ContentsLecture 2: Scattering and the Correlation FunctionScatteringApplication: Electron Energy Loss Spectroscopy (EELS)Application: Neutron ScatteringMIT OpenCourseWare http://ocw.mit.edu 8.512 Theory of Solids IISpring 2009For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture Notes: Theory of Solids II Patrick Lee Massachusetts Institute of Technology Cambridge, MA March 2, 2004Contents Contents 1 1 Lecture 2: Scattering and the Correlation Function 2 1.1 Scattering 2 1.2 Application: Electron Energy Loss Spectroscopy (EELS) 3 1.3 Application: Neutron Scattering 4| ��� � �� � � � � ��� � �� �� � � � � �Chapter 1 Lecture 2: Scattering and the Correlation Function We ended the last lecture with a brief discussion of the connection between scattering experiments and measurements of the correlation function S(�q, ω). In this lecture we will discuss scattering in more depth in terms of two concrete examples (electron and neutron scattering). After that, we will look at some more general properties of response functions. 1.1 Scattering The picture we have is of some blob of material, with a plane wave ki � coming in, and a different plane wave �kf � coming out. We define the momentum and energy transfer to the sample | Q = �ki − �kf (1.1) (1.2)ki − E�ω = E�kf Let �R be the coordinate of the scattering particle. Recall from last time that application of Fermi’s Golden Rule and the 1st order Born Approximation leads to the differential rate 2 φ0� ki )·R e−i� Rd�kf −�R e i(�q· →[f ] d3kf = 2π ρˆ†q | δ (ω − (En − E0)) d3kf (1.3)Wiv�q �n|qn = |v�2 2π |�n|ρˆ†2 δ(Ef − Ei)d3kf (1.4)QQ|�|φ0�|n = |v�2 S( �Q| Q, ω) d3kf (1.5) (1.6) Q, ω) = |v�2 S( �P ( �Q| Q, ω) (1.7) for scattering into a final state with momentum s omewhere in a volume element d3kf of momen-tum space centered on kf . Here, v�Q is the Fourier Transform of the interaction p otential. The key result here is that the rate of scattering with momentum transfer Q and energy loss ω is directly proportional to the correlation function S( �Q, ω). 2�(��(�� � �� Application: Electron Energy Los s Spectroscopy (EELS) 3 1.2 Application: Electron Energy Loss Spectroscopy (EELS) The experiment we imagine here is that of shooting high energy electrons ( 100 keV) at a thin film of material, and collecting them as they emerge with an energy-resolved detector. For this case, the interaction potential is just the Coulomb interaction between the electron and the sample’s charge density, so 4πe2 v�q | = (1.8)|q2 Recall the definition 1 UT ot = (1.9)q, ω) UExt Uscr= 1 + (1.10)UExt Remembering that Uscr (�q) = 4π e 2 δn(�q), where n(q�) are the Fourier components of the 2qdensity fluctuations, 1 4πe2 δn(�q, ω)= 1 + (1.11)�(� UExt(�q, ω) q2 q, ω) As defined in the previous lecture, the (linear) density response function χ(�q, ω) is defined by the ratio δn(�q, ω)χ(�q, ω) = (1.12)UExt(�q, ω) 1Substituting this into the relation for �(�q,ω) , we get 1 4πe2 = 1 + χ(�q, ω) (1.13)q, ω) q2 With χ��(�q, ω) defined as the imaginary part of χ, the relation S(� q, ω) (1.14)q, ω) = −2χ��(�combined with equation (1.7) for the scattering rate into momentum space volume d3kf gives the following relation for the scattering rate in terms of the dielectric function: 8πe2 1 q, ω) = q2P (� −Im �(�(1.15)q, ω) What useful information can we get out of this? For one, we are able to investigate the dielectric constant at finite values of �q (0 to kF ). In optical exp erime nts, the vanishingly small photon momentum in comparison with typical electron/nucleus momenta means that we are only able to investigate the �q ≈ 0 regime with photons. On the downside, the best energy resolution we can achieve today is around 0.1 eV, which is far too coarse to obtain much useful information. This energy resolution is already 1 : 106 when compared with the total electron energy of around 100 keV. To get around this, one might�(��(�� � Application: Neutron Scattering 4 consider trying lower energy experiments. However, the problem with low energy experiments is that the probability of multiple scattering events within the sample becomes significant, leading to complicated and messy results. With EELS, we can also look at high energy excitations of the electrons in a metal. Recall that there is a high energy collective mode of the sample e lectrons at a frequency equal to the plasma frequency ωpl. The plasma frequency is defined in terms of the zero of the dielectric function q, ωpl) = 0 (1.16) The situation where the dielectric function becomes zero is interesting, because it represents a singularity in the system’s response to an external perturbation: 1 UT ot = (1.17)q, ωpl) UExt Thus even a tiny perturbation at the plasma frequency results in a large response of the system . 1.3 Application: Neutron Scattering Since neutrons are uncharged, they do not see the electrons as they fly through a piece of mate-rial1 . The dominant scattering mechanism is through a contact potential with the nuclei of the sample 2πb V (�r) = δ(�r) (1.18)Mn where b is the scattering length and Mn is the mass of the neutron. Since the Fourier transform of a delta function in space has no �q dependence, the Fourier components of the interaction potential are all simply 2πb v�= (1.19)q Mn Inserting this into equation (1.7) for the scattering rate, we get � �22πb Q, ω) = S( �P ( �Q, ω) (1.20)Mn Here, S( �Q, ω) is the correlation for the nuclear positions (density) Q, ω) = dt e iω t� ˆQ(t) ˆQ(0) �T (1.21)S( �ρ�ρ − �with ρˆ�= e Q· �i�Ri (t) (1.22)Q i 1 The can interact, however t hrou gh spin-spin magnetic interactions.� � ��� � � �5 Application: Neutron Scattering where { �Ri(t)} are the coordinates of the nuclei at time t. Now we can substitute this in to the expression for S( �Q, ω): Q·Rj (t)e Q· ��i�R� (0) S( �Q, ω) = dt e iω t � e−i��T (1.23) j,� To m ake progress, we must put in a s pecific form for �Rj (t). We consider the case of