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. Neutral particle oscillations were first investigated in 1954 by Murray Gell-mann and Abraham Pais.
For example, a neutron cannot transmute into an antineutron as that would violate the conservation of baryon number. But in those hypothetical extensions of the Standard Model which include interactions that do not strictly conserve baryon number, neutron–antineutron oscillations are predicted to occur.
Such oscillations do regularly occur for other neutral particles, and are classified into two types:
- Particle–antiparticle oscillation (for example, oscillation).
- Flavor oscillation (for example, oscillation).
In those cases where the particles decay to some final product, then the system is not purely oscillatory, and an interference between oscillation and decay is observed.
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 symmetry that is conserved. However, in 1964 Cronin and Fitch reported CP violation in the neutral kaon system. Direct CP violation in the system was reported by both the labs by 2005.
The Kaon#Oscillation| and the systems can be studied as two state systems, considering the particle and its antiparticle as two states of a single particle.
Solar neutrino problem
The pp chain in the sun produces an abundance of . In 1968, R. Davis et al. first reported the results of the Homestake experiment. 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 to produce argon via the reaction
: <math>\mathrm{\nu_e + \ </math>
But this is physically same as <math>\ \left| 1 \right\rangle\ ,</math> 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 to be a basis in which the unperturbed Hamiltonian operator, <math> H_0 </math>, is diagonal:
: <math>\ H_0 = \begin{pmatrix}
E_1 & 0 \\
0 & E_2 \\
\end{pmatrix}\ =\ E_1\ \left| 1 \right\rangle\ +\ E_2\ \left| 2 \right\rangle
\ </math>
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 <math> W </math> imposed on <math> H_0 </math> such that the resultant Hamiltonian <math> H </math> is still Hermitian. Then
: <math>W = \begin{pmatrix}
W_{11} & W_{12} \\
W_{12}^* & W_{22} \\
\end{pmatrix}\ </math>
where <math> W_{11}, W_{22} \in \mathbb{R} </math> and <math> W_{12} \in \mathbb{C} </math> and
The eigenvalues of the perturbed Hamiltonian, , then change to <math>E_+</math> and , where
\\
&= I \\
\end{align} </math>
|-
|style="background:white;border-left:1px solid white;border-right:1px solid white;"| where the following results have been used:
|-
|
- <math> \sigma_j \sigma_k = \delta_{jk}\ I + i\ \sum\limits_{\ell=1}^3 {\varepsilon_{jk\ell} \sigma_\ell} </math>
- <math> \hat{n} </math> is a unit vector and hence <math> \sum\limits_{j=1}^3 \right\rangle</math> represent CP conjugate states (i.e. particle-antiparticle) of one another (i.e. <math>CP\left| P \right\rangle = e^{i\delta} \left| \bar{P} \right\rangle</math> and <math>CP\left| \bar{P} \right\rangle = e^{-i\delta} \left| P \right\rangle</math>), 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 <math>\left\{ \left| P \right\rangle, \left| \bar{P} \right\rangle \right\}</math> decay to final states <math>\left\{ \left| f \right\rangle, \left| \bar{f} \right\rangle \right\}</math>, where the barred and the unbarred kets of each set are CP conjugates of one another.
The probability of <math>\left| P \right\rangle</math> decaying to <math>\left| f \right\rangle</math> is given by,
: <math>
\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
</math>,
and that of its CP conjugate process by,
: <math>
\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
</math>
{| class="wikitable collapsible collapsed"
! where,
|-
| <math>\begin{align}
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
\end{align}</math>
|}
If there is no CP violation due to mixing, then <math>\left| \frac{q}{p} \right| = 1</math>.
Now, the above two probabilities are unequal if
{A_f} \right| \ne 1</math> and <math>\left| \frac{A_\bar{f{\bar{A_f \right| \ne 1</math>
| ref=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 <math>\left| \bar{P} \right\rangle</math> starting from <math>\left| P \right\rangle</math> is given by,
: <math>
\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
</math>,
and that of its CP conjugate process by,
: <math>
\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
</math>.
The above two probabilities are unequal if
Hence, the particle-antiparticle oscillation becomes a CP violating process as the particle and its antiparticle (say, <math>\left| P \right\rangle</math> and <math>\left| {\bar{P \right\rangle</math> respectively) are no longer equivalent eigenstates of CP.
CP violation through mixing-decay interference
Let <math>\left| f \right\rangle</math> be a final state (a CP eigenstate) that both <math>\left| P \right\rangle</math> and <math>\left| \bar{P} \right\rangle</math> can decay to. Then, the decay probabilities are given by,
: <math>\begin{align}
\operatorname{\mathbb P}_{P \to f} \left( t \right)
&= \Bigl| \left\langle f | P( t ) \right\rangle \Bigr|^2 \\
&= \Bigl| A_f \Bigr|^2 \tfrac{1}{2} e^{-\gamma t} \left[
\ \left( 1 + \left| \lambda_f \right|^2 \right) \cosh\!\left( \tfrac{1}{2} \Delta\gamma t \right)
+ 2\ \operatorname\mathcal{R_e}\!\left\{\ \lambda_f\ \right\}\ \sinh\!\left( \tfrac{1}{2}\Delta\gamma t \right)
+ \left( 1 - \left| \lambda_f \right|^2 \right) \cos\!\left( \Delta mt \right)
+ 2\ \operatorname\mathcal{I_m}\!\left\{\ \lambda_f\ \right\}\ \sin\!\left( \Delta mt \right)
\ \right] \\
\end{align}</math>
and,
:<math>\begin{align}
\operatorname{\mathbb P}_{\bar{P} \to f}( t )
&= \Bigl| \left\langle f | \bar{P}( t ) \right\rangle \Bigr|^2 \\
&= \Bigl| A_f \Bigr|^2 \left| \frac{p}{q} \right|^2 \tfrac{1}{2} e^{-\gamma t} \left[
\ \left( 1 + \left| \lambda_f \right|^2 \right) \cosh\!\left( \tfrac{1}{2}\Delta\gamma t \right)
+ 2\ \operatorname{\mathcal R_e}\!\left\{\ \lambda_f\ \right\}\ \sinh\!\left( \tfrac{1}{2}\Delta\gamma t \right)
- \left( 1 - \left| \lambda_f \right|^2 \right) \cos\left( \Delta mt \right)
- 2\ \operatorname{\mathcal I_m}\!\left\{\ \lambda_f\ \right\}\ \sin\left( \Delta mt \right)
\ \right] \\
\end{align}</math>
{| class="wikitable collapsible autocollapse"
! where,
|-
| <math>\begin{align}
\gamma &= \tfrac{1}{2} \left( \gamma_\mathsf H + \gamma_\mathsf L \right)\ \Delta\gamma = \gamma_\mathsf H - \gamma_\mathsf L \\
\Delta m &= m_\mathsf H - m_\mathsf 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
\end{align}</math>
|}
From the above two quantities, it can be seen that even when there is no CP violation through mixing alone (i.e. <math>\ \left| \tfrac{q}{p} \right| = 1\ </math>) and neither is there any CP violation through decay alone (i.e. <math>\ \left| \tfrac{\bar{A}_f}{A_f} \right| = 1\ </math>) and thus <math>\ \left| \lambda_f \right| = 1\ ,</math> the probabilities will still be unequal, provided that
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: 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.
<math>\left| K^0 \right\rangle</math> and <math>\left| \bar{K}^0 \right\rangle</math> being the strangeness eigenstates (with eigenvalues +1 and −1 respectively), the energy eigenstates are
: <math>\begin{align}
\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{align}</math>
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 <math>\left| K_{^1}^0 \right\rangle</math> 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.
- <math>\left| K_2^0 \right\rangle</math> 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, <math>\left| K_{^1}^0 \right\rangle</math> was referred to as the short-lived Kaon <math>\left| K_S^0 \right\rangle</math>, and <math>\left| K_2^0 \right\rangle</math> as the long-lived Kaon <math>\left| K_L^0 \right\rangle</math>. The 1964 experiment showed that contrary to what was expected, <math>\left| K_L^0 \right\rangle</math> could decay to two pions. This implied that the long lived Kaon cannot be purely the CP eigenstate <math>\left| K_2^0 \right\rangle</math>, but must contain a small admixture of <math>\left| K_{^1}^0 \right\rangle</math>, thereby no longer being a CP eigenstate.
Writing <math>\left| K_{^1}^0 \right\rangle</math> and <math>\left| K_2^0 \right\rangle</math> in terms of <math>\left| K^0 \right\rangle</math> and <math>\left| \bar{K}^0 \right\rangle</math>, we obtain (keeping in mind that <math>m_{K_L^0} > m_{K_S^0}</math>
: <math>\begin{align}
K^0 &\to f_{CP} \\
K^0 &\to \bar{K}^0 \to f_{CP}
\end{align}</math>.
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:
Mixing matrix - a brief introduction
If the system is a three state system (for example, three species of neutrinos , three species of quarks ), then, just like in the two state system, the flavor eigenstates (say <math>
\left| {\varphi_\alpha} \right\rangle</math>, <math>
\left| {\varphi_\beta} \right\rangle</math>, <math>
\left| {\varphi_\gamma} \right\rangle
</math>) are written as a linear combination of the energy (mass) eigenstates (say <math>
\left| \psi_1 \right\rangle</math>, <math>
\left| \psi_2 \right\rangle</math>, <math>
\left| \psi_3 \right\rangle
</math>). That is,
: <math>
\begin{pmatrix}
\left| {\varphi_\alpha} \right\rangle \\
\left| {\varphi_\beta} \right\rangle \\
\left| {\varphi_\gamma} \right\rangle \\
\end{pmatrix} = \begin{pmatrix}
\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{pmatrix}\begin{pmatrix}
\left| \psi_1 \right\rangle \\
\left| \psi_2 \right\rangle \\
\left| \psi_3 \right\rangle \\
\end{pmatrix}
</math>.
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.
See also
- CKM matrix
- CP violation
- CPT symmetry
- Kaon
- PMNS matrix
- Neutral current
- Flavor-changing neutral current
- Rabi cycle