CP violation edit

After the striking evidence for parity violation provided by Wu et al. in 1957, it was assumed that CP (charge conjugation-parity) is the quantity which is conserved.^[6] However, in 1964 Cronin and Fitch reported CP violation in the neutral Kaon system.^[7] They observed the long-lived K_L (with CP = −1 ) undergoing decays into two pions (with CP = [−1]·[−1] = +1 ) thereby violating CP conservation.

In 2001, CP violation in the
B⁰
⇄
B⁰
system was confirmed by the BaBar and the Belle experiments.^[8][9] Direct CP violation in the
B⁰
⇄
B⁰
system was reported by both the labs by 2005.^[10][11]

The
K⁰
⇄
K⁰
and the
B⁰
⇄
B⁰
systems can be studied as two state systems, considering the particle and its antiparticle as the two states.

The solar neutrino problem edit

The pp chain in the sun produces an abundance of
ν
_e. In 1968, R. Davis et al. first reported the results of the Homestake experiment.^[12][13] Also known as the Davis experiment, it used a huge tank of perchloroethylene in Homestake mine (it was deep underground to eliminate background from cosmic rays), South Dakota. Chlorine nuclei in the perchloroethylene absorb
ν
_e to produce argon via the reaction

${\displaystyle \mathrm {\nu _{e}+{{}_{17}^{37}Cl}\rightarrow {{}_{18}^{37}}Ar+e^{-}} }$ ,

which is essentially

${\displaystyle \mathrm {\nu _{e}+n\to p+e^{-}} }$ .^[14]

The experiment collected argon for several months. Because the neutrino interacts very weakly, only about one argon atom was collected every two days. The total accumulation was about one third of Bahcall's theoretical prediction.

In 1968, Bruno Pontecorvo showed that if neutrinos are not considered massless, then
ν
_e (produced in the sun) can transform into some other neutrino species (
ν
_μ or
ν
_τ), to which Homestake detector was insensitive. This explained the deficit in the results of the Homestake experiment. The final confirmation of this solution to the solar neutrino problem was provided in April 2002 by the SNO (Sudbury Neutrino Observatory) collaboration, which measured both
ν
_e flux and the total neutrino flux.^[15]

This 'oscillation' between the neutrino species can first be studied considering any two, and then generalized to the three known flavors.

A special case: considering mixing only edit

Caution: "mixing" discussed in this article is not the type obtained from mixed quantum states. Rather, "mixing" here refers to the superposition of "pure state" energy (mass) eigenstates, described by a "mixing matrix" (e.g. the CKM or PMNS matricies).

Let ${\displaystyle \,H_{0}\,}$  be the Hamiltonian of the two-state system, and ${\displaystyle \;\left|1\right\rangle \;}$  and ${\displaystyle \;\left|2\right\rangle \;}$  be its orthonormal eigenvectors with eigenvalues ${\displaystyle \,E_{1}\,}$  and ${\displaystyle \,E_{2}\,}$  respectively.

Let ${\displaystyle \,\left|\Psi \left(t\right)\right\rangle \,}$  be the state of the system at time ${\displaystyle \,t~.}$ 

If the system starts as an energy eigenstate of ${\displaystyle \,H_{0}\;,}$  i.e. say

${\displaystyle \left|\Psi \left(0\right)\right\rangle =\left|1\right\rangle }$ 

then, the time evolved state, which is the solution of the Schrödinger equation

will be,^[16]

${\displaystyle \left|\Psi \left(t\right)\right\rangle =\left|1\right\rangle e^{-i{\frac {E_{1}t}{\hbar }}}}$ 

But this is physically same as ${\displaystyle \left|1\right\rangle }$  as the exponential term is just a phase factor and does not produce a new state. In other words, energy eigenstates are stationary eigenstates, i.e. they do not yield physically new states under time evolution.

In the basis ${\displaystyle \,\left\{\left|1\right\rangle ,\left|2\right\rangle \right\}\;,}$  ${\displaystyle \,H_{0}\,}$  is diagonal. That is,

${\displaystyle H_{0}={\begin{pmatrix}E_{1}&0\\0&E_{2}\\\end{pmatrix}}}$ 

It can be shown, that oscillation between states will occur if and only if off-diagonal terms of the Hamiltonian are non-zero.

Hence let us introduce a general perturbation ${\displaystyle W}$  in ${\displaystyle H_{0}}$  such that the resultant Hamiltonian ${\displaystyle H}$  is still Hermitian. Then,

${\displaystyle W={\begin{pmatrix}W_{11}&W_{12}\\W_{12}^{*}&W_{22}\\\end{pmatrix}}}$  where ${\displaystyle W_{11},W_{22}\in \mathbb {R} }$  and ${\displaystyle W_{12}\in \mathbb {C} }$ 

and,

Then, the eigenvalues of ${\displaystyle H}$  are,^[17]

Since ${\displaystyle \,H\,}$  is a general Hamiltonian matrix, it can be written as,^[18]

${\displaystyle H=\sum \limits _{j=0}^{3}a_{j}\sigma _{j}=a_{0}\sigma _{0}+H'}$ 

The following two results are clear:

• ${\displaystyle \,\left[H,H'\right]=0\,}$ 

Proof
${\displaystyle {\begin{aligned}HH'&=a_{0}\sigma _{0}H'+H'H'=a_{0}\sigma _{0}+{H'}^{2}\\H'H&=a_{0}H'\sigma _{0}+H'H'=a_{0}\sigma _{0}+{H'}^{2}\\\therefore \left[H,H'\right]&=HH'-H'H=0\\\end{aligned}}}$

• ${\displaystyle \,{H'}^{2}=I\,}$ 

Proof

${\displaystyle {\begin{aligned}{H'}^{2}&=\sum \limits _{j=1}^{3}{n_{j}\sigma _{j}}\sum \limits _{k=1}^{3}{n_{k}\sigma _{k}}=\sum \limits _{j,k=1}^{3}{n_{j}n_{k}\sigma _{j}\sigma _{k}}\\&=\sum \limits _{j,k=1}^{3}{n_{j}n_{k}\left(\delta _{jk}I+i\sum \limits _{\ell =1}^{3}{\varepsilon _{jk\ell }\sigma _{\ell }}\right)}\\&=\left(\sum \limits _{j=1}^{3}{n_{j}}^{2}\right)I+i\sum \limits _{\ell =1}^{3}{\sigma _{l}\sum \limits _{j,k=1}^{3}\varepsilon _{jk\ell }}\\&=I\\\end{aligned}}}$  where the following results have been used: ${\displaystyle \sigma _{j}\sigma _{k}=\delta _{jk}I+i\sum \limits _{\ell =1}^{3}{\varepsilon _{jk\ell }\sigma _{\ell }}}$  ${\displaystyle {\hat {n}}}$  is a unit vector and hence ${\displaystyle \sum \limits _{j=1}^{3}{{n_{j}}^{2}}=\left|{\hat {n}}\right|^{2}=1}$  The Levi-Civita symbol ${\displaystyle \varepsilon _{jk\ell }}$  is antisymmetric in any two of its indices (${\displaystyle j}$  and ${\displaystyle k}$  in this case) and hence ${\displaystyle \sum \limits _{j,k=1}^{3}\varepsilon _{jk\ell }=0}$

With the following parametrization^[18] (this parametrization helps as it normalizes the eigenvectors and also introduces an arbitrary phase ${\displaystyle \phi }$  making the eigenvectors most general)

${\displaystyle {\hat {n}}=\left(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta \right)}$ ,

and using the above pair of results the orthonormal eigenvectors of ${\displaystyle H'}$  and hence of ${\displaystyle H}$  are obtained as,

Writing the eigenvectors of ${\displaystyle \,H_{0}\,}$  in terms of those of ${\displaystyle \,H\,}$  we get,

Now if the particle starts out as an eigenstate of ${\displaystyle \,H_{0}\,}$  (say, ${\displaystyle \,\left|1\right\rangle \,}$ ), that is,

${\displaystyle \left|\Psi \left(0\right)\right\rangle =\left|1\right\rangle }$ 

then under time evolution we get,^[17]

${\displaystyle \left|\Psi \left(t\right)\right\rangle =e^{i{\frac {\phi }{2}}}\left(\cos {\frac {\theta }{2}}\left|+\right\rangle e^{-i{\frac {E_{+}t}{\hbar }}}-\sin {\frac {\theta }{2}}\left|-\right\rangle e^{-i{\frac {E_{-}t}{\hbar }}}\right)}$ 

which unlike the previous case, is distinctly different from ${\displaystyle \;\left|1\right\rangle ~.}$ 

We can then obtain the probability of finding the system in state ${\displaystyle \;\left|2\right\rangle \;}$  at time ${\displaystyle \,t\,}$  as,^[17]

which is called Rabi's formula. Hence, starting from one eigenstate of the unperturbed Hamiltonian ${\displaystyle \,H_{0}\;,}$  the state of the system oscillates between the eigenstates of ${\displaystyle \,H_{0}\,}$  with a frequency (known as Rabi frequency),

From the expression of ${\displaystyle P_{21}(t)}$  we can infer that oscillation will exist only if ${\displaystyle \;\left|W_{12}\right|^{2}\neq 0~.}$  ${\displaystyle \,W_{12}\,}$  is thus known as the coupling term as it couples the two eigenstates of the unperturbed Hamiltonian ${\displaystyle H_{0}}$  and thereby facilitates oscillation between the two.

Oscillation will also cease if the eigenvalues of the perturbed Hamiltonian ${\displaystyle H}$  are degenerate, i.e. ${\displaystyle \;E_{+}=E_{-}~.}$  But this is a trivial case as in such a situation, the perturbation itself vanishes and ${\displaystyle H}$  takes the form (diagonal) of ${\displaystyle H_{0}}$  and we're back to square one.

Hence, the necessary conditions for oscillation are:

• Non-zero coupling, i.e. ${\displaystyle \;\left|W_{12}\right|^{2}\neq 0~.}$ 
• Non-degenerate eigenvalues of the perturbed Hamiltonian ${\displaystyle \,H\,}$ , i.e. ${\displaystyle \;E_{+}\neq E_{-}~.}$ 

The general case: considering mixing and decay edit

If the particle(s) under consideration undergoes decay, then the Hamiltonian describing the system is no longer Hermitian.^[19] Since any matrix can be written as a sum of its Hermitian and anti-Hermitian parts, ${\displaystyle H}$  can be written as,

${\displaystyle H=M-{\frac {i}{2}}\Gamma ={\begin{pmatrix}M_{11}&M_{12}\\M_{12}^{*}&M_{11}\\\end{pmatrix}}-{\frac {i}{2}}{\begin{pmatrix}\Gamma _{11}&\Gamma _{12}\\\Gamma _{12}^{*}&\Gamma _{11}\\\end{pmatrix}}}$ 

The eigenvalues of ${\displaystyle H}$  are,

The suffixes stand for Heavy and Light respectively (by convention) and this implies that ${\displaystyle \Delta m}$  is positive.

The normalized eigenstates corresponding to ${\displaystyle \mu _{L}}$  and ${\displaystyle \mu _{H}}$  respectively, in the natural basis ${\displaystyle \left\{\left|P\right\rangle ,\left|{\bar {P}}\right\rangle \right\}\equiv \left\{\left(1,0\right),\left(0,1\right)\right\}}$  are,

${\displaystyle p}$  and ${\displaystyle q}$  are the mixing terms. Note that these eigenstates are no longer orthogonal.

Let the system start in the state ${\displaystyle \left|P\right\rangle }$ . That is,

${\displaystyle \left|P\left(0\right)\right\rangle =\left|P\right\rangle ={\frac {1}{2p}}\left(\left|P_{L}\right\rangle +\left|P_{H}\right\rangle \right)}$ 

Under time evolution we then get,

${\displaystyle \left|P\left(t\right)\right\rangle ={\frac {1}{2p}}\left(\left|P_{L}\right\rangle e^{-{\frac {i}{\hbar }}\left(m_{L}-{\frac {i}{2}}\gamma _{L}\right)t}+\left|P_{H}\right\rangle e^{-{\frac {i}{\hbar }}\left(m_{H}-{\frac {i}{2}}\gamma _{H}\right)t}\right)=g_{+}\left(t\right)\left|P\right\rangle -{\frac {q}{p}}g_{-}\left(t\right)\left|{\bar {P}}\right\rangle }$ 

Similarly, if the system starts in the state ${\displaystyle \left|{\bar {P}}\right\rangle }$ , under time evolution we obtain,

${\displaystyle \left|{\bar {P}}(t)\right\rangle ={\frac {1}{2q}}\left(\left|P_{L}\right\rangle e^{-{\frac {i}{\hbar }}\left(m_{L}-{\frac {i}{2}}\gamma _{L}\right)t}-\left|P_{H}\right\rangle e^{-{\frac {i}{\hbar }}\left(m_{H}-{\frac {i}{2}}\gamma _{H}\right)t}\right)=-{\frac {p}{q}}g_{-}\left(t\right)\left|P\right\rangle +g_{+}\left(t\right)\left|{\bar {P}}\right\rangle }$ 

If in a system ${\displaystyle \left|P\right\rangle }$  and ${\displaystyle \left|{\bar {P}}\right\rangle }$  represent CP conjugate states (i.e. particle-antiparticle) of one another (i.e. ${\displaystyle CP\left|P\right\rangle =e^{i\delta }\left|{\bar {P}}\right\rangle }$  and ${\displaystyle CP\left|{\bar {P}}\right\rangle =e^{-i\delta }\left|P\right\rangle }$ ), and certain other conditions are met, then CP violation can be observed as a result of this phenomenon. Depending on the condition, CP violation can be classified into three types:^[19][21]

CP violation through decay only edit

Consider the processes where ${\displaystyle \left\{\left|P\right\rangle ,\left|{\bar {P}}\right\rangle \right\}}$  decay to final states ${\displaystyle \left\{\left|f\right\rangle ,\left|{\bar {f}}\right\rangle \right\}}$ , where the barred and the unbarred kets of each set are CP conjugates of one another.

The probability of ${\displaystyle \left|P\right\rangle }$  decaying to ${\displaystyle \left|f\right\rangle }$  is given by,

${\displaystyle \wp _{P\to f}\left(t\right)=\left|\left\langle f|P\left(t\right)\right\rangle \right|^{2}=\left|g_{+}\left(t\right)A_{f}-{\frac {q}{p}}g_{-}\left(t\right){\bar {A}}_{f}\right|^{2}}$ ,

and that of its CP conjugate process by,

${\displaystyle \wp _{{\bar {P}}\to {\bar {f}}}\left(t\right)=\left|\left\langle {\bar {f}}|{\bar {P}}\left(t\right)\right\rangle \right|^{2}=\left|g_{+}\left(t\right){\bar {A}}_{\bar {f}}-{\frac {p}{q}}g_{-}\left(t\right)A_{\bar {f}}\right|^{2}}$ 

If there is no CP violation due to mixing, then ${\displaystyle \left|{\frac {q}{p}}\right|=1}$ .

Now, the above two probabilities are unequal if,

.

Hence, the decay becomes a CP violating process as the probability of a decay and that of its CP conjugate process are not equal.

CP violation through mixing only edit

The probability (as a function of time) of observing ${\displaystyle \left|{\bar {P}}\right\rangle }$  starting from ${\displaystyle \left|P\right\rangle }$  is given by,

${\displaystyle \wp _{P\to {\bar {P}}}\left(t\right)=\left|\left\langle {\bar {P}}|P\left(t\right)\right\rangle \right|^{2}=\left|{\frac {q}{p}}g_{-}\left(t\right)\right|^{2}}$ ,

and that of its CP conjugate process by,

${\displaystyle \wp _{{\bar {P}}\to P}\left(t\right)=\left|\left\langle P|{\bar {P}}\left(t\right)\right\rangle \right|^{2}=\left|{\frac {p}{q}}g_{-}\left(t\right)\right|^{2}}$ .

The above two probabilities are unequal if,

Hence, the particle-antiparticle oscillation becomes a CP violating process as the particle and its antiparticle (say, ${\displaystyle \left|P\right\rangle }$  and ${\displaystyle \left|{\bar {P}}\right\rangle }$  respectively) are no longer equivalent eigenstates of CP.

CP violation through mixing-decay interference edit

Let ${\displaystyle \left|f\right\rangle }$  be a final state (a CP eigenstate) that both ${\displaystyle \left|P\right\rangle }$  and ${\displaystyle \left|{\bar {P}}\right\rangle }$  can decay to. Then, the decay probabilities are given by,

${\displaystyle {\begin{aligned}\wp _{P\to f}\left(t\right)&=\left|\left\langle f|P\left(t\right)\right\rangle \right|^{2}\\&=\left|A_{f}\right|^{2}{\frac {e^{-\gamma t}}{2}}\left[\left(1+\left|\lambda _{f}\right|^{2}\right)\cosh \left({\frac {\Delta \gamma }{2}}t\right)+2\operatorname {Re} \left(\lambda _{f}\right)\sinh \left({\frac {\Delta \gamma }{2}}t\right)+\left(1-\left|\lambda _{f}\right|^{2}\right)\cos \left(\Delta mt\right)+2\operatorname {Im} \left(\lambda _{f}\right)\sin \left(\Delta mt\right)\right]\\\end{aligned}}}$ 

and,

${\displaystyle {\begin{aligned}\wp _{{\bar {P}}\to f}\left(t\right)&=\left|\left\langle f|{\bar {P}}\left(t\right)\right\rangle \right|^{2}\\&=\left|A_{f}\right|^{2}\left|{\frac {p}{q}}\right|^{2}{\frac {e^{-\gamma t}}{2}}\left[\left(1+\left|\lambda _{f}\right|^{2}\right)\cosh \left({\frac {\Delta \gamma }{2}}t\right)+2\operatorname {Re} \left(\lambda _{f}\right)\sinh \left({\frac {\Delta \gamma }{2}}t\right)-\left(1-\left|\lambda _{f}\right|^{2}\right)\cos \left(\Delta mt\right)-2\operatorname {Im} \left(\lambda _{f}\right)\sin \left(\Delta mt\right)\right]\\\end{aligned}}}$ 

From the above two quantities, it can be seen that even when there is no CP violation through mixing alone (i.e. ${\displaystyle \left|q/p\right|=1}$ ) and neither is there any CP violation through decay alone (i.e. ${\displaystyle \left|{\bar {A}}_{f}/A_{f}\right|=1}$ ) and thus ${\displaystyle \left|\lambda _{f}\right|=1}$ , the probabilities will still be unequal provided,

The last terms in the above expressions for probability are thus associated with interference between mixing and decay.

An alternative classification edit

Usually, an alternative classification of CP violation is made:^[21]

Neutrino oscillation edit

Considering a strong coupling between two flavor eigenstates of neutrinos (for example,
ν
_e–
ν
_μ,
ν
_μ–
ν
_τ, etc.) and a very weak coupling between the third (that is, the third does not affect the interaction between the other two), equation (6) gives the probability of a neutrino of type ${\displaystyle \alpha }$  transmuting into type ${\displaystyle \beta }$  as,

${\displaystyle P_{\beta \alpha }\left(t\right)=\sin ^{2}\theta \sin ^{2}\left({\frac {E_{+}-E_{-}}{2\hbar }}t\right)}$ 

where, ${\displaystyle E_{+}}$  and ${\displaystyle E_{-}}$  are energy eigenstates.

The above can be written as,

Thus, a coupling between the energy (mass) eigenstates produces the phenomenon of oscillation between the flavor eigenstates. One important inference is that neutrinos have a finite mass, although very small. Hence, their speed is not exactly the same as that of light but slightly lower.

Neutrino mass splitting edit

With three flavors of neutrinos, there are three mass splittings:

${\displaystyle {\begin{aligned}\left(\Delta m^{2}\right)_{12}&={m_{1}}^{2}-{m_{2}}^{2}\\\left(\Delta m^{2}\right)_{23}&={m_{2}}^{2}-{m_{3}}^{2}\\\left(\Delta m^{2}\right)_{31}&={m_{3}}^{2}-{m_{1}}^{2}\end{aligned}}}$ 

But only two of them are independent, because ${\displaystyle \left(\Delta m^{2}\right)_{12}+\left(\Delta m^{2}\right)_{23}+\left(\Delta m^{2}\right)_{31}=0~}$ .

This implies that two of the three neutrinos have very closely placed masses. Since only two of the three ${\displaystyle \Delta m^{2}}$  are independent, and the expression for probability in equation (13) is not sensitive to the sign of ${\displaystyle \Delta m^{2}}$  (as sine squared is independent of the sign of its argument), it is not possible to determine the neutrino mass spectrum uniquely from the phenomenon of flavor oscillation. That is, any two out of the three can have closely spaced masses.

Moreover, since the oscillation is sensitive only to the differences (of the squares) of the masses, direct determination of neutrino mass is not possible from oscillation experiments.

Length scale of the system edit

Equation (13) indicates that an appropriate length scale of the system is the oscillation wavelength ${\displaystyle \lambda _{\text{osc}}}$ . We can draw the following inferences:

• If ${\displaystyle x/\lambda _{\text{osc}}\ll 1}$ , then ${\displaystyle P_{\beta \alpha }\simeq 0}$  and oscillation will not be observed. For example, production (say, by radioactive decay) and detection of neutrinos in a laboratory.
• If ${\displaystyle x/\lambda _{\text{osc}}\simeq n}$ , where ${\displaystyle n}$  is a whole number, then ${\displaystyle P_{\beta \alpha }\simeq 0}$  and oscillation will not be observed.
• In all other cases, oscillation will be observed. For example, ${\displaystyle x/\lambda _{\text{osc}}\gg 1}$  for solar neutrinos; ${\displaystyle x\sim \lambda _{\text{osc}}}$  for neutrinos from nuclear power plant detected in a laboratory few kilometers away.

Neutral kaon oscillation and decay edit

CP violation through mixing only edit

The 1964 paper by Christenson et al.^[7] provided experimental evidence of CP violation in the neutral Kaon system. The so-called long-lived Kaon (CP = −1) decayed into two pions (CP = (−1)(−1) = 1), thereby violating CP conservation.

${\displaystyle \left|K^{0}\right\rangle }$  and ${\displaystyle \left|{\bar {K}}^{0}\right\rangle }$  being the strangeness eigenstates (with eigenvalues +1 and −1 respectively), the energy eigenstates are,

${\displaystyle {\begin{aligned}\left|K_{^{1}}^{0}\right\rangle &={\frac {1}{\sqrt {2}}}\left(\left|K^{0}\right\rangle +\left|{\bar {K}}^{0}\right\rangle \right)\\\left|K_{2}^{0}\right\rangle &={\frac {1}{\sqrt {2}}}\left(\left|K^{0}\right\rangle -\left|{\bar {K}}^{0}\right\rangle \right)\end{aligned}}}$ 

These two are also CP eigenstates with eigenvalues +1 and −1 respectively. From the earlier notion of CP conservation (symmetry), the following were expected:

• Because ${\displaystyle \left|K_{^{1}}^{0}\right\rangle }$  has a CP eigenvalue of +1, it can decay to two pions or with a proper choice of angular momentum, to three pions. However, the two pion decay is a lot more frequent.
• ${\displaystyle \left|K_{2}^{0}\right\rangle }$  having a CP eigenvalue −1, can decay only to three pions and never to two.

Since the two pion decay is much faster than the three pion decay, ${\displaystyle \left|K_{^{1}}^{0}\right\rangle }$  was referred to as the short-lived Kaon ${\displaystyle \left|K_{S}^{0}\right\rangle }$ , and ${\displaystyle \left|K_{2}^{0}\right\rangle }$  as the long-lived Kaon ${\displaystyle \left|K_{L}^{0}\right\rangle }$ . The 1964 experiment showed that contrary to what was expected, ${\displaystyle \left|K_{L}^{0}\right\rangle }$  could decay to two pions. This implied that the long lived Kaon cannot be purely the CP eigenstate ${\displaystyle \left|K_{2}^{0}\right\rangle }$ , but must contain a small admixture of ${\displaystyle \left|K_{^{1}}^{0}\right\rangle }$ , thereby no longer being a CP eigenstate.^[22] Similarly, the short-lived Kaon was predicted to have a small admixture of ${\displaystyle \left|K_{2}^{0}\right\rangle }$ . That is,

${\displaystyle {\begin{aligned}\left|K_{L}^{0}\right\rangle &={\frac {1}{\sqrt {1+\left|\varepsilon \right|^{2}}}}\left(\left|K_{2}^{0}\right\rangle +\varepsilon \left|K_{1}^{0}\right\rangle \right)\\\left|K_{S}^{0}\right\rangle &={\frac {1}{\sqrt {1+\left|\varepsilon \right|^{2}}}}\left(\left|K_{1}^{0}\right\rangle +\varepsilon \left|K_{2}^{0}\right\rangle \right)\end{aligned}}}$ 

where, ${\displaystyle \varepsilon }$  is a complex quantity and is a measure of departure from CP invariance. Experimentally, ${\displaystyle \left|\varepsilon \right|=\left(2.228\pm 0.011\right)\times 10^{-3}}$ .^[23]

Writing ${\displaystyle \left|K_{^{1}}^{0}\right\rangle }$  and ${\displaystyle \left|K_{2}^{0}\right\rangle }$  in terms of ${\displaystyle \left|K^{0}\right\rangle }$  and ${\displaystyle \left|{\bar {K}}^{0}\right\rangle }$ , we obtain (keeping in mind that ${\displaystyle m_{K_{L}^{0}}>m_{K_{S}^{0}}}$ ^[23]) the form of equation (9):

${\displaystyle {\begin{aligned}\left|K_{L}^{0}\right\rangle &=\left(p\left|K^{0}\right\rangle -q\left|{\bar {K}}^{0}\right\rangle \right)\\\left|K_{S}^{0}\right\rangle &=\left(p\left|K^{0}\right\rangle +q\left|{\bar {K}}^{0}\right\rangle \right)\end{aligned}}}$ 

where, ${\displaystyle {\frac {q}{p}}={\frac {1-\varepsilon }{1+\varepsilon }}}$ .

Since ${\displaystyle \left|\varepsilon \right|\neq 0}$ , condition (11) is satisfied and there is a mixing between the strangeness eigenstates ${\displaystyle \left|K^{0}\right\rangle }$  and ${\displaystyle \left|{\bar {K}}^{0}\right\rangle }$  giving rise to a long-lived and a short-lived state.

CP violation through decay only edit

The
K⁰
_L and
K⁰
_S have two modes of two pion decay:
π⁰

π⁰
or
π⁺

π⁻
. Both of these final states are CP eigenstates of themselves. We can define the branching ratios as,^[21]

${\displaystyle {\begin{aligned}\eta _{+-}&={\frac {\left\langle \pi ^{+}\pi ^{-}|K_{L}^{0}\right\rangle }{\left\langle \pi ^{+}\pi ^{-}|K_{S}^{0}\right\rangle }}={\frac {pA_{\pi ^{+}\pi ^{-}}-q{\bar {A}}_{\pi ^{+}\pi ^{-}}}{pA_{\pi ^{+}\pi ^{-}}+q{\bar {A}}_{\pi ^{+}\pi ^{-}}}}={\frac {1-\lambda _{\pi ^{+}\pi ^{-}}}{1+\lambda _{\pi ^{+}\pi ^{-}}}}\\[3pt]\eta _{00}&={\frac {\left\langle \pi ^{0}\pi ^{0}|K_{L}^{0}\right\rangle }{\left\langle \pi ^{0}\pi ^{0}|K_{S}^{0}\right\rangle }}={\frac {pA_{\pi ^{0}\pi ^{0}}-q{\bar {A}}_{\pi ^{0}\pi ^{0}}}{pA_{\pi ^{0}\pi ^{0}}+q{\bar {A}}_{\pi ^{0}\pi ^{0}}}}={\frac {1-\lambda _{\pi ^{0}\pi ^{0}}}{1+\lambda _{\pi ^{0}\pi ^{0}}}}\end{aligned}}}$ .