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ucph>Tommy: /* Elastic scattering from one and two nuclei */

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Elastic scattering from one and two nuclei

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We will now go through the principles of neutron scattering from nuclei
in a way that is more strictly quantum mechanical than that of section~\ref{sect:scattering_simple}.
This section does not contain
new results, but may be more satisfactory for readers with
a physics background. Furthermore, the formalism developed here carries
on to the detailed treatment of magnetic scattering and to inelastic scattering in subsequent chapters.
This section is strongly inspired by the treatment of scattering in the famous textbook by
Squires <ref>G.L. Squires, Thermal Neutron Scattering, Cambridge University Press, 1978.</ref>.

== The initial and final states ==

We define the state of the incoming wave as

\begin{equation} %(S2.10)
| \psi_{\rm i} \rangle = \frac{1}{\sqrt{Y}}
\exp(i {\bf k}_{\rm i} \cdot {\bf r}) \, ,
\end{equation}

where \(Y=L^3\) can be identified
as the (large) normalization volume for the state which is assumed
enclosed in a cubic box with a side length \(L\).
The incoming neutron flux is given as \eqref{eq:flux_plane}

\begin{equation} %(S2.12)
\Psi = |\psi_{\rm i}|^2 v
= \frac{1}{Y} \frac{\hbar k_{\rm i}}{m_{\rm n}} .
\end{equation}

In contrast to the spherical outgoing wave from [[Wave description of nuclear scattering]],
we express the final state as a (superposition of) plane wave(s)

\begin{equation}
| \psi_{\rm f} \rangle = \frac{1}{\sqrt{Y}} \exp(i {\bf k}_{\rm f} \cdot {\bf r}) \, .
\end{equation}

We here ignore the spin state of the neutron,
which will be discussed in the later section on magnetic scattering.

== Density of states ==

For the spinless states, we calculate the number density in \({\bf k}\)-space:

\begin{equation} %(S2.7-2.9)
\frac{dn}{dV_k} = \left(\frac{2\pi}{L}\right)^{-3} = \frac{Y}{(2\pi)^3}.
\end{equation}

In order to describe the differential scattering cross sections,
we would like to describe the fraction of the wavefunction
which is emitted into directions of \({\bf k}_{\rm f}\), corresponding to
a solid angle \(d\Omega\). Here, the densities are given by

\begin{equation} %(S2.7-2.9)
\left.\frac{dn}{dV_k}\right|_{d\Omega}
= \frac{dn}{dV_k} \frac{d\Omega}{4\pi}
= \frac{Y}{(2\pi)^3}\frac{d\Omega}{4\pi} .
\end{equation}

We now consider a spherical shell in \({\bf k}\)-space to calculate
the (energy) density of states within the
scattering direction \(d\Omega\), using (\ref{eq:neutron_energy}):

\begin{equation} \label{eq:DOS} %(S2.7-2.9)
\left.\frac{dn}{dE_{\rm f}}\right|_{d\Omega} = \left.\frac{dn}{dV_k}\right|_{d\Omega} \frac{dV_k}{dk_{\rm f}}
\left(\frac{dE_{\rm f}}{dk_{\rm f}}\right)^{-1}
= \frac{Y}{(2 \pi)^3} 4\pi k_{\rm f}^2 \frac{m_{\rm n}}{k_{\rm f} \hbar^2}
= \frac{Y k_{\rm f} m_{\rm n}}{2 \pi^2 \hbar^2} .
\end{equation}

We will need this expression in the further calculations.

== The master equation for scattering ==

We describe the interaction responsible for the scattering by an
operator denoted \(\hat{V}\).
The scattering process itself is described by
the ''Fermi Golden Rule'' <ref>E. Merzbacher, Quantum Mechanics, Wiley, 1998.</ref>.

This gives the rate of change between the neutron in the single
incoming state, \(|\psi_{\rm i}\rangle\)
and a final state, \(|\psi_{\rm f}\rangle\),
where \(|\psi_{\rm f}\rangle\) resides in a continuum of possible states.

\begin{equation} \label{eq:goldenrule} %(S2.1-2.2)
W_{{\rm i} \rightarrow {\rm f}}
= \frac{2\pi}{\hbar} \frac{dn}{dE_{\rm f}}
\big| \big\langle \psi_{\rm i}\big| \hat{V} \big|
\psi_{\rm f}\big\rangle \big|^2 .
\end{equation}

We wish to consider only neutrons scattered into the solid angle \(d\Omega\).
Using (\ref{eq:DOS}) and (\ref{eq:goldenrule}), we reach

\begin{equation} %(S2.1-2.2)
W_{{\rm i} \rightarrow {\rm f},d\Omega}
= \frac{Y k_{\rm f} m_{\rm n}}{(2 \pi)^2 \hbar^3} d\Omega
\big| \big\langle \psi_{\rm i} \big| \hat{V} \big| \psi_{\rm f}\big\rangle \big|^2 .
\end{equation}

\(W_{{\rm i} \rightarrow {\rm f},d\Omega}\)
is the number of neutrons scattered into \(d\Omega\) per second.
We now only need the expression for the incoming flux (\ref{eq:flux_plane})

to reach the result for the differential scattering cross section (\ref{eq:dscs})

\begin{align} \label{eq:master_scatt} %(S2.1-2.2)
\frac{d\sigma}{d\Omega}
&= \frac{1}{\Psi} \frac{W_{{\rm i} \rightarrow {\rm f},d\Omega}}{d\Omega} \\ \nonumber
&= Y^2 \frac{k_{\rm f}}{k_{\rm i}} \left(\frac{m_{\rm n}}{2\pi \hbar^2}\right)^2
\big| \big\langle \psi_{\rm i} \big|\hat{V}\big| \psi_{\rm f}\big\rangle \big|^2 .
\end{align}

In this expression, the normalization volume, \(Y\), will eventually vanish
due to the factor \(1/\sqrt{Y}\) in the states \(|\psi_{\rm i}\rangle\) and \(|\psi_{\rm f}\rangle\),
since the interaction, \(\hat{V}\), is independent of \(Y\).
We will thus from now on neglect the \(Y\) dependence
in the states and in the cross sections.

The factor \(k_{\rm f}/k_{\rm i}\) in (\ref{eq:master_scatt})
is of importance only for inelastic neutron scattering, where it always
appears in the final expressions.
For elastic scattering, \(k_{\rm f} / k_{\rm i} = 1\) and is thus removed from the expression.

== Elastic scattering from one and two nuclei ==

The interaction between the neutron and the nuclei
is expressed by the Fermi pseudopotential

\begin{equation} %(S2.27,2.28,2.32,2.33) Squires do the derivation!
\hat{V}_j({\bf r}) = \frac{2 \pi \hbar^2}{m_{\rm n}} b_j \delta({\bf r}-{\bf r}_j) \, .
\end{equation}

Here, \(b_j\) has the unit of length and is of the order fm.
It is usually denoted the scattering length.
The spatial delta function represents the short range of the
strong nuclear forces and is a sufficient description for the
scattering of thermal neutrons.

It should here be noted that a strongly absorbing nucleus will have
a significant imaginary contribution to the scattering length.
We will, however, not deal with this complication here.
For a single nucleus, we can now calculate the scattering cross
section. We start by calculating
the matrix element

\begin{align} \label{eq:matrixelem} %(S2.3,2.11)
\big\langle \psi_{\rm f} \big|\hat{V}_j \big| \psi_{\rm i}\big\rangle &=
\frac{2\pi \hbar^2}{m_{\rm n}} \,b_j
\int \exp(-i {\bf k}_{\rm f} \cdot {\bf r}) \delta({\bf r}-{\bf r}_j)
\exp(i {\bf k}_{\rm i} \cdot {\bf r}) d^3{\bf r} \\ \nonumber
&= \frac{2\pi\hbar^2}{m_{\rm n}} \,b_j \exp(i {\bf q} \cdot {\bf r}_j) ,
\end{align}

where we have defined the scattering vector, as

\begin{equation} %(S2.24)
{\bf q} = {\bf k}_{\rm i} - {\bf k}_{\rm f} \, .
\end{equation}

Inserting (\ref{eq:matrixelem}) into (\ref{eq:master_scatt}), we reassuringly reach the same
result as found from the semi-classical calculation (\ref{eq:nuclear_diff_cross}):

\begin{equation} %(S2.31)

\frac{d\sigma}{d\Omega} = b_j^2\, .
\end{equation}

For a system of two nuclei, we obtain interference between the scattered waves.
We can write the scattering potential as a sum \(\hat{V}=\hat{V}_j+\hat{V}_{j'}\).
In this case, the matrix element becomes

\begin{equation} \label{eq:scatter_matrix_2}
\big\langle \psi_{\rm f} \big|\hat{V} \big| \psi_{\rm i}\big\rangle
= \frac{1}{Y}\frac{2\pi\hbar^2}{m_{\rm n}}
\big( b_j \exp(i {\bf q} \cdot {\bf r}_j) + b_{j'} \exp(i {\bf q} \cdot {\bf r}_{j'}) \big) .
\end{equation}

Inserting into (\ref{eq:master_scatt}),
we reach the same result (\ref{eq:interference2}) as found by the simpler approach in
section \ref{sect:scattering_simple}.

== The cross section for a system of particles ==
In analogy with (\ref{eq:scatter_matrix_2}), the scattering potential
for a system of particles can be considered as the sum of the single nuclear potentials,
\begin{equation} \label{eq:nuclear_potential}
\hat{V}=\frac{2\pi \hbar^2}{m_{\rm n}}\sum_j b_j \delta({\bf r}-{\bf r}_{j}) .
\end{equation}
By insertion into (\ref{eq:master_scatt}),
\begin{equation}
\frac{d\sigma}{d\Omega}
= Y^2 \frac{k_{\rm f}}{k_{\rm i}} \left(\frac{m_{\rm n}}{2\pi\hbar^2}\right)^2
\big| \big\langle \psi_{\rm i} \big|\hat{V}\big| \psi_{\rm f}\big\rangle \big|^2 .
\end{equation}
and performing the same calculations as above, we arrive directly at (\ref{eq:diffraction}),
\begin{equation} %Squires never writes this - strange!
\boxed{
\left.\frac{d\sigma}{d\Omega}\right|_{\rm coh}
= \biggr|\sum_j b_j \exp(i {\bf q} \cdot {\bf r}_j)\biggr|^2
} \, .
\end{equation}

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ucph>Tommy: /* Elastic scattering from one and two nuclei */