Question: The deuteron is a bound state of a proton and a neutron of total angular momentum $J=1$. It is known to be principally an $S(L=0)$ state with a small admixture of a $D(L=2)$ state.

Calculate the magnetic moment of the pure $D$ state $n-p$ system with $J=1$. Assume that the $n$ and $p$ spins are to be coupled to make the total spin $S$ which is then coupled to the orbital angular momentum $L$ to give the total angular momentum $J$. Express your result in nuclear magnetons. The proton and neutron magnetic moments are $2.79$ and $-1.91$ nuclear magnetons respectively.


Answer: The parities of the $S$ and $D$ states are positive, while the parity of the $P$ state is negative. Because of the conservation of parity in strong interaction, a quantum state that is initially an $S$ state cannot have a $P$ state component at any later moment.

The possible spin values for a system composed of a proton and a neutron are 1 and 0 . We are given $\mathbf{J}=\mathbf{L}+\mathbf{S}$ and $J=1$. If $S=0, L=1$, the system would be in a $P$ state, which must be excluded as we have seen in (a). The allowed values are then $S=1, L=2,1,0$. Therefore a $G$ state $(L=4)$ cannot contribute.

The total spin is $\mathbf{S}=\mathbf{s}_{p}+\mathbf{s}_{n}$. For a pure $D$ state with $J=1$, the orbital angular momentum (relative to the center of mass of the $n$ and $p$ ) is $L=2$ and the total spin must be $S=1$. The total magnetic moment arises from the coupling of the magnetic moment of the total spin, $\mu$, with that of the orbital angular momentum, $\boldsymbol{\mu}_{L}$, where $\boldsymbol{\mu}=\boldsymbol{\mu}_{p}+\boldsymbol{\mu}_{n}, \boldsymbol{\mu}_{p}, \boldsymbol{\mu}_{n}$ being the spin magnetic moments of $p$ and $n$ respectively.

The average value of the component of $\mu$ in the direction of the total spin $\mathbf{S}$ is

$$
\mu_{s}=\frac{\left(g_{p} \mu_{N} \mathbf{s}_{p}+g_{n} \mu_{N} \mathbf{s}_{n}\right) \cdot \mathbf{S}}{\mathbf{S}^{2}} \mathbf{S}=\frac{1}{2}\left(g_{p}+g_{n}\right) \mu_{N} \mathbf{S}
$$

where

$$
\mu_{N}=\frac{e \hbar}{2 m_{p} c}, \quad g_{p}=5.58, \quad g_{n}=-3.82,
$$

as $\mathbf{s}_{p}=\mathbf{s}_{n}=\frac{1}{2} \mathbf{S}$.

The motion of the proton relative to the center of mass gives rise to a magnetic moment, while the motion of the neutron does not as it is uncharged. Thus

$$
\mu_{L}=\mu_{N} \mathbf{L}_{p},
$$

where $\mathbf{L}_{p}$ is the angular momentum of the proton relative to the center of mass. As $\mathbf{L}_{p}+\mathbf{L}_{n}=\mathbf{L}$ and we may assume $\mathbf{L}_{p}=\mathbf{L}_{n}$, we have $\mathbf{L}_{p}=\mathbf{L} / 2$ (the center of mass is at the mid-point of the connecting line, taking $m_{p} \approx m_{n}$ ). Consequently, $\mu_{L}=\mu_{N} \mathbf{L} / 2$.

The total coupled magnetic moment along the direction of $\mathbf{J}$ is then

$$
\mu_{T}=\frac{\left[\frac{1}{2} \mu_{N} \mathbf{L} \cdot \mathbf{J}+\frac{1}{2}\left(g_{p}+g_{n}\right) \mu_{N} \mathbf{S} \cdot \mathbf{J}\right] \mathbf{J}}{J(J+1)} .
$$

Since $\mathbf{J}=\mathbf{L}+\mathbf{S}, \mathbf{S} \cdot \mathbf{L}=\frac{1}{2}\left(J^{2}-L^{2}-S^{2}\right)$. With $J=1, L=2, S=1$ and so $J^{2}=2, L^{2}=6, S^{2}=2$, we have $\mathbf{S} \cdot \mathbf{L}=-3$ and thus $\mathbf{L} \cdot \mathbf{J}=3$, $\mathbf{S} \cdot \mathbf{J}=-1$. Hence 

$$
\begin{aligned}
\boldsymbol{\mu}_{T} &=\left[\frac{1}{2} \mu_{N} \cdot 3+\frac{1}{2}\left(g_{p}+g_{n}\right) \mu_{N}(-1)\right] \mathbf{J} / 2 \\
&=\left[1.5-\frac{1}{2}\left(g_{p}+g_{n}\right)\right] \frac{1}{2} \mu_{N} \mathbf{J}=0.31 \mu_{N} \mathbf{J} .
\end{aligned}
$$

Taking the direction of $\mathbf{J}$ as the $z$-axis and letting $J_{z}$ take the maximum value $J_{z}=1$, we have $\mu_{T}=0.31 \mu_{N}$.