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.[1]
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.[2] [3] [4]
Such oscillations can be classified into two types:
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.
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. They observed the long-lived KL (with) undergoing decays into two pions (with) thereby violating CP conservation.
In 2001, CP violation in the system was confirmed by the BaBar and the Belle experiments.[7] [8] Direct CP violation in the system was reported by both the labs by 2005.[9] [10]
The and the systems can be studied as two state systems, considering the particle and its antiparticle as two states of a single particle.
The pp chain in the sun produces an abundance of . In 1968, R. Davis et al. first reported the results of the Homestake experiment.[11] [12] 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
37 | |
\nue+{{ | |
17 |
Cl} →
37 | |
{{} | |
18 |
which is essentially
\nue+n\top+e- |
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 (produced in the sun) can transform into some other neutrino species (or), 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 flux and the total neutrino flux.[14]
This 'oscillation' between the neutrino species can first be studied considering any two, and then generalized to the three known flavors.
See main article: Two-state quantum system.
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
H0
\left|1\right\rangle
\left|2\right\rangle
E1
E2
Let
\left|\Psi(t)\right\rangle
t~.
If the system starts as an energy eigenstate of
H0 ,
\left|\Psi(0)\right\rangle=\left|1\right\rangle ,
will be[15]
\left|\Psi(t)\right\rangle = \left|1\right\rangle
| ||||
e |
But this is physically same as
\left|1\right\rangle ,
Define
\left\{ \left|1\right\rangle , \left|2\right\rangle \right\} ,
H0
H0=\begin{pmatrix} E1&0\\ 0&E2\\ \end{pmatrix} = E1 \left|1\right\rangle + E2 \left|2\right\rangle
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
W
H 0
H
W=\begin{pmatrix} W11&W12\\
* | |
W | |
12 |
&W22\\ \end{pmatrix}
W11,W22\inR
W12\inC
The eigenvalues of the perturbed Hamiltonian,
H ,
E+
E- ,
Since
H
H=
3 | |
\sum\limits | |
j=0 |
aj\sigmaj=a0\sigma0+H'
The following two results are clear:
\left[H,H'\right]=0
Proof | |
---|---|
\begin{align} HH' &= a0\sigma0H'+H'H' = a0\sigma0+{H'}2\\ H'H &= a0H'\sigma0+H'H' = a0\sigma0+{H'}2\\ \end{align} | |
therefore | |
\left[H,H'\right]\equivHH'-H'H=0 |
{H'}2=I
Proof | ||||||||||||||||||||||||||||||||||||||||||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
\begin{align} {H'}2 &=
{nj\sigmaj}
{nk\sigmak} =
{njnk\sigmaj\sigmak}\\ &=
{njnk\left(\deltajkI+
{\varepsilonjk\ell\sigma\ell}\right)}\\ &=\left(
\right)I+
{\sigmal
\varepsilonjk\ell | ||||||||||||||||||||||||||||||||||||||||||||||||||||||
where the following results have been used: | ||||||||||||||||||||||||||||||||||||||||||||||||||||||
\sigmaj\sigmak=\deltajk I+
{\varepsilonjk\ell\sigma\ell} \hat{n}
=\left | \hat \right | ^2 = 1\ \varepsilonjk\ell j k
\varepsilonjk\ell=0~. |
With the following parametrization (this parametrization helps as it normalizes the eigenvectors and also introduces an arbitrary phase
\phi
\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'
H
Writing the eigenvectors of
H0
H
Now if the particle starts out as an eigenstate of
H0
\left|1\right\rangle
\left| \Psi(0) \right\rangle = \left|1\right\rangle
then under time evolution we get[16]
\left| \Psi(t
| ||||
) \right\rangle = e |
\left(\cos\tfrac{\theta}{2} \left|+
| ||||
\right\rangle e |
- \sin\tfrac{\theta}{2} \left|-
| ||||
\right\rangle e |
\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
t
which is called Rabi's formula. Hence, starting from one eigenstate of the unperturbed Hamiltonian
H0 ,
H0
From equation (6), for
P21(t) ,
\left|W12\right|2\ne0~.
W12
H0
Oscillation will also cease if the eigenvalues of the perturbed Hamiltonian
H
E+=E-~.
H
H0
Hence, the necessary conditions for oscillation are:
\left|W12\right|2\ne0~.
H ,
E+\neE-~.
If the particle(s) under consideration undergoes decay, then the Hamiltonian describing the system is no longer Hermitian.[18] Since any matrix can be written as a sum of its Hermitian and anti-Hermitian parts,
H
H = M-
i | |
2 |
\Gamma = \begin{pmatrix} M11&M12\\
* | |
M | |
12 |
&M11\\ \end{pmatrix}-
i | |
2 |
\begin{pmatrix} \Gamma11&\Gamma12\\
* | |
\Gamma | |
12 |
&\Gamma11\\ \end{pmatrix}
The eigenvalues of
H
The suffixes stand for Heavy and Light respectively (by convention) and this implies that
\Deltam
The normalized eigenstates corresponding to
\muL
\muH
l\{\left|P\right\rangle , \left|\bar{P}\right\rangler\}~\equiv~l\{ (1,0) , (0,1) r\}
p
q
Let the system start in the state
\left|P\right\rangle~.
\left| P(0) \right\rangle = \left|P\right\rangle =
1 | |
2 p |
l( \left|PL\right\rangle + \left|PH\right\rangle r)
Under time evolution we then get
\left| P(t) \right\rangle =
1 | |
2 p |
\left( \left|
-\tfrac{i | |
P | |
L\right\rangle e |
{\hbar} \left(mL-\tfrac{i}{2}\gammaL\right) t} + \left|
-\tfrac{i | |
P | |
H\right\rangle e |
{\hbar} \left(mH-\tfrac{i}{2}\gammaH\right) t} \right) = g+(t) \left|P\right\rangle -
q | |
p |
g-(t) \left|\bar{P}\right\rangle
Similarly, if the system starts in the state
\left|\bar{P}\right\rangle
\left| \bar{P}(t) \right\rangle=
1 | |
2 q |
\left(\left|
-\tfrac{i | |
P | |
L\right\rangle e |
{\hbar} \left(mL-\tfrac{i}{2}\gammaL\right) t}- \left|
-\tfrac{i | |
P | |
H\right\rangle e |
{\hbar} \left(mH-\tfrac{i}{2}\gammaH\right) t} \right) = -
p | |
q |
g-(t) \left|P\right\rangle + g+(t) \left|\bar{P}\right\rangle
If in a system
\left|P\right\rangle
\left|{\bar{P}}\right\rangle
CP\left|P\right\rangle=ei\delta\left|\bar{P}\right\rangle
CP\left|\bar{P}\right\rangle=e-i\delta\left|P\right\rangle
Consider the processes where
\left\{\left|P\right\rangle,\left|\bar{P}\right\rangle\right\}
\left\{\left|f\right\rangle,\left|\bar{f}\right\rangle\right\}
The probability of
\left|P\right\rangle
\left|f\right\rangle
\wpP\left(t\right)= \left|\left\langlef|P\left(t\right)\right\rangle\right|2= \left|g+\left(t\right)Af-
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}-
p | |
q |
g-\left(t\right)A\bar{f}\right|2
If there is no CP violation due to mixing, then
\left|
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.
The probability (as a function of time) of observing
\left|\bar{P}\right\rangle
\left|P\right\rangle
\wpP
and that of its CP conjugate process by,
\wp\bar{P\toP}\left(t\right)= \left|\left\langleP|\bar{P}\left(t\right)\right\rangle\right|2= \left|
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,
\left|P\right\rangle
\left|{\bar{P}}\right\rangle
Let
\left|f\right\rangle
\left|P\right\rangle
\left|\bar{P}\right\rangle
\begin{align} \operatorname{lP}P\left(t\right) &=l|\left\langlef|P(t)\right\rangler|2\\ &=l|Afr|2\tfrac{1}{2}e-\gamma\left[ \left(1+\left|λf\right|2\right)\cosh\left(\tfrac{1}{2}\Delta\gammat\right) +2 \operatornamel{Re}\left\{ λf \right\} \sinh\left(\tfrac{1}{2}\Delta\gammat\right)+\left(1-\left|λf\right|2\right)\cos\left(\Deltamt\right) +2 \operatornamel{Im}\left\{ λf \right\} \sin\left(\Deltamt\right) \right]\\ \end{align}
and,
\begin{align} \operatorname{lP}\bar{P\tof}(t) &=l|\left\langlef|\bar{P}(t)\right\rangler|2\\ &=l|Afr|2\left|
p | |
q |
\right|2\tfrac{1}{2}e-\gamma\left[ \left(1+\left|λf\right|2\right)\cosh\left(\tfrac{1}{2}\Delta\gammat\right) +2 \operatorname{lRe}\left\{ λf \right\} \sinh\left(\tfrac{1}{2}\Delta\gammat\right) -\left(1-\left|λf\right|2\right)\cos\left(\Deltamt\right) -2 \operatorname{lIm}\left\{ λf \right\} \sin\left(\Deltamt\right) \right]\\ \end{align}
From the above two quantities, it can be seen that even when there is no CP violation through mixing alone (i.e.
\left|\tfrac{q}{p}\right|=1
\left|\tfrac{\bar{A}f}{Af}\right|=1
\left|λf\right|=1 ,
The last terms in the above expressions for probability are thus associated with interference between mixing and decay.
Usually, an alternative classification of CP violation is made:
Direct CP violation | Direct CP violation is defined as, \left | \bar_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. |
See main article: Neutrino oscillation. Considering a strong coupling between two flavor eigenstates of neutrinos (for example, –, –, etc.) and a very weak coupling between the third (that is, the third does not affect the interaction between the other two), equation gives the probability of a neutrino of type
\alpha
\beta
P\beta\alpha\left(t\right)=\sin2\theta\sin2\left(
E+-E- | |
2\hbar |
t\right)
where,
E+
E-
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.
With three flavors of neutrinos, there are three mass splittings:
\begin{align} \left(\Deltam2\right)12&=
2 | |
{m | |
1} |
-
2 | |
{m | |
2} |
\\ \left(\Deltam2\right)23&=
2 | |
{m | |
2} |
-
2 | |
{m | |
3} |
\\ \left(\Deltam2\right)31&=
2 | |
{m | |
3} |
-
2 \end{align} | |
{m | |
1} |
But only two of them are independent, because
\left(\Deltam2\right)12+\left(\Deltam2\right)23+\left(\Deltam2\right)31=0~
For solar neutrinos | \left(\Deltam2\right)sol\simeq8 x 10-5\left(eV/c2\right)2 | |
For atmospheric neutrinos | \left(\Deltam2\right)atm\simeq3 x 10-3\left(eV/c2\right)2 |
This implies that two of the three neutrinos have very closely placed masses. Since only two of the three
\Deltam2
\Deltam2
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.
Equation indicates that an appropriate length scale of the system is the oscillation wavelength
λosc
x/λosc\ll1
P\beta\alpha\simeq0
x/λosc\simeqn
n
P\beta\alpha\simeq0
x/λosc\gg1
x\simλosc
See main article: Kaon.
The 1964 paper by Christenson et al.[21] 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|K0\right\rangle
\left|\bar{K}0\right\rangle
\begin{align} \left|
0 | |
K | |
1 |
\right\rangle&=
1 | |
\sqrt{2 |
These two are also CP eigenstates with eigenvalues +1 and −1 respectively. From the earlier notion of CP conservation (symmetry), the following were expected:
\left|
0 | |
K | |
1 |
\right\rangle
\left|
0 | |
K | |
2 |
\right\rangle
Since the two pion decay is much faster than the three pion decay,
\left|
0 | |
K | |
1 |
\right\rangle
\left|
0 | |
K | |
S |
\right\rangle
\left|
0 | |
K | |
2 |
\right\rangle
\left|
0 | |
K | |
L |
\right\rangle
\left|
0 | |
K | |
L |
\right\rangle
\left|
0 | |
K | |
2 |
\right\rangle
\left|
0 | |
K | |
1 |
\right\rangle
\left|
0 | |
K | |
2 |
\right\rangle
\begin{align} \left|
0 | |
K | |
L |
\right\rangle&=
1 | |
\sqrt{1+\left|\varepsilon\right|2 |
where,
\varepsilon
\left|\varepsilon\right|=\left(2.228\pm0.011\right) x 10-3
Writing
\left|
0 | |
K | |
1 |
\right\rangle
\left|
0 | |
K | |
2 |
\right\rangle
\left|K0\right\rangle
\left|\bar{K}0\right\rangle
m | |||||||
|
>
m | |||||||
|
\begin{align} \left|
0 | |
K | |
L |
\right\rangle&=\left(p\left|K0\right\rangle-q\left|\bar{K}0\right\rangle\right)\\ \left|
0 | |
K | |
S |
\right\rangle&=\left(p\left|K0\right\rangle+q\left|\bar{K}0\right\rangle\right) \end{align}
where,
q | |
p |
=
1-\varepsilon | |
1+\varepsilon |
Since
\left|\varepsilon\right|\ne0
\left|K0\right\rangle
\left|\bar{K}0\right\rangle
The and have two modes of two pion decay: or . Both of these final states are CP eigenstates of themselves. We can define the branching ratios as,[20]
\begin{align} η+-&=
| |||||||||
|
=
| |||||||
|
Experimentally,
η+-=\left(2.232\pm0.011\right) x 10-3
η00=\left(2.220\pm0.011\right) x 10-3
η+-\neη00
\left|
A | |
\pi+\pi- |
/\bar{A} | |
\pi+\pi- |
\right|\ne1
\left|
A | |
\pi0\pi0 |
/\bar{A} | |
\pi0\pi0 |
\right|\ne1
In other words, direct CP violation is observed in the asymmetry between the two modes of decay.
If the final state (say
fCP
\begin{align} K0&\tofCP\\ K0&\to\bar{K}0\tofCP\end{align}
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.
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:[24]
See main article: Cabibbo–Kobayashi–Maskawa matrix and Pontecorvo–Maki–Nakagawa–Sakata matrix.
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
\left|{\varphi\alpha}\right\rangle
\left|{\varphi\beta}\right\rangle
\left|{\varphi\gamma}\right\rangle
\left|\psi1\right\rangle
\left|\psi2\right\rangle
\left|\psi3\right\rangle
\begin{pmatrix} \left|{\varphi\alpha}\right\rangle\\ \left|{\varphi\beta}\right\rangle\\ \left|{\varphi\gamma}\right\rangle\\ \end{pmatrix}=\begin{pmatrix} \Omega\alpha&\Omega\alpha&\Omega\alpha\\ \Omega\beta&\Omega\beta&\Omega\beta\\ \Omega\gamma&\Omega\gamma&\Omega\gamma\\ \end{pmatrix}\begin{pmatrix} \left|\psi1\right\rangle\\ \left|\psi2\right\rangle\\ \left|\psi3\right\rangle\\ \end{pmatrix}
In case of leptons (neutrinos for example) the transformation matrix is the PMNS matrix, and for quarks it is the CKM matrix.[25]
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.