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  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  %% Indranil's LaTeX macros
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  \def\ek{\epsilon_{k}}
  \def\xk{\xi_k} 
  \def\la{{\langle\, }}
  \def\ra{{\,\rangle }}
  \def\p{\,|\,}
  \def\infintd3k{ \int_{-\infty}^\infty d^3r\,}
  \def\pm{\prime}
  \def\vm{\vec{m}}
  \def\vs{\vec{S}}
  \def \vr{\vec{r}} 
   \def\vnab{\vec{\nabla}}
  % Equation environments
  \def\bea{\begin{eqnarray}}
  \def\eea{\end{eqnarray}}
  \def\ben{\begin{equation}}
  \def\een{\end{equation}}
  \def\benu{\begin{enumerate}}
  \def\enu{\end{enumerate}}

  \def\bdq{b^{\dagger}_q}
  \def\bq{b_q} 
  \def\e{\epsilon}
  \def\p{\partial}
  \def\vk{\vec{k}}
  \def\vRa{\vec{R}_{\alpha}} 
  \def\vq{\vec{q}}
  \def\s{\sigma}
  \def\hs{\hat{S}}
  \def\ua{\uparrow}
  \def\da{\downarrow} 
  \def\a{\alpha}

   \begin{document} 

   \begin{center}
   Solution to Problem Set 3
   \end{center}


First we will consider the scattering of a neutron by a single spin. If the 
spin is due to an electron, the magnetic dipole moment is $\vec{m} = 
2 \mu_B \vec{S}$, where $\vec{S}$ is the spin operator and can be represented
by Pauli matrices as $\vec{S} = \vec{\sigma}/2$. The vector potential due to 
a magnetic dipole moment $\vm$ is,
\ben
\vec{A}(\vr) = - \frac{\mu_0}{4 \pi} \left [ \vm \times \vec{\nabla} 
\left ( \frac{1}{r} \right ) \right ],
\een
from which the magnetic field can be calculated as,
\bea
\vec{B}(\vr) &=& \vnab \times \vec{A}(\vr)  \nonumber \\
       &=& \frac{\mu_0}{4 \pi} \left [ \frac{3 \left( \hat{r} \cdot \vm 
\right)\hat{r} - \vm}{r^3} + 4 \pi \vm \delta^3(\vr) \right ]. 
\eea
The magnetic moment of the neutron is $\vm_N = \gamma_N \mu_N \vs_N$, and the 
magnetic interaction between the two is given by the Hamiltonian,
\bea
\mathcal{H} &=& -\vm_N \cdot \vec{B} \nonumber \\
    &=& \lambda \vs_N \cdot \left[ \vnab \times \left \{ \vs_e \times 
\vnab \left(\frac{1}{r} \right) \right\} \right] \nonumber \\
    &=& \lambda \left(S_N \right)_i \e_{imk} \p_m \e_{kjl} \left( S_e 
\right)_j \p_l \left( \frac{1}{r} \right),
\eea
where $\lambda = 2\mu_B \gamma_N \mu_N \mu_0 /4 \pi$.

To calculate the scattering cross-section for the neutron with incident 
wave-vector $\vk$ and final wave-vector $\vk^{\pm}$, we have to evaluate 
the matrix element $\la k | \e_{imk} \e_{kjl} \p_m \p_l \left( \frac{1}{r}
\right) | k^{\pm} \ra$. This is the Fourier transform of the potential that 
the neutron experiences. The calculation is done using the trick 
(thanks to Chaikin \& Lubensky),
\ben
\frac{1}{r} = \int \frac{d^3q}{(2 \pi)^3} \frac{4 \pi}{q^2} e^{-i \vq \cdot 
\vr}.
\een
Then,
\bea
\la k | \e_{imk} \e_{kjl} \p_m \p_l \left( \frac{1}{r} \right) | k^{\pm} \ra 
&=& \e_{imk} \e_{kjl} \int d^3r e^{i \vq \cdot \vr} \p_m \p_l 
\left( \frac{1}{r} \right) \nonumber \\
&=& \e_{imk} \e_{kjl} \int d^3r e^{i \vq \cdot \vr} \p_m \p_l \int 
\frac{d^3 k}{(2 \pi)^3} \frac{4 \pi}{k^2} e^{-i \vk \cdot \vr} \nonumber \\
&=& - 4 \pi \e_{imk} \e_{kjl} \int \frac{d^3 k}{(2 \pi)^3} \frac{k_m k_l}
{k^2} \int d^3r e^{i(\vq - \vk) \cdot \vr} \nonumber \\
&=& 4 \pi \left( q_i q_j - q^2 \delta_{ij} \right) \frac{1}{q^2},
\eea
where $\vq = \vk - \vk^{\pm}$. 

The matrix element for the transition of the neutron from an initial
state of wave-vector $\vk$ and spin state $|s \ra$ and the elctron in the
initial spin state $| \s \ra$  to corresponding final states $\vk^{\pm}$ and
$|s^{\pm} \ra$ for the neutron and $|\s^{\pm} \ra$ for the electron is,
\ben
\la k s; \s | \mathcal{H} | k^{\pm} s^{\pm}; \s^{\pm} \ra
= 4 \pi \lambda \la s| (\hs_N)_i | s^{\pm} \ra \la \s | (\hs_e)_j | 
\s^{\pm} \ra (q_i q_j - q^2 \delta_{ij} ) \frac{1}{q^2}.
\een
In Born approximation the differential scattering cross-section for the 
neutron is given by,
\ben 
\left( \frac{d \s}{d \Omega} \right)_{\vk,s \rightarrow \vk^{\pm},s^{\pm}}
= \left( \frac{m_N}{2 \pi \hbar^2} \right)^2 \sum_{\s, \s^{\pm}} P(\s)
\left| \la k s; \s | \mathcal{H} | k^{\pm} s^{\pm}; \s^{\pm} \ra \right|^2.
\een
Here $P(\s)$ is the probability of the electron to be in the spin state 
$|\s \ra$. The summation over the electron spins can be done easily and we 
have,
\ben
\sum_{\s, \s^{\pm}} P(\s) \la \s | (\hs_e)_j | \s^{\pm} \ra \la \s^{\pm} |
(\hs_e)_k | \s \ra = (\frac{1}{4}) \delta_{jk},
\een
if the electron has an equal probability to be in up and down states. Then,
combining equations (6), (7) and (8) we have,
\ben
\left( \frac{d \s}{d \Omega} \right)_{\vk,s \rightarrow \vk^{\pm},s^{\pm}}
= \left( \frac{m_N}{2 \pi \hbar^2} \right)^2 (4 \pi \lambda )^2 (\frac{1}{4}) 
\left(q^2 \delta_{ij} - q_i q_j \right) \frac{1}{q^2} \la s | (\hs_N)_i |
s^{\pm} \ra \la s^{\pm} | (\hs_N)_j | s \ra.
\een
If the neutron is initially in the state $|\ua \ra $ and finally in 
$|\da \ra $, then the scattering cross-section is,
\bea
\left( \frac{d \s}{d \Omega} \right)_{\vk,\ua \rightarrow \vk^{\pm},\da}
&=& \left( \frac{m_N}{2 \pi \hbar^2} \right)^2 (4 \pi \lambda )^2 \frac{1}
{q^2} \left\{ \la \ua | (\hs_N)_x | \da \ra \la \da | (\hs_N)_x | \ua \ra 
(q^2 -q_x^2) \right. \nonumber \\ 
&+& \la \ua | (\hs_N)_x | \da \ra \la \da | (\hs_N)_y | \ua \ra 
(- q_x q_y) 
+ \la \ua | (\hs_N)_y | \da \ra \la \da | (\hs_N)_x | \ua \ra (- q_y q_x)
\nonumber \\
&+& \left. \la \ua | (\hs_N)_y | \da \ra \la \da | (\hs_N)_y | \ua \ra 
(q^2- q_y^2) \right\} \nonumber \\
&=& \left( \frac{m_N}{2 \pi \hbar^2} \right)^2 (4 \pi \lambda )^2 (\frac{1}
{4})^2 \left(1+ \cos^2\theta \right).
\eea
Similarly, we can show that,
\ben
\left( \frac{d \s}{d \Omega} \right)_{\vk,\ua \rightarrow \vk^{\pm},\ua}
= \left( \frac{m_N}{2 \pi \hbar^2} \right)^2 (4 \pi \lambda )^2 
(\frac{1}{4})^2 \sin^2\theta.
\een

Now we consider, instead of a single spin, a regular array of spins in a 
lattice. Let the position of the spins be $\vRa$. The magnetic moment per
unit volume of the sample is defined as,
\ben
\vm(\vr^{\pm}) = \sum_{\alpha} \vm_{\alpha} \delta(\vr^{\pm} - \vRa),
\een
where $\vm_{\a}$ is the magnetic moment of the spin at $\vRa$. The magnetic 
field created by the array of spins is,
\bea
\vec{B}(\vr) &=& - \left( \frac{\mu_0}{4 \pi} \right) \vnab \times 
\left[ \vm_{\a} \times \vnab \left( \frac
{1}{\left| \vr - \vRa \right|} \right) \right] \nonumber \\
&=& - \left( \frac{\mu_0}{4 \pi} \right) \int d^3 r^{\pm} \ \vnab 
\times \left[ \vm(\vr^{\pm}) \times 
\vnab \left( \frac {1}{\left| \vr - \vr^{\pm}  \right|} \right) \right],
\eea
where we have used equation (12) in the last step. The magnetic interaction 
between the neutron and the spins is given by,
\bea
\mathcal{H} &=& \lambda \vs_N \cdot \left[ \int d^3 r^{\pm} \ \vnab
\times  \vs(\vr^{\pm}) \times 
\vnab \left( \frac {1}{\left| \vr - \vr^{\pm}  \right|} \right) \right]
\nonumber \\
&=& \lambda \int d^3 r^{\pm} (\hs_N)_i  \e_{imk} \p_m \e_{kjl} 
\hs_j (\vr^{\pm}) \ \p_l \left( \frac {1}{\left| \vr - \vr^{\pm}  \right|} 
\right),
\eea
where $\vs(\vr)$ is the total spin per unit volume of the array. The Fourier 
transform of the potential is similar to what we did before, and we have,
\ben
\la k | \e_{imk} \e_{kjl} \p_m \p_l \left( \frac{1}{\left| \vr - \vr^{\pm}  
\right|} \right) | k^{\pm} \ra = 4 \pi \left( q_i q_j - q^2 \delta_{ij} 
\right) \frac{1}{q^2}e^{i \vq \cdot \vr^{\pm}}.
\een
The relevant matrix element is,
\ben
\la k s; S | \mathcal{H} | k^{\pm} s^{\pm}; S^{\pm} \ra
= 4 \pi \lambda \la s| (\hs_N)_i | s^{\pm} \ra \la S | \hs_j(\vq) | 
S^{\pm} \ra (q_i q_j - q^2 \delta_{ij} ) \frac{1}{q^2}.
\een
Here $|S \ra$ and $|S^{\pm} \ra$ are the initial and final spin states of the
array. $\vs(\vq)$ is the Fourier transform of the spin of the array and is 
defined as, 
\[
\vs(\vq) = \int d^3 r \vs(\vr) e^{i \vr \cdot \vq}.
\]

In the previous case, where there was a single spin, we considered elastic 
scattering of the neutron. This is usually the case, unless the spin has
other degrees of freedom that the neutron can excite. However, now the 
neutron can exchange energy with the array. This is what happens in inelastic
neutron scattering experiments. The scattering cross-section per unit energy,
called the partial scattering cross-section is defined as,
\ben
\left( \frac{d^2 \s}{ d\Omega d E^{\pm}}\right)_{\vk,s,E \rightarrow 
\vk^{\pm},s^{\pm},E^{\pm}} = \frac{k}{k^{\pm}} 
\left( \frac{m_N}{2 \pi \hbar^2} \right)^2
\sum_{S,S^{\pm}} P(S) \left| \la k s; S| \mathcal{H} | k^{\pm} s^{\pm}; 
S^{\pm} \ra \right|^2 \delta ( E_S -E_{S^{\pm}} + \hbar \omega),
\een
where $E_S$ is the energy of the array in the spin state $|S \ra$ and $\hbar
\omega = E - E^{\pm}$, is the energy transferred from the neutron to the 
array. The delta function, which ensures total energy conservation during 
the scattering, can be represented as,
\ben
 \delta ( E_S -E_{S^{\pm}} + \hbar \omega) = \int \frac{dt}{2 \pi \hbar}
e^{(i/\hbar)(E_S -E_{S^{\pm}} + \hbar \omega)t}.
\een
The summation over the spin of the array gives,
\bea
\sum_{S,S^{\pm}} P(S) \la S |\hs_j(\vq) | S^{\pm} \ra \la S^{\pm}| \hs_l(-\vq)
| S \ra e^{(i/\hbar) (E_S -E_{S^{\pm}})t} &=&
\sum_{S} P(S) \la S |\hs_j(\vq,t) \hs_l(-\vq,0) |S \ra \nonumber \\
&=& \chi_{jl}(\vq,t),
\eea
where $\chi_{jl}(\vq,t)$ is the dynamic spin-spin correlation function. The 
time integral from (18) Fourier transforms the correlation function, 
\ben
\int dt \chi_{jl}(\vq,t)e^{i \omega t} = \chi_{jl}(\vq,\omega).
\een
Using equations (16), (17), (19) and (20) we can write a general expression
for the partial differential scattering cross-section as,
\bea
\left( \frac{d^2 \s}{ d\Omega d E^{\pm}}\right)_{\vk,s,E \rightarrow 
\vk^{\pm},s^{\pm},E^{\pm}} &=& \frac{k}{k^{\pm}}
\left( \frac{m_N}{2 \pi \hbar^2} \right)^2 \frac{(4 \pi \lambda)^2}{2 \pi 
\hbar} \frac{1}{q^4} \left( q_i q_j - q^2 \delta_{i,j} \right) 
\left( q_k q_l - q^2 \delta_{k,l} \right) \nonumber \\
&& \times \la s| (\hs_N)_i | s^{\pm} \ra \la s^{\pm} | (\hs_N)_k |s \ra 
\chi_{jl}(\vq,\omega).
\eea

Suppose the neutron is initially in the spin up state, and the detector can
detect only down spins. Then, $|s \ra = | \ua \ra $, and 
$|s^{\pm} \ra = | \da \ra$.In this case it is easy to see that in equation 
(21) indices $i,k \neq z $. Doing a summation over the appropriate indices
in equation (21), and after a bit of algebra we get the result,
\bea
\left( \frac{d^2 \s}{ d\Omega d E^{\pm}}\right)_{\vk,\ua,E \rightarrow 
\vk^{\pm},\da,E^{\pm}}  &=& \frac{k}{k^{\pm}}
\left( \frac{m_N}{2 \pi \hbar^2} \right)^2 \frac{(4 \pi \lambda)^2}{2 \pi 
\hbar} (\frac{1}{4}) \frac{1}{q^4} \left( q^4 \chi_{-+}(\vq,\omega) +
q_-q_+q_jq_l  \chi_{jl}(\vq,\omega) \right. \nonumber \\
&& \left. - q^2q_+q_l \chi_{-l}(\vq,\omega)
-q^2q_-q_j \chi_{j+}(\vq,\omega) \right),
\eea
where, $q_+ = q_x + i q_y$ and $q_- = q_x - i q_y$.

If the neutron is in spin up state both initially and finally after the
scattering, then $|s \ra = |s^{\pm} \ra = | \ua \ra$. In this case, we have
$i=k= \hat{z}$ in equation (21).

If both the initial and final neutron beam are unpolarized, then we have to
do a sum over the neutron spins. Since,
\[
\sum_{s, s^{\pm}} P(s) \la s| (\hs_N)_i | s^{\pm} \ra \la s^{\pm} | 
(\hs_N)_k |s \ra = (\frac{1}{4}) \delta_{ik},
\]
now equation (21) can be simplified as,
\ben
\left( \frac{d^2 \s}{ d\Omega d E^{\pm}}\right)_{\vk,E \rightarrow 
\vk^{\pm},E^{\pm}} = \frac{k}{k^{\pm}}\left( \frac{m_N}{2 \pi \hbar^2} 
\right)^2 \frac{(4 \pi \lambda)^2}{2 \pi \hbar} (\frac{1}{4}) \frac{1}{q^2}
\left(q^2 \delta_{jl} - q_j q_l \right) \chi_{jl}(\vq,\omega).
\een

  

Suppose the spins are ordered in the $\hat{z}$ direction, with an ordering 
wavevector $\vec{Q}$ (for ferromagnetic ordering $\vec{Q}= 0$, while $\vec{Q}$
is finite for antiferromagnetic ordering). Then $\chi_{zz}(\vec{Q})$ is 
peaked and the corresponding elastic scattering cross-section is large.

\end{document}