## 物理代写|粒子物理代写particle physics代考|Magnetic Dipole Moments

Since nuclei with an odd number of protons and/or neutrons have intrinsic spin, they also generally possess a magnetic dipole moment.

The unit of the magnetic dipole moment for a nucleus, the “nuclear magneton”, is defined as
$$\mu_N=\frac{e \hbar}{2 m_p},$$
which is analogous to the Bohr magneton but with the electron mass replaced by the proton mass. It is defined such that the magnetic moment of a proton due to its orbital angular momentum $\ell$ is $\mu_N \ell$. Experimentally it is found that the magnetic moment of the proton due to its spin is
$$\mu_p=2.79 \mu_N=5.58 \mu_N s, \quad\left(s=\frac{1}{2}\right)$$
and that of the neutron is
$$\mu_n=-1.91 \mu_N=-3.82 \mu_N s, \quad\left(s=\frac{1}{2}\right) .$$
If we apply a magnetic field in the $z$-direction to a nucleus, then the unpaired proton with orbital angular momentum $\boldsymbol{\ell}$, spin $\boldsymbol{s}$ and total angular momentum $\boldsymbol{j}$ will give a contribution to the $z$-component of the nuclear magnetic moment
$$\mu_z=\left(5.58 s_z+\ell_z\right) \mu_N .$$

As in the case of the “Zeeman effect”, we can use the vector model approach to project $s$ and $\ell$ onto the total momentum, $\boldsymbol{j}$, and then project $\boldsymbol{j}$ onto the $z$-axis yielding
$$\mu_z=\frac{(5.58\langle\boldsymbol{s} \cdot \boldsymbol{j}\rangle+\langle\boldsymbol{\ell} \cdot \boldsymbol{j}\rangle)}{\left\langle\boldsymbol{j}^2\right\rangle} j_z \mu_N .$$

## 物理代写|粒子物理代写particle physics代考|The Collective Model

The Shell Model has its shortcomings. In spite of its great success, the usefulness of the Shell Model should not be overstated. It has a limited range of validity – it can explain phenomena mainly relevant to the light spherical nuclei, but even in this case one observes discrepancies between the predictions of the model and experiment. These discrepancies are even larger for heavier nuclei. We have already seen that the Shell Model does not predict magnetic dipole moments or the spectra of excited states very well.

One further failing of the Shell Model is the prediction of electric quadrupole moments. The Shell Model predicts very small values for thesese. Ilowever, for heavier nuclei with $A$ in the range of 150-190 and for $A>220$, these electric quadrupole moments are found to be rather large. The failure of the Shell Model to correctly predict electric quadrupole moments arises from the assumption that the nucleons move in a spherically symmetric potential.

A model that generalizes the Shell Model is the Collective Model, which considers the effect of a non-spherically symmetric potential (leading to substantial deformations for large nuclei and consequently large electric quadrupole moments) and takes into account interactions between nucleons. One of the most striking consequences of the Collective Model is the explanation of low-lying excited states of heavy nuclei. These excitations are of two types:

• Rotational States. A nucleus whose nucleon density distributions are spherically symmetric (zero quadrupole moment) cannot have rotational excitations (this is analogous to the application of the principle of equipartition of energy to monatomic molecules for which there are no degrees of freedom associated with rotation). On the other hand, a nucleus with a non-zero quadrupole moment can have excited levels due to rotation perpendicular to the (rotational) axis of symmetry.

For an even-even nucleus whose ground state has zero spin, these states have energies
$$E_{\mathrm{rot}}=\frac{I(I+1) \hbar^2}{2 \mathcal{I}},$$
where $\mathcal{I}$ is the moment of inertia of the nucleus about an axis through the centre perpendicular to the axis of rotational symmetry as shown in Fig. 4.6, and the integer $I$ is the quantum number of rotational angular momentum.

## 物理代写|粒子物理代写particle physics代考|Magnetic Dipole Moments

$$\mu_N=\frac{e \hbar}{2 m_p},$$

$$\mu_p=2.79 \mu_N=5.58 \mu_N s, \quad\left(s=\frac{1}{2}\right)$$

$$\mu_n=-1.91 \mu_N=-3.82 \mu_N s, \quad\left(s=\frac{1}{2}\right) .$$

$$\mu_z=\left(5.58 s_z+\ell_z\right) \mu_N .$$

$$\mu_z=\frac{(5.58\langle\boldsymbol{s} \cdot \boldsymbol{j}\rangle+\langle\boldsymbol{\ell} \cdot \boldsymbol{j}\rangle)}{\left\langle\boldsymbol{j}^2\right\rangle} j_z \mu_N .$$

## 物理代写|粒子物理代写particle physics代考|The Collective Model

• 旋转状态。核子密度分布为球对称 (零四极矩) 的原子核不能具有旋转激发 (这类似于将能量均分原理 应用于不存在与旋转相关联的自由度的单原子分子)。另一方面，具有非零四极矩的原子核可以由于垂 直于 (旋转) 对称轴的旋转而具有激发能级。
对于基态自旋为零的偶偶核，这些状态具有能量
$$E_{\text {rot }}=\frac{I(I+1) \hbar^2}{2 \mathcal{I}},$$
在哪里 $\mathcal{I}$ 是原子核绕通过中心的轴的转动愢量，该轴垂直于如图 $4.6$ 所示的旋转对称轴，整数 $I$ 是旋转角动 量的量子数。

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