# 物理代写|核物理代写nuclear physics代考|PHYSICS404

## 物理代写|核物理代写nuclear physics代考|Density of States

We have shown in Appendix 3 that the number of allowed momentum states for a particle confined to a box of volume $V$, with magnitude of momentum between $p$ and $p+d p$, is given by

$$n(p)=V \frac{p^2 d p}{2 \pi^2 \hbar^3},$$
so the number of allowed states with an electron momentum between $p_e$ and $p_e+$ $d p_e$ and an antineutrino with momentum between $q_{\bar{v}}$ and $q_{\bar{v}}+d q_{\bar{v}}$ is given by
$$n_{e \bar{v}}\left(p_e, q_{\bar{v}}\right)=n\left(p_e\right) n\left(q_{\bar{v}}\right)=\left(\frac{V}{2 \pi^2 \hbar^3}\right)^2 p_e^2 d p_e q_{\bar{\nu}}^2 d q_{\bar{v}} .$$
Let us now convert this into the number of states, $n_{e \bar{v}}\left(T_e, E_{\bar{v}}\right)$, with electron kinetic energy between $T_e$ and $T_e+d T_e$ and antineutrino energy between $E_{\bar{v}}$ and $E_{\bar{v}}+d E_{\bar{v}}$. Neglecting the neutrino mass, the antineutrino energy and momentum are simply related by
$$q_{\bar{v}}=\frac{E_{\overline{\bar{v}}}}{c}, \quad d q_{\bar{v}}=\frac{d E_{\bar{v}}}{c} .$$
For electrons and neutrinos we need to use the relativistic relation between momentum and kinetic energy, since very often the electron will have kinetic energy of the order of or greater than its rest energy, $m_e c^2$ :
$$p_e=\frac{1}{c} \sqrt{T_e\left(T_e+2 m_e c^2\right)}, \quad d p_e=\frac{\left(T_e+m_e c^2\right)}{c^2 p_e} d T_e .$$
Inserting (7.17) and (7.18) into (7.16) we get the expression for $n_{e \bar{\nu}}\left(T_e, E_{\bar{v}}\right)$,
$$n_{e \bar{v}}\left(T_e, E_{\bar{v}}\right)=\frac{V^2}{4 \pi^4 \hbar^6 c^5} E_{\bar{v}}^2\left(T_e+m_e c^2\right) p_e d T_e d E_{\bar{v}} .$$

## 物理代写|核物理代写nuclear physics代考|Decay Matrix Element

Let us consider now the matrix element of the weak-interaction Hamiltonian
$$\left\langle f\left|H_{\mathrm{WK}}\right| i\right\rangle=\int \Psi_{\bar{v}}^(\boldsymbol{r}) \Psi_e^(\boldsymbol{r}) \Psi_D^*(\boldsymbol{r}) H_{\mathrm{WK}} \Psi_P(\boldsymbol{r}) d^3 \boldsymbol{r},$$
where we have used the fact that the interactions are very short range so that we can take the wavefunctions for all the particles at the same point, $\boldsymbol{r}$.

The wavefunctions (normalized within a volume $V$ ) of the leptons are outgoing plane-wave functions
$$\Psi_e(r)=\frac{e^{i p_e \cdot r / h}}{\sqrt{V}} \text { and } \Psi_{\bar{\nu}}(r)=\frac{e^{i q_{\bar{v}} \cdot r / h}}{\sqrt{V}} .$$
Since the nuclear wavefunctions become negligible for $r \gg R$, where $R$ is the nuclear radius and the momenta of the leptons are such that
$$\frac{p_e \cdot \boldsymbol{r}}{\hbar}, \frac{\boldsymbol{q}{\bar{v}} \cdot \boldsymbol{r}}{\hbar} \ll 1 \text { for } r \leq R,$$ we may approximate these wavefunctions by $$\Psi_e(\boldsymbol{r}), \Psi{\bar{\nu}}(\boldsymbol{r}) \approx \frac{1}{\sqrt{V}} .$$
The weak Hamiltonian can be written as
$$H_{\mathrm{WK}}=G_F \hbar^3 c^3 \mathcal{O}{\mathrm{WK}}(\boldsymbol{r}),$$ where the operator $\mathcal{O}{\mathrm{WK}}$ acts only on the nuclear wavefunctions, $\Psi_P$ of the parent nucleus and $\Psi_D$ of the daughter nucleus.

## 物理代写|核物理代写nuclear physics代考|Density of States

$$n(p)=V \frac{p^2 d p}{2 \pi^2 \hbar^3},$$

$$n_{e \bar{v}}\left(p_e, q_{\bar{v}}\right)=n\left(p_e\right) n\left(q_{\bar{v}}\right)=\left(\frac{V}{2 \pi^2 \hbar^3}\right)^2 p_e^2 d p_e q_{\bar{\nu}}^2 d q_{\bar{v}} .$$

$$q_{\bar{v}}=\frac{E_{\bar{v}}}{c}, \quad d q_{\bar{v}}=\frac{d E_{\bar{v}}}{c} .$$

$$p_e=\frac{1}{c} \sqrt{T_e\left(T_e+2 m_e c^2\right)}, \quad d p_e=\frac{\left(T_e+m_e c^2\right)}{c^2 p_e} d T_e .$$

$$n_{e \bar{v}}\left(T_e, E_{\bar{v}}\right)=\frac{V^2}{4 \pi^4 \hbar^6 c^5} E_{\bar{v}}^2\left(T_e+m_e c^2\right) p_e d T_e d E_{\bar{v}}$$

## 物理代写|核物理代写nuclear physics代考|Decay Matrix Element

$$\left.\left.\left\langle f\left|H_{\mathrm{WK}}\right| i\right\rangle=\int \Psi_{\bar{v}}^{(} \boldsymbol{r}\right) \Psi_e^{(} \boldsymbol{r}\right) \Psi_D^*(\boldsymbol{r}) H_{\mathrm{WK}} \Psi_P(\boldsymbol{r}) d^3 \boldsymbol{r},$$

$$\Psi_e(r)=\frac{e^{i p_e \cdot r / h}}{\sqrt{V}} \text { and } \Psi_{\bar{\nu}}(r)=\frac{e^{i q_{\bar{v}} \cdot r / h}}{\sqrt{V}} .$$

$$\frac{p_e \cdot \boldsymbol{r}}{\hbar}, \frac{\boldsymbol{q} \bar{v} \cdot \boldsymbol{r}}{\hbar} \ll 1 \text { for } r \leq R,$$

$$\Psi_e(\boldsymbol{r}), \Psi \bar{\nu}(\boldsymbol{r}) \approx \frac{1}{\sqrt{V}}$$

$$H_{\mathrm{WK}}=G_F \hbar^3 c^3 \mathcal{O W K}(r),$$

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