# Quantum treatment of elastic neutron scattering

This chapter contains the quantum mechanical formulation of neutron scattering theory. The chapter should be seen as a complementary to Basics of neutron scattering in the way that the present chapter contains the proper derivations of the expressions discussed from a semi-classical point of view in the previous chapter.

The contents of this chapter form the basis for the understanding of the quantum mechanical derivations of the expressions for inelastic neutron scattering, to be presented later in these notes.

# 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 [1].

## The initial and final states

We define the state of the incoming wave as

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

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}

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

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)

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

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:

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

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

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

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}):

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

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 [2].

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.

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

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

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

$$W_{{\rm i} \rightarrow {\rm f},d\Omega}$$ 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

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

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

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

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}):

$$%(S2.31) \frac{d\sigma}{d\Omega} = b_j^2\, .$$

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

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

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, $$\label{eq:nuclear_potential} \hat{V}=\frac{2\pi \hbar^2}{m_{\rm n}}\sum_j b_j \delta({\bf r}-{\bf r}_{j}) .$$ By insertion into (\ref{eq:master_scatt}), $$\frac{d\sigma}{d\Omega} = Y^2 \frac{k_{\rm f}}{k_{\rm i}} \left(\frac{m_{\rm n}}{2\pi\hbar^2}\right)^2 \big| \big\langle \psi_{\rm i} \big|\hat{V}\big| \psi_{\rm f}\big\rangle \big|^2 .$$ and performing the same calculations as above, we arrive directly at (\ref{eq:diffraction}), $$%Squires never writes this - strange! \boxed{ \left.\frac{d\sigma}{d\Omega}\right|_{\rm coh} = \biggr|\sum_j b_j \exp(i {\bf q} \cdot {\bf r}_j)\biggr|^2 } \, .$$

# Quantum mechanics of magnetic diffraction

We now develop the quantum mechanical formalism for elastic magnetic neutron scattering. The formalism for inelastic magnetic scattering is developed in Scattering theory for magnetic dynamics.

## The magnetic interaction

The interaction responsible for magnetic neutron scattering is the nuclear Zeeman term for a neutron in an external magnetic field:

$$\label{eq:nuclear_Zeeman} H_{\rm Z} = - {\bf \mu} \cdot {\bf B} = - \gamma \mu_{\rm N} \hat{\bf \sigma} \cdot {\bf B} ,$$

where $$\hat{\bf \sigma}$$ represents the spin of the neutron and is composed by the Pauli matrices.

The external field that scatters the neutron comes from the individial electrons on the atoms.

The magnetic moment of one electronic spin, $${\bf s}_j$$, is given by (\ref{eq:emoment})

The field from a dipole placed at the Origin can be described as

$$B = \frac{\mu_0}{(4\pi)} \nabla \times ({\bf \mu} \times {\bf r}/r^3).$$

We now position the spin at position $${\bf r}_j$$, and transform (\ref{eq:nuclear_Zeeman}) into

$$H_{{\rm Z},j} = \frac{\mu_0}{4\pi} g \mu_{\rm B} \gamma \mu_{\rm N} \hat{\bf \sigma} \cdot \nabla\times\left(\frac{{\bf s}_j\times({\bf r}-{\bf r}_j)}{|{\bf r}-{\bf r}_j|^3}\right) .$$

The neutron interaction with the magnetic ions is given as the total nuclear Zeeman interaction, summed over all magnetic sites, $$j$$. This we use as the scattering potential,

$$\label{eq:magnetic_potential} \hat{V} = \sum_j H_{{\rm Z},j} ,$$

that we insert in the master equation for magnetic diffraction (\ref{eq:master_scatt}).

$$\frac{d\sigma}{d\Omega} = \frac{k_{\rm f}}{k_{\rm i}} \left(\frac{m_{\rm n}}{2\pi \hbar^2}\right)^2 \big| \big\langle \psi_{\rm i} \big|\hat{V}\big| \psi_{\rm f}\big\rangle \big|^2 .$$

By substitution, the resulting equation for the magnetic diffraction cross section becomes

\begin{align} \label{eq:diffract_spinonly} \left. \frac{d \sigma}{d\Omega} \right|_{\sigma_{\rm i} \rightarrow \sigma_{\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 \\ &\quad\times \biggr| \biggr\langle {\bf k}_{\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} \sigma_{\rm i} \biggr\rangle \biggr|^2 . \nonumber \end{align}

We here note that we have explicitly added the spin value to the initial and final state of the neutron.

## The magnetic matrix element

We now turn to the calculation of the complicated matrix element in (\ref{eq:diffract_spinonly}).

We utilize a mathematical identity \cite{squires}, $$\nabla \times \left( \frac{{\bf s} \times {\bf r}}{r^3} \right) = \frac{1}{2\pi^2} \int \hat{\bf q}' \times ({\bf s} \times \hat{\bf q}') \exp(i {\bf q}' \cdot {\bf r}) d^3{\bf q}'$$ to reach \begin{align} &\biggr\langle {\bf k}_{\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} \sigma_{\rm i} \biggr\rangle \nonumber \\ &\quad= \frac{1}{2\pi^2} \biggr\langle {\bf k}_{\rm f} \sigma_{\rm f} \biggr| \sum_j \int d^3{\bf q'} \hat{\bf \sigma} \cdot (\hat{\bf q}' \times ({\bf s}_j \times \hat{\bf q}')) \exp(i {\bf q}' \cdot ({\bf r}-{\bf r}_j)) \biggr| {\bf k}_{\rm i} \sigma_{\rm i} \biggr\rangle \nonumber\\ &\quad= \frac{1}{2\pi^2} \biggr\langle \sigma_{\rm f} \biggr| \sum_j \int d^3{\bf r} d^3{\bf q'} \exp(i {\bf q}\cdot {\bf r}) \nonumber\\ &\quad\quad\times \exp(i {\bf q}' \cdot ({\bf r}-{\bf r}_j)) \hat{\bf \sigma} \cdot (\hat{\bf q}' \times ({\bf s}_j \times \hat{\bf q}')) \biggr| \sigma_{\rm i} \biggr\rangle \nonumber \\ &\quad= 4\pi \biggr\langle \sigma_{\rm f} \biggr| \sum_j \exp(i {\bf q} \cdot {\bf r}_j) \hat{\bf \sigma} \cdot (\hat{\bf q} \times ({\bf s}_j \times \hat{\bf q})) \biggr| \sigma_{\rm i} \biggr\rangle . \label{eq:magn_matrix} \end{align}

To perform the last step we used that $$\int \exp(i ({\bf q}+{\bf q'})\cdot {\bf r}) d^3 {\bf r} = (2\pi)^3 \delta({\bf q}+{\bf q}')$$. The equation contains a term

$$\label{eq:spinperp} \hat{\bf q} \times ({\bf s}_j \times \hat{\bf q}) \equiv {\bf s}_{j,\perp},$$

which is simply the component of the spin on site $$j$$ perpendicular to the scattering vector. Equation (\ref{eq:magn_matrix}) therefore proves that the spin component parallel to $${\bf q}$$ is invisible to neutrons, as mentioned in The magnetic scattering length.

## Matrix element for unpolarized neutrons

For the remainder of this notes, we assume that the neutrons are unpolarized, $$p_\uparrow = p_\downarrow = 1/2$$. We also assume that we do not observe the final spin state, $$\sigma_{\rm f}$$, of the neutron. To obtain the cross section for unpolarised neutrons, we therefore sum over $$\sigma_{\rm f}$$ and average over the initial spin state, $$\sigma_{\rm i}$$.

We save this derivation for later and just state the result here:

$$\sum_{\sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \left| \left\langle \sigma_{\rm f} \left| {\bf \sigma} \cdot {\bf s}_{\perp} \right| \sigma_{\rm i} \right\rangle \right|^2 = \left\langle \left| {\bf s}_{\perp} \cdot {\bf s}_{\perp} \right| \right\rangle.$$

When this expression is used in the calculation for the cross section, we will encounter terms of the general type $${\bf s}_{j \perp} \cdot {\bf s}_{j' \perp}$$. We here utilize that the perpendicular projection is defined as $${\bf s}_{j \perp} \equiv {\bf s}_j - ({\bf s}_j \cdot \hat{\bf q}) \hat{\bf q}$$, where $$\hat{\bf q}$$ is a unit vector in the direction of $${\bf q}$$, to reach

$$\label{eq:cartesian_perp} {\bf s}_{j \perp} \cdot {\bf s}_{j' \perp} = {\bf s}_j \cdot {\bf s}_{j'} - 2({\bf s}_j \cdot \hat{\bf q})({\bf s}_{j'} \cdot \hat{\bf q}) + ({\bf s}_j \cdot \hat{\bf q})({\bf s}_{j'} \cdot \hat{\bf q}) = \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) s_j^\alpha s_{j'}^\beta ,$$

where the indices $$\alpha$$ and $$\beta$$ run over the Cartesian coordinates ($$x$$, $$y$$, and $$z$$), and $$\hat{q}^{\alpha}$$ and $$s_j^{\alpha}$$ etc. are now scalar variables.

## The master equation for magnetic diffraction

We now collect the prefactors from the calculations above, assuming that the proton and neutron masses are identical: $$\frac{m_{\rm N}}{2\pi\hbar^2} g \mu_{\rm B} \gamma \mu_{\rm N} \mu_0 = \gamma \frac{\mu_0}{4\pi}\frac{e^2}{m_{\rm e}}\frac{g}{2} = \gamma r_0 \frac{g}{2},$$

where $$r_0$$ is the classical electron radius $$r_0=e^2\mu_0/(4\pi m_{\rm e})=2.8179$$~fm.

Collecting all equations, we end up with the master equation for the magnetic differential scattering cross section for unpolarized neutrons[3]:

$$\label{eq:magnetic_master_diffract} \frac{d \sigma}{d\Omega } = \left(\gamma r_0 \right)^2 \left(\frac{g}{2}\right)^2 \frac{k_{\rm f}}{k_{\rm i}} \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) \left\langle Q_\alpha \right\rangle \left\langle Q_\beta \right\rangle ,$$

where we have defined $${\bf Q}$$ as the Fourier transform of the spins $${\bf s}_j$$

positioned at $${\bf r}_j$$, with respect to the scattering vector, $${\bf q}$$[3]:

$${\bf Q}({\bf q}) = \sum_j \exp(i {\bf q} \cdot {\bf r}_j) {\bf s}_j .$$

## The magnetic form factor

We assume the electrons causing the magnetism to be located in orbitals around particular ions as discussed in Elastic magnetic scattering. The electron coordinates are therefore replaced by the nuclear positions, $${\bf r}_j$$, plus a small deviation from this, $${\bf r}$$, representing the extension of the particular electron orbital. We thus make the substitution

\begin{eqnarray} {\bf Q}({\bf q}) &=& \sum_{j}\int \exp(i {\bf q} \cdot ({\bf r}_j+{\bf r})) {\bf s}_j d^3{\bf r} \nonumber \\ &=& \sum_{j} \exp(i {\bf q} \cdot {\bf r}_{j}) {\bf s}_{j} F({\bf q}) \nonumber \\ &=& {\bf M}({\bf q}) F({\bf q}), \end{eqnarray} where $${\bf M}({\bf q})$$ and $$F({\bf q})$$ are defined as in Correlation between nuclear and magnetic scattering.

## Orbital contributions

When taking contributions from orbital magnetism into account, e.g. from rare-earth ions, the term $$g{\bf s}$$ is replaced by $$g_{\rm L}{\bf J}$$, where $$g_{\rm L}$$ is the Land\'e factor:

$$g_{\rm L} = 1 + \frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)} ,$$

which is a number between 1 and 2, and $${\bf J}$$ is the total angular momentum However, in the expressions to follow we keep the notation $${\bf s}$$ for simplicity.

The derivation of the contribution from orbital moment is lengthy and adds nothing to the general understanding of magnetic neutron scattering, so we simply omit it here. Details of this derivation are found in Theory of Thermal Neutron Scattering[3].

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1. G.L. Squires, Thermal Neutron Scattering, Cambridge University Press, 1978.
2. E. Merzbacher, Quantum Mechanics, Wiley, 1998.
3. W. Marshall and S.W. Lovesey, Theory of Thermal Neutron Scattering, Oxford, 1971.