Scattering in Quantum Mechanics

1 1 Paul Avery PHZ4390 Sep. 25, 2013 Scattering in Quantum Mechanics 1 Rutherford Scattering .. 1 Derivation of the classical Rutherford Scattering formula .. 1 Using the momentum transfer q .. 3 2 General Formula for Scattering in Nonrelativistic QM .. 3 3 QM Scattering from a Coulomb Potential .. 5 Derivation of Scattering cross section .. 5 Matrix element and Feynman diagram for coulomb Scattering .. 5 Example: large angle Scattering in the Rutherford experiment .. 6 Coulomb Scattering of two finite mass particles .. 7 4 QM Scattering from a Potential with Massive Particle Exchange .. 8 Fixed potential .. 8 Two interacting particles: weak interactions.

2 9 5 Appendix: Density of States .. 10 Derivation of density of states formula .. 10 Example 1: Total density of states for p pmax .. 10 Example 2: Density of states and Maxwell velocity distribution .. 10 Example 3: Density of states and Planck photon energy distribution .. 11 Example 4: Fermi momentum at low temperature .. 12 1 Rutherford Scattering Our understanding of Scattering goes back to the Rutherford-Geiger-Marsden Scattering experi-ments of 1909 which aimed to measure the structure of atoms. The experimenters used radium and a collimation system to produce a narrow alpha particle beam that struck a thin gold foil tar-get in an evacuated chamber. Although most alphas were gently deflected as expected (the pre-vailing atomic model assumed smooth charges embedded throughout the atom), the experiment-ers were astonished to find that a small fraction were scattered at large angles by what they later surmised was a massive, positively charged nucleus occupying a region less than 1/4000 the di-ameter of the atom.

3 In their 1911 paper they derived an equation, known today as the Rutherford Scattering formula, that explained their results in terms of the Scattering of charged particles by a heavy point charge. Derivation of the classical Rutherford Scattering formula Consider a beam of particles, each of mass m and charge e, incident on a heavy nucleus of charge Ze, where the nucleus is assumed not to move ( M m). Each beam particle will be deflected, depending on its momentum p and impact parameter b, through an angle , as shown in Figure 1. 2 Figure 1: Scattering of a particle of charge e by a heavy nucleus of charge Ze. The force acting on the beam particle is F=Ze2/4 0r2=Z /r2 in natural units, where r is the distance between the particles.

4 We can determine the angle of Scattering from the following argument. After the Scattering the beam particle momentum p is unchanged (elastic Scattering ), but it is deflected by an angle . Thus py=psin . But we can also calculate py by integrating the force along the y direction. Let be the angle measured from the nucleus to the particle start-ing from the x-axis, with . Then using py=Fsin dt , we obtain py=Z r2sin dt =Z r2sin dtd d =Z bvcos =Z bv1+cos () where r2 =bv from angular momentum conservation. We thus obtain a relation between the Scattering angle , momentum p and impact parameter b: psin =Z bv1+cos () tan12 =Z pvb We are interested in finding the cross sectional area d corresponding to a particle Scattering into the small angular region to + d.

5 This will happen if the particle has an impact parameter be-tween b and b + db, if it lies in an annulus of radius b centered on the nucleus. The area of the annulus is d =2 bdb and using the relation just derived for tan12 b=Z pvcot12 db=Z 2pv1sin212 we obtain d =2 bdb=2 Z ()22p2v2cos12 sin312 =Z ()2p2v22 sin d 1 cos ()2 !bp 3 The right-most expression was obtained by multiplying numerator and denominator by sin12 and applying the half-angle trig identities sin =2sin12 cos12 and 1 cos =2sin212 . Since the Scattering process does not depend on (axially symmetric Scattering ), d =2 and we can write the numerator of the rightmost fraction as 2 sin d d , the differential of solid angle.

6 We thus obtain the well-known Rutherford differential cross section: d d =Z c()2p2v211 cos ()2 where I have restored the c term. Two things should be noted: (1) c=e2/4 0 so this clas-sical result does not depend on the QM-related quantity ; (2) integrating over d =2 dcos d gives an infinite total cross section, a consequence of the infinite range of the coulomb force. Using the momentum transfer q It s conventional to write the cross section in terms of the momentum transfer q=pi pf (ini-tial minus final momentum), where q2=2p21 cos (). The Rutherford formula then becomes d d =4Z cm()2q4 In QM Scattering theory we normally express the differential cross section in terms of q2 or the Lorentz invariant quantity q2.

7 These are the same up to a sign for elastic Scattering ( Ei Ef). 2 General Formula for Scattering in Nonrelativistic QM We can compute cross section formulas in nonrelativistic QM. The rate dJ for a process to pro-ceed from an initial Quantum state i to a final state f is given by the Fermi Golden Rule: dJ= fM i2 dNf 2 Ef Ei() where fM i= fM id3x describes the QM transition, dNf is the differential number of states for final state f (derived in the Appendix) and Ef Ei() is a delta function enforcing energy conservation. The term phase space is used for the dNf term. In relativistic Scattering phase space generalizes to multiple particles in the final state and has a delta function for 4-momentum conservation.

8 For Scattering processes, we assume that we have a free particle in the initial state before scatter-ing and a free particle in the final state after Scattering . Thus, for a Scattering potential U(x), 4 ix()=1 Veipi x fx()=1 Veipf x fM i=1 VUx()eiq xd3x dNf=Vd3pf2 ()3 where q=pi pf is the momentum transfer and I have used natural units. To get a cross sec-tion, we use the standard expression dJ=jiNtd (derived in a previous note) with one target particle ( Nt=1) and a plane wave flux density ji=nivi=vi/V (a single beam particle in the volume moving at velocity vi). When we divide the rate by the initial flux the volumes cancel and we obtain: d =dJjiNt=1viMfiq()2d3pf2 ()32 Ef Ei()=1viMfiq()2pf2d f4 2dpf Ef Ei() Mfiq()=eiq xUx()d3x , the Fourier transform of the potential, is called the matrix element.

9 We can integrate out the delta function using a standard variable change dpf Ef Ei() =dpfdEfdEf Ef Ei() =dpfdEf=1vf using dpf/dEf=1/vf, which holds true even for relativistic momenta. We finally obtain a general formula for the differential cross section: d d =Mfiq()2pf24 2vivf This formula allows the scattered particle to change species with different masses, though nor-mally vi=vf. Note that all the physics is in the matrix element Mfiq(); everything else is kin-ematics and phase space. 5 3 QM Scattering from a Coulomb Potential Derivation of Scattering cross section Here the potential is Ux()=Z /r. We calculate Mfiq()eiq xZ /r()d3x =4 Z /q2 (left as an exercise1) and get the cross section (using vi=vf in the final step) d d =Mfiq()2pf24 2vivf=4 Z ()2q4pf24 2vivf=4Z2 2m2q4 which agrees exactly with the classical Rutherford formula.

10 The graph in Figure 2 shows the Rutherford differential cross section for particles on gold for 1 cos Figure 2: Rutherford differential cross section (barn/steradian) vs cos Matrix element and Feynman diagram for coulomb Scattering Scattering from a fixed coulomb potential is represented by the Feynman diagram in Figure 3. The diagram gives the essential elements of the matrix element. The upper and lower vertices have coupling constants e and Ze, respectively, while the massless photon propagator brings in a factor 1/q2. Putting it all together yields Mfi Ze2/q2=4 Z /q2, which happens to be the 1 This can most easily be shown using spherical coordinates, with eip x=eiprcos and d3x=2 r2drdcos.