Chapter 15, Problem 22P

(a)

To determine

The expression for ${\nu }_{\tau }\left(t\right)$.

Answer to Problem 22P

The expression for ${\nu }_{\tau }\left(t\right)$ is ${\nu }_{\tau }\left(t\right)=\sin \theta {\nu }_2{e}^{-i{E}_{2}t/\hslash }+\cos \theta {\nu }_3{e}^{-i{E}_{3}t/\hslash }$.

Explanation of Solution

Write the expression for ${\nu }_{\tau }\left(t\right)$.

  ${\nu }_{\tau }\left(t\right)=\sin \theta {\nu }_2{e}^{-i{E}_{2}t/\hslash }+\cos \theta {\nu }_3{e}^{-i{E}_{3}t/\hslash }$        (I)

Here, ${\nu }_{\tau }\left(t\right)$ the time evolution of ${\nu }_{\tau }$, $\theta$ is the mixing angle, $E_2$ is the energy of the stationary state ${\nu }_2$ , $E_3$ is the energy of the stationary state ${\nu }_3$ , $t$ is the time and $\hslash$ is the reduced Planck’s constant.

Conclusion:

Therefore, the expression for ${\nu }_{\tau }\left(t\right)$ is ${\nu }_{\tau }\left(t\right)=\sin \theta {\nu }_2{e}^{-i{E}_{2}t/\hslash }+\cos \theta {\nu }_3{e}^{-i{E}_{3}t/\hslash }$.

(b)

To determine

The mathematical expression to show that the neutrino created at $t=0$ is entirely a $\mu$ neutrino.

Answer to Problem 22P

The mathematical expressions to show that the neutrino created at $t=0$ is entirely a $\mu$ neutrino are $\begin{array}{c}{\left(\cos \theta {\nu }_2-sin\theta {\nu }_3\right)}^{*}\left(\cos \theta {\nu }_2-\sin \theta {\nu }_3\right)=1\\ {\left(sin\theta {\nu }_2+\cos \theta {\nu }_3\right)}^{*}\left(\sin \theta {\nu }_2+\cos \theta {\nu }_3\right)=0\end{array}$ .

Explanation of Solution

Write the conditions for the neutrino created at $t=0$ to be entirely a $\mu$ neutrino.

  ${\left|v_{\mu }\left(0\right)\right|}^2=1$        (II)

Here, ${\left|v_{\mu }\left(0\right)\right|}^2$ is the probability for the particle to be found in ${\nu }_{\mu }$ state at $t=0$ .

  ${\left|v_{\tau }\left(0\right)\right|}^2=0$        (III)

Here, ${\left|v_{\tau }\left(0\right)\right|}^2$ is the probability for the particle to be found in ${\nu }_{\tau }$ state at $t=0$ .

Write the expression for ${\nu }_{\mu }\left(t\right)$ .

  ${\nu }_{\mu }\left(t\right)=\cos \theta {\nu }_2{e}^{-i{E}_{2}t/\hslash }-\sin \theta {\nu }_3{e}^{-i{E}_{3}t/\hslash }$        (IV)

Substitute $0$ for $t$ in equation (IV).

  $\begin{array}{c}{\nu }_{\mu }\left(0\right)=\cos \theta {\nu }_2{e}^0-\sin \theta {\nu }_3{e}^0\\ =\cos \theta {\nu }_2-\sin \theta {\nu }_3\end{array}$        (V)

Put equation (V) in equation (II).

  ${\left(\cos \theta {\nu }_2-\sin \theta {\nu }_3\right)}^{*}\left(\cos \theta {\nu }_2-\sin \theta {\nu }_3\right)=1$        (VI)

Substitute $0$ for $t$ in equation (I).

  $\begin{array}{c}{\nu }_{\tau }\left(0\right)=\sin \theta {\nu }_2{e}^0+\cos \theta {\nu }_3{e}^0\\ =\sin \theta {\nu }_2+\cos \theta {\nu }_3\end{array}$        (VII)

Put equation (VII) in equation (III).

  ${\left(\sin \theta {\nu }_2+\cos \theta {\nu }_3\right)}^{*}\left(\sin \theta {\nu }_2+\cos \theta {\nu }_3\right)=0$

Conclusion:

Therefore, the mathematical expressions to show that the neutrino created at $t=0$ is entirely a $\mu$ neutrino are $\begin{array}{c}{\left(\cos \theta {\nu }_2-sin\theta {\nu }_3\right)}^{*}\left(\cos \theta {\nu }_2-\sin \theta {\nu }_3\right)=1\\ {\left(sin\theta {\nu }_2+\cos \theta {\nu }_3\right)}^{*}\left(\sin \theta {\nu }_2+\cos \theta {\nu }_3\right)=0\end{array}$ .

(c)

To determine

To show that the probability of finding a $\tau$ neutrino at time $t$ is $P\left(v_{\mu }\to v_{\tau }\right)=\frac{{\sin }^{2}2\theta }{2}\left(1-\cos \frac{E_2-E_3}{\hslash }t\right)$ .

Answer to Problem 22P

It is showed that the probability of finding a $\tau$ neutrino at time $t$ is $P\left(v_{\mu }\to v_{\tau }\right)=\frac{{\sin }^{2}2\theta }{2}\left(1-\cos \frac{E_2-E_3}{\hslash }t\right)$ .

Explanation of Solution

Write the equation for the probability of finding a $\tau$ neutrino at time $t$ .

  $P\left(v_{\mu }\to v_{\tau }\right)={\left|v_{\tau }\left(t\right)\right|}^2$

Here, $P\left(v_{\mu }\to v_{\tau }\right)$ is the probability of finding a $\tau$ neutrino at time $t$ .

Put equation (I) in the above equation.

  $P\left(v_{\mu }\to v_{\tau }\right)={\left(\sin \theta {\nu }_2{e}^{-i{E}_{2}t/\hslash }+\cos \theta {\nu }_3{e}^{-i{E}_{3}t/\hslash }\right)}^{*}\left(\sin \theta {\nu }_2{e}^{-i{E}_{2}t/\hslash }+\cos \theta {\nu }_3{e}^{-i{E}_{3}t/\hslash }\right)$

After some algebra, it can be showed that

  $P\left(v_{\mu }\to v_{\tau }\right)=\frac{{\sin }^{2}2\theta }{2}\left(1-\cos \frac{E_2-E_3}{\hslash }t\right)$

Conclusion:

Therefore, it is showed that the probability of finding a $\tau$ neutrino at time $t$ is $P\left(v_{\mu }\to v_{\tau }\right)=\frac{{\sin }^{2}2\theta }{2}\left(1-\cos \frac{E_2-E_3}{\hslash }t\right)$ .

(d)

To determine

To show that for a fixed momentum $p$ and small masses $m_2$ and $m_3$ , $E_2-E_3\cong \frac{\left(m_2^2-m_3^2\right)c^3}{2p}$ .

Answer to Problem 22P

It is showed that for a fixed momentum $p$ and small masses $m_2$ and $m_3$ , $E_2-E_3\cong \frac{\left(m_2^2-m_3^2\right)c^3}{2p}$ .

Explanation of Solution

In this case, the experiment involves setting up a $\mu$-beam at $t=0$ with a definite momentum $p$ .

Write the expression for $E_2$ .

  $E_2=pc\sqrt{1+\frac{{\left(m_2{c}^2\right)}^2}{{\left(pc\right)}^2}}$

Here, $m_2$ is the mass of ${\nu }_2$ .

Rewrite the above equation for very small masses such that $m_2{c}^2<<pc$ .

  $E_2\cong pc\left(1+\frac{{\left(m_2{c}^2\right)}^2}{2{\left(pc\right)}^2}\right)$        (VIII)

Write the expression for $E_3$ .

  $E_3=pc\sqrt{1+\frac{{\left(m_3{c}^2\right)}^2}{{\left(pc\right)}^2}}$

Here, $m_3$ is the mass of ${\nu }_3$ .

Rewrite the above equation for very small masses such that $m_3{c}^2<<pc$ .

  $E_3\cong pc\left(1+\frac{{\left(m_3{c}^2\right)}^2}{2{\left(pc\right)}^2}\right)$        (IX)

Find $E_2-E_3$ using equations (VIII) and (IX).

  $\begin{array}{c}{E}_2-E_3=pc\left(1+\frac{{\left(m_2{c}^2\right)}^2}{2{\left(pc\right)}^2}\right)-pc\left(1+\frac{{\left(m_3{c}^2\right)}^2}{2{\left(pc\right)}^2}\right)\\ =\frac{{\left(m_2{c}^2\right)}^2}{2{\left(pc\right)}^2}-\frac{{\left(m_3{c}^2\right)}^2}{2{\left(pc\right)}^2}\\ =\frac{\left(m_2^2-m_3^2\right)c^3}{2p}\end{array}$

Conclusion:

Therefore, it is showed that for a fixed momentum $p$ and small masses $m_2$ and $m_3$ , $E_2-E_3\cong \frac{\left(m_2^2-m_3^2\right)c^3}{2p}$ .