16.451 Lecture 6: Cross Section for Electron Scattering 23/9/2003 Recall from last time: where V n is the normalization volume for the plane wave electron.

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Lecture 6: Cross Section for Electron Scattering 23/9/2003 Recall from last time: where V n is the normalization volume for the plane wave electron states, and if is the transition rate from the initial to final state, which we calculate using a standard result from quantum mechanics known as “Fermi’s Golden Rule:” (transitions / sec) We will first calculate the matrix element M if and then the density of states  f 1

Matrix Element for Scattering, M if 3 – momenta: p i, p f, p R ( Nonrelativistic QM treatment!!! ) Use plane wave states to represent the incoming and outgoing electrons, and let p i = ħ k i, p f = ħ k f, p R = ħ q, and normalization volume = V n (Note: slight change of notation here from last class to make sure we don’t miss any factors of ħ. The recoil momentum of the proton in MeV/c is p R ; the momentum transfer in fm -1 is q = p R / ħ ) 2

Matrix element, continued... Insight #1: RHS is the Fourier transform of the scattering potential V(r), and it only depends on the momentum transfer q ! Next, proceed with caution: V(r) is the Coulomb potential of the extended charge distribution of the target atom that our electron is scattering from... at large distances, the atom is electrically neutral, so V(r)  0 faster than 1/r at short distances, we have to keep track of geometry carefully, accounting for the details of the nuclear charge distribution.... 3

Matrix element, continued... Insight #1: RHS is the Fourier transform of the scattering potential V(r), and it only depends on the momentum transfer q ! Next, proceed with caution: V(r) is the Coulomb potential of the extended charge distribution of the target atom that our electron is scattering from... at large distances, the atom is electrically neutral, so V(r)  0 faster than 1/r at short distances, we have to keep track of geometry carefully, accounting for the details of the nuclear charge distribution.... 4

x y z Screened Coulomb potential: For the atom, where Z is the atomic number, and  is a distance scale of order Å, the atomic radius. But the electron interacts with charge elements dQ inside the nucleus: dQ Bottom line: 5

Now do the integral... normalization: [  ] = m -3 substitute for dQ: Finally, for the matrix element: 6

How to deal with the variables: problem: r, R and s in here! 1. inside the nucleus, where  (R) differs from zero, (where the screening factor really matters is at large r, and there r  s to an even better approximation!) 2. there is a one to one mapping between all electron positions r and all displacements from the charge element dQ, so: ( this expression can be factored into 2 parts... ) Solution: 7

Finishing the integral for M if : use the relation: to simplify... Fourier transform of the nuclear charge density  F(q 2 ) Exact integral: (Eureka!) 8

What does this mean? Consider: If the scattering object is a point charge,, i.e. the normalized charge density is a Dirac delta function, with the property: F(q 2 ) = 1 for a point charge (Really important result!) 9

Finally, work out the density of states factor: from slide 1... for the cross-section: All we have left to calculate is the “density of states” factor, where E f is the total energy in the final state when the electron scatters at angle , and this factor accounts for the number of ways it can do that. 10

Consider the total final state energy: (electron)(recoil) (being careful with the factor of c !) This is useful because the momentum states are quantized – we have our electrons in a normalization volume, and we can “count the states” inside... 11

Counting the scattered electron momentum states: Recall the wave function: The normalization volume is arbitrary, but we have to be consistent.... let V n = L 3, i.e. the electron wave function is contained in a cubical box, so its wave function must be identically zero on all 6 faces of the cube. Since: Then it follows that: 12

So, momentum is quantized on a 3-d lattice: ħ 2  /L For a relativistic electron beam, the quantum numbers n x etc. are very large, but finite. We use the quantization relation not to calculate the allowed momentum, but rather to calculate the density of states! Allowed states are dots, 1 per cube of volume  p = (2  ħ/L) 3 13

Finally, consider the scattered momentum into d  at  : shaded area: number of momentum points in the shaded ring: 14

End of the calculation: We want the density of states factor: FINALLY, from slide 9: 15