Neutral particle oscillation

Description as a two-state system

Special case that only considers mixing

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, prescribed by a "mixing matrix" (e.g. the CKM or PMNS matricies).

Let ${\displaystyle H_0}$ be the Hamiltonian of the two-state system, and ${\displaystyle |1⟩}$ and ${\displaystyle |2⟩}$ be its orthonormal eigenvectors with eigenvalues ${\displaystyle E_1}$ and ${\displaystyle E_2}$ respectively.

Let ${\displaystyle |\mathrm{\Psi }(t)⟩}$ be the state of the system at time ⁠${\displaystyle t}$⁠.

If the system starts as an energy eigenstate of ⁠${\displaystyle H_0}$⁠, for example, say

${\displaystyle |\mathrm{\Psi }(0)⟩=|1⟩,}$

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

${\displaystyle {\hat{H}}_0|\mathrm{\Psi }(t)⟩=i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }t}|\mathrm{\Psi }\left(t\right)⟩}$ ()

will be

${\displaystyle |\mathrm{\Psi }(t)⟩=|1⟩e^{-i\frac{E_{1}t}{\hslash }}}$

But this is physically same as ${\displaystyle |1⟩,}$ since the exponential term is just a phase factor: It does not produce an observable new state. In other words, energy eigenstates are stationary eigenstates, that is, they do not yield observably distinct new states under time evolution.

Define ⁠${\displaystyle \left\{|1⟩,|2⟩\right\}}$⁠ to be a basis in which the unperturbed Hamiltonian operator, ${\displaystyle H_0}$, is diagonal:

${\displaystyle H_0=\left(\begin{array}{cc}{E}_1& 0\\ 0& E_2\end{array}\right)=E_1|1⟩+E_2|2⟩}$

It can be shown, that oscillation between states will occur if and only if off-diagonal terms of the Hamiltonian are not zero.

Hence let us introduce a general perturbation ${\displaystyle W}$ imposed on ${\displaystyle H_0}$ such that the resultant Hamiltonian ${\displaystyle H}$ is still Hermitian. Then

${\displaystyle W=\left(\begin{array}{cc}{W}_{11}& W_{12}\\ W_{12}^{\ast }& W_{22}\end{array}\right)}$

where ${\displaystyle W_{11},W_{22}\in \mathbb{R}}$ and ${\displaystyle W_{12}\in \mathbb{C}}$ and

${\displaystyle H=H_0+W=\left(\begin{array}{cc}{E}_1+W_{11}& W_{12}\\ W_{12}^{\ast }& E_2+W_{22}\end{array}\right)}$ ()

The eigenvalues of the perturbed Hamiltonian, ⁠${\displaystyle H}$⁠, then change to ${\displaystyle E_{+}}$ and ⁠${\displaystyle E_{-}}$⁠, where

${\displaystyle E_{\pm }=\frac{1}{2}\left[E_1+W_{11}+E_2+W_{22}\pm \sqrt{{\left(E_1+W_{11}-E_{{}^2}-W_{22}\right)}^2+4{\left|W_{12}\right|}^2}\right]}$ ()

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

${\displaystyle H=\sum _{j=0}^3{a}_j{\sigma }_j=a_0{\sigma }_0+H^{\prime }}$

The following two results are clear:

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

• ${\displaystyle {H^{\prime }}^2=I}$

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

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

and using the above pair of results the orthonormal eigenvectors of ${\displaystyle H^{\prime }}$ and consequently those of ${\displaystyle H}$ are obtained as

${\displaystyle \begin{array}{cc}|+⟩& =\left(\begin{array}{c}\cos \frac{\theta }{2}{e}^{-i\frac{\varphi }{2}}\\ \sin \frac{\theta }{2}{e}^{+i\frac{\varphi }{2}}\end{array}\right)\equiv \cos \frac{\theta }{2}{e}^{-i\frac{\varphi }{2}}|1⟩+\sin \frac{\theta }{2}{e}^{+i\frac{\varphi }{2}}|2⟩\\ |-⟩& =\left(\begin{array}{c}-\sin \frac{\theta }{2}{e}^{+i\frac{\varphi }{2}}\\ \cos \frac{\theta }{2}{e}^{-i\frac{\varphi }{2}}\end{array}\right)\equiv -\sin \frac{\theta }{2}{e}^{-i\frac{\varphi }{2}}|1⟩+\cos \frac{\theta }{2}{e}^{+i\frac{\varphi }{2}}|2⟩\end{array}}$ ()

Writing the eigenvectors of ${\displaystyle H_0}$ in terms of those of ${\displaystyle H}$ we get

${\displaystyle \begin{array}{cc}|1⟩& =e^{i\frac{\varphi }{2}}\left(\cos \frac{\theta }{2}|+⟩-\sin \frac{\theta }{2}|-⟩\right)\\ |2⟩& =e^{-i\frac{\varphi }{2}}\left(\sin \frac{\theta }{2}|+⟩+\cos \frac{\theta }{2}|-⟩\right)\end{array}}$ ()

Now if the particle starts out as an eigenstate of ${\displaystyle H_0}$ (say, ⁠${\displaystyle |1⟩}$⁠), that is

${\displaystyle |\mathrm{\Psi }(0)⟩=|1⟩}$

then under time evolution we get

${\displaystyle |\mathrm{\Psi }(t)⟩=e^{i\frac{\varphi }{2}}\left(\cos \frac{\theta }{2}|+⟩e^{-i\frac{E_{+}t}{\hslash }}-\sin \frac{\theta }{2}|-⟩e^{-i\frac{E_{-}t}{\hslash }}\right)}$

which unlike the previous case, is distinctly different from ⁠${\displaystyle |1⟩}$⁠.

We can then obtain the probability of finding the system in state ${\displaystyle |2⟩}$ at time ${\displaystyle t}$ as

${\displaystyle \begin{array}{cc}{P}_{21}(t)& =|⟨2|\mathrm{\Psi }(t)⟩{|}^2={\sin }^2\theta {\sin }^2\left(\frac{E_{+}-E_{-}}{2\hslash }t\right)\\ & =\frac{4{\left|W_{12}\right|}^2}{4{\left|W_{12}\right|}^2+{\left(E_1-E_2\right)}^2}{\sin }^2\left(\frac{\sqrt{4{\left|W_{12}\right|}^2+{\left(E_1-E_2\right)}^2}}{2\hslash }t\right)\end{array}}$ ()

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),

${\displaystyle \omega =\frac{E_1-E_2}{2\hslash }=\frac{\sqrt{4{\left|W_{12}\right|}^2+{\left(E_1-E_2\right)}^2}}{2\hslash }}$ ()

From equation (6), for ⁠${\displaystyle P_{21}(t)}$⁠, we can conclude that oscillation will exist only if ⁠${\displaystyle {\left|W_{12}\right|}^2\ne 0}$⁠. So ${\displaystyle W_{12}}$ is known as the coupling term as it connects 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. ⁠${\displaystyle {\left|W_{12}\right|}^2\ne 0}$⁠.
• Non-degenerate eigenvalues of the perturbed Hamiltonian ⁠${\displaystyle H}$⁠, i.e. ⁠${\displaystyle E_{+}\ne E_{-}}$⁠.

General case: considering mixing and decay

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

${\displaystyle H=M-\frac{i}{2}\mathrm{\Gamma }=\left(\begin{array}{cc}{M}_{11}& M_{12}\\ M_{12}^{\ast }& M_{11}\end{array}\right)-\frac{i}{2}\left(\begin{array}{cc}{\mathrm{\Gamma }}_{11}& {\mathrm{\Gamma }}_{12}\\ {\mathrm{\Gamma }}_{12}^{\ast }& {\mathrm{\Gamma }}_{11}\end{array}\right)}$

The eigenvalues of ${\displaystyle H}$ are

${\displaystyle \begin{array}{cc}{\mu }_{\mathsf{H}}& =M_{11}-\frac{i}{2}{\mathrm{\Gamma }}_{11}+\frac{1}{2}\left(\mathrm{\Delta }m-\frac{i}{2}\mathrm{\Delta }\mathrm{\Gamma }\right),\\ {\mu }_{\mathsf{L}}& =M_{11}-\frac{i}{2}{\mathrm{\Gamma }}_{11}-\frac{1}{2}\left(\mathrm{\Delta }m-\frac{i}{2}\mathrm{\Delta }\mathrm{\Gamma }\right)\end{array}}$ ()

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

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

${\displaystyle \begin{array}{cc}|P_{\mathsf{L}}⟩& =p|P⟩+q|\overline{P}⟩\\ |P_{\mathsf{H}}⟩& =p|P⟩-q|\overline{P}⟩\end{array}}$ ()

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

Let the system start in the state ⁠${\displaystyle |P⟩}$⁠.That is

${\displaystyle |P(0)⟩=|P⟩=\frac{1}{2p}(|P_{\mathsf{L}}⟩+|P_{\mathsf{H}}⟩)}$

Under time evolution we then get

${\displaystyle |P(t)⟩=\frac{1}{2p}\left(|P_{\mathsf{L}}⟩e^{-\frac{i}{\hslash }\left(m_L-\frac{i}{2}{\gamma }_L\right)t}+|P_{\mathsf{H}}⟩e^{-\frac{i}{\hslash }\left(m_H-\frac{i}{2}{\gamma }_H\right)t}\right)=g_{+}(t)|P⟩-\frac{q}{p}{g}_{-}(t)|\overline{P}⟩}$

Similarly, if the system starts in the state ${\displaystyle |\overline{P}⟩}$, under time evolution we obtain

${\displaystyle |\overline{P}(t)⟩=\frac{1}{2q}\left(|P_{\mathsf{L}}⟩e^{-\frac{i}{\hslash }\left(m_{\mathsf{L}}-\frac{i}{2}{\gamma }_{\mathsf{L}}\right)t}-|P_{\mathsf{H}}⟩e^{-\frac{i}{\hslash }\left(m_{\mathsf{H}}-\frac{i}{2}{\gamma }_{\mathsf{H}}\right)t}\right)=-\frac{p}{q}{g}_{-}(t)|P⟩+g_{+}(t)|\overline{P}⟩}$

CP violation as a consequence

If in a system ${\displaystyle |P⟩}$ and ${\displaystyle |\overline{P}⟩}$ represent CP conjugate states (i.e. particle-antiparticle) of one another (i.e. ${\displaystyle CP|P⟩=e^{i\delta }|\overline{P}⟩}$ and ${\displaystyle CP|\overline{P}⟩=e^{-i\delta }|P⟩}$), 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:

CP violation through decay only

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

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

${\displaystyle {\mathrm{\wp }}_{P\to f}\left(t\right)={\left|⟨f|P\left(t\right)⟩\right|}^2={\left|g_{+}\left(t\right)A_f-\frac{q}{p}{g}_{-}\left(t\right){\overline{A}}_f\right|}^2}$,

and that of its CP conjugate process by,

${\displaystyle {\mathrm{\wp }}_{\overline{P}\to \overline{f}}\left(t\right)={\left|⟨\overline{f}|\overline{P}\left(t\right)⟩\right|}^2={\left|g_{+}\left(t\right){\overline{A}}_{\overline{f}}-\frac{p}{q}{g}_{-}\left(t\right)A_{\overline{f}}\right|}^2}$

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

${\displaystyle \left|\frac{{\overline{A}}_{\overline{f}}}{A_f}\right|\ne 1}$ and ${\displaystyle \left|\frac{A_{\overline{f}}}{\overline{A_f}}\right|\ne 1}$ ()

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

The probability (as a function of time) of observing ${\displaystyle |\overline{P}⟩}$ starting from ${\displaystyle |P⟩}$ is given by,

${\displaystyle {\mathrm{\wp }}_{P\to \overline{P}}\left(t\right)={\left|⟨\overline{P}|P\left(t\right)⟩\right|}^2={\left|\frac{q}{p}{g}_{-}\left(t\right)\right|}^2}$,

and that of its CP conjugate process by,

${\displaystyle {\mathrm{\wp }}_{\overline{P}\to P}\left(t\right)={\left|⟨P|\overline{P}\left(t\right)⟩\right|}^2={\left|\frac{p}{q}{g}_{-}\left(t\right)\right|}^2}$.

The above two probabilities are unequal if

${\displaystyle \left|\frac{q}{p}\right|\ne 1}$ ()

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

CP violation through mixing-decay interference

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

${\displaystyle \begin{array}{cc}{\mathbb{P}}_{P\to f}\left(t\right)& =|⟨f|P(t)⟩{|}^2\\ & =|A_f{|}^2\frac{1}{2}{e}^{-\gamma t}\left[\left(1+{\left|{\lambda }_f\right|}^2\right)\cosh \left(\frac{1}{2}\mathrm{\Delta }\gamma t\right)+2{\mathcal{R}}_{\mathcal{e}}\left\{{\lambda }_f\right\}\sinh \left(\frac{1}{2}\mathrm{\Delta }\gamma t\right)+\left(1-{\left|{\lambda }_f\right|}^2\right)\cos \left(\mathrm{\Delta }mt\right)+2{\mathcal{I}}_{\mathcal{m}}\left\{{\lambda }_f\right\}\sin \left(\mathrm{\Delta }mt\right)\right]\end{array}}$

and,

${\displaystyle \begin{array}{cc}{\mathbb{P}}_{\overline{P}\to f}(t)& =|⟨f|\overline{P}(t)⟩{|}^2\\ & =|A_f{|}^2{\left|\frac{p}{q}\right|}^2\frac{1}{2}{e}^{-\gamma t}\left[\left(1+{\left|{\lambda }_f\right|}^2\right)\cosh \left(\frac{1}{2}\mathrm{\Delta }\gamma t\right)+2{\mathcal{R}}_{\mathrm{e}}\left\{{\lambda }_f\right\}\sinh \left(\frac{1}{2}\mathrm{\Delta }\gamma t\right)-\left(1-{\left|{\lambda }_f\right|}^2\right)\cos \left(\mathrm{\Delta }mt\right)-2{\mathcal{I}}_{\mathrm{m}}\left\{{\lambda }_f\right\}\sin \left(\mathrm{\Delta }mt\right)\right]\end{array}}$

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

${\displaystyle {\mathcal{I}}_{\mathrm{m}}\left\{{\lambda }_f\right\}={\mathcal{I}}_{\mathrm{m}}\left\{\frac{q}{p}\frac{{\overline{A}}_f}{A_f}\right\}\ne 0}$ ()

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

An alternative classification

Usually, an alternative classification of CP violation is made:

Specific cases

Neutrino oscillation

Considering a strong coupling between two pairs of flavor eigenstates of neutrinos (for example, ν
_e–ν
_μ, ν
_μ–ν
_τ, etc.) and a very weak coupling between either pair and the excluded 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

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

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

The above can be written as

${\displaystyle P_{\beta \alpha }\left(x\right)={\sin }^2\theta {\sin }^2\left(\frac{\mathrm{\Delta }{m}^2{c}^3}{4E\hslash }x\right)={\sin }^2\theta {\sin }^2\left(\frac{2\pi }{{\lambda }_{\text{osc}}}x\right)}$ ()

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

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

${\displaystyle \begin{array}{cc}{\left(\mathrm{\Delta }{m}^2\right)}_{12}& ={m_1}^2-{m_2}^2\\ {\left(\mathrm{\Delta }{m}^2\right)}_{23}& ={m_2}^2-{m_3}^2\\ {\left(\mathrm{\Delta }{m}^2\right)}_{31}& ={m_3}^2-{m_1}^2\end{array}}$

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

This implies that two of the three neutrinos have very closely placed masses. Since only two of the three ${\displaystyle \mathrm{\Delta }{m}^2}$ are independent, and the expression for probability in equation (13) is not sensitive to the sign of ${\displaystyle \mathrm{\Delta }{m}^2}$ (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

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

CP violation through mixing only

The 1964 paper by Christenson et al. 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.

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

${\displaystyle \begin{array}{cc}|K_{{}^1}^0⟩& =\frac{1}{\sqrt{2}}\left(|K^0⟩+|{\overline{K}}^0⟩\right)\\ |K_2^0⟩& =\frac{1}{\sqrt{2}}\left(|K^0⟩-|{\overline{K}}^0⟩\right)\end{array}}$

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 ${\displaystyle |K_{{}^1}^0⟩}$ 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.
• ${\displaystyle |K_2^0⟩}$ 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, ${\displaystyle |K_{{}^1}^0⟩}$ was referred to as the short-lived Kaon ${\displaystyle |K_S^0⟩}$, and ${\displaystyle |K_2^0⟩}$ as the long-lived Kaon ${\displaystyle |K_L^0⟩}$. The 1964 experiment showed that contrary to what was expected, ${\displaystyle |K_L^0⟩}$ could decay to two pions. This implied that the long lived Kaon cannot be purely the CP eigenstate ${\displaystyle |K_2^0⟩}$, but must contain a small admixture of ${\displaystyle |K_{{}^1}^0⟩}$, thereby no longer being a CP eigenstate. Similarly, the short-lived Kaon was predicted to have a small admixture of ${\displaystyle |K_2^0⟩}$. That is,

${\displaystyle \begin{array}{cc}|K_L^0⟩& =\frac{1}{\sqrt{1+{\left|\epsilon \right|}^2}}\left(|K_2^0⟩+\epsilon |K_1^0⟩\right)\\ |K_S^0⟩& =\frac{1}{\sqrt{1+{\left|\epsilon \right|}^2}}\left(|K_1^0⟩+\epsilon |K_2^0⟩\right)\end{array}}$

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

Writing ${\displaystyle |K_{{}^1}^0⟩}$ and ${\displaystyle |K_2^0⟩}$ in terms of ${\displaystyle |K^0⟩}$ and ${\displaystyle |{\overline{K}}^0⟩}$, we obtain (keeping in mind that ${\displaystyle m_{K_L^0}>m_{K_S^0}}$) the form of equation (9):

${\displaystyle \begin{array}{cc}|K_L^0⟩& =\left(p|K^0⟩-q|{\overline{K}}^0⟩\right)\\ |K_S^0⟩& =\left(p|K^0⟩+q|{\overline{K}}^0⟩\right)\end{array}}$

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

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

CP violation through decay only

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,

${\displaystyle \begin{array}{cc}{\eta }_{+-}& =\frac{⟨{\pi }^{+}{\pi }^{-}|K_L^0⟩}{⟨{\pi }^{+}{\pi }^{-}|K_S^0⟩}=\frac{p{A}_{{\pi }^{+}{\pi }^{-}}-q{\overline{A}}_{{\pi }^{+}{\pi }^{-}}}{p{A}_{{\pi }^{+}{\pi }^{-}}+q{\overline{A}}_{{\pi }^{+}{\pi }^{-}}}=\frac{1-{\lambda }_{{\pi }^{+}{\pi }^{-}}}{1+{\lambda }_{{\pi }^{+}{\pi }^{-}}}\\ {\eta }_{00}& =\frac{⟨{\pi }^0{\pi }^0|K_L^0⟩}{⟨{\pi }^0{\pi }^0|K_S^0⟩}=\frac{p{A}_{{\pi }^0{\pi }^0}-q{\overline{A}}_{{\pi }^0{\pi }^0}}{p{A}_{{\pi }^0{\pi }^0}+q{\overline{A}}_{{\pi }^0{\pi }^0}}=\frac{1-{\lambda }_{{\pi }^0{\pi }^0}}{1+{\lambda }_{{\pi }^0{\pi }^0}}\end{array}}$.

Experimentally, ${\displaystyle {\eta }_{+-}=\left(2.232\pm 0.011\right)\times {10}^{-3}}$ and ${\displaystyle {\eta }_{00}=\left(2.220\pm 0.011\right)\times {10}^{-3}}$. That is ${\displaystyle {\eta }_{+-}\ne {\eta }_{00}}$, implying ${\displaystyle \left|A_{{\pi }^{+}{\pi }^{-}}/{\overline{A}}_{{\pi }^{+}{\pi }^{-}}\right|\ne 1}$ and ${\displaystyle \left|A_{{\pi }^0{\pi }^0}/{\overline{A}}_{{\pi }^0{\pi }^0}\right|\ne 1}$, and thereby satisfying condition (10).

In other words, direct CP violation is observed in the asymmetry between the two modes of decay.

CP violation through mixing-decay interference

If the final state (say ${\displaystyle f_{CP}}$) is a CP eigenstate (for example π⁺
π⁻
), then there are two different decay amplitudes corresponding to two different decay paths:

${\displaystyle \begin{array}{cc}{K}^0& \to f_{CP}\\ K^0& \to {\overline{K}}^0\to f_{CP}\end{array}}$.

CP violation can then result from the interference of these two contributions to the decay as one mode involves only decay and the other oscillation and decay.

Which then is the "real" particle

The above description refers to flavor (or strangeness) eigenstates and energy (or CP) eigenstates. But which of them represents the "real" particle? What do we really detect in a laboratory? Quoting David J. Griffiths:

The neutral Kaon system adds a subtle twist to the old question, 'What is a particle?' Kaons are typically produced by the strong interactions, in eigenstates of strangeness (K⁰
and K⁰
), but they decay by the weak interactions, as eigenstates of CP (K₁ and K₂). Which, then, is the 'real' particle? If we hold that a 'particle' must have a unique lifetime, then the 'true' particles are K₁ and K₂. But we need not be so dogmatic. In practice, it is sometimes more convenient to use one set, and sometimes, the other. The situation is in many ways analogous to polarized light. Linear polarization can be regarded as a superposition of left-circular polarization and right-circular polarization. If you imagine a medium that preferentially absorbs right-circularly polarized light, and shine on it a linearly polarized beam, it will become progressively more left-circularly polarized as it passes through the material, just as a K⁰
beam turns into a K₂ beam. But whether you choose to analyze the process in terms of states of linear or circular polarization is largely a matter of taste.

Mixing matrix - a brief introduction

If the system is a three state system (for example, three species of neutrinos ν
_e ⇄ ν
_μ ⇄ ν
_τ, three species of quarks d ⇄ s ⇄ b ), then, just like in the two state system, the flavor eigenstates (say ${\displaystyle |{\phi }_{\alpha }⟩}$, ${\displaystyle |{\phi }_{\beta }⟩}$, ${\displaystyle |{\phi }_{\gamma }⟩}$) are written as a linear combination of the energy (mass) eigenstates (say ${\displaystyle |{\psi }_1⟩}$, ${\displaystyle |{\psi }_2⟩}$, ${\displaystyle |{\psi }_3⟩}$). That is,

${\displaystyle \left(\begin{array}{c}|{\phi }_{\alpha }⟩\\ |{\phi }_{\beta }⟩\\ |{\phi }_{\gamma }⟩\end{array}\right)=\left(\begin{array}{ccc}{\mathrm{\Omega }}_{\alpha 1}& {\mathrm{\Omega }}_{\alpha 2}& {\mathrm{\Omega }}_{\alpha 3}\\ {\mathrm{\Omega }}_{\beta 1}& {\mathrm{\Omega }}_{\beta 2}& {\mathrm{\Omega }}_{\beta 3}\\ {\mathrm{\Omega }}_{\gamma 1}& {\mathrm{\Omega }}_{\gamma 2}& {\mathrm{\Omega }}_{\gamma 3}\end{array}\right)\left(\begin{array}{c}|{\psi }_1⟩\\ |{\psi }_2⟩\\ |{\psi }_3⟩\end{array}\right)}$.

In case of leptons (neutrinos for example) the transformation matrix is the PMNS matrix, and for quarks it is the CKM matrix.

The off diagonal terms of the transformation matrix represent coupling, and unequal diagonal terms imply mixing between the three states.

The transformation matrix is unitary and appropriate parameterization (depending on whether it is the CKM or PMNS matrix) is done and the values of the parameters determined experimentally.