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Non-Relativistic Scattering

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Non-Relativistic Scattering Non-relativistic scattering Contents 1 Scattering theory 2 1.1 Scattering amplitudes . 3 1.2 The Born approximation . 5 2 Virtual Particles 5 3 The Yukawa Potential 6 Appendices 10 .A Beyond Bornx 10 1 Non-relativistic scattering 1 Scattering theory We are interested in a theory that can describe the scattering of a particle from a potential V (x). Our Hamiltonian is H = H0 + V: where H0 is the free-particle kinetic energy operator p2 H0 = : 2m In the absence of the potential V the solutions of the Hamiltonian could be written as the free-particle states satisfying H0jφi = Ejφi: These free-particle eigenstates could be written as momentum eigenstates jpi, but since that isn't the only possibility we hold off writing an explicit form for jφi for now. The full Schr¨odingerequation is (H0 + V )j i = Ej i: We define the eigenstates of H such that in the limit where the potential disappears (V !0), we have j i!jφi, where jφi and j i are states with the same energy eigenvalue. (We are able to do this since the spectra of both H and H + V are continuous.) A possible solution is1 1 j i = V j i + jφi: (1) E − H0 By multiplying by (E − H0) we can show that this agrees with the definitions above. There is, however the problem of the operator 1=(E − H0) being singular. The singular behaviour in (1) can be fixed by making E slightly complex and defining ( ) 1 ( ) j ± i = jφi + V j ± i : (2) E − H0 ± i This is the Lippmann-Schwinger equation. We will find the physical meeting of ( ) the (±) in the j ± i shortly. 1 ^ P Functions of operators are defined by f(A) = i f(ai)jaiihaij. The reciprocal of an operator is well defined provided that its eigenvalues are non-zero. January 18, 2014 2 c A.J.Barr 2009-2013. 1.1 Scattering amplitudes 1.1 Scattering amplitudes To calculate scattering amplitudes we are going to have to use both the position and the momentum basis, the incoming beam is (almost) a momentum eigenstate, and V is a function of position x. If jφi stands for a plane wave with momentum ~k then the wavefunction can be written ik x e · hxjφi = 3 : (2π) 2 We can express (2) in the position basis by bra-ing through with hxj and inserting R 3 the identity operator d x0 jx0ihx0j Z ( ) 3 D 1 E ( ) hxj ± i = hxjφi + d x0 x x0 hx0jV j ± i: (3) E − H0 ± i The solution to the Greens function defined by ~2 D 1 E G (x; x0) ≡ x x0 ± 2m E − H0 ± i is ik x x0 1 e± j − j G (x; x0) = − : ± 4π jx − x0j Using this result we can see that the amplitude of interest simplifies to Z ik x x0 ( ) 1 2m 3 e± j − j ( ) hxj ± i = hxjφi − 2 d x0 V (x0)hx0j ± i (4) 4π ~ jx − x0j where we have also assumed that the potential is local in the sense that it can be written as 3 hx0jV jx00i = V (x0)δ (x0 − x00): The wave function (4) is a sum of two terms. The first is the incoming plane wave. ikr For large r = jxj the spatial dependence of the second term is e± =r. We can now ( ) understand the physical meaning of the j ± i states; they represent outgoing (+) and incoming (−) spherical waves respectively. We are interested in the outgoing (+) spherical waves { the ones which have been scattered from the potential. We want to know the amplitude of the outgoing wave at a point x. For practical experiments the detector must be far from the scattering centre, so we may assume jxj jx0j. x We define a unit vector ^r in the direction of the observation point x ^r = jxj x´ and also a wave-vector k0 for particles travelling in the direction ^x of the observer, k0 = k^r: January 18, 2014 3 c A.J.Barr 2009-2013. 1.1 Scattering amplitudes Far from the scattering centre we can write p 2 2 jx − x0j = r − 2rr0 cos α + r0 s r r 2 = r 1 − 2 0 cos α + 0 r r2 ≈ r − ^r · x0 where α is the angle between the x and the x0 directions. It's safe to replace the jx − x0j in the denominator in the integrand of (4) with just r, but the phase term will need to be replaced by r − ^r · x0. So we can simplify the wave function to ikr Z (+) r large 1 2m e 3 ik0 x0 (+) hxj i −−−−!hxjki − d x0 e− · V (x0)hx0j i 4π ~2 r which we can write as ikr (+) 1 ik x e hxj i = 3 e · + f(k0; k) : (2π) 2 r This makes it clear that we have a sum of an incoming plane wave and an outgoing spherical wave with amplitude f(k0; k) given by 1 3 2m ( ) f(k0; k) = − (2π) hk0jV j ± i: (5) 4π ~2 We will ignore the interference between the first term which represents the orig- inal `plane' wave and the second term which represents the outgoing `scattered' wave, which is equivalent to assuming that the incoming beam of particles is only approximately a plane wave over a region of dimension much smaller than r. We then find that the partial cross-section dσ | the number of particles scattered into a particular region of solid angle per unit time divided by the incident flux2 | is given by 2 r jjscattj 2 dσ = dΩ = jf(k0; k)j dΩ: jjincidj This means that the differential cross section is given by the simple result dσ 2 = jf(k0; k)j : dΩ The differential cross section is simply the mod-squared value of the scattering amplitude. 2 ~ ∗ ∗ Remember that the flux density is given by j = 2im [ r − r ]. January 18, 2014 4 c A.J.Barr 2009-2013. 1.2 The Born approximation 1.2 The Born approximation If the potential is weak we can assume that the eigenstates are only slightly modified ( ) by V , and so we can replace j ± i in (5) by jki. (1) 1 3 2m f (k0; k) = − (2π) hk0jV jki: (6) 4π ~2 This is known as the Born approximation. Within this approximation we have the simple result that (1) f (k0; k) / hk0jV jki: Up to some constant factors, the scattering amplitude is found by squeezing the perturbing potential V between incoming and the outgoing momentum eigenstates of the free-particle Hamiltonian. Expanding out (6) in the position representation (by insertion of a couple of com- R 3 pleteness relations d x0 jx0ihx0j) we can write Z (1) 1 2m 3 i(k k0) x0 f (k0; k) = − d x0e − · V (x0): 4π ~2 This result is telling us that scattering amplitude is proportional to the 3d Fourier transform of the potential. By scattering particles from targets we can measure dσ dΩ , and hence infer the functional form of V (r). This result is used, for example, in the nuclear form factor (Section ??). 2 Virtual Particles One of the insights of subatomic physics is that at the microscopic level forces are caused by the exchange of force-carrying particles. For example the Coulomb force between two electrons is mediated by excitations of the electromagnetic field { i.e. photons. There is no real `action at a distance'. Instead the force is transmitted between the two scattering particles by the exchange of some unobserved photon or photons. The mediating photons are emitted by one electron and absorbed by the other. It's generally not possible to tell which electron emitted and which absorbed the mediating photons { all one can observe is the net effect on the electrons. e− Other forces are mediated by other force-carrying particles. In each case the mes- senger particles are known as virtual particles. Virtual particles are not directly e− γ observed, and have properties different from `real particles' which are free to prop- agate. e− To illustrate why virtual particles have unusual properties, consider the elastic scat- e− tering of an electron from a nucleus, mediated by a single virtual photon. We can assume the nucleus to be much more massive than the electron so that it is approx- imately stationary. Let the incoming electron have momentum p and the outgoing, scattered electron have momentum p0. For elastic scattering, the energy of the January 18, 2014 5 c A.J.Barr 2009-2013. electron is unchanged E0 = E. The electron has picked up a change of momentum ∆p = p0 − p from absorbing the virtual photon, but absorbed no energy. So the photon must have energy and momentum Eγ = 0 e− e− pγ = ∆p = p0 − p: γ The exchanged photon carries momentum, but no energy. This sounds odd, but is 2 2 nevertheless correct. What we have found is that for this virtual photon, Eγ 6= pγ . The particular value Eγ = 0 is special to the case we have chosen, but the general result is that for any virtual particle there is an energy-momentum invariant3 which is not equal to the square of its mass P · P = E2 − p · p 6= m2: Such virtual particles do not satisfy the usual energy-momentum invariant and are said to be `off mass shell'. Note that we would not have been able to escape this conclusion if we had taken the alternative viewpoint that the electron had emitted the photon and the nucleus had absorbed it. In that case the photon's momentum would have been pγ = −∆p.
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