trying to understand neutrino oscillations

Neutrinos only interact Weakly. Er go, neutrinos are generated into Weak eigenstates, which are "mixtures" of the canonical mass eigenstates; and neutrinos remain in those Weak eigenstates, until detection. For example, electron neutrinos $\nu_e$ are a mixed Weak eigenstate, representing a mixture of the neutrino mass eigenstates $\nu_1, \nu_2, \nu_3$ , specifically

$|\nu_e> \approx 0.9 |\nu_1> + 0.5 |\nu_2>$

The following analysis employs normalized units $c \rightarrow 1$ ; and assumes that the neutrino mass eigenstates, are also eigenstates, of mass $\left( \hat{m} \right)$ , momentum $\left( \hat{p} \right)$ , & squared-energy $\left( \hat{E}^2 \right)$ .

take 1

How can such a Weak eigenstate possess a well-defined energy & momentum ? For, if

$m \equiv \alpha m_1 + \beta m_2$

$E \equiv \alpha E_1 + \beta E_2$

$p \equiv \alpha p_1 + \beta p_2$

then

$E^2 = m^2 + p^2$

$\left( \alpha E_1 + \beta E_2 \right)^2 = \left( \alpha m_1 + \beta m_2 \right)^2 + \left( \alpha p_1 + \beta p_2 \right)^2$

$\begin{array}{c}\alpha^2 E_1^2 + \beta^2 E_2^2 \\+ 2 \alpha \beta E_1 E_2 \end{array} = \begin{array}{c}\alpha^2 m_1^2 + \beta^2 m_2^2 \\+ 2 \alpha \beta m_1 m_2 \end{array} + \begin{array}{c}\alpha^2 p_1^2 + \beta^2 p_2^2 \\+ 2 \alpha \beta p_1 p_2 \end{array}$

$\therefore E_1 E_2 = m_1 m_2 + p_1 p_2$

But w.h.t.

$E = \gamma m$
$p = \gamma m \beta = \beta E$

$E_1 E_2 = m_1 m_2 + p_1 p_2$

$\gamma_1 \gamma_2 = 1 + \gamma_1 \beta_1 \gamma_2 \beta_2$
$\gamma_1 \gamma_2 \left(1 - \beta_1 \beta_2 \right) = 1$
$\left(1 - \beta_1 \beta_2 \right)^2 = \left(1 - \beta_1^2 \right) \left( 1 - \beta_2^2 \right)$
$\begin{array}{c}1 + \beta_1^2 \beta_2^2 \\- 2 \beta_1 \beta_2 \end{array} = \begin{array}{c}1 + \beta_1^2 \beta_2^2 \\- 2 \beta_1^2 \beta_2^2 \end{array}$
$1 = \beta_1 \beta_2$

$\therefore 1 = \beta_1 = \beta_2$

But, neutrinos have mass, i.e. $\beta_{1,2} < 1$ .

take 2

Are "Weak neutrinos", i.e. neutrinos in Weak eigenstates, "off mass shell" ? For, if:

$|\nu_e> = \alpha |\nu_1> + \beta |\nu_2>$

then is the mass expectation value:

$<m_{\nu_e}> = \left( \alpha^* <\nu_1| + \beta^* <\nu_2| \right) \hat{m} \left( \alpha |\nu_1> + \beta |\nu_2> \right)$

$= |\alpha|^2 <\nu_1|\hat{m}|\nu_1> + |\beta|^2 <\nu_2|\hat{m}|\nu_2>$

$= |\alpha|^2 <m_1> +|\beta|^2 <m_2>$

$\equiv |\alpha|^2 m_1 +|\beta|^2 m_2$

$\therefore m_{\nu_e} \approx \frac{3}{4} m_1 + \frac{1}{4} m_2$

or is the momentum expectation value:

$<p_{\nu_e}> = \left( \alpha^* <\nu_1| + \beta^* <\nu_2| \right) \hat{p} \left( \alpha |\nu_1> + \beta |\nu_2> \right)$

$= |\alpha|^2 <\nu_1|\hat{p}|\nu_1> + |\beta|^2 <\nu_2|\hat{p}|\nu_2>$

$= |\alpha|^2 <p_1> +|\beta|^2 <p_2>$

$\equiv |\alpha|^2 p_1 +|\beta|^2 p_2$

$\therefore p_{\nu_e} \approx \frac{3}{4} p_1 + \frac{1}{4} p_2$

or is the squared-energy expectation value:

$<E_{\nu_e}^2> = \left( \alpha^* <\nu_1| + \beta^* <\nu_2| \right) \left[ \hat{p}^2 + \hat{m}^2 \right] \left( \alpha |\nu_1> + \beta |\nu_2> \right)$

$= |\alpha|^2 <\nu_1|\left[ \hat{p}^2 + \hat{m}^2 \right]|\nu_1> + |\beta|^2 <\nu_2|\left[ \hat{p}^2 + \hat{m}^2 \right]|\nu_2>$

$= |\alpha|^2 <E_1^2> +|\beta|^2 <E_2^2>$

$\equiv |\alpha|^2 E_1 +|\beta|^2 E_2$

$\therefore E_{\nu_e}^2 \approx\frac{3}{4} E_1^2 + \frac{1}{4} E_2^2$

??? If so, then Weak neutrinos are "off mass shell", i.e. $E_{\nu_e}^2 \ne m_{\nu_e}^2 + p_{\nu_e}^2$ , i.e.

$m_{\nu_e}^2 + p_{\nu_e}^2 = \left( |\alpha|^2 m_1 +|\beta|^2 m_2 \right)^2 + \left( |\alpha|^2 p_1 +|\beta|^2 p_2 \right)^2$

$= \begin{array}{c}|\alpha|^4 m_1^2 +|\beta|^4 m_2^2 \\+ 2 |\alpha|^2 |\beta|^2 m_1 m_2 \end{array} + \begin{array}{c}|\alpha|^4 p_1^2 +|\beta|^4 p_2^2 \\+ 2 |\alpha|^2 |\beta|^2 p_1 p_2 \end{array}$

$= |\alpha|^4 E_1^2 +|\beta|^4 E_2^2 + 2 |\alpha|^2 |\beta|^2 \left( m_1 m_2 + p_1 p_2 \right)$

$= |\alpha|^2 \left(|\alpha|^2 \right)^2 E_1^2 +|\beta|^2 \left( |\beta|^2 \right)^2 E_2^2 + 2 |\alpha|^2 |\beta|^2 \left( m_1 m_2 + p_1 p_2 \right)$

$= |\alpha|^2 \left(1 - |\beta|^2 \right)^2 E_1^2 +|\beta|^2 \left(1 - |\alpha|^2 \right)^2 E_2^2 + 2 |\alpha|^2 |\beta|^2 \left( m_1 m_2 + p_1 p_2 \right)$

$= \left( |\alpha|^2 E_1^2 +|\beta|^2 E_2 \right) - |\alpha|^2 |\beta|^2 \left( \left( E_1^2 + E_2^2 \right) - 2 \left( m_1 m_2 + p_1 p_2 \right)\right)$

$= E_{\nu_e}^2 - |\alpha|^2 |\beta|^2 \left( \left( E_1^2 + E_2^2 \right) - 2 \left( m_1 m_2 + p_1 p_2 \right)\right)$

$= E_{\nu_e}^2 - |\alpha|^2 |\beta|^2 \left( \left( m_1^2 + p_1^2 + m_2^2 + p_2^2 \right) - 2 \left( m_1 m_2 + p_1 p_2 \right)\right)$

$= E_{\nu_e}^2 - |\alpha|^2 |\beta|^2 \left( \left(m_1-m_2 \right)^2 + \left(p_1-p_2 \right)^2 \right)$

$\approx E_{\nu_e}^2 - \frac{3}{16} \left( \left(m_2-m_1 \right)^2 + \left(p_2-p_1 \right)^2 \right)$

If so, then $E_{\nu_e}^2 > m_{\nu_e}^2 + p_{\nu_e}^2$ , i.e. "electron neutrinos are energy rich" (by a few eV ?).

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