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I was reading various reviews on neutrino physics but I couldn't understand the following to complete satisfaction.

How is $\theta_{12}$ identified with the Solar mixing angle and $\Delta m^2_{21}$ the Solar mass-squared difference?

The formulae for three-flavor oscillation probabilities, in general, depends on all three mixing angles $\theta_{12},\theta_{23}$ and $\theta_{13}$:

$$P_{\alpha\rightarrow\beta}=\delta_{\alpha\beta} - 4 \sum_{i>j} {\rm Re}(U_{\alpha i}^{*} U_{\beta i} U_{\alpha j} U_{\beta j}^{*}) \sin^2 \left(\frac{\Delta m_{ij}^2 L}{4E}\right) \\ + 2\sum_{i>j}{\rm Im}(U_{\alpha i}^{*}U_{\beta i}U_{\alpha j} U_{\beta j}^{*}) \sin\left(\frac{\Delta m_{ij}^2 L}{2E}\right).$$

It's not clear to me why only one mixing angle be associated with solar neutrino oscillation. An answer starting from the three-flavour oscillation formulae will be helpful for my understanding.

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    $\begingroup$ Related: physics.stackexchange.com/q/270727 $\endgroup$ – dmckee --- ex-moderator kitten Jun 28 '17 at 16:54
  • $\begingroup$ Too bad there is not official mechanism to combine good answers into a comprehensive one because with all the answers to that very same question, each stressing particularly well a specific aspect of it, we would have the mother of all answers on the subject!!! $\endgroup$ – user154997 Jul 1 '17 at 19:17
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The most important effect in estimating the anticipated production of neutrino types out of the sun is the MSW effect, resonant conversion in matter, and the estimate is a complicated business, but apparently you are not interested in that, in this question.

The point, well made before in @PascExchange 's answer, is sheer numerical accident. Since L, the distance from the sun, is so enormous, the oscillatory factors (see below) in the transition formula in vacuum are dominated by the smallest $\Delta m^2$s: in this case, the 1-2 mass-squared-difference is two orders of magnitude smaller than the other two.

This fortunate accident makes 1-3 and 2-3 oscillation frequencies a hundred times higher, so the corresponding trig functions average out to zero. An intermediate expression reflecting this fact is oscillation equations

Proceed to neglect the very small $\theta_{13}$, and you are there, as only $S_{12}$ survives.

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  • $\begingroup$ "Proceed to neglect the very small $\theta_{13}$, and you are there, as only $S_{12}$ survives." Why do you say that? Even when $\theta_{13}$ is neglected, your equations (30)-(40) still contain $\theta_{23}$. $\endgroup$ – SRS Jul 14 '18 at 13:00
  • $\begingroup$ Adding (39) to (40) in that limit to subtract from 1, yields the persistence probability characteristically depending on $\theta_{12}$, and not $\theta_{23}$, as implied. $\endgroup$ – Cosmas Zachos Jul 14 '18 at 13:51
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In order to answer this question let me go back a bit in the discussion about solar neutrinos. The process, which keeps our sun alive can basically be described as $$p+p\to\ ^2_1\text{H}+e^++\nu_e$$ $$D+p\to\ ^3\text{He}+\gamma$$ The important fact for our current discussion nonetheless is, that during this process we produce electron neutrinos $\nu_e$.

Using the Luminosity, the distance between the sun and the earth and as the energy being released by one single such fusion process we can calculate the expected neutrino flux on earth. Various experiments, covering different Neutrino energies and allowing to detect $\nu_\mu$ and $\nu_e$ now showed, that the there is only half of the expected amount measured. The Sudbury Neutrino Observatory (SNO) now included the capability to measure $\nu_\tau$ by using heavy water instead of normal water, where-after the expected flux was measured. Thus we can conclude that the lack is actually an observation of $\nu_e \to \nu_\mu\to \nu_\tau$ aswell as $\nu_e\to \nu_\tau$ oscillation. To understand why we still claim that we get one squared mass difference out of that can be understood when consulting the particle physics booklet. Here we find the following values for the squared mass differences between the two generations $\Delta m_{12}^2=(7.53\pm 0.18)\cdot 10^{-5}$ and $\Delta m_{23}^2=(2.52\pm 0.05)\cdot 10^{-3}$. We realize that the difference $\Delta m_{12}$ is quite small compared to $\Delta m_{23}$ and therefore we can only distinguish between the first two, respectively the second and third generation and therefore we can only regard the $\mu_e\to \mu_\tau$ part.

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We offer detailed consideration of the applicable physics, for those who would like to gain a quantitative as well as qualitative understanding of solar neutrino flavor change.

quibbling

Not to quibble, but there are not any solar neutrino flavor oscillations, at least no one on Earth has detected periodic $L/E$ dependendent flavor change of the $\nu_e$ produced in the solar fusion chains Solar neutrinos: Oscillations or No-oscillations?.

solar deficit

However, there is flavor change, since fewer solar electron neutrinos $\nu_e$ are detected than predicted by the SSM Standard Solar Model and other explanations have been ruled out (2017 Neutrino Physics discusses the experimental history, as well as all aspects of neutrino physics). About $0.55$ of the expected low energy $\nu_e$ (say below $1 \, \mathrm{MeV}$, think $pp$ and $^7\mathrm{Be}$ neutrinos) are detected. On the other hand, only about $0.3$ of the higher energy solar neutrinos, primarily $^8\mathrm{B}$ above $5 \, \mathrm{MeV}$, are observed. You can see this in the following Fig. 84 from a 2014 Borexino experiment report:

2014 Borexino graph

how to know if oscillations are observable

When you consider detecting neutrino flavor change you first need to calculate the distance scale ($\lambda_{osc}$) over which oscillation effects could be observable.

The characteristic wavelength $\lambda_{osc}$, also known as oscillation length $L_{osc}$, can by calculated by setting the oscillation term (phase difference), often denoted $\Delta_{jk}$, (the argument to the $\sin^2$ term in your posted equation) equal to $\pi$ and solving for $L$: $$ \begin{align} \pi &= \frac{\Delta m_{jk}^2 L}{4E} \quad \quad \text{set osc factor } \Delta_{jk} \text{ equal to }\pi \\ 4\pi E &= \Delta m_{jk}^2 L \quad \quad \quad \text{solve for }L\\ \frac{4\pi E}{ \Delta m_{jk}^2} &= L_{osc} = \lambda_{osc} \quad \quad \text{osc wavelength} \end{align} $$ For more information, see Section 6.4 and Chapter 8 in 2017 Neutrino Physics 2018 PDG Review of Particle Physics Chapter 14 (in particular section 14.7 and Eqs $(14.36)$, $(14.39)$, and $(14.50)$ and surrounding text).

If you apply the correct $\hbar c$ factors to the $\lambda_{osc}$ equation in natural units you can obtain a convenient formula for the vacuum oscillation length in $\mathrm{km}$ for a given neutrino energy $E_{\nu}$ in $\mathrm{GeV}$ and mass split difference in $\mathrm{eV}^2$:

$$ \begin{align} \lambda_{osc} &= \frac{4\pi E}{ \Delta m_{jk}^2} = \frac{ (12.57) \, E \, (\mathrm{GeV}) }{ \Delta m_{jk}^2 \, (\mathrm{eV}^2 )}\\ \lambda_{osc}(\mathrm{km}) &= \frac{ 12.57 \, E \,\times 10^9 \, ( \mathrm{eV}) } { \Delta m_{jk}^2 \, (\mathrm{eV}^2 ) }= \frac{ 12.57 \,\times 10^9 \, (\hbar c) } { (\mathrm{eV} ) } \\ & \text{select }\hbar c\text{ above in units }(\mathrm{eV \, km}) \text{ to cancel denom remaining }\,\mathrm{eV}\text{ and leave }\mathrm{km}\\ \lambda_{osc}(\mathrm{km}) &= \frac{ 12.57 \,\times 10^9 \, \left(\, 1.97 \, \times 10^{-10}) \, [\mathrm{eV \, km}] \,\right)} { (\mathrm{eV} ) } = 2.47 \, \frac{E \, (\mathrm{GeV})} { \Delta m_{jk}^2 \, (\mathrm{eV}^2 )} \end{align} $$

For example, the oscillation length is $\lambda_{osc} \sim 8.8 \, \mathrm{km}$ for the $\Delta m_{12}^2 = 7.5 \, \times 10^{-5} \, \mathrm{eV}^2$ mass split with neutrino energy $267 \, \mathrm{keV} = 267 \, \times 10^3 \times 10^{-9} = 267 \, \times 10^{-6} \, \mathrm{GeV}$:

$$ \lambda_{osc}= 2.47 \, \times \frac{(267 \, \times 10^{-6} \, \mathrm{GeV})}{7.5 \, \times 10^{-5} \mathrm{eV}^2 } = 8.8 \, \mathrm{km} $$

In order to detect oscillations (see works cited), the oscillation length $\lambda_{jk}$ of the squared mass splitting(s) being probed should be close to or slightly less than the baseline $L$ (distance from source to detector). Clearly, the distance from the Sun to Earth, $L\sim 1.50 \, \times 10^8 \, \mathrm{km}$ ($1 \, \mathrm{au}$), is far greater than the $\lambda_{12}$ associated with $\Delta m_{12}^2 = 7.5 \, \times 10^{-5} \, \mathrm{eV}^2$. The other mass splitting produces a similarly small oscillation length, $\lambda_{32} \approx \lambda_{31} \sim 0.3 \, \mathrm{km}$ for the $\Delta m_{31}^2 = 2.457 \, \times 10^{-3} \, \mathrm{eV}$ squared mass splitting.

Put another way, an optimal neutrino oscillation experiment is one in which ratio of the neutrino energy and baseline are of the same order as the mass splitting, i.e.,

$$ \begin{align} \Delta_{ij} &= \frac{\Delta m_{ij}^2 L}{4E} \gtrsim 1 \quad \text{ oscillation experiment} \\ \end{align} $$

You require $\Delta_{ij}\gtrsim 1$ for at least one $\Delta m_{jk}^2$, i.e., the neutrino oscillation length should be of the order of the source-detector distance. If you keep in mind that we are talking about the argument of the $\sin^2$ term it should be obvious that making that argument $\pi/2=1.57 \, \mathrm{rad}$ (or a few odd integer multiples of $\pi/2$), for example, would make the $\sin^2$ function maximum (and that $\pi$ multiples would result in zero).

For the distance between the Earth and the Sun $L \sim 1.50 \, \times 10^8 \, \mathrm{km}$ you have, for a low energy neutrino:

$$ \begin{align} \Delta_{ij} &= 1.27 \frac{ \Delta m^2 (\mathrm{eV}^2) L(\mathrm{km}) }{ E_{\nu} (\mathrm{GeV})}\\ \Delta_{31} = \frac{\Delta m_{31}^2 L}{4E} &= \mathcal{O}(10^{9}) \gg 1, \\[1ex] \Delta_{21} = \frac{\Delta m_{21}^2 L}{4E} &= \mathcal{O}(10^{7}) \gg 1,\\ \end{align} $$

where $E= 267 \, \mathrm{keV}$ ($267 \, \times 10^{-6} \, \mathrm{GeV}$), $\Delta m_{32}^2 \sim \Delta m_{31}^2 = 2.457 \, \times 10^{-3} \, \mathrm{eV}^2$, and $ \Delta m_{21}^2 = 7.5 \, \times 10^{-5} \, \mathrm{eV}^2$.

That tells you that an oscillation experiment is out of the question with solar neutrinos. The required energy resolution (see Section 2.3 of Solar neutrinos and neutrino physics) would be

$$ \Delta E_{\nu} \approx \frac{E_{\nu}}{N_{osc}} $$

For the distance from the Sun to Earth, $L\sim 1.50 \, \times 10^8 \, \mathrm{km}$, that implies $N_{osc} \approx L/\lambda_{osc} \simeq 17 \, \times 10^6$ oscillations of the $267 \, \mathrm{keV} $ neutrino with $\lambda_{osc} \sim 8.8 \, \mathrm{km}$ we are considering (setting aside other factors like whether the mass states of the neutrino packets would have separated over that distance and the averaging that occurs over the region of neutrino production in the Sun, estimated to be $\Delta R \simeq (0.04 \text{-} 0.20)R_{\odot}$ or $\simeq 10^5 \, \mathrm{km}$).

Using the $\Delta E_{\nu}$ equation above, that would require a detector with $\mathcal{O}(0.02 \, \mathrm{eV})$ resolution. For comparison, the Borexino experiment, using a liquid scintillator detector, has a neutrino energy threshold $E_{min}\sim150 \, \mathrm{keV}$, roughly seven orders of magnitude from the required resolution to detect oscillation as a function of neutrino energy in this case.

what about the solar mixing angle and mass squared splitting

Yes, you say, but why are $\theta_{12}$ and $\Delta m_{12}^2$ associated with solar neutrino flavor change when the general equations involve three mixing angles, three square mass splittings?

As we mentioned above, there are two regimes in the Borexino graph (Fig 84 above) of experimental observations of solar $\nu_e$ survival fractions (three if you count the transition region between the two, the predicted upturn from the low $P_{ee} \approx 0.3$ around $10 \, \mathrm{MeV}$ neutrino energy, to nearer $P_{ee} \gtrsim 0.4$ at $5 \, \mathrm{MeV}$ as the MSW effect begins to turn off, apparently reported for the first time by Super-Kamiokande at Neutrino2020).

the low energy region

We detect about $0.55$ of the expected low energy (expectation $\langle E_{\nu}\rangle \sim 267 \, \mathrm{keV}$) electron neutrinos from $pp$ fusion and subsequent $\beta^+$-decay: $$p + p \rightarrow \; ^2_1\mathrm{H} + e^+ + \nu_e $$ ($^2_1\mathrm{H}$ is deuterium, i.e., one proton $+$ one neutron, resulting from $\beta^+$ decay of the initial diproton $p+p$ fusion.)

The $\nu_e$ electron neutrino vacuum mix (probabilities, hence square the mixing matrix elements; see $(G11)$ $U$ mixing matrix farther below) of mass states $\nu_j$ is:

$$ \lvert \nu_e \rangle= U_{e1}^2\lvert \nu_{1} \rangle + U_{e2}^2\lvert \nu_{2} \rangle + U_{e3}^2 \lvert \nu_{3} \rangle\\ = [\cos{\theta_{12} } \cos{\theta_{13} }]^2 \lvert \nu_{1} \rangle+ [\sin{\theta_{12} } \cos{\theta_{13} }]^2\lvert \nu_{2} \rangle + [\sin{\theta_{13} }\lvert]^2 \nu_{3} \rangle \\ =0.6805\lvert \nu_{1} \rangle + 0.2977\lvert \nu_{2} \rangle + 0.0218 \lvert \nu_{3} \rangle \quad \quad \text{for }\theta_{12}=33.48^{\circ}, \, \theta_{13}=8.50^{\circ} \tag{G99} $$

The electron neutrino is therefore $68\%$ $\nu_1$ and $\sim 30\%$ $\nu_2$ (almost no $\nu_3$ at $2\%$), so it is natural to associate the $\theta_{12}$ angle that specifies the $\nu_1$ and $\nu_2$ mixing with the solar electron neutrino, and the mass split difference between them, $\Delta m_{12}^2$.

back on the trail of low energy solar neutrinos

Resuming the analysis of the lower energy solar neutrinos, at these low energies there is not much matter potential. Using a PDG formula for the electron number density $N_e \, \mathrm{cm}^{-3}$ required for MSW resonance, we can quantify this:

$$ N_e^{res} = \frac{\Delta m_{jk}^2 \cos 2\theta_{jk}}{2 E_{\nu} \sqrt{2} G_F} \simeq 6.56 \, \times 10^6 \, \frac{\Delta m_{jk}^2 [\mathrm{eV}^2]}{E_{\nu}[\mathrm{MeV}]} \, \cos 2 \theta_{jk} \, \mathrm{cm}^{-3} \, \mathrm{N_A} $$ where $\mathrm{N_A}= 6.022140857 \, \times 10^{23}$ is Avogadro's number and $G_F = 1.1664 \, \times 10^{-5} \, \mathrm{GeV}^{-2}$ is the Fermi coupling constant.

For the $\Delta m_{12}^2 =7.5 \, \times 10^{-5} \, \mathrm{eV}^2$ mass split and associated mixing angle $\theta_{12} = 33.48^{\circ}$ (we are using 2016 global fit values), a $pp$ neutrino with $E_{\nu} = 0.267 \, \mathrm{MeV}$ would require an electron number density of

$$ N_e^{res} \simeq 6.56 \, \times 10^6 \, \frac{ 7.5 \, \times 10^{-5} \, [\mathrm{eV}^2]}{ 0.267 \,[\mathrm{MeV}]} \, \cos [(2) (0.5843 \, \mathrm{rad})] \, \mathrm{cm}^{-3} \, 6.022140857 \, \times 10^{23}\\ = 4.3431\, \times 10^{26}\, \mathrm{cm}^{-3} $$

Using the above formula, we find that a $E_{\nu} = 0.267 \, \mathrm{MeV}$ neutrino would require $N_e(t=0)=4.3431\, \times 10^{26}\, \mathrm{cm}^{-3}$ $e^-$ number density at birth for MSW resonance (our criterion here for significant matter effect). Consulting the B16(GS98) SSM (Standard Solar Model) data we find the radial location of the maximum emission point of the $pp$ neutrinos is $r_{\odot} = 0.0990$ ($r_{\odot} = 0$ is the center of the Sun, $r_{\odot} = 1$ the normalized radius) and the electron number density there is $N_e(r_{\odot} = 0.0990)=4.0249\, \times 10^{25}\, \mathrm{cm}^{-3}$. Therefore, for significant matter effect on the $pp$ neutrino flavor evolution we require approximately an order of magnitude more electron number density than is present at the production location in the solar core.

how do we end up with survival probability dependent on only the $\theta_{12}$ angle?

$pp$ neutrino flavor oscillations therefore occur in the Sun as in vacuum (no significant level jump probability either, but we will not get into that here). Now, since we cannot observe solar neutrino oscillations at Earth (for the reasons discussed earlier), we may only observe average flavor change. For the solar $pp$ $\nu_e$, PDG offers a two-neutrino model survival probability equation: $$ \bar{P}^{2 \nu} (\nu_e \rightarrow \nu_e) \simeq 1 - \frac{1}{2} \sin^2 2 \theta_{12} \tag{14.89} $$

We obtain $\bar{P}^{2 \nu} (\nu_e \rightarrow \nu_e) \simeq 0.5766$ for $\theta_{12} = 33.48^{\circ}$ with this equation.

Aha! We see the low energy solar neutrino survival probability (the left side of the Borexino graph above, $P_{ee} \approx 0.55$) is dependent solely on the $\theta_{12}$ mixing angle. Why is this true when there are three mass splittings and three mixing angles?

The amplitude for finding a $\nu_{l^{\prime}}$ in an original $\nu_l$ neutrino flux is

$$ \langle \nu_{l^{\prime}} \mid \nu_l \rangle = \sum_{j} \langle \nu_{j,l^{\prime}} \lvert U^{\dagger}_{j l^{\prime}} e^{-iE_jt} U^{}_{j l^{}} \rvert \nu_{j,l} \rangle \\ = \sum_{j} e^{-iE_jt} U^{}_{ l^{} j} U^{\ast}_{l^{\prime}j} \tag{5.4} $$ where we used the orthogonality relation $\langle \nu_{j} \mid \nu_k \rangle = \delta_{jk}$.

If the experimental conditions permit coherent detection, then in usual quantum mechanical fashion you take the squared modulus of the transition amplitude to obtain the probability for the transition:

$$ P_{} (t) = \lvert \langle \nu_{l^{\prime}} \mid \nu_l(t) \rangle \rvert^2 = \left \lvert \sum_{j} e^{-iE_jt} U^{}_{ l^{} j} U^{\ast}_{l^{\prime}j} \right \rvert^2 \tag{G300} $$

Your posted equation is derived from that statement $(G300)$. However, since we can detect only an average, then you cannot sum amplitudes, but instead must sum probabilities, i.e., sum the squared amplitudes:

$$ \bar{P}(\nu_l \rightarrow \nu_{l^{\prime}})= \sum_j \left\lvert \, \langle \nu_l^{\prime} \mid \nu_j \rangle e^{-iE_jt} \langle \nu_j\mid \nu_l \rangle \, \right\rvert^2 =\sum_j \lvert U_{l^{\prime}j} \rvert^2 \lvert U_{l j} \rvert^2 \tag{G301} $$

In $G301$the $\bar{P}$ indicates the average probability. Notice the difference between $(G300)$, where the amplitudes are summed inside the squared modulus, and $(G301)$, where the squared modulus of the product of the production braket $\langle \nu_j\mid \nu_l \rangle$ and detection braket $\langle \nu_l^{\prime} \mid \nu_j \rangle$ is summed.

Note that the exponential phase evolution factor vanishes in a squared modulus, i.e., $$ \left \lvert \, e^{-iE_jt} \, \right \rvert^2 = \left( \sqrt{ \, e^{-iE_jt} \, \, e^{+iE_jt} } \right)^2 = \left( \sqrt{ \, e^{E_jt-iE_jt} } \right)^2 = \left( \sqrt{ \, e^{0} } \right)^2 = 1 $$

Then we take the squared moduli of the $U$ transition elements in $(G301)$, $$ \lvert U_{l^{\prime}j} \rvert^2 \lvert U_{l j} \rvert^2 = \left( \sqrt{ U_{l^{\prime}j}U_{l^{\prime}j}^{\ast} } \right)^2 \left( \sqrt{ U_{l^{}j}U_{l^{}j}^{\ast} } \right)^2 = U_{l^{\prime}j}U_{l^{\prime}j}^{\ast} U_{l^{}j}U_{l^{}j}^{\ast} $$

For $l = l^{\prime} =e$, $(G301)$ expands to $$ \bar{P}(\nu_{e} \rightarrow \nu_{e} ) = \lvert U_{e1}U_{e 1}^* \rvert^2 + \lvert U_{e2}U_{e 2}^* \rvert^2 + \lvert U_{e3} U_{e 3}^*\rvert^2 \\ = U_{e1}U_{e 1}^*U_{e1}U_{e 1}^* + U_{e2}U_{e 2}^*U_{e2}U_{e 2}^* + U_{e3} U_{e 3}^*U_{e3} U_{e 3}^* \tag{G399} $$

Replace the $U_{\alpha j}$ in $(G399)$ with their values from the standard parameterization of the PMNS matrix:

$$ \begin{pmatrix} \nu_{e} \\ \nu_{\mu} \\ \nu_{\tau} \end{pmatrix} = \begin{pmatrix} U_{e1} & U_{e2} & U_{e3} \\ U_{\mu 1} & U_{\mu 2} & U_{\mu 3} \\ U_{\tau 1} & U_{\tau 2} & U_{\tau 3} \end{pmatrix} \, \begin{pmatrix} \nu_1 \\ \nu_2 \\ \nu_3 \end{pmatrix} \\ \begin{pmatrix} \nu_{e} \\ \nu_{\mu} \\ \nu_{\tau} \end{pmatrix}= \begin{pmatrix} c_{12}c_{13} & s_{12}c_{13} & s_{13}e^{-i\delta_{CP}} \\ - s_{12}c_{23} - c_{12}s_{13}s_{23}e^{i \delta_{CP}} & c_{12}c_{23} - s_{12}s_{13}s_{23} e^{i \delta_{CP}} & c_{13}s_{23} \\ s_{12}s_{23}- c_{12}s_{13}c_{23}e^{i \delta_{CP}} & - c_{12}s_{23} - s_{12}s_{13}c_{23}e^{i \delta_{CP}} & c_{13}c_{23} \end{pmatrix} \begin{pmatrix} \nu_1 \\ \nu_2 \\ \nu_3 \end{pmatrix} \tag{G11} $$

$c_{12}$ denotes $\cos \theta_{12}$, $s_{13}$ is $\sin \theta_{13}$ and so on. We are not interested in the possibility of CP violation in the solar context, so just let $\delta_{CP}=0$ (as of 2020 LBS long baseline experiemtns NO normal ordering still favors CP conservation $\delta_{CP}= \pi$).

We obtain: $$ \bar{P}(\nu_{e} \rightarrow \nu_{e} ) = U_{e1}U_{e 1}^*U_{e1}U_{e 1}^* + U_{e2}U_{e 2}^*U_{e2}U_{e 2}^* + U_{e3} U_{e 3}^*U_{e3} U_{e 3}^* \\ = \cos^{4}{\left(\theta_{12} \right)} \cos^{4}{\left(\theta_{13} \right)}+ \sin^{4}{\left(\theta_{12} \right)} \cos^{4}{\left(\theta_{13} \right)} + \sin^{4}{\left(\theta_{13} \right)} \tag{G401} $$

$(G401)$ simplifies to $$ \bar{P}(\nu_{e} \rightarrow \nu_{e} ) = \cos^{4}\theta_{13} \left[ \cos^{4}{\theta_{12}} + \sin^{4}\theta_{12} \right] + \sin^{4}{\left(\theta_{13} \right)}\\ = \cos^{4}\theta_{13} \left[ 1 - \frac{1}{2} \sin^2 2\theta_{12} \right] + \sin^{4}{\left(\theta_{13} \right)} \tag{G777} $$

where we made the substitution: $$ \left[ \cos^{4}{\left(\theta_{12} \right)} + \sin^{4}\theta_{12} \right] \Longrightarrow \left[ 1 - \frac{1}{2} \sin^2 2\theta_{12} \right] $$

With $\theta_{13} = 8.5^{\circ}$, $\sin^{4}{\left(\theta_{13} \right)} = 0.00048$ and may reasonably be dropped from $(G777)$. With that value of mixing angle, $\cos^{4}\theta_{13}= 0.96 \approx 1$ and may be omitted also in $(G777)$, leading to the two-neutrino PDG low energy solar neutrino survival probability $$ \bar{P}^{2 \nu} (\nu_e \rightarrow \nu_e) = \underbrace{ \cos^{4}\theta_{13} }_{\sim 1} \left[ 1 - \frac{1}{2} \sin^2 2\theta_{12} \right] + \underbrace{ \sin^{4}{\left(\theta_{13} \right)} }_{\approx 0}\\ \simeq 1 - \frac{1}{2} \sin^2 2 \theta_{12} \tag{G778} $$

$\Longrightarrow$So, for the LER low energy range solar neutrinos, we might say that the reason the $\theta_{12}$ mixing angle is associated with solar neutrinos here is that the $\theta_{13}$ mixing angle is very small (which means that the fraction of $\nu_3$ mass state is very small in the electron neutrino, since $U_{e3} = \sin \theta_{13}$, referring to $(G11)$ matrix above) and so drops out of the above averaged (incoherent) survival probability equation.

the high energy region

On the right side of the Borexino graph above we see the survival probability $P_{ee}$ observed by various experiments is $\sim 0.3$ in the $10 \, \mathrm{MeV}$ area (this is the sweet spot for the SNO experiment, though the threshold is lower). The solar neutrinos in this range are predominantly produced in the $\beta^+$ decay from $^8_5\mathrm{B} \rightarrow \; ^8_4\mathrm{Be}^* + e^+ + \nu_e$ with energy expectation $\langle E_{\nu}\rangle = 6.735 \, \mathrm{MeV}$.

This $0.3$ survival probability is very close to simply $\sin^2 \theta_{12} = 0.304$ for $\theta_{12}=33.48^{\circ}$. Why is $\theta_{12}$ again so closely connected to an observed survival probability of solar neutrinos?

The $\Delta m_{12}^2$ oscillation length for a $10 \, \mathrm{MeV}$ neutrino is $\sim 330 \, \mathrm{km}$ (about $\sim 10 \, \mathrm{km}$ for the $\Delta m_{13}^2$ split), again much smaller than the distance from the Sun to Earth, so detecting oscillations would again be impossible on that basis alone (and the averaging over the region of neutrino production in the Sun, $\Delta R \simeq (0.04 \text{-} 0.20)R_{\odot}\simeq 10^5 \, \mathrm{km}$ as we mentioned earlier). However, there is a much more significant feature for a $10 \, \mathrm{MeV}$ solar neutrino, i.e., the matter potential.

matter matters

Consulting the B16(GS98) SSM (Standard Solar Model) again, we find that the $^8\mathrm{B}$ neutrino production peaks around $r = 0.0460 \, r_{\odot}$ (see discussion above about normalized solar radius) with electron number density at that location $N_e \sim 5.467 \, \times 10^{25} \, \mathrm{cm^{-3}}$. Using the PDG equation above, we see that a $10 \, \mathrm{MeV}$ neutrino reaches MSW resonance for the $\Delta m_{12}^2$ mass split at $N_e^{res} = 1.1596 \, \times 10^{25} \mathrm{cm}^{-3}$. The $^8\mathrm{B}$ solar neutrino is produced at almost $5$ times the required electron number density for MSW resonance, so we expect the matter potential to be highly significant.

For the $\Delta m_{13}^2=2.457 \, \times 10^{-3} \, \mathrm{eV}^2$ mass split, on the other hand, the required electron density for resonance, $N_e^{res}= 9.2823 \, \times 10^{26} \mathrm{cm}^{-3}$ is more than ten times the available electron number density. We understand, then, why $\Delta m_{13}^2$ is not associated with solar neutrino flavor change.

Consider a $10 \, \mathrm{MeV}$ $^8\mathrm{B}$ electron neutrino weak (flavor) state $\lvert \nu_e \rangle$ produced at $t=0$. Read off the $ \nu_e$ weak state description in terms of mass states from the top row $(U_{e1} ,U_{e2} , U_{e3})$ of $U$ $(G11)$ applied to the mass state column on rhs $(\nu_1, \nu_2, \nu_3)^T$ (this notation means to write the elements of this row as a column, i.e., $T$ transpose, since is typographically inconvenient to include a column inline):

$$ \lvert \nu_e \rangle= U_{e1}\lvert \nu_{1m} \rangle + U_{e2}\lvert \nu_{2m} \rangle + U_{e3} \lvert \nu_{3m} \rangle\\ = \cos{\theta_{12m} } \cos{\theta_{13m} } \lvert \nu_{1m} \rangle+ \sin{\theta_{12m} } \cos{\theta_{13m} }\lvert \nu_{2m} \rangle + \sin{\theta_{13m} }\lvert \nu_{3m} \rangle \\ \tag{G14} $$

If you are not familiar with the procedure (matrix vector multiplication) for assigning the $U$ elements to mass states for a particular flavor state as we did in $(G14)$ (we are using the same mixing matrix, we have simply transformed the parameters under the influence of the matter potential, denoted with $_m$ subscripts), informally put, think of $\nu_e$ in row $1$of the column vector on the lhs in $(G11)$ "owning" row $1$ of the $U$ matrix on the rhs. To see the composition of $\nu_e$ in terms of the $\nu_1,\nu_2,\nu_3$ mass states on the rhs, just pick up that row $1$ from the $U$ matrix and place it as a column to the left of the $\nu_1,\nu_2,\nu_3$ column like this:

$$ \begin{pmatrix} U_{e1} \\ U_{e2} \\ U_{e3} \end{pmatrix} \begin{pmatrix} \nu_1 \\ \nu_2 \\ \nu_3 \end{pmatrix} $$

Then read off each row of the two columns as a product, placing a plus sign between each row so constructed:

$$ \nu_e = U_{e1}\nu_1 + U_{e2}\nu_2+ U_{e3}\nu_3 $$

Similarly, $\nu_{\mu}$ owns row $2$ of $U$, so you would pick up that second row of $U$, placing it to the left of the mass state column and then read off the components as we did with the $\nu_e$.

Again, the $_m$ subscripts indicate that these are states and values in solar plasma, i.e., modified by matter effect (increased coherent forward elastic scattering potential, i.e., from exchange of a virtual $W$-boson, for $\nu_e$ vs $\nu_{\mu,\tau}$). The mixing angles with matter effect here are (we are working from Solar neutrinos and neutrino physics), who did include the small correction from the $\nu_1 \nu_3$ mixing):

$$ \cos 2 \theta_{12}^m = \frac{\cos 2 \theta_{12} - \cos^2 \theta_{13} \epsilon_{12}} { \sqrt{ \left(\cos 2 \theta_{12} - \cos^2 \theta_{13}\epsilon_{12}\right)^2 + \sin^2 2 \theta_{12}}} \tag{16} $$

where $$ V_e = \sqrt{2}G_F N_e $$ and $$ \epsilon_{12} \equiv\frac{2V_eE}{\Delta m_{12}^2} \tag{17} $$

The matter potential $V_e$ is calculated using our test case $^8\mathrm{B}$ $N_e$ and $E$: $$ V_e = \sqrt{2}G_F N_e (\hbar c)^3 = [\sqrt{2}] [1.1664 \, \times 10^{-5} \, \mathrm{GeV}^{-2}] [5.467 \, \times 10^{25} \, \mathrm{cm^{-3}}] [1.97 \, \times 10^{-14} \,\mathrm{GeV \, cm}]^3 $$

We obtain $V_e = 6.93 \, \times 10^{-21} \, \mathrm{ GeV }$ (or $V_e = 6.93 \, \times 10^{-12} \, \mathrm{ eV }$ if you are comparing with the general potential in solar core given in the literature, i.e., $\mathcal{O}(10^{-12} \, \mathrm{ eV })$).

$\epsilon_{12}$ is then (you see factors converting all energy units to $ \mathrm{ GeV }$) $$ \epsilon_{12} =\frac{2V_eE}{\Delta m_{12}^2} = \frac{(2)(6.93 \, \times 10^{-21} \, \mathrm{ GeV }) (10 \, \times 10^{-3}\mathrm{GeV})} {(7.5 \, \times 10^{-5} \,\times 10^{-18} \mathrm{GeV}^2)} = 1.848 $$

We see the energy units cancel, making $\epsilon_{12}$ a dimensionless ratio of the matter potential of the $10 \, \mathrm{MeV}$ neutrino in $5.467 \, \times 10^{25} \, \mathrm{cm^{-3}}$ electron number density to the $\Delta m_{12}^2$ mass splitting, $1.848$ indicating that the matter potential is nearly twice the mass splitting here.

Plug that value of $\epsilon_{12}$ into Eq. $(16)$ and obtain the cosine of $2\theta_{12m}$ mixing angle in matter: $$ \cos 2 \theta_{12}^m = \frac{\cos 2 \theta_{12} - \cos^2 \theta_{13} \epsilon_{12}} { \sqrt{ \left(\cos 2 \theta_{12} - \cos^2 \theta_{13}\epsilon_{12}\right)^2 + \sin^2 2 \theta_{12}}} \\[2ex] = \frac{(0.3914 \,\mathrm{rad}) - (0.9782 \, \mathrm{rad}) (1.848)} { \sqrt{ \left[ (0.3914 \,\mathrm{rad}) - (0.9782 \, \mathrm{rad}) (1.848)\right]^2 + (0.8468 \, \mathrm{rad})}} = −0.83854 $$

The $\arccos$ of $-0.83854$ is $2.56540 \, \mathrm{rad}$. Divide by $2$ to obtain the $\theta_{12m}$ mixing angle with the calculated matter potential, $1.28270 \, \mathrm{rad}$, or $\sim 74^{\circ}$. That is significant rotation (positive, counterclockwise) from the vacuum $\theta_{12}$ angle of $33.48^{\circ}$.

We obtain the $\theta_{13m}$ mixing angle through $$ \sin^2 \theta_{13}^m = \sin^2 \theta_{13} (1 + 2 \epsilon_{13}) + \mathcal{O}(s_{13}^2 \epsilon_{13}^2, s^4_{13},\epsilon_{13}) $$

We already have the $V_e$ potential and merely require $\epsilon_{13}$:

$$ \epsilon_{13} \equiv\frac{2V_eE}{\Delta m_{13}^2} \tag{19} $$

Inserting the values, with $\Delta m_{13}^2=2.457 \, \times 10^{-3} \, \mathrm{eV}^2$ the only change from our previous $\epsilon$ calculation: $$ \epsilon_{13} \equiv\frac{2V_eE}{\Delta m_{13}^2} = \frac{(2)(6.93 \, \times 10^{-21} \, \mathrm{ GeV }) (10 \, \times 10^{-3}\mathrm{GeV})} {(2.457 \, \times 10^{-3} \,\times 10^{-18} \mathrm{GeV}^2)} = 0.0564 $$

Drop the higher order tems and calculate $\sin^2 \theta_{13}^m $ using our $\epsilon_{13}$ value: $$ \sin^2 \theta_{13}^m = \sin^2 \theta_{13} (1 + 2 \epsilon_{13}) = 0.0218[1 + 2 (0.0564)] = 0.024259 $$

$\theta_{13}^m$ is then

$$ \arcsin(\sqrt{0.024259}) = 0.1566 \, \mathrm{rad} $$ or $8.9704^{\circ}$. As expected, the $\theta_{13}^m$ mix angle has not changed much from the vacuum value $8.5^{\circ}$, the larger $\Delta m_{13}^2$ mass split requiring more electron density than is available in order to produce significant rotation.

Referring to Eq. $(G14)$ above, we can look at the mass state mix at production for our test neutrino: $$ \lvert \nu_e \rangle= U_{e1}\lvert \nu_{1m} \rangle + U_{e2}\lvert \nu_{2m} \rangle + U_{e3} \lvert \nu_{3m} \rangle\\ = \cos{\theta_{12m} } \cos{\theta_{13m} } \lvert \nu_{1m} \rangle+ \sin{\theta_{12m} } \cos{\theta_{13m} }\lvert \nu_{2m} \rangle + \sin{\theta_{13m} }\lvert \nu_{3m} \rangle \\ \tag{G14} $$

The square of the mixing elements $U_{e1m},U_{e1m},U_{e1m}$ with mixing angle in matter give the probability fraction for each mass state in the electron neutrino produced under the specified conditions in the solar plasma (the crowded ket $\lvert \nu_e (N_e,E,t=0) \rangle$ is intended to remind you that the state of this neutrino at production $t=0$ is dependent on its energy and the local electron density in solar plasma): $$ \lvert \nu_e (N_e,E,t=0) \rangle =\\ [ \cos \theta_{12m} \cos \theta_{13m} ]^2 = [ \cos(74^{\circ}) \cos(8.9704^{\circ})]^2 = 0.0788 \lvert \nu_{1m} \rangle +\\ [ \sin{\theta_{12m} } \cos{\theta_{13m} } ]^2 = [\sin(74^{\circ}) \cos(8.9704^{\circ})]^2 =0.8969 \lvert \nu_{2m} \rangle \, + \quad \Longleftarrow\\ [\sin{\theta_{13m} } ]^2 = [\sin(8.9704^{\circ})]^2= 0.0243 \lvert \nu_{3m} \rangle $$

We find the single $\nu_2$ state detected at Earth produces closely the observed survival probability $P_{ee} \sim 0.3$ of $^8\mathrm{B}$ electron neutrinos on the right side of the Borexino graph above. With $\cos^2 \theta_{13} = 0.9781 \approx 1$, this survival probability is produced largely by simply $\sin^2 \theta_{12} = 0.304$.

$\Longrightarrow$We see the $10 \, \mathrm{MeV}$ $^8\mathrm{B}$ electron neutrino consists almost entirely ($\sim 90\%$)) of the $\lvert \nu_{2m}\rangle$ mass state. Because the solar plasma electron number density decreases smoothly (approximately as an exponential), the $\lvert \nu_{2m}\rangle$ evolves continuously (without level jumps) into the state $\lvert \nu_{2}\rangle$ at the surface of the Sun and is ultimately detected at Earth as that $\lvert \nu_{2}\rangle$ mass state, with the vacuum flavor probabilities assigned to $\nu_2$.

Well, we had a little more to present, but we are apparently at the limit of content allowed $30000$ characters), so will close here.

$\endgroup$
  • $\begingroup$ Since we ran out of space above, we prepared a formal pdf document with the entire article (with some historical material and a complete derivation of the standard three neutrino oscillation posted in the original question): sites.google.com/view/bentleyphysics/solsplit $\endgroup$ – Dalton Bentley Nov 10 '20 at 21:39

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