# Quantum treatment of inelastic neutron scattering

We here present the quantum mechanical derivations of the expression for the inelastic nuclear and magnetic scattering.

# *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

$$\label{eq:master_scatt_inel} %(S2.13-2.16), he does the derivation \left. \frac{d^2\sigma}{d\Omega dE_{\rm f}} \right|_{\lambda_{\rm i}\rightarrow \lambda_{\rm f}} = \frac{k_{\rm f}}{k_{\rm i}} \left(\frac{m_{\rm n}}{2\pi\hbar^2}\right)^2 \big| \big\langle \lambda_{\rm i} {\bf k}_{\rm i} \big|\hat{V}\big| {\bf k}_{\rm f} \lambda_{\rm f}\big\rangle \big|^2 \delta(E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}+\hbar\omega) \, ,$$ 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:

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

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[1] 2.3):

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

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

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

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:

$$\sum_{\rm f} |\lambda_{\rm f}\rangle \langle \lambda_{\rm f}| = 1,$$

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

$$\langle A \, \rangle \equiv \sum_i p_i \langle \lambda_i|A |\lambda_i\rangle,$$

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

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

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

1. G.L. Squires. Thermal Neutron Scattering. Cambridge University Press, 1978.

# *Scattering theory for magnetic dynamics

We now develop the formalism for inelastic magnetic neutron scattering.

### The magnetic master equation for inelastic scattering

First, we lift the requirement from Quantum mechanics of magnetic diffraction that the state of the sample is unchanged in the scattering process. We insert the scattering potential (\ref{eq:magnetic_potential}) in the master equation for inelastic neutron scattering (\ref{eq:master_scatt_inel}).

We note that we do not measure the final state of the sample. Furthermore, we assume that the sample is initially in thermal equilibrium. Therefore, we perform a thermal average over the initial states of the sample, $$|\lambda_{\rm i} \rangle$$, and sum over those final states, $$|\lambda_{\rm f}\rangle$$, which are consistent with the observed momentum transfer, $${\bf q}$$ and energy transfer, $$\hbar \omega$$. This treatment is similar to the derivation of the inelastic nuclear cross section, Scattering theory for nuclear dynamics.

By substitution, the resulting equation for the scattering cross section becomes \begin{align} \label{eq:cross_spinonly} \left. \frac{d^2 \sigma}{d\Omega dE_{\rm f}} \right|_{\sigma_{\rm i} \rightarrow \sigma_{\rm f} } &= \frac{k_{\rm i}}{k_{\rm f}} \left( \frac{\mu_0}{4\pi} \right)^2 \left( \frac{m_{\rm N}}{2\pi\hbar^2} \right)^2 \left( g \mu_{\rm B} \gamma \mu_{\rm N} \right)^2 \sum_{\lambda_{\rm i},\lambda_{\rm f}} p_{\lambda_{\rm i}} \\ &\quad\times \biggr| \biggr\langle {\bf k}_{\rm f} \lambda_{\rm f} \sigma_{\rm f} \biggr| \sum_j \hat{\bf \sigma} \cdot \nabla\times\left(\frac{{\bf s}_j \times({\bf r}-{\bf r}_j)}{|{\bf r}-{\bf r}_j|^3}\right) \biggr| {\bf k}_{\rm i} \lambda_{\rm i} \sigma_{\rm i} \biggr\rangle \biggr|^2 \nonumber \\ &\quad\times \delta\left( \hbar\omega+E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}} \right) . \nonumber \end{align}

### Matrix element for unpolarized neutrons

We start with the spin part of the matrix element (\ref{eq:cross_spinonly}), transformed using (\ref{eq:magn_matrix}) to reach $$\label{eq:magn_matrix_spin} \sum_{\sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \left| \left\langle \sigma_{\rm f} \lambda_{\rm f} \left| \hat{\bf \sigma} \cdot {\bf s}_{\perp} \right| \sigma_{\rm i} \lambda_{\rm i}\right\rangle \right|^2 .$$ Now, the dot product will contain terms of the type $$\sigma^x s_{\perp}^x$$, where the first factor depends only on the neutron spin coordinate, $$\sigma$$, and the second only on the sample coordinate, $$\lambda$$. The initial neutron state cannot be correlated with the initial state of the sample. Hence, we can factorize the two inner products: \begin{align} \label{eq:magn_matrix_spin2} &\sum_{\sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \left| \left\langle \sigma_{\rm f} \lambda_{\rm f} \left| \hat{\bf \sigma} \cdot {\bf s}_{\perp} \right| \sigma_{\rm i} \lambda_{\rm i}\right\rangle \right|^2 \\ &\quad= \sum_{\sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \biggr| \sum_\alpha \left\langle \sigma_{\rm f} \left| \sigma^\alpha \right| \sigma_{\rm i} \right\rangle \left\langle \lambda_{\rm f} \left| {\bf s}_{\perp}^\alpha \right| \lambda_{\rm i}\right\rangle \biggr|^2 \nonumber \\ &\quad= \sum_{\alpha, \beta, \sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \langle \sigma_{\rm i} | \sigma^\beta | \sigma_{\rm f} \rangle \langle \sigma_{\rm f} | \sigma^\alpha | \sigma_{\rm i} \rangle \langle \lambda_{\rm i} | {\bf s}_{\perp}^\beta | \lambda_{\rm f}\rangle \langle \lambda_{\rm f} | {\bf s}_{\perp}^\alpha | \lambda_{\rm i}\rangle \nonumber \\ &\quad= \sum_{\alpha, \beta, \sigma_{\rm i}} p_{\sigma_{\rm i}} \langle \sigma_{\rm i} | \sigma^\beta \sigma^\alpha | \sigma_{\rm i} \rangle \langle \lambda_{\rm i} | {\bf s}_{\perp}^\beta | \lambda_{\rm f}\rangle \langle \lambda_{\rm f} | {\bf s}_{\perp}^\alpha | \lambda_{\rm i}\rangle , \nonumber \end{align} where we in the last step have used the completeness relation in the form, $$\sum_{\sigma_{\rm f}} |\sigma_{\rm f}\rangle \langle \sigma_{\rm f} | = 1$$. For unpolarized neutrons, $$\alpha = \beta$$ leads to $$\sum_{\sigma_{\rm i}} p_{\sigma_{\rm i}} \langle \sigma_{\rm i} | \sigma^\alpha \sigma^\alpha | \sigma_{\rm i} \rangle = 1.$$

Likewise, if $$\alpha \neq \beta$$, we have that $$\sum_{\sigma_{\rm i}} p_{\sigma_{\rm i}} \langle \sigma_{\rm i} | \sigma^\alpha \sigma^\beta | \sigma_{\rm i} \rangle = 0$$. Using this to perform the sum over $$\sigma_{\rm i}$$, we obtain $$\label{eq:magn_matrix_spin3} \sum_{\sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \left| \left\langle \sigma_{\rm f} \lambda_{\rm f} \left| {\bf \sigma} \cdot {\bf s}_{\perp} \right| \sigma_{\rm i} \lambda_{\rm i}\right\rangle \right|^2 = \sum_\alpha \left\langle \lambda_{\rm i}| s_\perp^\alpha | \lambda_{\rm f}\right\rangle \left\langle \lambda_{\rm f}| s_\perp^\alpha | \lambda_{\rm i}\right\rangle .$$ Summing over the final states, $$|\lambda_{\rm f} \rangle$$, we again use the completeness relation to obtain $$\sum_{\sigma_{\rm i}, \sigma_{\rm f}, \lambda_{\rm f}} p_{\sigma_{\rm i}} \left| \left\langle \sigma_{\rm f} \lambda_{\rm f} \left| {\bf \sigma} \cdot {\bf s}_{\perp} \right| \sigma_{\rm i} \lambda_{\rm i}\right\rangle \right|^2 = \left\langle \lambda_{\rm i} \left| {\bf s}_{\perp} \cdot {\bf s}_{\perp} \right| \lambda_{\rm i}\right\rangle.$$

### The master equation for magnetic scattering

Collecting the prefactors as in the previous section and all other equations from above, we end up with {\em the master equation} for the partial differential magnetic scattering cross section for unpolarized neutrons [1]:

\begin{align} \frac{d^2 \sigma}{d\Omega dE_{\rm f}} &= \left(\gamma r_0 \right)^2 \frac{k_{\rm f}}{k_{\rm i}} \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) \nonumber \\ &\times \sum_{\lambda_{\rm i} \lambda_{\rm f}} p_{\lambda_{\rm i}} \left\langle \lambda_{\rm i}|Q_\alpha|\lambda_{\rm f}\right\rangle \left\langle \lambda_{\rm f}|Q_\beta|\lambda_{\rm i}\right\rangle \delta\left( \hbar\omega + E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}\right) , \end{align}

Performing all replacements above, the final cross section for magnetic neutron scattering reads

\begin{align} \label{eq:magnetic_master_final} \frac{d^2\sigma}{d\Omega dE_{\rm f}} &= \left(\gamma r_0 \right)^2 \frac{k_{\rm f}}{k_{\rm i}} \left[ \frac{g}{2} F(q)\right]^2 \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) \\ &\quad \times \sum_{\lambda_{\rm i} \lambda_{\rm f}} p_{\lambda_{\rm i}} \sum_{j,j'} \big\langle \lambda_{\rm i}| \exp(-i {\bf q} \cdot {\bf r}_{j}) {\bf s}_{j}^\alpha | \lambda_{\rm f}\big\rangle \big\langle \lambda_{\rm f}\big| \exp(i {\bf q} \cdot {\bf r}_{j'}) {\bf s}_{j'}^\beta \big| \lambda_{\rm i}\big\rangle \nonumber \\ &\quad \times \delta\left( \hbar\omega + E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}\right) . \nonumber \end{align}

### Inelastic magnetic neutron scattering from a lattice

As a final part of this derivation, we will now specialize the master equation for magnetic scattering on a lattice, starting from (\ref{eq:magnetic_master_final}):

The next step closely follows the treatment of phonon scattering. First, we rewrite the energy-conserving $$\delta$$-function in terms of a time integral. To repeat (\ref{eq:delta_hw}): $$\label{eq:delta_transform} \delta(\hbar\omega+E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}) = \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} dt \exp\left(-\frac{i(\hbar\omega+E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}})t}{\hbar} \right) .$$ Subsequently, we transform the stationary operators, here denoted $$\hat{A}$$, to time-dependent Heisenberg operators, $$\hat{A}(t)$$, through \begin{align} &\biggr\langle \lambda_{\rm i} \biggr| \exp\biggr( \frac{i E_{\lambda_{\rm i}} t}{\hbar} \biggr) \hat{A} \exp\biggr( -\frac{i E_{\lambda_{\rm f}} t}{\hbar} \biggr) \biggr| \lambda_{\rm f} \biggr\rangle \\ &\quad= \biggr\langle \lambda_{\rm i} \biggr| \exp\biggr( \frac{i \hat{H}_\lambda t}{\hbar} \biggr) \hat{A} \exp\biggr( -\frac{i \hat{H}_\lambda t}{\hbar} \biggr) \biggr| \lambda_{\rm f} \biggr\rangle \nonumber \\ &\quad= \langle \lambda_{\rm i} | \hat{A}(t) | \lambda_{\rm f} \rangle . \nonumber \end{align} Making this transformation and performing the completeness sum over the final states, ($$\sum_{\lambda_{\rm f}}|\lambda_{\rm f}\rangle\langle\lambda_{\rm f}| = 1$$), we reach the equation for the magnetic scattering cross section in the Heisenberg picture:

\begin{align} \frac{d^2\sigma}{d\Omega dE_{\rm f}} &= \left(\gamma r_0 \right)^2 \frac{k_{\rm f}}{k_{\rm i}} \left[ \frac{g}{2} F(q)\right]^2 \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) \\ &\quad\times \frac{1}{2\pi\hbar} \sum_{j,j'} \int_{-\infty}^{\infty} dt \exp(-i \omega t) \nonumber\\ &\quad\times \left\langle \exp(-i {\bf q} \cdot {\bf R}_{j}(0)) {\bf s}_{j}^\alpha(0) \exp(i {\bf q} \cdot {\bf R}_{j'}(t)) {\bf s}_{j'}^\beta(t) \right\rangle , \nonumber \end{align} where $$\langle \hat{A} \rangle$$ means the thermal average of the expectation value of an operator $$\hat{A}$$, in the same way as the calculations leading to the phonon expression (\ref{eq:sum_lambda_f}).

The nuclear positions $${\bf R}(t)$$ and the atomic spin $${\bf s}(t)$$ are both operators. Each of these gives rise to both an elastic and an inelastic contribution; in total the cross section becomes a sum of four terms. The one that is elastic in both channels is the magnetic diffraction signal, while the contribution that is elastic in the nuclear positions (the phonon channel) and inelastic in the spins (the magnetic channel) is the true magnetic inelastic signal. The two other contributions involve creation/annihilation of phonons via the magnetic interactions, and are not discussed further here; we refer to Ref.~[1].

The two contributions that are elastic in the phonon channel are reached in analogy with Scattering from lattice vibrations. We replace the general nuclear position operators $${\bf R}_j(t)$$ with the nuclear equilibrium positions in a lattice, $${\bf r}_j$$

and utilize the periodicity of the lattice. Furthermore, we multiply with the Debye-Waller factor, $$\exp(-2W)$$:

\begin{align} \left(\frac{d^2\sigma}{d\Omega dE_{\rm f}}\right)_{\rm magn.} &= \left(\gamma r_0 \right)^2 \frac{k_{\rm f}}{k_{\rm i}} \left[ \frac{g}{2} F(q)\right]^2 \exp(-2W) \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{\bf q}_\alpha\hat{\bf q}_\beta\right) \\ &\quad\times \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} \sum_{j,j'} \exp(i {\bf q} \cdot ({\bf r}_{j'}-{\bf r}_j)) \left\langle {\bf s}_{j}^\alpha(0) {\bf s}_{j'}^\beta(t) \right\rangle \exp(-i \omega t) dt. \nonumber \end{align} As a final step, we utilize the translational symmetry of the lattice to see that the double sum of $$j$$ and $$j'$$ equals $$N$$ times the sum over $$j-j'$$. The final result becomes: \begin{align} \label{eq:magn_cross_master} \left(\frac{d^2\sigma}{d\Omega dE_{\rm f}}\right)_{\rm magn.} &= \left(\gamma r_0 \right)^2 \frac{k_{\rm f}}{k_{\rm i}} \left[ \frac{g}{2} F(q)\right]^2 \exp(-2W) \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{\bf q}_\alpha\hat{\bf q}_\beta\right) \\ &\quad\times \frac{N}{2\pi\hbar} \int_{-\infty}^{\infty} \sum_{j'} \exp(i {\bf q} \cdot {\bf r}_{j'}) \left\langle {\bf s}_{0}^\alpha(0) {\bf s}_{j'}^\beta(t) \right\rangle \exp(-i \omega t) dt. \nonumber \end{align}

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1. W. Marshall and S.W. Lovesey. Theory of Thermal Neutron Scattering. Oxford, 1971 (Wiley, 2001)