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# 物理代写|粒子物理学作业代写Nuclear and Particle Physics代考|Leptons

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## 物理代写|粒子物理学作业代写Nuclear and Particle Physics代考|Lepton Number

From experimental observations as well as symmetries of the underlying theory, we can deduce a conserved quantity called Lepton number. Negatively charged Leptons and the corresponding neutrino carry a Lepton number of 1 . Antiparticles carry the opposite Lepton Number of $-1$. For states with more than one Lepton, the Lepton number is additive.
$$L=N\left(l^{-}\right)-N\left(l^{+}\right)+N\left(\nu_{l}\right)-N\left(\bar{\nu}{l}\right) ; l=e, \mu, \tau$$ The lepton number $L$ as well as the individual lepton numbers $\left(L{e}, L_{\mu}, L_{\tau}\right)$ are conserved in all interactions.

## 物理代写|粒子物理学作业代写Nuclear and Particle Physics代考|Lepton decays

As neutrinos are the lightest particles of a generation, they are stable. The lightest electrically charged lepton, the electron, is stable. The heavier leptons can decay through the exchange of a weak W boson.
Lepton Universality means that the coupling of the $\mathrm{W}$ boson to any Lepton Doublet (a charged lepton and a neutrino) are the same for all three generations. However decay rates etc. will also depend on the “phase space”, e.g. the available energy and hence we need to take the mass differences between lepton species into account. We can make a simple dimensional argument to understand this dependence. The decay width $\Gamma$ has units of [mass]. The Feynman diagram for charged lepton decay (and hence the decay amplitude) contains a $\mathrm{W}$ propagator $\frac{1}{m_{W}^{2}}$, which has to be squared to calculate the decay rate, proving a term with dimensions $\sim[\text { mass }]^{-4}$. In order to arrive at units of $[$ mass $]$ for the decay width $\Gamma$, a term with dimensions $\sim[\text { mass }]^{5}$ is needed. The relevant Q-value is the mass of the decaying Lepton, hence:
$$\frac{1}{\tau_{\mu}}=\Gamma\left(\mu^{-}\right) \sim \Gamma\left(\mu^{-} \rightarrow e^{-} \nu_{\mu} \bar{\nu}{e}\right) \sim\left|\frac{1}{m{W}^{2}}\right|^{2}\left(m_{\mu}\right)^{5} \rightarrow[\text { mass }]$$

## 物理代写|粒子物理学作业代写NUCLEAR AND PARTICLE PHYSICS代考|Neutrino Oscillations

In the SM neutrinos are massless. However we now know that this description is not adequate and indeed neutrinos do posses mass. This has some important implications. In particular the mass eigenstates could be different from the flavour eigenstates of the weak interaction. The flavour eigenstates would than be superpositions of the mass eigenstates as shown in Eq.2.8.
\begin{aligned} &\left|\nu_{e}\right\rangle=\left|\nu_{1}\right\rangle \cos \theta+\left|\nu_{2}\right\rangle \sin \theta \ &\left|\nu_{\mu}\right\rangle=-\left|\nu_{1}\right\rangle \sin \theta+\left|\nu_{2}\right\rangle \cos \theta \end{aligned}
This essentially corresponds to a rotation between two sets of basis state vectors.
$$\left|\begin{array}{l} \nu_{e} \ \nu_{\mu} \end{array}\right\rangle=\left(\begin{array}{cc} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{array}\right)\left|\begin{array}{l} \nu_{1} \ \nu_{2} \end{array}\right\rangle$$

## 物理代写|粒子物理学作业代写NUCLEAR AND PARTICLE PHYSICS代考|LEPTON NUMBER

$$L=N\left(l^{-}\right)-N\left(l^{+}\right)+N\left(\nu_{l}\right)-N\left(\bar{\nu}{l}\right) ; l=e, \mu, \tau$$ The lepton number $L$ as well as the individual lepton numbers $\left(L{e}, L_{\mu}, L_{\tau}\right)$ 在所有相互作用中都是守恒的。

## 物理代写|粒子物理学作业代写NUCLEAR AND PARTICLE PHYSICS代考|LEPTON DECAYS

$$\frac{1}{\tau_{\mu}}=\Gamma\left(\mu^{-}\right) \sim \Gamma\left(\mu^{-} \rightarrow e^{-} \nu_{\mu} \bar{\nu}{e}\right) \sim\left|\frac{1}{m{W}^{2}}\right|^{2}\left(m_{\mu}\right)^{5} \rightarrow[\text { mass }]$$

## 物理代写|粒子物理学作业代写NUCLEAR AND PARTICLE PHYSICS代考|NEUTRINO OSCILLATIONS

\begin{aligned} &\left|\nu_{e}\right\rangle=\left|\nu_{1}\right\rangle \cos \theta+\left|\nu_{2}\right\rangle \sin \theta \ &\left|\nu_{\mu}\right\rangle=-\left|\nu_{1}\right\rangle \sin \theta+\left|\nu_{2}\right\rangle \cos \theta \end{aligned}
This essentially corresponds to a rotation between two sets of basis state vectors.
$$\left|\begin{array}{l} \nu_{e} \ \nu_{\mu} \end{array}\right\rangle=\left(\begin{array}{cc} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{array}\right)\left|\begin{array}{l} \nu_{1} \ \nu_{2} \end{array}\right\rangle$$

## Matlab代写

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