The angular momentum of the planar loop rotating around its perpendicular axis is given by
$$
\vec{L} = M R^2 \vec{\omega},
$$while the current generated by the rotation is $I = Q/T = \omega Q/2\pi$.
then the magnetic dipole moment of the planar loop is given by
$$
\vec{\mu} = I A \vec{\omega} = \frac{\omega Q}{2\pi} \pi R^2\hat{\omega} = \frac{1}{2} Q R^2 \vec{\omega}
$$It follows that
$$
\vec{\mu} = \frac{Q}{2M} \vec{L}
$$and

The magnetic field due to the first dipole on the second is
$$
\vec{B}_{1\, on \, 2} = - \frac{\mu_0}{4 \pi} \frac{\mu_1}{d^3}
$$Then
$$
U = - \vec{\mu}_2 \cdot \vec{B}_{1\, on \, 2} = - \frac{\mu_0}{4 \pi d^3}\mu_1 \mu_2 \cos(\pi -\theta),
$$leading to
$$
U = \frac{\mu_0 \gamma^2}{4 \pi d^3} \vec{L}_1 \cdot \vec{L}_2
$$

The $i$th spin interacts with the $i − 1$ and $i + 1$ spins, with an energy $E_i = −J(\vec{S}_{i−1} +\vec{S}_{i+1}) \cdot \vec{S}_i$ which is analogous to $E_i = −\vec{B}_i \cdot \vec{\mu}_i$. Using ⃗$\mu_i = \gamma S_i$, we find

Given the effective magnetic field

$$
\begin{aligned}
\frac{\mathrm{d} \vec{S}_{i}}{\mathrm{~d} t} & =\vec{\tau}=\vec{\mu}_{i} \times \vec{B}_{i, \text { eff }} =J \vec{S}_{i} \times\left(\vec{S}_{i-1}+\vec{S}_{i+1}\right)
\end{aligned}
$$

None

Find the relationship between $\omega$ and $k$ (known as the dispersion relation, $\omega(k)$) for the spin waves in terms of $J, S$ and $a$.

Hint: express the position of the $i$th spin as $x=a\cdot i$.

We can write the rate of change of the $x$ and $y$ components of $\vec{S}_{i}$, keeping in mind the approximation $S_{i, z} \approx S$ for all $i$.

$$
\begin{aligned}
\frac{\mathrm{d} S_{i, x}}{\mathrm{~d} t} & =J\left[S_{i, y}\left(S_{i-1, z}+S_{i+1, z}\right)\right. \\
& \left.-S_{i, z}\left(S_{i-1, y}+S_{i+1, y}\right)\right] \\
& \approx J S\left(2 S_{i, y}-S_{i-1, y}-S_{i+1, y}\right)
\end{aligned}
$$
and

$$
\begin{aligned}
\frac{\mathrm{d} S_{i, y}}{\mathrm{~d} t} & =J\left[-S_{i, x}\left(S_{i-1, z}+S_{i+1, z}\right)\right. \\
& \left.-S_{i, z}\left(S_{i-1, x}+S_{i+1, x}\right)\right] \\
& \approx-J S\left(2 S_{i, x}-S_{i-1, x}-S_{i+1, x}\right)
\end{aligned}
$$
The structure of these equations along with the traveling wave behavior leads us to the ansatz

$$
\begin{aligned}
S_{i, x} & =\delta S \cos (k x-\omega t) \\
S_{i, y} & =\delta S \sin (k x-\omega t)
\end{aligned}
$$
Where $\delta S$ is the amplitude. This yields

$$
\begin{aligned}
\omega \delta S \sin (k x-\omega t) & =J S \delta S \cdot[2 \sin (k x-\omega t) \\
& -\sin (k x-\omega t-k a) \\
& -\sin (k x-\omega t+k a)]
\end{aligned}
$$
Using $\sin (A)+\sin (B)=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$, we get

$$
\omega(k)=2 J S[1-\cos (k a)]
$$
Note: We can show that the amplitude for $S_{x}$ equals that of $S_{y}$, but substituting predictions with $\delta S_{x}$ and $\delta S_{y}$, which leads to

$$
\begin{aligned}
\omega \delta S_{x} \sin (k x-\omega t) & =J S \delta S_{y} \cdot[2 \sin (k x-\omega t) \\
& -\sin (k x-\omega t-k a) \\
& -\sin (k x-\omega t+k a)] \\
\omega \delta S_{y} \cos (k x-\omega t) & =J S \delta S_{x} \cdot[2 \cos (k x-\omega t) \\
& -\cos (k x-\omega t-k a) \\
& -\cos (k x-\omega t+k a)]
\end{aligned}
$$
which can only be satisfied given $\delta S_{x}=\delta S_{y}$.

None

For small $k$,

$$
\omega(k) \approx 2 J S\left[1-1+\frac{1}{2}(k a)^{2}\right]=J S a^{2} k^{2}
$$
The de Broglie relations are $E=\hbar \omega$ and $p=\hbar k$, plugging these into the expression for $\omega(k)$, we find

$$
E=\hbar \omega=\frac{J S a^{2}}{\hbar} p^{2} \equiv \frac{p^{2}}{2 m_{\mathrm{eff}}}
$$
where

$$
m_{\mathrm{eff}} \equiv \frac{\hbar}{2 J S a^{2}}
$$

In this inelastic scattering, energy and momentum for the entire system (including the chain) is conserved. In particular, the spin wave does not have a momentum along the $y$-axis. Therefore

$$
p_{\text {in }} \cos \theta_{\text {in }}=p_{\text {out }} \cos \theta_{\text {out }}
$$
Combining this with $E_{n}=p_{n}^{2} / 2 m_{n}$ valid for the neutron, we find

$$
E_{\text {out }}=\left(\frac{\cos \theta_{\text {in }}}{\cos \theta_{\text {out }}}\right)^{2} E_{\text {in }}
$$
Using energy conservation, the energy $E_{s}$ of the spin wave is $E_{s}=E_{\text {in }}-E_{\text {out }}$, so we get

$$
E_{s}=\frac{\cos ^{2} \theta_{\mathrm{out}}-\cos ^{2} \theta_{\mathrm{in}}}{\cos ^{2} \theta_{\mathrm{out}}} E_{\mathrm{in}}
$$
Conservation of momentum along the $x$-axis gives

$$
\begin{aligned}
p_{s} & =p_{\text {in }} \sin \theta_{\text {in }}-p_{\text {out }} \sin \theta_{\text {out }} \\
& =\sqrt{2 m_{n} E_{\text {in }}}\left(\sin \theta_{\text {in }}-\frac{\cos \theta_{\text {in }}}{\cos \theta_{\text {out }}} \sin \theta_{\text {out }}\right)
\end{aligned}
$$
Then from $m_{\text {eff }}=p_{s}^{2} / 2 E_{s}$, we find

$$
m_{\text {eff }}=\frac{\cos ^{2} \theta_{\text {out }}\left(\sin \theta_{\text {in }}-\frac{\cos \theta_{\text {in }}}{\cos \theta_{\text {out }}} \sin \theta_{\text {out }}\right)^{2}}{\cos ^{2} \theta_{\text {out }}-\cos ^{2} \theta_{\text {in }}} m_{n}
$$
which after simplifying gives

$$
m_{\mathrm{eff}}=\frac{\sin ^{2}\left(\theta_{\mathrm{in}}-\theta_{\mathrm{out}}\right)}{\cos ^{2} \theta_{\mathrm{out}}-\cos ^{2} \theta_{\mathrm{in}}} m_{n}=\frac{\sin \left(\theta_{\mathrm{in}}-\theta_{\mathrm{out}}\right)}{\sin \left(\theta_{\mathrm{in}}+\theta_{\mathrm{out}}\right)} m_{n}
$$

In a Boltzmann distribution, given the system's temperature $T$, the probability to find a system in a given state with energy $\varepsilon_{i}$ is

$$
p_{i} \propto \exp \left(-\frac{\varepsilon_{i}}{k_{B} T}\right)
$$
In this case, the probability to find a spin up state of energy $\varepsilon_{\uparrow}=-h s_{\uparrow}=-h$ is

$$
p_{\uparrow} \sim \mathrm{e}^{h / k_{B} T}
$$
Therefore,

$$
\frac{p_{\uparrow}}{p_{\downarrow}}=\frac{\mathrm{e}^{h / k_{B} T}}{\mathrm{e}^{-h / k_{B} T}}=\mathrm{e}^{2 h / k_{B} T}
$$
We also note that, normalization requires, $p_{\uparrow}+p_{\downarrow}=1$. Thus, a full solution for $p_{\uparrow}$ and $p_{\downarrow}$ is


\begin{align*}
& p_{\uparrow}=\frac{\mathrm{e}^{h / k_{B} T}}{\mathrm{e}^{h / k_{B} T}+\mathrm{e}^{-h / k_{B} T}}, \tag{1}\\
& p_{\downarrow}=\frac{\mathrm{e}^{-h / k_{B} T}}{\mathrm{e}^{h / k_{B} T}+\mathrm{e}^{-h / k_{B} T}} \tag{2}
\end{align*}

The average polarization of the system $\bar{s}$ can be written as

$$
\begin{aligned}
\bar{s} & =\frac{1}{N} \sum_{i} s_{i} \\
& =\frac{1}{N}\left[N p_{\uparrow} \cdot 1+N p_{\downarrow} \cdot(-1)\right] \\
& =p_{\uparrow}-p_{\downarrow}
\end{aligned}
$$
where $N p_{\uparrow}$ and $N p_{\downarrow}$ are the number of spin vectors pointing up and down, respectively. Substituting for probabilities, we find


\begin{equation*}
\bar{s}=\frac{\mathrm{e}^{h / k_{B} T}-\mathrm{e}^{-h / k_{B} T}}{\mathrm{e}^{h / k_{B} T}+\mathrm{e}^{-h / k_{B} T}}=\tanh \left(\frac{h}{k_{B} T}\right) \tag{3}
\end{equation*}

None

The energy of the system is minimized when all the spins are aligned, so


\begin{equation*}
E_{g}=-\tilde{J} \sum_{i} 1=-\tilde{J}(N-1) \approx-\tilde{J} N \tag{4}
\end{equation*}


where we used $N \gg 1$.

$$
E=-\tilde{J} \sum_{i} s_{i} s_{i+1}=-\tilde{J} \sum_{i} s_{i} \bar{s}=-\tilde{J}_{\mathrm{eff}} \sum_{i} s_{i}
$$
where we define $\tilde{J}_{\text {eff }}=\tilde{J} \bar{s}$. Using $2 \bar{s}$ is double counting the energy.

None

By looking at the earlier equation given in part C4, and comparing it to what we had earlier in C2, we find that the average polarization $\bar{s}$ should satisfy the following transcendental equation: $$ \bar{s}=\tanh \left(\frac{\tilde{J}_{\mathrm{eff}}}{k_{B} T}\right)=\tanh \left(\frac{\tilde{J} \bar{s}}{k_{B} T}\right) $$ With the help of the plots we did earlier in C2, one can see that for $\tilde{J} \ll k_{B} T$, there exists only one simple solution, $\bar{s}=0$. Where as for $\tilde{J} \gg k_{B} T$, there exist two non-trivial solutions. As clarified by the figure below, this crossing behavior occurs when the tangent line for $\tanh \left(\tilde{J} \bar{s} / k_{B} T\right)$ at $\bar{s}=0$ equals the slope of $\bar{s}$: $$ \begin{aligned} \left.\frac{d}{d \bar{s}} \tanh \left(\frac{\tilde{J} \bar{s}}{k_{B} T_{c}}\right)\right|_{\bar{s}=0} & =\left.\frac{d}{d \bar{s}} \bar{s}\right|_{\bar{s}=0} \\ \left.\implies\frac{\tilde{J}}{k_{B} T_{c}} \frac{1}{\cosh ^{2}\left(\tilde{J} \bar{s} / k_{B} T_{c}\right)}\right|_{\bar{s}=0} & =1 \\ \implies\frac{\tilde{J}}{k_{B} T_{c}} & =1, \end{aligned} $$ leading to $T_{c}=\tilde{J} / k_{B}$.

Near the critical temperature, the average polarization is small. Thus, we can approximate the transcendental equation as $$ \bar{s}=\tanh \left(\frac{T_{c}}{T} \bar{s}\right) \approx \frac{T_{c}}{T} \bar{s}-\frac{1}{3}\left(\frac{T_{c}}{T} \bar{s}\right)^{3} $$ We see that $\bar{s}=0$ is still a solution. After rearranging for $\bar{s} \neq 0$, we get $$ \begin{aligned} \bar{s} & = \pm \sqrt{3\left[\left(\frac{T}{T_{c}}\right)^{2}-\left(\frac{T}{T_{c}}\right)^{3}\right]} \\ & = \pm \sqrt{3\left(\frac{T}{T_{c}}\right)^{2}\left(1-\frac{T}{T_{c}}\right)} \\ & \approx \pm \sqrt{3 \frac{T_{c}-T}{T_{c}}} \end{aligned} $$ where we have used $\left(T / T_{c}\right)^{2} \approx 1$.

None

In the absence of magnetic fields, when $T > T_{c}$, there is only one solution at $\bar{s}=0$, which means that the system cannot maintain a net polarization. However, applying a magnetic field leads to a net magnetization, which is a characteristic of paramagnets. On the other hand, when $T < T_{c}$, the system can support a net magnetization even in the absence of magnetic fields, which is the characteristic of ferromagnets.