# 物理代写|凝聚态物理代写condensed matter physics代考|PHYSICS754

## 物理代写|凝聚态物理代写condensed matter physics代考|Introduction to Neutron Scattering

Radioactive decay produces high-energy neutrons. Here we will be concerned with thermal neutrons (i.e. neutrons that have diffused around inside a solid long enough to come to thermal equilibrium with it). These low-energy neutrons can be made mono-energetic by using a crystal as a 3D Bragg diffraction grating (to select a single momentum, much as in X-ray scattering). This mono-energetic beam can then be scattered from a sample and the resulting energy and momentum transfer analyzed very accurately [24]. Hence one can measure the spectrum of lattice and electronic excitations in a solid much more precisely than with X-rays. Crudely speaking, since the speed of thermal neutrons is comparable to the speed of atoms vibrating in the sample, we cannot make the approximation of neglecting atomic motion during the scattering event (as we did for the case of $\mathrm{X}$-ray scattering). ${ }^2$

More precisely, the very good energy resolution available with neutrons means that the neutron wave packets are long and take many vibrational periods to pass by any given atom. As a result, motion of the atom (either classical or quantum mechanical) has significant effects on the scattering process. Hence we must develop a theory of the dynamical correlations being probed by the neutrons. This is quite different from the X-ray scattering, which takes a “snapshot” of the instantaneous configuration of the atoms, and is not sensitive to the dynamics. The source of the difference lies in the energy and, in particular, the energy resolution of the different probes, as we will discuss in more detail later.

Neutrons, like X-rays, interact only weakly, and hence have great penetration depths. They therefore undergo bulk diffraction but will give an appreciable signal only if large enough samples can be obtained. A microscopic crystallite will not suffice! Neutrons, being electrically neutral, interact primarily with the nuclei (via the strong force) and to a much lesser extent with the magnetic moments of the electrons. If a neutron experiences a magnetic interaction which causes its spin to flip, the scattering is called incoherent, because diffraction does not occur in this case. The reason is the following: in principle, we can look back in the sample and see which nucleus or electron had its spin flipped by the neutron. Thus we know where in the crystal the neutron scattered. The situation is very much like what happens in the famous two-slit diffraction experiment. When detectors are inserted to see which slit the particle went through, the interference pattern is destroyed. Crystal diffraction comes from not knowing which atom does the scattering so that we have a coherent superposition of all possibilities,
$$F(\vec{q})=f(\vec{q}) \sum_j e^{-i \vec{q} \cdot \vec{r}_j}$$

## 物理代写|凝聚态物理代写condensed matter physics代考|Inelastic Neutron Scattering

The total Hamiltonian describing the system being studied and the neutron used to probe it is
$$H_{\mathrm{tot}}=H_{\mathrm{n}}+H+H_{\text {int }},$$
where $H$ is the Hamiltonian of the system, $H_{\mathrm{n}}=P_{\mathrm{n}}^2 / 2 M_{\mathrm{n}}$ is the neutron kinetic energy, and $H_{\text {int }}$ describes the interaction (to be specified later) between the neutron and the system.
Let the initial state of the neutron plus crystal system be
$$\Psi_i(\vec{R},{\vec{r}})=\psi_{\vec{p}}(\vec{R}) \Phi_i({\vec{r}}),$$
where $\vec{R}$ is the neutron coordinate and ${\vec{r}}$ stands for $\left(\vec{r}1, \vec{r}_2, \ldots, \vec{r}_N\right)$, the set of nuclear coordinates (atom positions). For simplicity we will focus mostly on the case of lattice vibrations and hence ignore the nuclear and electronic spin degrees of freedom; our results, however, are quite general (and apply to liquid states), and can be generalized to spin-dependent interactions straightforwardly (see Exercise 4.2). We assume that the only effect of the electrons is to produce chemical bonds, which give the effective Hamiltonian for the atoms a potential-energy term $V\left(\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_N\right)$. We otherwise ignore the electrons insofar as degrees of freedom are concerned. We take the initial state of the neutron to be a plane wave, $$\psi{\vec{p}}(\vec{R})=\frac{1}{\sqrt{L^3}} e^{i \vec{p} \cdot \vec{R}},$$
and the initial energy to be the neutron kinetic energy plus the initial crystal energy:
$$\mathcal{E}i=\frac{\hbar^2 p^2}{2 M{\mathrm{n}}}+E_i .$$
After the scattering, the final state is
$$\Psi_f=\psi_{\vec{p}^{\prime}}(\vec{R}) \Phi_f({\vec{r}}),$$
with energy
$$\mathcal{E}f=\frac{\hbar^2 p^{\prime 2}}{2 M{\mathrm{n}}}+E_f .$$
We note that the $E_{i, f}$ are eigenvalues of the system (or crystal) Hamiltonian $H$.

## 物理代写|凝聚态物理代写condensed matter physics代考|Introduction to Neutron Scattering

$$F(\vec{q})=f(\vec{q}) \sum_j e^{-i \vec{q} \cdot \vec{r}_j}$$

## 物理代写|凝聚态物理代写condensed matter physics代考|Inelastic Neutron Scattering

$$H_{\text {tot }}=H_{\mathrm{n}}+H+H_{\text {int }},$$

$$\Psi_i(\vec{R}, \vec{r})=\psi_{\vec{p}}(\vec{R}) \Phi_i(\vec{r}),$$

$$\mathcal{E} f=\frac{\hbar^2 p^{\prime 2}}{2 M \mathrm{n}}+E_f .$$

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