Quantum Physics III Chapter 7: Scattering

Chapter 7 Scattering c B. Zwiebach In high energy physics experiments a beam of particles hits a target composed of particles. By detecting the by-products one aims to study the interactions that occur during the collision. Collisions can be rather intricate. For example, the particles involved may be not elementary (protons) or they may be elementary (electrons and positrons) p+p → p + p + π0 p+p → p + n + π+ e+ + e− → µ + + µ− The final products may not be the same as the initial particles. Some particles may be created. Here collisions are like reactions in which a+b → c + d + e + ... We have scattering when the particles in the initial and final state are the same a+b → a+b 133 CHAPTER 7. SCATTERING 134 The scattering is elastic if none of the particles internal states change in the collision1 . We will focus on elastic scattering of particles without spin in the nonrelativistic approximation. We will also assume that the interaction potential is translational invariant, that is, V (r1 − r2 ). It follows that in the CM frame the problem reduces, as we did for hydrogen atom, to scattering of a single particle of reduced mass off a potential V (r). We will work with energy eigenstates and we will not attempt to justify steps using wave-packets. 7.1 The Schrödinger equation for elastic scattering We are interested in energy eigenstates p2 + V (r) 2M iEt ψ(r, t) = ψ(r)e− ~   ~2 2 − ∇ + V (r) ψ(r) = Eψ(r) 2M H= Figure 7.1: Potential is finite range, or vanishes faster than 1 (7.1.1) (7.1.2) (7.1.3) r as r → ∞ Will be consider solutions with positive energy. The energy is the energy of the particle 2 k2 far away from the potential, E = ~2M . The Schrödinger equation then takes the form  
~2 2 2 ∇ + k + V (r) ψ(r) = 0 − 2M (7.1.4) Now we must set up the waves! Recall the 1D case. Physics dictates the existence of three waves: an incoming one, a reflected one, and a transmitted one. We can think of the reflected and transmitted waves as the scattered wave, the waves produced given the incoming wave. Setting up the waves is necessary in order to eventually solve the problem by looking into the details in the region where the potential is non-zero. When the potential has finite range the incoming and scattered waves are simple plane waves that are easily 1 Frank-Hertz experiment involved inelastic collisions of electrons with mercury atoms in which the mercury atoms are excited. 7.1. THE SCHRÖDINGER EQUATION FOR ELASTIC SCATTERING 135 written. Equation (7.1.4) has an inifinite degeneracy of energy eigenstates. When V (r) ≡ 0, for example, eik·x for any k such that k · k = k 2 is a solution. Assume an incident wave moving towards +ẑ. Then the wavefunction will look like ϕ(r) = eikz (7.1.5) If we assume V (r) has a finite range a, this ϕ(r) satisfies (7.1.4) for any r > a. For r < a, however, it does not satisfy the equation; ϕ(r) is a solution everywhere only if the potential vanishes. Given an incident wave we will also have a scattered wave. Could it be an ψ(r) = eikr that propagates radially out?
fails badly for r 6= 0!! (7.1.6) ∇2 + k 2 eikr = 6 0 On the other hand
eikr =0 for r 6= 0 . (7.1.7) r This is consistent with the radial equation having a solution u(r) = eirk in the region where the potential vanishes. Recall that the full radial solution takes the form u(r)/r. ikr Can the scattered wave therefore be e r ? Yes, but this is not general enough. We need to introduce some angular dependence. Hence our ansatz for the scattered wave is ∇2 + k 2 ψs (r) = fk (θ, φ) eikr r ( 7.1.8) We expect the intensity of the scattered wave to depend on direction and the function fk (θ, φ) does that. We will see that ψs is only a solution for r  a, arbitrarily far. Both the incident and the scattered wave must be present, hence the wavefunction is ψ(r) = ψs (r) + ϕ(r) ' eikz + fk (θ, φ) eikr , r r  a. (7.1.9) CHAPTER 7. SCATTERING 136 As indicated this expression is only true far away from the scattering center. We physically expect fk (θ, φ) to be determined by V (r). fk (θ, φ) is called the scattering amplitude. We now relate fk (θ, φ) to cross section!! h # particles scattered per unit time into solid angle dΩ about (θ, φ) i i dσ = h particles flux of incident particles = # area · time (7.1.10) dσ is called differential cross section, it’s the area that removes from the incident beam the particles to be scattered into the solid angle dΩ. Let us calculate the numerator and the denominator. First the denominator, which is the probability current: Incident flux = ~ Im [ψ ∗ ∇ψ] = ~k ẑ (7.1.11) in eikz m m Intuitively, this can be calculated by multiplying the probability density |eikz |2 = 1, by the p ~k = ~k velocity m m . The result is again an incident flux equal in magnitude to m . To calculate the numerator we first find the number of particles in the little volume of thickness dr and area r2 dΩ dn = number of particles in this little volume dn = |ψ(r)|2 d3 r = fk (θ, φ) eikr r 2 r2 dΩ dr = |fk (θ, φ)|2 dΩ dr (7.1.12) With velocity v = ~k m all these particles in the little volume will go cross out in time dt = therefore the number of particles per unit time reads dn dΩ dr ~k = |fk (θ, φ)|2 dr = |fk (θ, φ)|2 dΩ dt m v dr v , Back in the formula for the cross section we get ~k  |fk (θ, φ)|2 dΩ dσ = m ~k  m  hence dσ = |fk (θ, φ)|2 dΩ Z Z Total cross section: σ = dσ = |fk (θ, φ)|2 dΩ Differential cross section: (7.1.13) (7.1.14) 7.2. PHASE SHIFTS 7.2 137 Phase shifts Assume V (r) = V (r), so that we are dealing with a central potential. First recall the description of a free particle in spherical coordinates. With V = 0, E= ~2 k 2 , 2m ψ(r) = uE` (r) Y`m (Ω) r the Schrödinger equation for uE`   ~2 `(` + 1) ~2 k 2 ~2 d2 + u uE` (r) (r) = − E` 2m dr2 2m r2 2m   d2 `(` + 1) − 2+ uE` (r) = k 2 uE` (r) . dr r2 Now take ρ = kr, then (7.2.2) reads   d2 `(` + 1) − 2+ uE` (ρ) = uE` (ρ) dρ ρ2 (7.2.1) (7.2.2) (7.2.3) Since k2 disappeared from the equation, the energy is not quantized. The solution to (7.2.3) is uE` (ρ) = A` ρj` (ρ) + B` ρ n` (ρ) or uE` (r) = A` rj` (kr) + B` r n` (kr) (7.2.4) where j` (ρ) is the spherical Bessel function j` (ρ) is non singular at the origin n` (ρ) is the spherical Neumann function n` (ρ) is singular at the origin Both have finite limits as ρ → ∞   `π ρ j` (ρ) → sin ρ − 2   `π ρ n` (ρ) → − cos ρ − 2 (7.2.5) (7.2.6) Now, with the plane wave a solution, we must have eikz = eikr cos θ = ∞ X a` P` (cos θ)j` (kr) (7.2.7) `=0 for some coefficients a` . Using r 2` + 1 Y`,0 (θ) = P` (cos θ) 4π and 1 j` (x) = ` 2i Z 1 −1 eixu P` (u)du (7.2.8) CHAPTER 7. SCATTERING 138 it can be shown that ikz e = √ 4π ∞ X √ 2` + 1 i` Y`,0 (θ)j` (kr) . (7.2.9) `=0 This is an incredible relation in which a plane wave is built by a linear superposition of spherical waves with all possible values of angular momentum! Each ` contribution is a partial wave. Each partial wave is an exact solution when V = 0. We can see the spherical ingoing and outgoing waves in each partial wave by expanding (7.2.9) for large r: " #   `π `π 1 1 ei(kr− 2 ) e−i(kr− 2 ) `π j` (kr) → = sin kr − − kr 2 2ik r r (7.2.10) Thus ikz e √ `π `π ∞ 4π X √ 1 h ei(kr− 2 ) e−i(kr− 2 ) i ` = 2` + 1 i Y`,0 (θ) − , k 2i | {z r } | {z r } ra (7.2.11) `=0 outgoing 7.2.1 incoming Calculating the scattering amplitude in terms of phase shifts Recall 1D case ϕ(x) = sin(kx) =  1  ikx e − e|−ikx {z } 2i solution if V = 0 ingoing (7.2.12) Exact solution ψ(x) =  1  ikx 2iδk e e − e|−ikx {z } 2i same for x > a (7.2.13) ingoing wave where the outgoing wave can only differ from the ingoing one by a phase, so that probability is conserved. Finally we defined ψ(x) = ψs (x) + ϕ(x) (7.2.14) So do a similar transformation to write a consistent ansatz for ψ(r). We have from (7.1.9) ψ(r) ' eikz + fk (θ) eikr , r r  a. (7.2.15) 7.2. PHASE SHIFTS 139 The incoming partial waves in the left-hand side must be equal to the incoming partial waves in eikz since the scattered wave is outgoing. Introducing the phase shifts on the outgoing waves of the left-hand side we get √ ψ(r) = `π `π ∞ 4π X √ 1 h ei(kr− 2 ) e2iδk e−i(kr− 2 ) i eikr ` 2` + 1 i Y`,0 (θ) − = eikz + fk (θ) k 2i | r r r {z } | {z } `=0 outgoing incoming The incoming partial waves in eikz cancel in between the two sides of the equation, and moving the outgoing partial waves in eikz into the other side we get √ ∞  eikr e− i`π 2 4π X √ 1  2iδ` eikr ` 2` + 1 i Y`,0 (θ) −1 e = fk (θ) k r r |2i {z } `=0 eiδ` sin δ` √ = (7.2.16) ∞ eikr 4π X √ 2` + 1 Y`,0 (θ)eiδ` sin δ` , k r (7.2.17) `=0 where we noted that e− i`π 2 = (−i)` and i` (−i)` = 1. Therefore we get √ fk (θ) = ∞ 4π X √ 2` + 1 Y`,0 (θ)eiδ` sin δ` . k (7.2.18) `=0 This is our desired expression for the scattering amplitude in terms of phase shifts. We had (7.2.19) dσ = |fk (θ)|2 dΩ and this differential cross section exhibits θ dependence. On the other hand for the full cross section the angular dependence, which is integrated over, must vanish Z Z 2 σ = |fk (θ)| dΩ = fk∗ (θ)fk (θ)dΩ Z √ 4π X √ −iδ` iδ`0 ∗ 0 = 2 (Ω)Y`0 ,0 (Ω) 2` + 1 2` + 1 e sin δ` e sin δ`0 dΩ Y`,0 k `,`0 | {z } δ``0 Hence σ= ∞ 4π X (2` + 1) sin2 δ` . k2 `=0 Now let us explore the form of f (θ) in the forward direction θ = 0. Given that r r 2` + 1 2` + 1 Y`,0 (θ) = P` (cos θ) =⇒ Y`,0 (θ = 0) = 4π 4π (7.2.20) CHAPTER 7. SCATTERING 140 then √ fk (θ = 0) = ∞ 4π X √ 2` + 1 k r `=0 ∞ 1X 2` + 1 iδ` e sin δ` = (2` + 1)eiδ` sin δ` 4π k `=0 Hence Im(f (0)) = ∞ 1X 1 k2 (2` + 1)eiδ` sin2 δ` = σ k k 4π (7.2.21) `=0 Therefore we have found out a remarkable relation between the total elastic cross section and the imaginary part of the forward (θ = 0) scattering amplitude. This relation is known as the Optical theorem and reads σ= 4π Im(f (0)) . k Optical Theorem Let us consider ways in which we can identify the phase shifts δk . Consider a solution, restricted to a fixed ` ψ(x) = (A` j` (kr) + B` n` (kr)) Y`,0 (θ) ` x>a (7.2.22) If B 6= 0 then V = 6 0. As a matter of fact, if V = 0 the solution should be valid everywhere and n` is singular at the origin, thus B` = 0. Now, let’s expand ψ(x)|` for large kr as in (7.2.10)      A` `π B` `π ψ(x) ' sin kr − − cos kr − Y`,0 (θ) (7.2.23) kr 2 kr 2 ` Define B` ( 7.2.24) A` We must now confirm that this agrees with the postulated definition as a relative phase between outgoing and ingoing spherical waves. We thus calculate      1 `π sin δ` `π ψ(x) ' sin kr − + cos kr − Y`,0 (θ) kr 2 cos δ` 2 `   1 `π sin kr − + δ` Y`,0 (θ) ' (7.2.25) kr 2 tan δ` ≡ − i `π 1 1 h i(kr− `π +δ` ) 2 − e−i(kr− 2 +δ` ) Y`,0 (θ) e kr 2i i `π 1 1 h i(kr− `π +2δ` ) 2 ' e−iδ e − e−i(kr− 2 ) Y`,0 (θ) kr 2i = (7.2.26) 7.2. PHASE SHIFTS 141 This shows that our definition of δk above is indeed consistent the expansion in (7.2.16). Finally we can read the phase shift from (7.2.25)   1 `π ψ(x) ' (7.2.27) sin kr − + δ` Y`,0 (θ) ra kr 2 ` in the partial wave solution. As an aside note that both the outgoing and ingoing components of a partial wave are separately acceptable asymptotic solutions obtained as: • If A` = −iB`      `π Y`,0 (θ) ei(kr− 2 ) `π `π ψ(x) ∼ i sin kr − + cos kr − ∼ Y`,0(θ) 2 2 kr kr ` • If A` = iB` (7.2.28) `π e−i(kr− 2 ) ψ(x) ∼ Y`,0 (θ) kr ` (7.2.29) Each wave is a solution, but not a scattering one. 7.2.2 Example: hard sphere ( ∞ V (r) = 0 Radial solution u = A` j` (kr) + B` n` (kr) r X ψ(a, θ) = R` (a)P` (cos θ) ≡ 0 R` (r) = r≤a r>a (7.2.30) (7.2.31) (7.2.32) ` The P` (cos θ) are complete, meaning that 0 = R` (a) = A` j` (ka) + B` n` (ka) ∀ ` (7.2.33) Recalling tan δ` ≡ − B` j` (ka) = A` n` (ka) (7.2.34) this implies all phase shifts are determined tan δ` = j` (ka) n` (ka) (7.2.35) CHAPTER 7. SCATTERING 142 We can now easily compute the cross section σ, which, recalling (7.2.20) is proportional to sin2 δ` j`2 (ka) tan2 δ` = 2 (7.2.36) sin2 δ` = 2 1 + tan δ` j` (ka) + n2` (ka) hence ∞ ∞ j`2 (ka) 4π X 4π X 2 σ= 2 (2` + 1) sin δ` = 2 (2` + 1) 2 k k j` (ka) + n2` (ka) `=0 (7.2.37) `=0 Griffiths gives you the low energy ka  1 expansion  ` 4 ∞ 2 `! 4π X 1 σ' 2 (ka)4`+2 k 2` + 1 (2`)! (7.2.38) `=0 At low energy the dominant contribution is from ` = 0 σ' 4π (ka)2 = 4πa2 k2 Full area of the sphere! Not the cross section! (7.2.39) One more calculation! i tan δ = eiδ − e−iδ e2iδ − 1 = eiδ + eiδ e2iδ + 1 Therefore e2iδ` = e2iδ` = 7.2.3 1 + i nj``(ka) (ka) 1− i nj``(ka) (ka) =⇒ = e2iδ = 1 + i tan δ 1 − i tan δ n + ij i(j − in) = n − ij −i(j + in) j` (ka) − in` (ka) , j` (ka) + in` (ka) Hard sphere phase shifts (7.2.40) (7.2.41) (7.2.42) General computation of the phase shift Suppose you have a radial solution R` (r) known for r ≤ a (V (r ≥ a) = 0). This must be matched to the general solution that holds for V = 0, as it holds for r > a: At r = a must match the function and its derivative R` (a) = A` j` (ka) + B` n` (ka) (7.2.43) 7.2. PHASE SHIFTS 143   aR`0 (a) = ka A` j`0 (ka) + B` n0` (ka) (7.2.44) Let’s form the ratio and define β` as the logarithmic derivative of the radial solution ` 0 j` (ka) + B j` (ka) − tan δ` η`0 (ka) aR`0 (a) A` j` (ka) + B` η`0 (ka) A` η` (ka) = ka = ka = ka ` R` (a) A` j` (ka) + B` n` (ka) j` (ka) − tan δ` n` (ka) j` (ka) + B A` n` (ka) (7.2.45) In principle we can use eq. (7.2.45) to get tan δ`, but let’s push the calculation further to extract the phase shift from e2iδ`

j` eiδ` + e−iδ` + iη`0 eiδ` − e−iδ` j` cos δ` − η`0 sin δ` β` = ka = ka j` cos δ` − n` sin δ` j` (eiδ` + e−iδ` ) + in` (eiδ` − e−iδ` ) β` ≡ = ka e2iδ` (j` + iη`0 ) + (j` − iη`0 ) eiδ` (j` + iη`0 ) + e−iδ` (j` − iη`0 ) = ka eiδ` (j` + in` ) + e−iδ` (j` − in` ) e2iδ` (j` + in` ) + (j` − in` ) Solve for e2iδ` e2iδ` = − ka(j` − iη`0 ) − β` (j` − in` ) ka(j` + iη`0 ) − β` (j` + in` ) (7.2.46) (7.2.47) which can be rewritten as e2iδ` = −      β − ka j` −iη`0 j −in j` − in`  `  ` 0`   j +iη j` + in` β` − ka j``+in`` (7.2.48) which we can also write as  e2iδ` = 2iξ` e|{z} Hard sphere phase shift  β` − ka  j` −iη`0 j` −in`  β` − ka  j` +iη`0 j` +in`  (7.2.49) Phase shifts are useful when a few of them dominate the cross section. This happens when ka < 1 with a the range of the potential. So short range and/or low energy. 1. Angular momentum of the incident particle is L = b · p with b impact parameter and p momentum L ' bp → ~` ' b~k → b' ` k (7.2.50) CHAPTER 7. SCATTERING 144 When b > a there is no scattering ` >a k Expect no scattering for ` > ka (7.2.51) Thus only values ` ≤ ka contribute to σ. 2. Confirm the impact parameter intuition from partial wave expression. Consider the free partial wave ∼ j` (kr)Y`,0 . The impact parameter b` of such a wave would be estimated to be `/k. Recall that j` (kr) is a solution of the V = 0 radial equation with angular momentum ` and energy ~2 k 2 /(2m). Setting the effective potential equal to the energy: ~2 `(` + 1) ~2 k 2 = 2mr2 2m (7.2.52) we find that the turning point for the solution is at k 2 r2 = `(` + 1) . Thus we expect j` (kr) to be exponentially small beyond the turning point p kr ≤ `(` + 1) ' ` (7.2.53) (7.2.54) confirming that the wave is negligible for r less than the impact parameter `/k! Figure 7.2: Plot of ρ2J 2 ` (ρ) 7.3 Integral scattering equation Some useful approximations can be made when we reformulate the time-independent Schrödinger equation as an integral equation. These approximations, as opposed to the partial waves method, allow us to deal with potentials that are not spherically symmetric. We consider the Schrödinger equation   ~2 2 − ∇ + V (r) ψ(r) = Eψ(r) , (7.3.1) 2M 7.3. INTEGRAL SCATTERING EQUATION 145 set the energy equal to that of a plane wave of wavenumber k and rewrite the potential in terms of a rescaled version U (r) that simplifies the units: E= ~2 k 2 2M and V (r) = ~2 U (r) 2M (7.3.2) then the Schrödinger equation reads   −∇2 + U (r) ψ(r) = k 2 ψ(r) (7.3.3)
∇2 + k 2 ψ(r) = U (r)ψ(r) (7.3.4) rewrite as This is the equation we want to solve. Let us introduce G(r − r0 ), a Green function for the operator ∇2 + k 2 , i.e.
∇2 + k 2 G(r − r0 ) = δ (3) (r − r0 ) (7.3.5) Then we claim that any solution of the integral equation Z ψ(r) = ψ0 (r) + d3 r0 G(r − r0 )U (r0 )ψ(r0 ) (7.3.6) where ψ0 (r) is a solution of the homogeneus equation
∇2 + k 2 ψ0 (r) = 0 (7.3.7) is a solution of equation (7.3.4). Let’s check that this is true Z

2 2 2 2 ∇r + k ψ(r) = ∇r + k d3 r0 G(r − r0 )U (r0 )ψ(r0 ) Z = d3 r0 δ (3) (r − r0 )U (r0 )ψ(r0 ) = U (r)ψ(r) To find G we first recall that  ±ikr  Ce for r = 6 0 r G(r) ∼ − 1 for r → 0 4πr (7.3.8) X
e±ikr as a matter of fact ∇2r + k 2 as a matter of fact
1 ∇2 − 4πr = δ 3 (r) r = 0 ∀ r 6= 0 (7.3.9) Thus try G± (r) = − 1 e±ikr 4π r (7.3.10) CHAPTER 7. SCATTERING 146 and we are going to refer to G+ as the outgoing wave Green function and to G− as the ingoing wave Green function.
1 Let’s verify that this works, write G as a product: G± (r) = e±ikr − 4πr , so that           1 1 1 2 2 ±ikr ±ikr 2 ±ikr ~ ~ − ∇ G± (r) = ∇ e + e ∇ − + 2 ∇e ·∇ − 4πr 4πr 4πr (7.3.11) Recall that r ∇2 = ∇ · ∇ , ∇r = , ∇ · (f A) = ∇f · A + f ∇ · A (7.3.12) r Therefore   r ±ikr 2ik ±ikr 2 ±ikr 2 e±ikr ∇e = ±ik e , ∇ e = −k ± (7.3.13) r r then      1 r  r  2ik ±ikr 2 2 e − + e±ikr δ 3 (r) + 2 ±ike±ikr · ∇ G± (r) = −k ± r 4πr r 4πr3     2ik ±ikr 2ik ±ikr = −k 2 G± (r) ∓ e + δ 3 (r) ± e  4πr  4πr = −k 2 G± (r) + δ 3 (r) X (7.3.14) We will use ψ0 (r) = eikz and G = G+ (7.3.15) Thus, with these choices, (7.3.6) takes the form Z ψ(r) = eikz + d3 r0 G+ (r − r0 )U (r0 )ψ(r0 ) where (7.3.16) 0 1 eik|r−r | (7.3.17) 4π |r − r0 | We now want to show that this is consistent with our asymptotic expansion for the energy eigenstates. For that we can make the following approximations G+ (r − r0 ) = − For the G+ denominator: 0 (7.3.18) 0 |r − r | ' r − n · r For the G+ numerator: where n≡ |r − r0 | ' r r r (7.3.19) 7.3. INTEGRAL SCATTERING EQUATION 147 In this way G+ (r − r0 ) = − Thus 1 ikr −ikn·r0 e e 4πr (7.3.20)  ikr Z 1 e 3 0 −ikn·r0 0 0 ψ(r) = e + − d re U (r )ψ(r ) (7.3.21) 4π r The object in brackets is a function of the unit vector n in the direction of r. This shows that the integral equation, through the choice of G, incorporates both the Schrödinger equation and the asymptotic conditions. By definition, the object in brackets is fk (θ, φ), i.e. Z 1 0 fk (θ, φ) = − (7.3.22) d3 r0 e−ikn·r U (r0 )ψ(r0 ) . 4π ikz  Of course, this does not yet determine fk as the undetermined wavefunction ψ(r) still appears under the integral. The incident wave has the form of eiki ·r with ki the incident wave number, |ki | = k. For the outgoing wave we define ks ≡ nk (in the direction of n), the scattered wave vector. The expression for ψ(r) then becomes   ikr Z 1 e iki ·r 3 0 −iks ·r0 0 0 (7.3.23) ψ(r) = e + − d re U (r )ψ(r ) 4π r We will do better with the Born approximation. 7.3.1 The Born approximation Consider the original integral expression: Z iki ·r ψ(r) = e + d3 r0 G+ (r − r0 )U (r0 )ψ(r0 ) (7.3.24) Rewrite by just relabeling r → r0 0 ψ(r0 ) = eiki ·r + Z d3 r00 G+ (r0 − r00 )U (r00 )ψ(r00 ) . (7.3.25) Now plug (7.3.25) under the integral in (7.3.24) to get Z Z Z iki ·r 0 0 3 0 0 0 iki ·r0 3 0 ψ(r) = e + d r G+ (r−r )U (r )e + d r G+ (r−r )U (r ) d3 r00 G+ (r0 −r00 )U (r00 )ψ(r00 ) (7.3.26) Repeat the trick once more to find Z 0 ψ(r) = eiki ·r + d3 r0 G+ (r − r0 ) U (r0 )eiki ·r Z Z 00 + d3 r0 G+ (r − r0 )U (r0 ) d3 r00 G+ (r0 − r00 ) U (r00 )eiki ·r Z Z Z + d3 r0 G+ (r − r0 )U (r0 ) d3 r00 G+ (r0 − r00 )U (r00 ) d3 r000 G+ (r000 − r00 )U (r000 )ψ(r000 ) (7.3.27) CHAPTER 7. SCATTERING 148 By iterating this procedure we can form an infinite series which schematically looks like ψ = eiki ·r + Z G U eiki r + Z Z GU G U eiki r + Z Z GU Z GU G U eikir + . . . (7.3.28) The approximation in which we keep the first integral in this series and set to zero all others is called the first Born approximation. The preparatory work was done in (7.3.23), so we have now Z  ikr 1 e Born iki ·r 3 0 −iks ·r0 0 iki ·r0 − ψ (r) = e d re U (r )e (7.3.29) 4π r hence fkBorn (θ, φ) = − 1 4π Z d3 re−iK·r U (r) . (7.3.30) Here we defined the wave-number transfer K: |ki | = |ks | , K ≡ ks − ki (7.3.31) Note that we eliminated the primes on the variable of integration – it is a dummy variable after all. The wave-number transfer is the momentum that must be added to the incident one to get the scattered one. We call θ the angle between ki and ks (it is the spherical angle θ if ki is along the positive z-axis). Note that K = |K| = 2k sin 2θ (7.3.32) In the Born approximation the scattering amplitude fk (θ, φ) is simply the Fourier transform of U (r) evaluated at the momentum transfer K! fk (θ, φ) captures some information of V (r). If we have a central potential V (r) = V (r), we can simplify the expression for the Born scattering amplitude further by performing the radial integration. We have fkBorn (θ) = − 1 2m 4π ~2 Z d3 re−iK·r V (r) . (7.3.33) By spherical symmetry this integral just depends on the norm K of the vector K. This is why we have a result that only depends on θ: while K is a vector that depends on both θ and φ, its magnitude only depends on θ. To do the integral think of K fixed and let Θ be 7.3. INTEGRAL SCATTERING EQUATION 149 the angle with r fk (θ) = − m 2π~2 m =− 2 ~ Z 0 ∞ Z 2πdr r2 1 d(cos Θ)e−iKr cos Θ V (r) −1 0 ∞ Z e−iKr − eiKr m dr r V (r) =− 2 −iKr ~ 2 Z ∞  2 dr r V (r) 0 2 sin(Kr) Kr  (7.3.34) hence fk (θ) = − 2m K~2 ∞ Z dr rV (r) sin(Kr) (7.3.35) 0 with K = 2k sin 2θ . The Born approximation treats the potential as a perturbation of the free particle waves. This wave must therefore have kinetic energies larger than the potential. So most naturally, this is a good high-energy approximation. Example: Yukawa potential Given V (r) = V (r) = β e−µr r β, µ > 0 (7.3.36) from (7.3.35), one gets 2mβ fk (θ) = − K~2 Z 0 ∞ dr e−µr sin(Kr) = − 2mβ . + K 2) ~2 (µ2 (7.3.37) We can give a graphical representation of the Born series. Two waves reach the desired point r. The first is the direct incident wave. The second is a secondary wave originating at the scattering “material” at a point r0 . The amplitude of a secondary source at r0 is given by the value of the incident wave times the density U (r0 ) of scattering material at r0 . In the second figure again an incident wave hits r. The secondary wave now takes two steps: the incident wave hits scattering material at r00 which then propagates and hits scattering material at r0 , from which it travels to point r. 150 CHAPTER 7. SCATTERING Figure 7.3: Pictorial view of Born approximation. Wave at r is the sum of the free incident wave at r, plus an infinite number of waves coming from the secondary source at r0, induced by the incident wave. Figure 7.4: Pictorial view of second order Born term Figure 7.5: Given (θ, φ) and k (∼ energy), K is determined.