The question asks about the probability that the measurement of the muon spin along the direction of its momentum is positive in the decay of a charged pion (π −) into a muon (μ −) and an anti-muon neutrino (v̅ μ).

The decay process is:

\begin{equation*} \pi^- \rightarrow \mu^- + \bar{\nu}_\mu \end{equation*}

Pion Decay and Conservation Laws

Let's consider this decay in the rest frame of the pion.

• The initial particle, the pion ($\pi^-$), is a spin-0 particle. Its total angular momentum is 0.
• The decay products are a muon ($\mu^-$) and an anti-muon neutrino ($\bar{\nu}_\mu$). Both are spin-1/2 particles.
• In the pion's rest frame, the muon and the anti-muon neutrino are emitted back-to-back to conserve linear momentum.
• Angular momentum must also be conserved during the decay. The total angular momentum of the final state (muon + anti-neutrino) must equal the initial angular momentum of the pion, which is 0.

Neutrino Helicity

Neutrinos and anti-neutrinos are produced in weak interactions and have a special property called helicity. Helicity is the projection of the particle's spin onto its direction of momentum.

• Neutrinos ($\nu$) are observed to be left-handed, meaning their spin is oriented opposite to their direction of momentum. Their helicity is negative (-1/2).
• Anti-neutrinos ($\bar{\nu}$) are observed to be right-handed, meaning their spin is oriented along their direction of momentum. Their helicity is positive (+1/2).

In this decay, we have an anti-muon neutrino ($\bar{\nu}_\mu$), which is a type of anti-neutrino. Therefore, the anti-muon neutrino must be right-handed. This means its spin is along its direction of momentum, and its spin projection along its momentum is +1/2.

Muon Spin Direction

Let's denote the spin of the muon as $\vec{S}_\mu$ and the spin of the anti-muon neutrino as $\vec{S}_{\bar{\nu}_\mu}$. Let the momentum of the muon be $\vec{p}_\mu$ and the momentum of the anti-muon neutrino be $\vec{p}_{\bar{\nu}_\mu}$.

In the pion rest frame, $\vec{p}_\mu = -\vec{p}_{\bar{\nu}_\mu}$. The direction of the muon's momentum is opposite to the direction of the anti-neutrino's momentum. Let $\hat{p}_\mu$ be the unit vector in the direction of the muon's momentum, and $\hat{p}_{\bar{\nu}_\mu}$ be the unit vector in the direction of the anti-neutrino's momentum. Thus, $\hat{p}_\mu = -\hat{p}_{\bar{\nu}_\mu}$.

The anti-muon neutrino is right-handed, so its spin along its momentum direction is positive:

\begin{equation*} \vec{S}_{\bar{\nu}_\mu} \cdot \hat{p}_{\bar{\nu}_\mu} = +\frac{1}{2} \end{equation*}

Now consider angular momentum conservation. In the center-of-momentum frame, the orbital angular momentum contribution along the direction of motion is zero. Therefore, the sum of the spins of the muon and anti-neutrino must add up to the initial angular momentum along this direction, which is zero. The total angular momentum vector must be zero, so $\vec{S}_\mu + \vec{S}_{\bar{\nu}_\mu} = \vec{0}$, which means $\vec{S}_\mu = -\vec{S}_{\bar{\nu}_\mu}$.

We are interested in the probability that the muon's spin along its momentum direction ($\hat{p}_\mu$) is positive. This is given by $\vec{S}_\mu \cdot \hat{p}_\mu$.

Substitute $\vec{S}_\mu = -\vec{S}_{\bar{\nu}_\mu}$:

\begin{equation*} \vec{S}_\mu \cdot \hat{p}_\mu = (-\vec{S}_{\bar{\nu}_\mu}) \cdot \hat{p}_\mu \end{equation*}

Substitute $\hat{p}_\mu = -\hat{p}_{\bar{\nu}_\mu}$:

\begin{equation*} (-\vec{S}_{\bar{\nu}_\mu}) \cdot (-\hat{p}_{\bar{\nu}_\mu}) = \vec{S}_{\bar{\nu}_\mu} \cdot \hat{p}_{\bar{\nu}_\mu} \end{equation*}

From the anti-neutrino helicity, we know $\vec{S}_{\bar{\nu}_\mu} \cdot \hat{p}_{\bar{\nu}_\mu} = +\frac{1}{2}$.

Therefore, the muon's spin along its momentum direction must be:

\begin{equation*} \vec{S}_\mu \cdot \hat{p}_\mu = +\frac{1}{2} \end{equation*}

This shows that the muon produced in this decay is always right-handed, meaning its spin is aligned with its momentum direction.

Since the muon's spin along the direction of its momentum is always +1/2 (positive), the probability of this measurement being positive is 1.

The rest masses of the muon and pion (105 MeV and 140 MeV) are relevant for the energy-momentum kinematics of the decay, but they do not affect the spin and helicity conservation principles that determine the spin orientation of the decay products.

Probability Result

Based on the conservation of angular momentum and the right-handed nature of the anti-muon neutrino, the muon produced in the $\pi^- \rightarrow \mu^- + \bar{\nu}_\mu$ decay must be right-handed. Its spin projection along its momentum direction is therefore always positive (+1/2). The probability of this occurring is 1.