristic time scale of hadron resonances). The weak interaction is supposed to violate strangeness and let strange parti- cles decay. In modern language, we need the decay of the strange quark s → ue−^ ν¯e, very analogous to the nuclear beta decay that is caused by d → ue−^ ¯νe. If you look at the lifetime of K+, Λ, etc, the universal strength of the weak interaction predicts too short lifetimes. The strange particles are long-lived even in the standard of the weak interaction! However, given the success of the universality in nuclear beta decay and the muon decay, we do not want to give up the idea of universality. Cabibbo in 1960 proposed a slightly extended notion of universality. He proposed that the weak interaction acts on the linear combination of the down and strange quarks (he didn’t use the language of quarks back then; this is the modern translation of what he said),
d′^ = d cos θC + s sin θC. (5)
The strength of the weak interaction on d′^ is still the same as in the muon decay. But if you specialize to the d, the vertex is smaller by a factor of cos θC , which is still close to one. However, the vertex for the strange quark is much more suppressed, by a factor of sin θC ' 0 .22. This makes the lifetime of the strange particles longer by a factor of 1/ sin^2 θC ' 20. At this point, the fact that cos θC < 1 could not be established because of the experimental uncertainties. Nowadays, however, we know that in- deed the nuclear β-decay is slightly weaker than the muon decay. In fact, the best test of universality is done using the nuclear beta decay. It gives that the nuclear beta decay is slightly weaker than the muon decay by
a factor of cos θC = 0. 9740 ± 0 .0005. On the other hand the decay of strange particles, especially K+^ → π^0 e+νe and K^0 → π−e+νe, determine sin θC = 0. 2196 ± 0 .0026. Putting them together, we find the total strength of the quark weak interaction sin^2 θC + cos^2 θC = 0. 9969 ± 0 .0015. It is consistent with universality (one) at the two sigma level.
4 CP Violation
The fact that neutrinos are all left-handed and anti-neutrinos right-handed explicitly break parity, but also charge conjugation. But the product CP still appears to be a symmetry. Under parity, a left-handed neutrino becomes a right-handed neutrino, a state that does not exist. Under charge conjugation, a right-handed neutrino becomes a right-handed anti-neutrino, a state that does exist. Therefore it was hoped that CP is still a symmetry of nature. It is believed that the combination CPT is an exact symmetry. The CPT theorem states that it is a symmetry to perform charge conjugation C, parity P, and time reversal T at the same time. The assumptions to prove this theorem is very reasonable: (1) Lorentz invariance, (2) hermiticity of the Hamiltonian, and (3) locality. The third one means there is no “action at a distance,” but all forces must be mediated by virtual particles which are produced where the sources are. The CPT theorem implies that the particle and anti-particle must have the same mass and the same lifetime (if unstable). Despite the hope that CP may be preserved by the weak interaction, it fell also shortly after parity violation was discovered. The violation of CP is a fascinating subject, as it is a necessary condition for us to understand why there exists only matter in our Universe but no anti-matter.
4.1 Neutral Kaon System
Neutral kaon system is a very strange system, well beyond the fact that they are strange particles anyway. There are two neutral kaons, K^0 (d¯s) and
K 0 (s d¯), anti-particle to each other. Because the strangeness is violated in weak decays, both of them can decay into common final states, π+π−^ and π^0 π^0. Therefore, at the second order in the weak interaction, K^0 can go to
K 0
. A particle and its anti-particle can mix! Read Cahn–Goldhaber Chapter 7 for how this mixing was proposed, seen and studied.
The notation refers to “K-short” and “K-long” because their lifetimes are very different as we will see below.
Note that CP interchanges K^0 and K 0 , and hence the 12 and 21 elements. Therefore, the complex M 12 and Γ 12 break CP invariance. We will see below the difference between the real and complex cases.
4.1.1 CP Conserving Case
Let us suppose for the moment that M 12 , Γ 12 are real. This corresponds to the case where CP is conserved. In this case, p = q and the two eigenstates are
|K 1 〉 =
(|K^0 〉 + |K
0 〉), |K 2 〉 =
(|K^0 〉 − |K
0 〉). (10)
The eigenvalues are
H|K 1 〉 = (m 1 − i
Γ 1 )|K 1 〉 = (m +
∆m 2
i 2
i 4
∆Γ)|K 1 〉, (11)
H|K 2 〉 = (m 2 − i
Γ 2 )|K 2 〉 = (m −
∆m 2
i 2
i 4
∆Γ)|K 2 〉. (12)
The notation ∆m and ∆Γ therefore refers to the difference in the mass and the width between two states. The convention commonly used in the literature is that the charge con-
jugation interchanges K^0 and K 0 with a minus sign (I don’t know why),
C|K^0 〉 = −|K 0 〉, C|K 0 〉 = −|K^0 〉. (13)
Because they are pseudoscalar 0−^ mesons, the parity changes their signs, too,
P |K^0 〉 = −|K^0 〉, P |K 0 〉 = −|K 0 〉. (14)
Putting them together, CP interchanges them without an additional sign,
CP |K^0 〉 = +|K 0 〉, CP |K 0 〉 = +|K^0 〉. (15)
The states |K 1 , 2 〉 therefore are CP eigenstates,
CP |K 1 〉 = +|K 1 〉, CP |K 2 〉 = −|K 2 〉. (16)
When they decay, for example into π^0 π^0 , the two pions must be in L = 0 state because neither the initial state particle (kaon) nor the final state
particles (pions) have spins and the angular momentum must be conserved. Therefore the parity of the final state is (−1)L(−1)^2 = +1, where two factors of (−1) come from the intrinsic parity of the pion. Under charge conjugation, π^0 is an eigenstate of C = +1 (remember it decays into two photons and we could also derive this eigenvalue from the quark model) and hence π^0 π^0 state is also a state with C = +1. Therefore, the CP eigenvalue of the final state is CP = +1. If CP is conserved, only K 1 can decay into this final state. On the other hand, K 2 has odd CP eigenvalue, and (if CP is conserved) must decay into a state with odd CP eigenvalue. 3π^0 and π+π−π^0 state have odd CP and the decay of K 2 into these final states are allowed. However such a decay is suppressed because of much smaller phase space. The kaon mass is 497. 672 ± 0 .031 MeV, the charged pion mass 139. 57018 ± 0 .00035 MeV and the neutral pion mass 134. 9766 ± 0 .0006 MeV. The sum of masses for π+π−π^0 state is 414.117 MeV, quite close to the mass of the kaon. Therefore this decay is barely allowed, and it makes the lifetime of K 2 much longer than that of K 1. In fact, the lifetimes are 0.8935(8) × 10 −^10 sec and 5.17(4) × 10 −^8 sec, respectively, with almost three orders of magnitude difference. Suppose K^0 is produced in a collision of a proton on a nucleus target, with the final state pp → K^0 Σ+p. Because the strangeness is conserved in the strong interaction, the existence of Σ+^ implies that the kaon must be K^0 ,
not K 0
. Once it is produced, however, K^0 starts to mix with K 0 . The time evolution is given simply by the Schr¨odinger equation, with the phase factor e−iEt^ on Hamiltonian eigenstates. Therefore,
|K^0 (t)〉 = e−iHt^
(|K 1 〉+|K 2 〉) =
(|K 1 〉e−i(m^1 −iΓ^1 /2)t+|K 2 〉e−i(m^2 −iΓ^2 /2)t). (17)
Given this we can calculate the probability of finding K^0 or K 0 at a given moment,
P (K^0 → K^0 , t) = |〈K^0 |K^0 (t)〉|^2 =
∣∣ ∣∣^1 2
( e−i(m^1 −iΓ^1 /2)t^ + e−i(m^2 −iΓ^2 /2)t
)∣∣ ∣∣
2
( e−Γ^1 t^ + e−Γ^2 t^ + 2(cos ∆mt)e−(Γ^1 +Γ^2 )t/^2
) , (18)
P (K^0 → K
0 , t) = |〈K 0 |K^0 (t)〉|^2 =
∣∣ ∣∣^1 2
( e−i(m^1 −iΓ^1 /2)t^ − e−i(m^2 −iΓ^2 /2)t
)∣∣ ∣∣
2
( e−Γ^1 t^ + e−Γ^2 t^ − 2(cos ∆mt)e−(Γ^1 +Γ^2 )t/^2
) , (19)
The last term comes from the interference of two Hamiltonian eigenstates
Figure 2: CPLEAR measurement of the asymmetry in K^0 - K 0 oscillation. Taken from Phys. Lett. B 444, 38 (1998).
In order to explain this small admixture of the “wrong” CP state in the Hamiltonian eigenstate, we go back to the general case of complex M 12 and Γ 12. The eigenstates are given by
|KS 〉 = p|K^0 〉 + q|K 0 〉, |KL〉 = p|K^0 〉 − q|K 0 〉, (22)
with the complex coefficients p, q
q p
√√ √√ M 12 ∗ − i 2 Γ∗ 12 M 12 − i 2 Γ 12
Note that the phase of M 12 and Γ 12 can be changed simultaneously by re- defining the phase of states K^0 and K 0 , but the relative phase between M 12 and Γ 12 is convention independent. For instance, we can always choose our convention that Γ 12 is real. Then M 12 has an imaginary part M 12 = <e(M 12 ) + i=m(M 12 ). Assume =m(M 12 ) <e(M 12 ) because it turns out to be true. Then,
q p
√√ √√ <e(M 12 ) − i=m(M 12 ) − i 2 Γ 12 <e(M 12 ) + i=m(M 12 ) − i 2 Γ 12
−i=m(M 12 ) <e(M 12 ) − 2 i Γ 12
With this definition of , we find q = √^12 (1 − ), p = √^12 (1 + ) (all quantities
valid only up to the first order in ), and
|KL〉 = p|K^0 〉−q|K 0 〉 =
(1+)|K^0 〉−
(1−)|K
0 〉 = |K 2 〉+|K 1 〉, (25)
as desired. The data on KL → ππ determines the magnitude of as || = 2 .287(17) × 10 −^3 a very small number but non-vanishing. Another consequence of 6 = 0 is that KL → π−e+νe and KL → π+e−^ ν¯e have different rates. For these decays, K^0 or K 0 are singled out. For instance, K^0 (ds¯) can decay into π−e+νe because ¯s → ue¯−^ ¯νe, but cannot decay into π+e−^ ¯νe. The same is true for muon final states. Therefore, the decays into these states “measure” |q|/|p| directly. People define the observable
δL =
Γ(KL → π−+ν) − Γ(KL → π+−^ ν¯)) Γ(KL → π−+ν) + Γ(KL → π+−^ ν¯))
where ` = e, μ. Experimentally, δL = 3.27(12) × 10 −^3. It follows from the above discussions that δL = 2<e(). Using the observed values of ∆m and |Γ 12 | = ΓS − ΓL, you can verify that this observed value of δL is consistent with || mentioned above. Therefore, matter and anti-matter mix, but not quite 50:50. The world of anti-matter is not an exact mirror of the world of matter. This phenomenon is called “indirect CP Violation” because the violation of CP symmetry is in the mixing =m(M 12 ) 6 = 0 but not in the decay of the particles.
4.1.3 Direct CP Violation
Much more recently, even more direct but subtle difference between K^0 and
K 0 became known. This phenomenon is called “direct CP violation,” because it is the difference in the decays or particle and its anti-particle counterpart. This phenomenon was established by 2001. The direction CP violation is the difference
Γ(K^0 → π+π−) − Γ(K 0 → π+π−), (27)
and similarly for π^0 π^0. Because of the CPT theorem, the total width Γ(K^0 →
everything) and Γ(K 0 → everything) are equal. However the individual contributions can be different if CP is violated. This difference is extremely
Theoretically, we start with
|K^0 〉 =
2 p
(|KS 〉 + |KL〉). (33)
The time evolution gives
|K^0 (t)〉 =
2 p
(|KS 〉e−imS^ t−Γst/s^ + |KL〉e−imLt−ΓLt/s)
2 p
((p|K^0 〉 + q|K 0 〉)e−imS^ t−Γst/s^ + (p|K^0 〉 − q|K 0 〉)e−imLt−ΓLt/s)
= |K^0 〉
(e−imS^ t−ΓS^ t/s^ + e−imLt−ΓLt/s)
+|K 0 〉
q 2 p
(e−imS^ t−ΓS^ t/s^ − e−imLt−ΓLt/s). (34)
Therefore, for K^0 produced at t = 0 to decay as K 0 at τ is
R(K^0 (0) → e−π+^ ¯νe(τ ))
|q|^2 4 |p|^2
(e−ΓS^ τ^ + e−ΓLτ^ − (cos ∆mτ )e−(ΓS^ +ΓL)τ /^2 ). (35)
Similarly,
R(K
0 (0) → e+π−νe(τ ))
|p|^2 4 |q|^2
(e−ΓS^ τ^ + e−ΓLτ^ − (cos ∆mτ )e−(ΓS^ +ΓL)τ /^2 ). (36)
Therefore,
R(K 0 (0) → e+π−νe(τ )) − R(K^0 (0) → e−π+^ ν¯e(τ )) R(K 0 (0) → e+π−νe(τ )) + R(K^0 (0) → e−π+^ ¯νe(τ ))
|p|^4 − |q|^4 |p|^4 + |q|^4
= 4<e().
(37) Indeed, the observed T-violation is consistent with this expectation.
4.2 Neutral B-Meson System
The formalism used in the neutral kaon system can be used also in the
neutral B-meson system, B^0 (d¯b) and B 0 (b d¯). The main difference in this case is that the lifetime difference is very small. Therefore, the observation of CP violation would use a different technique. The idea is very simple. You first produce Υ(4S) resonance in an e+e−
annihilation, which decays into B^0 - B 0 pair. The point is that Υ is a spin one
resonance, and hence there must be L = 1 angular momentum in the final state (B^0 is a pseudo-scalar mesons just like the pion). But L = 1 means that the interchange of two particles gives −1, which is not possible for identical bosons. Therefore, even though B^0 and B 0 states evolve in time, you can
never have a moment when both of them are simultaneously B^0 (or B 0 ). In other words, if you are sure that one of them is B^0 at a given moment t, the other particle must be B 0 at the same moment t. This way, you can identify a particle to be 100% B^0 at the moment t, and see how it evolves in time. If you look for a specific final state such as J/ψKS , it is a CP eigenstate (CP = −1 to the extent we ignore mixture of CP-odd state in KS ) and
both B^0 and B 0 can decay into this state. This final state is particularly easy to identify when J/ψ → μ+μ−^ or e+e−^ and KS → π+π−^ with the well-known masses. This way, you can define a time-dependent asymmetry,
af (t) =
Γ(B
0 (t) → f ) − Γ(B^0 (t) → f ) Γ(B 0 (t) → f ) + Γ(B^0 (t) → f )
A problem to measure this quantity is that you need to resolve “time” when B-meson decays. Pier Oddone at Lawrence Berkeley Laboratory came up with the idea that you can collide two beams asymmetrically, so that the produce B and B mesons move in the laboratory frame. Then you can determine the vertex where they decay and translate it to the timing. This way, you can see if there is any time-dependent difference between the B^0 and B 0 decay. The expectation is easily derived by using the same type of equation as in the kaon case,
|B 1 〉 = p|B^0 〉 + q|B 0 〉, |B 2 〉 = p|B^0 〉 − q|B 0 〉, (39)
where the complex coefficients p, q are normalized |p|^2 + |q|^2 = 1 and
q p
√√ √√ M 12 ∗ − i 2 Γ∗ 12 M 12 − i 2 Γ 12
We don’t use “short” or “long” because the lifetime difference is small. (Un- like in the case of kaons, where K 1 , 2 were reserved for CP eigenstate, B 1 , 2 are not CP eigenstates but rather Hamiltonian eigenstates.) When you start with B^0 at t = 0,
|B^0 (0)〉 =
2 p
(|B 1 〉 + |B 2 〉), (41)
