Neutral particle oscillation

In particle physics, neutral particle oscillation is the transmutation of a particle with zero electric charge into another neutral particle due to a change of a non-zero internal quantum number via an interaction that does not conserve that quantum number. For example, a neutron cannot transmute into an antineutron as that would violate the conservation of baryon number.

Such oscillations can be classified into two types:

• Particle-antiparticle oscillation (for example, K0-K0 oscillation, B0-B0 oscillation, D0-D0 oscillation^[1]).
• Flavor oscillation (for example, ν
  e-ν
  μ oscillation).

In case the particles decay to some final product, then the system is not purely oscillatory, and an interference between oscillation and decay is observed.

N.B. It is the probability (of detecting any one of the two particles) that oscillates as a function of distance traveled (or, as a function of time of flight). It is not that the particle itself transmutes periodically as it travels. What propagates is a mixed state and on detection we measure the energy (mass) associated with any one of its pure states; and what oscillates is the probability of obtaining a particular energy (mass) as a result of measurement on the mixed state.

History and motivation

CP violation

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.^[2] However, in 1964 Cronin and Fitch reported CP violation in the neutral Kaon system.^[3] They observed the long-lived K₂ (CP = -1) undergoing two pion decays (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.^[4][5] Direct CP violation in the B0-B0 system was reported by both the labs by 2005.^[6][7]

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

The pp chain in the sun produces an abundance of ν
e. In 1968, Raymond Davis et al. first reported the results of the Homestake experiment.^[8][9] 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, USA. Chlorine nuclei in the perchloroethylene absorb ν
e to produce argon via the reaction

${{\nu }_{e}}+C{{l}^{37}}\to A{{r}^{38}}+{{e}^{-}}$

which is essentially

${{\nu }_{e}}+n\to p+e^{-}$.^[10]

The experiment collected argon for several months. Because 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.^[11] This 'oscillation' between the neutrino species can first be studied considering any two, and then generalized to the three known flavors.

Description as a two-state system

A special case: considering mixing only

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

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

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

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

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

$\hat H_{0}\left| {\Psi \left( t \right)} \right\rangle = {i \hbar }{\partial \over {\partial t}}\left| {\Psi \left( t \right)} \right\rangle$   (1)

will be,^[12]

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

But this is physically same as $\left| 1 \right\rangle$ 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 $\left\{ {\left| 1 \right\rangle ,\left| 2 \right\rangle } \right\}$, ${H_{0}}$ is diagonal. That is,

${{H}_{0}}=\left( \begin{matrix} {{E}_{1}} & 0 \\ 0 & {{E}_{2}} \\ \end{matrix} \right)$

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

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

$W=\left( \begin{matrix} {{W}_{11}} & {{W}_{12}} \\ {{W}_{12}}^{*} & {{W}_{22}} \\ \end{matrix} \right)$ where, ${{W}_{11}},{{W}_{22}}\in \mathbb{R}$ and ${{W}_{12}}\in \mathbb{C}$

and,

$H={{H}_{0}}+W=\left( \begin{matrix} {{E}_{1}}+{{W}_{11}} & {{W}_{12}} \\ {{W}_{12}}^{*} & {{E}_{2}}+{{W}_{22}} \\ \end{matrix} \right)$   (2)

Then, the eigenvalues of $H$ are,^[13]

${{E}_{\pm }}=\frac{1}{2}\left[ {{E}_{1}}+{{W}_{11}}+{{E}_{2}}+{{W}_{22}}\pm \sqrt{{{\left( {{E}_{1}}+{{W}_{11}}-{{E}_{^{2}}}-{{W}_{22}} \right)}^{2}}+4{{\left| {{W}_{12}} \right|}^{2}}} \right]$   (3)

Since $H$ is a general Hamiltonian matrix, it can be written as,^[14]

$H=\sum\limits_{j=0}^{3}{{{a}_{j}}{{\sigma }_{j}}}={{a}_{0}}{{\sigma }_{0}}+H'$

where,

$H'=\vec{a}.\vec{\sigma }=\left| a \right|\hat{n}.\vec{\sigma }$, $\vec{a}=\left( {{a}_{1}},{{a}_{2}},{{a}_{3}} \right)$, ${\hat{n}}$ is a real unit vector in 3 dimensions in the direction of ${\vec{a}}$, ${{\sigma }_{0}}=I=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right)$, and ${{\sigma }_{1}}={{\sigma }_{x}}=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right)$, ${{\sigma }_{2}}={{\sigma }_{y}}=\left( \begin{matrix} 0 & -i \\ i & 0 \\ \end{matrix} \right)$, ${{\sigma }_{3}}={{\sigma }_{z}}=\left( \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right)$ are the Pauli matrices.

The following two results are clear:

• $\left[ H,H' \right]=0$

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

• $H{{'}^{2}}=I$

Proof

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

With the following parametrization^[14] (this parametrization helps as it normalizes the eigenvectors and also introduces an arbitrary phase $\phi$ making the eigenvectors most general)

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

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

$\begin{align} & \left| + \right\rangle =\left( \begin{matrix} \cos \frac{\theta }{2}{{e}^{-i\frac{\phi }{2}}} \\ {} \\ \sin \frac{\theta }{2}{{e}^{i\frac{\phi }{2}}} \\ \end{matrix} \right)\equiv \cos \frac{\theta }{2}{{e}^{-i\frac{\phi }{2}}}\left| 1 \right\rangle +\sin \frac{\theta }{2}{{e}^{i\frac{\phi }{2}}}\left| 2 \right\rangle \\ & \left| - \right\rangle =\left( \begin{matrix} -\sin \frac{\theta }{2}{{e}^{i\frac{\phi }{2}}} \\ {} \\ \cos \frac{\theta }{2}{{e}^{-i\frac{\phi }{2}}} \\ \end{matrix} \right)\equiv -\sin \frac{\theta }{2}{{e}^{-i\frac{\phi }{2}}}\left| 1 \right\rangle +\cos \frac{\theta }{2}{{e}^{i\frac{\phi }{2}}}\left| 2 \right\rangle \\ \end{align}$   (4)

where,
$\tan \theta =\frac{2\left| {{W}_{12}} \right|}{{{E}_{1}}+{{W}_{11}}-{{E}_{2}}-{{W}_{22}}}$ and, ${{W}_{12}}=\left| {{W}_{12}} \right|{{e}^{i\phi }}$

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

$\begin{align} & \left| 1 \right\rangle ={{e}^{i\frac{\phi }{2}}}\left( \cos \frac{\theta }{2}\left| + \right\rangle -\sin \frac{\theta }{2}\left| - \right\rangle \right) \\ & \left| 2 \right\rangle ={{e}^{-i\frac{\phi }{2}}}\left( \sin \frac{\theta }{2}\left| + \right\rangle +\cos \frac{\theta }{2}\left| - \right\rangle \right) \\ \end{align}$   (5)

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

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

then under time evolution we get,^[13]

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

which unlike the previous case, is distinctly different from $\left| 1 \right\rangle$.

We can then obtain the probability of finding the system in state $\left| 2 \right\rangle$ at time $t$ as,^[13]

$\begin{align} {{P}_{21}}\left( t \right)&={{\left| \left\langle 2 | \Psi \left( t \right) \right\rangle \right|}^{2}}={{\sin }^{2}}\theta {{\sin }^{2}}\left( \frac{{{E}_{+}}-{{E}_{-}}}{2\hbar }t \right) \\ & =\frac{4{{\left| {{W}_{12}} \right|}^{2}}}{4{{\left| {{W}_{12}} \right|}^{2}}+{{\left( {{E}_{1}}-{{E}_{2}} \right)}^{2}}}{{\sin }^{2}}\left( \frac{\sqrt{4{{\left| {{W}_{12}} \right|}^{2}}+{{\left( {{E}_{1}}-{{E}_{2}} \right)}^{2}}}}{2\hbar }t \right) \\ \end{align}$   (6)

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

$\omega =\frac{{{E}_{+}}-{{E}_{-}}}{2\hbar }=\frac{\sqrt{4{{\left| {{W}_{12}} \right|}^{2}}+{{\left( {{E}_{1}}-{{E}_{2}} \right)}^{2}}}}{2\hbar }$   (7)

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

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

Hence, the necessary conditions for oscillation are:

• Non-zero coupling, i.e. ${{\left| {{W}_{12}} \right|}^{2}}\ne 0$.
• Non-degenerate eigenvalues of the perturbed Hamiltonian $H$, i.e. ${{E}_{+}}\ne {{E}_{-}}$.

The general case: considering mixing and decay

If the particle(s) under consideration undergoes decay, then the Hamiltonian describing the system is no longer Hermitian.^[15] Since any matrix can be written as a sum of its Hermitian and anti-Hermitian parts, $H$ can be written as,

$H=M-\frac{i}{2}\Gamma =\left( \begin{matrix} {{M}_{11}} & {{M}_{12}} \\ {{M}_{12}}^{*} & {{M}_{11}} \\ \end{matrix} \right)-\frac{i}{2}\left( \begin{matrix} {{\Gamma }_{11}} & {{\Gamma }_{12}} \\ {{\Gamma }_{12}}^{*} & {{\Gamma }_{11}} \\ \end{matrix} \right)$

where,

$M=\left( \begin{matrix} {{M}_{11}} & {{M}_{12}} \\ {{M}_{21}} & {{M}_{22}} \\ \end{matrix} \right)$ and, $\Gamma =\left( \begin{matrix} {{\Gamma }_{11}} & {{\Gamma }_{12}} \\ {{\Gamma }_{12}}^{*} & {{\Gamma }_{11}} \\ \end{matrix} \right)$ $M$ and $\Gamma$ are Hermitian. Hence, ${{M}_{21}}={{M}_{12}}^{*}$ and ${{\Gamma}_{21}}={{\Gamma}_{12}}^{*}$ CPT conservation (symmetry) implies, ${{M}_{22}}={{M}_{11}}$ and ${{\Gamma}_{22}}={{\Gamma}_{11}}$ Proof Let, $\Theta =CPT$$\Theta$ changes a particle to its antiparticle. That is, $\Theta \left| 1 \right\rangle =\left| 2 \right\rangle$ and $\Theta \left| 2 \right\rangle =\left| 1 \right\rangle$ CPT conservation implies that the Hamiltonian $H$ and hence $M$ and $\Gamma$ are invariant under the following transformation: ${{\Theta }^{-1}}M\Theta=M$ and ${{\Theta }^{-1}}\Gamma \Theta =\Gamma$ $\Theta$ is an anti-Unitary operator[16] and satisfies the relation ${{\Theta }^{\dagger }}\Theta =I$ Hence, ${{M}_{22}}=\left\langle 2 \right|M\left| 2 \right\rangle =\left\langle 2 \right|{{\Theta }^{-1}}M\Theta \left| 2 \right\rangle =\left\langle 2 \right|{{\Theta }^{\dagger }}M\Theta \left| 2 \right\rangle =\left\langle 1 \right|M\left| 1 \right\rangle ={{M}_{11}}$ and similarly for the diagonal elements of $\Gamma$. Hermiticity of $M$ and $\Gamma$ also implies that their diagonal elements are real.

The eigenvalues of $H$ are,

${{\mu }_{H}}={{M}_{11}}-\frac{i}{2}{{\Gamma }_{11}}+\frac{1}{2}\left( \Delta m-\frac{i}{2}\Delta \Gamma \right)$ and, ${{\mu }_{L}}={{M}_{11}}-\frac{i}{2}{{\Gamma }_{11}}-\frac{1}{2}\left( \Delta m-\frac{i}{2}\Delta \Gamma \right)$   (8)

where,
$\Delta m$ and $\Delta \Gamma$ satisfy, ${{\left( \Delta m \right)}^{2}}-{{\left( \frac{\Delta \Gamma }{2} \right)}^{2}}=4{{\left| {{M}_{12}} \right|}^{2}}-{{\left| {{\Gamma }_{12}} \right|}^{2}}$ and, $\Delta m\Delta \Gamma =4\operatorname{Re}\left( {{M}_{12}}{{\Gamma }_{12}}^{*} \right)$

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

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

$\left| {{P}_{L}} \right\rangle =p\left| P \right\rangle +q\left| \bar{P} \right\rangle$ and, $\left| {{P}_{H}} \right\rangle =p\left| P \right\rangle -q\left| \bar{P} \right\rangle$   (9)

where,
${{\left| p \right|}^{2}}+{{\left| q \right|}^{2}}=1$ and, ${{\left( \frac{p}{q} \right)}^{2}}=\frac{{{M}_{12}}^{*}-\frac{i}{2}{{\Gamma }_{12}}^{*}}{{{M}_{12}}-\frac{i}{2}{{\Gamma }_{12}}}$

$p$ and $q$ are the mixing terms. Note that the eigenstates are no longer orthogonal.

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

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

Under time evolution we then get,

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

where,
${{g}_{\pm }}\left( t \right)=\frac{1}{2}\left( {{e}^{-\frac{i}{\hbar }\left( {{m}_{H}}-\frac{i}{2}{{\gamma }_{H}} \right)t}}\pm {{e}^{-\frac{i}{\hbar }\left( {{m}_{L}}-\frac{i}{2}{{\gamma }_{L}} \right)t}} \right)$

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

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

CP violation as a consequence

If in a system $\left| P \right\rangle$ and $\left| {\bar{P}} \right\rangle$ represent CP conjugate states (i.e. particle-antiparticle) of one another (i.e. $CP\left| P \right\rangle ={{e}^{i\delta }}\left| {\bar{P}} \right\rangle$ and $CP\left| {\bar{P}} \right\rangle ={{e}^{-i\delta }}\left| P \right\rangle$), 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:^[15][17]

CP violation through decay only

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

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

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

and that of its CP conjugate process by,

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

where,
${{A}_{f}}=\left\langle f | P \right\rangle$ ${{{\bar{A}}}_{f}}=\left\langle f | {\bar{P}} \right\rangle$ ${{A}_{{\bar{f}}}}=\left\langle {\bar{f}} | P \right\rangle$ ${{{\bar{A}}}_{{\bar{f}}}}=\left\langle {\bar{f}} | {\bar{P}} \right\rangle$

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

Now, the above two probabilities are unequal if,

$\left| \frac{{{{\bar{A}}}_{{\bar{f}}}}}{{{A}_{f}}} \right|\ne 1$ and $\left| \frac{{{A}_{{\bar{f}}}}}{{{{\bar{A}}}_{f}}} \right|\ne 1$   (10)

.

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 $\left| {\bar{P}} \right\rangle$ starting from $\left| P \right\rangle$ is given by,

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

and that of its CP conjugate process by,

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

The above two probabilities are unequal if,

$\left| \frac{q}{p} \right|\ne 1$   (11)

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

CP violation through mixing-decay interference

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

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

and,

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

where,
$\gamma =\frac{{{\gamma }_{H}}+{{\gamma }_{L}}}{2}$$\Delta \gamma ={{\gamma }_{H}}-{{\gamma }_{L}}$ $\Delta m={{m}_{H}}-{{m}_{L}}$ ${{\lambda }_{f}}=\frac{q}{p}\frac{{{{\bar{A}}}_{f}}}{{{A}_{f}}}$ ${{A}_{f}}=\left\langle f | P \right\rangle$ ${{{\bar{A}}}_{f}}=\left\langle f | {\bar{P}} \right\rangle$

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

$\operatorname{Im}\left( {{\lambda }_{f}} \right)=\operatorname{Im}\left( \frac{q}{p}\frac{{{{\bar{A}}}_{f}}}{{{A}_{f}}} \right)\ne 0$   (12)

.

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:^[17]

Direct CP violation

Direct CP violation is defined as, $\left| {{{\bar{A}}}_{f}}/{{A}_{f}} \right| \ne 1$. In terms of the above categories, direct CP violation occurs in CP violation through decay only.

Indirect CP violation

Indirect CP violation is the type of CP violation that involves mixing. In terms of the above classification, indirect CP violation occurs through mixing only, or through mixing-decay interference, or both.

Specific cases

Neutrino Oscillation

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 $\alpha$ transmuting into type $\beta$ as,

${{P}_{\beta \alpha }}\left( t \right)={{\sin }^{2}}\theta {{\sin }^{2}}\left( \frac{{{E}_{+}}-{{E}_{-}}}{2\hbar }t \right)$

where, ${{E}_{+}}$ and ${{E}_{-}}$ are energy eigenstates.

The above can be written as,

${{P}_{\beta \alpha }}\left( x \right)={{\sin }^{2}}\theta {{\sin }^{2}}\left( \frac{\Delta {{m}^{2}}{{c}^{3}}}{4E\hbar }x \right)={{\sin }^{2}}\theta {{\sin }^{2}}\left( \frac{2\pi }{{{\lambda }_{osc}}}x \right)$   (13)

where,
$\Delta {{m}^{2}}={{m}_{+}}^{2}-{{m}_{-}}^{2}$, i.e. the difference between the squares of the masses of the energy eigenstates, $c$ is the speed of light in vacuum, $x$ is the distance traveled by the neutrino after creation, $E$ is the energy with which the neutrino was created, and ${{\lambda }_{osc}}$ is the oscillation wavelength.

Proof

${{E}_{\pm }}=\sqrt{{{p}^{2}}{{c}^{2}}+{{m}_{\pm }}^{2}{{c}^{4}}}\simeq pc\left( 1+\frac{{{m}_{\pm }}^{2}{{c}^{2}}}{2{{p}^{2}}} \right)\left[ \because \frac{{{m}_{\pm }}c}{p}\ll 1 \right]$ where, $p$ is the momentum with which the neutrino was created. Now, $E\simeq pc$ and $t\simeq x/c$. Hence, $\frac{{{E}_{+}}-{{E}_{-}}}{2\hbar }t\simeq \frac{\left( {{m}_{+}}^{2}-{{m}_{-}}^{2} \right){{c}^{3}}}{2p\hbar }t\simeq \frac{\Delta {{m}^{2}}{{c}^{3}}}{4E\hbar }x=\frac{2\pi }{{{\lambda }_{osc}}}x$ where, ${{\lambda }_{osc}}=\frac{8\pi E\hbar }{\Delta {{m}^{2}}{{c}^{3}}}$

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:

${{\left( \Delta {{m}^{2}} \right)}_{12}}={{m}_{1}}^{2}-{{m}_{2}}^{2}$

${{\left( \Delta {{m}^{2}} \right)}_{23}}={{m}_{2}}^{2}-{{m}_{3}}^{2}$

${{\left( \Delta {{m}^{2}} \right)}_{31}}={{m}_{3}}^{2}-{{m}_{1}}^{2}$

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

For solar neutrinos, ${{\left( \Delta {{m}^{2}} \right)}_{sol}}\simeq 8\times {{10}^{-5}}{{\left( eV/{{c}^{2}} \right)}^{2}}$.

For atmospheric neutrinos, ${{\left( \Delta {{m}^{2}} \right)}_{atm}}\simeq 3\times {{10}^{-3}}{{\left( eV/{{c}^{2}} \right)}^{2}}$.

This implies that two of the three neutrinos have very closely placed masses. Since only two of the three $\Delta {{m}^{2}}$ are independent, and the expression for probability in equation (13) is not sensitive to the sign of $\Delta {{m}^{2}}$ (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 ${{\lambda }_{osc}}$. We can draw the following inferences:

• If $x/{{\lambda }_{osc}}\ll 1$, then ${{P}_{\beta \alpha }}\simeq 0$ and oscillation will not be observed. For example, production (say, by radioactive decay) and detection of neutrinos in a laboratory.
• If $x/{{\lambda }_{osc}}\simeq n$, where $n$ is a whole number, then ${{P}_{\beta \alpha }}\simeq 0$ and oscillation will not be observed.
• In all other cases, oscillation will be observed. For example, $x/{{\lambda }_{osc}}\gg 1$ for solar neutrinos; $x\sim {{\lambda }_{osc}}$ 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.^[3] 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.

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

$\left| K_{^{1}}^{0} \right\rangle =\frac{1}{\sqrt{2}}\left( \left| {{K}^{0}} \right\rangle +\left| {{{\bar{K}}}^{0}} \right\rangle \right)$ and,

$\left| K_{2}^{0} \right\rangle =\frac{1}{\sqrt{2}}\left( \left| {{K}^{0}} \right\rangle -\left| {{{\bar{K}}}^{0}} \right\rangle \right)$.

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 $\left| K_{^{1}}^{0} \right\rangle$ 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.
• $\left| K_{2}^{0} \right\rangle$ having a CP eigenvalue -1, can decay only two three pions and never to two.

Since the two pion decay is much faster than the three pion decay, $\left| K_{^{1}}^{0} \right\rangle$ was referred to as the short-lived Kaon $\left| K_{S}^{0} \right\rangle$, and $\left| K_{2}^{0} \right\rangle$ as the long-lived Kaon $\left| K_{L}^{0} \right\rangle$. The 1964 experiment showed that contrary to what was expected, $\left| K_{L}^{0} \right\rangle$ could decay to two pions. This implied that the long lived Kaon cannot be purely the CP eigenstate $\left| K_{2}^{0} \right\rangle$, but must contain a small admixture of $\left| K_{^{1}}^{0} \right\rangle$, thereby no longer being a CP eigenstate.^[18] Similarly, the short-lived Kaon was predicted to have a small admixture of $\left| K_{2}^{0} \right\rangle$. That is,

$\left| K_{L}^{0} \right\rangle =\frac{1}{\sqrt{1+{{\left| \varepsilon \right|}^{2}}}}\left( \left| K_{2}^{0} \right\rangle +\varepsilon \left| K_{1}^{0} \right\rangle \right)$ and,

$\left| K_{S}^{0} \right\rangle =\frac{1}{\sqrt{1+{{\left| \varepsilon \right|}^{2}}}}\left( \left| K_{1}^{0} \right\rangle +\varepsilon \left| K_{2}^{0} \right\rangle \right)$

where, $\varepsilon$ is a complex quantity and is a measure of departure from CP invariance. Experimentally, $\left| \varepsilon \right|=\left( 2.228\pm 0.011 \right)\times {{10}^{-3}}$.^[19]

Writing $\left| K_{^{1}}^{0} \right\rangle$ and $\left| K_{2}^{0} \right\rangle$ in terms of $\left| {{K}^{0}} \right\rangle$ and $\left| {{{\bar{K}}}^{0}} \right\rangle$, we obtain (keeping in mind that ${{m}_{K_{_{L}}^{0}}}>{{m}_{K_{S}^{0}}}$^[19]) the form of equation (9):

$\left| K_{L}^{0} \right\rangle =\left( p\left| {{K}^{0}} \right\rangle -q\left| {{{\bar{K}}}^{0}} \right\rangle \right)$ and,

$\left| K_{S}^{0} \right\rangle =\left( p\left| {{K}^{0}} \right\rangle +q\left| {{{\bar{K}}}^{0}} \right\rangle \right)$

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

Since $\left| \varepsilon \right|\ne 0$, condition (11) is satisfied and there is a mixing between the strangeness eigenstates $\left| {{K}^{0}} \right\rangle$ and $\left| {{{\bar{K}}}^{0}} \right\rangle$ 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: π0π0 or π+π−. Both of these final states are CP eigenstates of themselves. We can define the branching ratios as,^[17]

${{\eta }_{+-}}=\frac{\left\langle {{\pi }^{+}}{{\pi }^{-}} | K_{L}^{0} \right\rangle }{\left\langle {{\pi }^{+}}{{\pi }^{-}} | K_{S}^{0} \right\rangle }=\frac{p{{A}_{{{\pi }^{+}}{{\pi }^{-}}}}-q{{{\bar{A}}}_{{{\pi }^{+}}{{\pi }^{-}}}}}{p{{A}_{{{\pi }^{+}}{{\pi }^{-}}}}+q{{{\bar{A}}}_{{{\pi }^{+}}{{\pi }^{-}}}}}=\frac{1-{{\lambda }_{{{\pi }^{+}}{{\pi }^{-}}}}}{1+{{\lambda }_{{{\pi }^{+}}{{\pi }^{-}}}}}$ and,

${{\eta }_{00}}=\frac{\left\langle {{\pi }^{0}}{{\pi }^{0}} | K_{L}^{0} \right\rangle }{\left\langle {{\pi }^{0}}{{\pi }^{0}} | K_{S}^{0} \right\rangle }=\frac{p{{A}_{{{\pi }^{0}}{{\pi }^{0}}}}-q{{{\bar{A}}}_{{{\pi }^{0}}{{\pi }^{0}}}}}{p{{A}_{{{\pi }^{0}}{{\pi }^{0}}}}+q{{{\bar{A}}}_{{{\pi }^{0}}{{\pi }^{0}}}}}=\frac{1-{{\lambda }_{{{\pi }^{0}}{{\pi }^{0}}}}}{1+{{\lambda }_{{{\pi }^{0}}{{\pi }^{0}}}}}$.

Experimentally, ${{\eta }_{+-}}=\left( 2.232\pm 0.011 \right)\times {{10}^{-3}}$^[19] and ${{\eta }_{00}}=\left( 2.220\pm 0.011 \right)\times {{10}^{-3}}$. That is ${{\eta }_{+-}}\ne {{\eta }_{00}}$, implying $\left| {{A}_{{{\pi }^{+}}{{\pi }^{-}}}}/{{{\bar{A}}}_{{{\pi }^{+}}{{\pi }^{-}}}} \right|\ne 1$ and $\left| {{A}_{{{\pi }^{0}}{{\pi }^{0}}}}/{{{\bar{A}}}_{{{\pi }^{0}}{{\pi }^{0}}}} \right|\ne 1$, 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 ${{f}_{CP}}$) is a CP eigenstate (for example π+π−), then there are two different decay amplitudes corresponding to two different decay paths:^[20]

${{K}^{0}}\to {{f}_{CP}}$ and,

${{K}^{0}}\to {{{\bar{K}}}^{0}}\to {{f}_{CP}}$.

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:^[18]

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.

The 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 $\left| {{\varphi }_{\alpha }} \right\rangle$, $\left| {{\varphi }_{\beta}} \right\rangle$, $\left| {{\varphi }_{\gamma}} \right\rangle$) are written as a linear combination of the energy (mass) eigenstates (say $\left| {{\psi }_{1}} \right\rangle$, $\left| {{\psi }_{2}} \right\rangle$, $\left| {{\psi }_{3}} \right\rangle$). That is,

$\left( \begin{matrix} \left| {{\varphi }_{\alpha }} \right\rangle \\ \left| {{\varphi }_{\beta }} \right\rangle \\ \left| {{\varphi }_{\gamma }} \right\rangle \\ \end{matrix} \right)=\left( \begin{matrix} {{\Omega }_{\alpha 1}} & {{\Omega }_{\alpha 2}} & {{\Omega }_{\alpha 3}} \\ {{\Omega }_{\beta 1}} & {{\Omega }_{\beta 2}} & {{\Omega }_{\beta 3}} \\ {{\Omega }_{\gamma 1}} & {{\Omega }_{\gamma 2}} & {{\Omega }_{\gamma 3}} \\ \end{matrix} \right)\left( \begin{matrix} \left| {{\psi }_{1}} \right\rangle \\ \left| {{\psi }_{2}} \right\rangle \\ \left| {{\psi }_{3}} \right\rangle \\ \end{matrix} \right)$.

In case of leptons (neutrinos for example) the transformation matrix is the PMNS matrix, and for quarks it is the CKM matrix.^[21]

N.B. The three familiar neutrino species ν
e-ν
μ-ν
τ are flavor eigenstates, whereas the three familiar quarks species d-s-b are energy eigenstates.

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.

See also

• CKM matrix
• CP violation
• CPT symmetry
• Kaon
• PMNS matrix
• Neutrino oscillation
• Rabi cycle

References

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