Scattering theory for nuclear dynamics

Formalism for inelastic scattering

When describing the quantum mechanics of the inelastic scattering process, it is important to keep track of the quantum state of the scattering system (the sample), since it changes during the scattering process (for \(\hbar \omega \neq 0\)). The initial and final sample states are denoted \(|\lambda_{\rm i}\rangle\) and \(|\lambda_{\rm f}\rangle\), respectively. The partial differential cross section for scattering from \(|\lambda_{\rm i},{\bf k}_{\rm i}\rangle\) to \(|\lambda_{\rm f},{\bf k}_{\rm f}\rangle\) is given in analogy with master scattering equation by

\begin{equation} \label{eq:master_scatt_inel} %(S2.13-2.16), he does the derivation \left. \frac{d^2\sigma}{d\Omega dE_{\rm f}} \right|_{\lambda_{\rm i}\rightarrow \lambda_{\rm f}} = \frac{k_{\rm f}}{k_{\rm i}} \left(\frac{m_{\rm n}}{2\pi\hbar^2}\right)^2 \big| \big\langle \lambda_{\rm i} {\bf k}_{\rm i} \big|\hat{V}\big| {\bf k}_{\rm f} \lambda_{\rm f}\big\rangle \big|^2 \delta(E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}+\hbar\omega) \, , \end{equation} where the \(\delta\)-function expresses explicit energy conservation and the normalization factor \(Y^2\) is omitted.

Scattering from initial to final state

We begin by expanding the expression for the nuclear potential on the Small angle neutron scattering page:

\begin{equation} \label{dummy1389712319} \hat{V}=\frac{2\pi \hbar^2}{m_{\rm n}}\sum_j b_j \delta({\bf r}-{\bf R}_j) , \end{equation}

where \({\bf R}_j\) is now the operator for the position of the \(j\)'th nucleus. We use this to expand the matrix element in the inelastic cross section:

\begin{align} \big| \big\langle & \lambda_{\rm i}\psi_{\rm i} \big|\hat{V}\big| \psi_{\rm f} \lambda_{\rm f}\big\rangle \big|^2 \label{dummy348279381263912}\\ &= \biggr( \frac{2\pi \hbar^2}{m_{\rm n}} \biggr)^2 \biggr[ \sum_{j} b_j \left\langle \lambda_{\rm i} \left| \int \psi_{\rm i}^* \delta({\bf r}-{\bf R}_j) \psi_{\rm f} d^3{\bf r} \right| \lambda_{\rm f} \right\rangle \biggr]^2 \nonumber \\ &= \biggr( \frac{2\pi \hbar^2}{m_{\rm n}} \biggr)^2 \sum_{j,j'} b_j b_{j'} \left\langle \lambda_{\rm i} \left| \exp(-i {\bf q} \cdot {\bf R}_j) \right| \lambda_{\rm f} \right\rangle \left\langle \lambda_{\rm f} \left| \exp(i {\bf q} \cdot {\bf R}_{j'}) \right| \lambda_{\rm i} \right\rangle . \nonumber \end{align}

If all nuclei were fixed in position, we would now reach the diffraction cross section by summing over the (in practice unmeasurable) finite states of the lattice, \(|\lambda_{\rm f}\rangle\), since the \(\delta\)-function in \eqref{eq:master_scatt_inel2} would factorize out and vanish by integration. However, we cannot do this simple calculation now, so we need to take a more difficult path. We rewrite the troublesome delta-function in \eqref{eq:master_scatt_inel2}, using \(2 \pi \delta(a) = \int_{-\infty}^{\infty} \exp(iax) dx \) (following Squires 2.3):

\begin{equation} \label{eq:delta_hw} \delta(E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}+\hbar\omega) = \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} \exp\biggr( \frac{i(E_{\lambda_{\rm f}}-E_{\lambda_{\rm i}})t}{\hbar} \biggr) \exp(-i\omega t) dt \, . \end{equation}

Now, we utilize a rather intuitive identity from quantum mechanics, valid when \(|\lambda\rangle\) is an eigenstate of the Hamiltonian \(H\) with eigenvalue \(E_\lambda\):

\begin{equation} \exp\left( \frac{i H t}{\hbar} \right) |\lambda\rangle = \exp\left( \frac{i E_{\lambda} t}{\hbar} \right) |\lambda\rangle . \end{equation}

We can then rewrite the inelastic scattering cross section \eqref{eq:master_scatt_inel2} into

\begin{align} \label{eq:master_scatt_phonon} &\frac{d^2\sigma}{d\Omega dE_{\rm f}} \biggr|_{\lambda_{\rm i}\rightarrow \lambda_{\rm f}} \\ &= \frac{k_{\rm f}}{k_{\rm i}} \sum_{j,j'} \frac{b_j b_{j'}}{2 \pi \hbar} \int_{-\infty}^{\infty} \left\langle \lambda_{\rm i} \left| \exp(-i {\bf q} \cdot {\bf R}_j) \right| \lambda_{\rm f} \right\rangle \left\langle \lambda_{\rm f} \left| \exp(i {\bf q} \cdot {\bf R}_{j'}) \right| \lambda_{\rm i} \right\rangle \nonumber \\ &\quad \times \exp\left(\frac{ iE_{\lambda_{\rm f}}t}{\hbar} \right) \exp\left(-\frac{iE_{\lambda_{\rm i}}t}{\hbar} \right) \exp(-i\omega t) dt \nonumber \\ &= \frac{k_{\rm f}}{k_{\rm i}} \sum_{j,j'} \frac{b_j b_{j'}}{2 \pi \hbar} \int_{-\infty}^{\infty} \left\langle \lambda_{\rm i} \left| \exp(-i {\bf q} \cdot {\bf R}_j) \right| \lambda_{\rm f} \right\rangle \nonumber \\ &\quad \times \left\langle \lambda_{\rm f} \left| \exp\left( \frac{i H t}{\hbar} \right) \exp(i {\bf q} \cdot {\bf R}_{j'}) \exp\left( -\frac{i H t}{\hbar} \right) \right| \lambda_{\rm i} \right\rangle \exp(-i\omega t) dt \nonumber \\ &= \frac{k_{\rm f}}{k_{\rm i}} \sum_{j,j'} \frac{b_j b_{j'}}{2 \pi \hbar} \int_{-\infty}^{\infty} \left\langle \lambda_{\rm i} \left| \exp(-i {\bf q} \cdot {\bf R}_j(0)) \right| \lambda_{\rm f} \right\rangle \nonumber \\ &\quad \times \left\langle \lambda_{\rm f} \left| \exp(i {\bf q} \cdot {\bf R}_{j'}(t)) \right| \lambda_{\rm i} \right\rangle \exp(-i\omega t) dt \nonumber . \end{align}

In the last step, we have employed the time-dependent Heisenberg operators, \({\bf R}(t) = \exp(i H t / \hbar) {\bf R} \exp(-i H t / \hbar)\); in particular, \({\bf R}(0) = {\bf R}\).

The observable nuclear cross section

In an experiment, we observe only the final state of the neutron, not of the sample. Thus, we can find the total cross section by summing over all final states of the system, using the completeness rule:

\begin{equation} \sum_{\rm f} |\lambda_{\rm f}\rangle \langle \lambda_{\rm f}| = 1, \end{equation}

We consider systems in thermal equilibrium and hence the contribution from a thermal average of initial states. We denote the thermal average by the notation

\begin{equation} \langle A \, \rangle \equiv \sum_i p_i \langle \lambda_i|A |\lambda_i\rangle, \end{equation}

where \(p_i\) is the Boltzmann probability for the individual states. After these observations, we can specify the observable cross section for nuclear scattering:

\begin{equation} \label{eq:sum_lambda_f} \frac{d^2\sigma}{d\Omega dE_{\rm f}} = \frac{k_{\rm f}}{k_{\rm i}} \sum_{j,j'} \frac{b_j b_{j'}}{2 \pi \hbar} \times \int_{-\infty}^{\infty} \big\langle \exp(-i {\bf q} \cdot {\bf R}_j(0)) \exp(i {\bf q} \cdot {\bf R}_{j'}(t)) \big\rangle \exp(-i\omega t) dt . \end{equation}

This final equation naturally covers both elastic and inelastic nuclear scattering.