Quantum mechanics of nuclear diffraction

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 .

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 .

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}