Electron Atom Scattering

Transcription

1 SOCRATES Intensive Programme Calculating Atomic Data for Astrophysics using New Technologies Electron Atom Scattering Kevin Dunseath Mariko Dunseath-Terao Laboratoire de Physique des Atomes, Lasers, Molécules et Surfaces Université de Rennes Rennes, April 4

2 c K M Dunseath, M Dunseath-Terao Chapter Introduction Collision processes are of fundamental importance in quantum physics. Scattering experiments allow us to investigate the structure and properties of atoms, nuclei, elementary particles, and the interactions between them. The existence of the atomic nucleus, for example, was demonstrated by Rutherford from the scattering of α particles by a thin metal foil, while at CERN many new elementary particles have been found by colliding very energetic electrons, protons, etc... An understanding of collision processes is also important in basic chemistry and in materials science. In this and following chapters, we will consider the collision of a projectile with a single, isolated atom or ion, i.e. collisional processes occuring in low density gases such as interstellar media, planetary atmospheres and plasmas. Collisions play and important role in determining the physical and chemical properties of such environments. For example, the kinetic energy of the constituent particles determines the temperature of the gas. Collisions between these particles can result in the loss of kinetic energy, and hence the cooling of the gas. Atoms can be excited by collisions, subsequently decaying by emitting photons. This gives rise to bright emission lines in the spectra of the gas. The gas may also absorb any radiation traversing it at characteristic wavelengths. This gives rise to dark absorption lines in the spectra at characteristic wavelengths depending on the constituents. The intensity of a particular line depends on the number of atoms or ions of the associated element. This number can be enhanced or depleted by collisions between the particles. These processes in turn depend on the temperature and on electron and atom densities. The intensities and shapes of observed spectral lines can therefore yield information on the temperature of the gas, its composition, the abundance and density of various elements,... In order to interpret the observed spectra, and also to model the behaviour of the gas, it is therefore necessary to have accurate data concerning all the various collision processes that can occur. This data may be obtained through experiments or by calculation. A typical scattering experiment is shown below figure.). It consists of a beam of particles P directed at a target T, which scatters them through various angles. Detectors are used to count the number of particles of a particular type scattered in a particular direction. Other parameters such as kinetic energy, the internal state of the scattered particles etc, may also be determined. The collimator shields the detectors from the source of incident particles by producing a narrow beam directed at the target. The angular divergence of this beam should be as small as possible. The results of a scattering experiment will vary with the energy of the incident beam. In order to simplify the analysis and interpretation of the measurements, the energy spectrum of the incident

3 Chapter. Introduction 3 Detector Incident beam d Ω θ φ Target Z Collimator Figure.: A typical scattering experiment beam should be sharply peaked, so that it may be considered monochromatic, i.e. the collision may be considered to occur at one particular energy. The densities of the particles in the incident beam and the target should be low, so that the effect of multiple collisions can be neglected collisions between particles in the incident beam, scattering of an incident particle by several target particles,... ). A number of different processes may occur during the collision: - Elastic scattering: the projectile P is scattered by the target T, without any change in the internal structure of the two particles. - Inelastic scattering: the internal state of one or both particles P and T changes during the collision. Denoting the new states by P and T we may have P + T P + T P + T P + T An important example is the excitation of an atom by electron impact. - Rearrangement collisions: the composite system P + T) splits into two or more particles: P + T R + R +... Important examples include ionisation, where the target loses one or more electrons; charge exchange, where the projectile captures one or more electrons from the target. Inverse processes are recombination processes, in which a neutral atom or molecule is formed by the combination of a positive ion and a negative ion or electron. The neutral species is often in an excited state, which may decay by emitting a photon or, in the case of molecules, dissociate. A channel is defined as a mode of fragmentation of the composite system P+T). It is characterised by the number and nature of the fragments into which the system can decompose. For example, in the scattering of electrons by hydrogen, we may distinguish different channels according to the state

4 Chapter. Introduction 4 of the hydrogen atom, as well as the ionisation channel e + H +. A channel is said to be open if the corresponding collision is allowed by known conservation laws usually energy conservation), otherwise it is said to be closed. In general, the number and complexity of the processes that can occur increases as the collision energy increases. A scattering experiment usually determines the cross section for a particular process. The differential cross section is defined as the ratio of the number of scattered particles corresponding to this processes into the solid angle dω per unit time and per unit scatterer, to the flux of incident particles relative to the target. The flux of a beam is defined as the number of particles crossing a surface of unit area perpendicular to the beam direction per unit time. Consider a typical experiment as shown in figure.. The target is placed at the origin. We denote the incident flux by F i. The detector is placed far from the target, and subtends a solid angle dω at the origin around the direction θ, φ). Let n be the number of particles counted in unit time by the detector. The differential cross section is then given by dσ n/dω θ, φ) =. dω F i The integral or total cross section is obtained by integrating the differential cross section over dω: σ = dσ dω dω = π dθ sin θ π dφ dσ θ, φ). dω The integral cross section has the dimensions of an area. It is usually given in units of a or πa, where a is the Bohr radius a = m, so that a =.88 m and πa = m ).

5 c K M Dunseath, M Dunseath-Terao 5 Chapter Scattering by a central potential We consider the collision of two particles interacting through a potential that depends only on their relative coordinate r. By separating out the motion of the center of mass, the problem reduces to that of the relative motion in the center of mass frame of the two particles, which in turn may be considered as the motion in a central potential of a particle of mass equal to the reduced mass µ of the two colliding particles. A rigorous treatment of this problem would be based on the study of the evolution in time of a wave packet describing the motion of the particle in the potential. This evolution is described by the timedependent Schrödinger equation. The wave packet however may be considered as a superposition of stationary scattering states Φr), each of which are solutions to the time-independent Schrödinger equation [ ] µ + V r) Φr) = EΦr),.) where E is the collision energy in the center of mass frame. It can be shown that this approach gives the same results as the more rigorous time-dependent approach based on wave packets.. Asymptotic boundary conditions We suppose that beyond some distance a from the origin the interaction potential V vanishes. The Schrödinger equation.) for r > a becomes + k ) Φr) =.) where we have written E = k /µ. One possible solution is the plane wave e ik r.3) which represents an incident flux of particles moving from left to right along the z-axis. possible solution is the outgoing spherical wave Another r eikr.4) which represents a scattered wave emanating from the target. In general, this scattered wave is not isotropic, and should be modulated by multiplying by an angular factor fθ, φ) called the scattering

6 Chapter. Scattering by a central potential 6 amplitude. We may therefore impose the boundary condition Φr) eik r + fθ, φ) r r eikr.5) to be satisfied by the stationary scattering state Φr). The form of the scattering amplitude fθ, φ) depends on the interaction potential V r). The central problem in scattering theory is the determination of this scattering amplitude. It should be stressed that the form.5) is only valid asymptotically, i.e. at very large distances from the origin. It is not valid in the region where the potential V is non-vanishing. Furthermore, this form does not correspond with the typical experimental situation given in figure.: the incident wave is not a plane wave but a narrow beam. Outside this beam we expect to see only the scattered wave. Using a wave packet description however, it can be shown that the form.5) does in fact give a correct description of the scattering process. In essence, that part of the incident beam which does not strike the target has no effect on the scattering process, and no error is introduced by treating this part of the incident beam incorrectly.. The cross section The probability current density is defined as j = Re For a plane wave of the form.3), the probability current density is [ ) ] Φ Φ..6) iµ j inc = k µ ẑ = v ẑ where v is the speed of the particle and ẑ is a unit vector along the z-axis. The probability current density for the scattered wave.4) is j sc = fθ, φ) r k µ ˆr +... = j inc fθ, φ) r ˆr +... where we have neglected terms of order /r 3 and higher since we are considering only the asymptotic form of the wave function. In order to calculate the cross section, we must calculate the number of particles per unit time that will be scattered towards the detector. This is located along a ray specified by the angles θ, φ) and subtends a solid angle dω = da/r, where da is the projected area of the detector perpendicular to the ray. The number of particles N counted per unit time is given by the scattered flux along the ray θ, φ) multiplied by this area: N = j sc ds = j sc da fθ, φ) = j inc da r = j inc fθ, φ) dω.

7 Chapter. Scattering by a central potential 7 The differential cross section is the ratio of this number to the incident flux j inc : dσθ, φ) = fθ, φ) dω or dσ dω θ, φ) = fθ, φ)..7) This is the fundamental relation between scattering theory and scattering experiments: it relates the quantity measured in the experiment the differential cross section), to the scattering amplitude which characterises the wave function at large distances..3 The optical theorem In a conservative potential, only elastic scattering is possible: the incident particle can only be deflected by the target. We therefore expect that the current density will be conserved. This may be expressed mathematically as j ds = surface where the integration is over the surface of an infinitely large sphere centered on the target. The current density for the asymptotic wave function.5) is j = k fθ, φ) {ẑ + µ r ˆr + [ r ẑ + ˆr)Re fθ, φ)e ikr z)]}. Using this, the surface integral is dω r j ˆr = ) { k dω fθ, φ) + Re dcos θ)dφ rfθ, φ) µ + cos θ) exp [ikr cos θ)]}. The first integral on the right-hand side is simply the total cross section. Thus if the conservation condition is to hold, σ TOT = Re dcos θ)dφ rfθ, φ) + cos θ) exp [ikr cos θ)]. The integral over θ on the right may be evaluated by integration by parts: dcos θ) fθ, φ) + cos θ) exp [ikr cos θ)] = i fθ, φ) + cos θ) exp [ikr cos θ)] kr = i kr fθ =, φ) + O k r ). π + O k r ) Since Φr) is a single-valued function, the scattering amplitude must be independent of φ in the forward direction. We therefore have the final result that σ TOT = 4π k Imfθ = )..8) This is known as the optical theorem. We have derived it in the case of potential scattering, but in fact it is also valid in more general cases. Physically, it may be interpreted as follows: the target casts a shadow in the forward direction where the intensity of the beam is reduced. The forward current is reduced by just the amount that appears in the scattered wave.

8 Chapter. Scattering by a central potential 8.4 The partial wave expansion We have shown in the previous section that the differential cross section is given by equation.7) dσ θ, φ) = fθ, φ) dω where the scattering amplitude fθ, φ) is related to the asymptotic behaviour of the wave function by Φr) Φr) is solution of the Schrödinger equation HΦr) = eik r + fθ, φ) r r eikr. ] [ µ + V r) Φr) = EΦr). In what follows, we shall consider the particular case where V r) is a central potential i.e. is spherically symmetric: V r) = V r)..9) It is usual to choose the z-axis parallel to k, the momentum of the incoming projectile. Both V r) and e ikz are then invariant under rotation about the z-axis. Φr) is thus independent of the azimuthal angle φ and so therefore is the scattering amplitude f: Φr) Φr, θ), fθ, φ) = fθ)..) We also know that H, l and l z commute. It is therefore possible to define a basis of common eigenfunctions of these three operators. Since the eigenfunctions of l and l z are the spherical harmonics it is possible to write the solutions as l Y lm θ, φ) = ll + ) Y lm θ, φ).) l z Y lm θ, φ) = m Y lm θ, φ),.) Φ Elm r) = R Elm r)y lm θ, φ)..3) By substituting Φ Elm r) into the Schrödinger equation, we find that the radial function must satisfy a second-order differential equation, called the radial equation, [ d µ dr + ) ] d ll + ) + r dr µr + V r) R Elm r) = ER Elm r)..4) All the coefficients in this equation are independent of m, and so therefore is the radial function. We can thus drop the index m, and from now on we shall also drop the index E and simply write R l r). The stationary scattering wave function, satisfying the boundary condition.5), cannot be written as R l r)y lm θ, φ) but can however be expanded in the complete basis of spherical harmonics as Φr) = lm c lm R l r)y lm θ, φ)..5) This is called the partial wave expansion of Φr). It allows us to reduce the problem of solving the full Schrödinger equation.) to solving the radial equation.4). To define the coefficients c lm, it is necessary to study the asymptotic behaviour of the solutions of.4).

9 Chapter. Scattering by a central potential 9.4. Partial wave expansion of the free particle wave function Let us first consider the case V r) = and write E = k /µ. Equation.4) simplifies to [ d dr + r ] d ll + ) dr r + k R l r) =..6) We set ρ = kr ρ is a dimensionless quantity). The previous equation becomes where we have used [ d dρ + d ρ dρ ] ll + ) ρ + R l ρ) =.7) d dρ = d k dr and where we have set R l r) = R l ρ). Equation.7) is called the spherical Bessel differential equation. It has two linearly independent solutions: the regular solution is the spherical Bessel function j l ρ) = while the irregular solution is the spherical Neumann function ) π / J ρ l+/ρ),.8) n l ρ) = ) l+ π ρ) / J l / ρ),.9) where J ν ρ) is the cylindrical Bessel function. Writing d dρ + d ρ dρ = ρ dρ ρ. d and substituting R l ρ) = ρ u lρ) into.7), we obtain [ ] d ll + ) dρ ρ + u l ρ) =..) As ρ, the regular and irregular solutions of this simpler equation behave as ρ l+ and ρ l respectively. The coefficients of these terms are determined from the definitions.8) and.9) and the properties of the cylindrical Bessel functions. We find that j l ρ) n l ρ) r r ρ l l + )!!.) l )!! ρ l+..) The behaviour of these functions for r is j l ρ) n l ρ) r sinρ lπ/).3) ρ cosρ lπ/)..4) r ρ

10 Chapter. Scattering by a central potential Specific examples of spherical Bessel functions are j ρ) = sin ρ ρ n ρ) = cos ρ ρ j ρ) = sin ρ ρ n ρ) = cos ρ j ρ) = n ρ) = ρ 3 ρ 3 ρ cos ρ ρ sin ρ ρ ) sin ρ 3 ρ cos ρ 3 ρ 3 ρ ) cos ρ 3 sin ρ. ρ.5).8 l = l = l = l = 3.6 j l ρ) ρ Figure.: Regular Bessel functions j l ρ).4. n l ρ) l = l = l = l = ρ Figure.: Irregular Bessel functions n l ρ)

11 Chapter. Scattering by a central potential The plane wave e ikz is obviously a regular function. It can therefore be expanded in the basis of regular spherical Bessel functions and spherical harmonics: e ikz = l l= m= l c lm j l kr)y lm θ, φ)..6) Since e ikz is invariant under rotation around the z-axis, it must be independent of φ and only terms with m = can appear in expansion.6). Using the explicit expression l + Y l θ, φ) = 4π P lcos θ), expansion.6) can be rewritten as e ikz = a l j l kr)p l cos θ)..7) l= The value of a l can be obtained by projecting.7) on a particular P l cos θ). Using the orthogonality of the Legendre polynomials we obtain dθ sin θp l cos θ)p l cos θ) = l + δ ll.8) π l + a lj l kr) = e ikr cos θ P l cos θ) sin θdθ..9) Evaluating this integral for r and comparing with the asymptotic behaviour of j l kr), we find that a l = l + )i l. We therefore have e ikz = l + )i l j l kr)p l cos θ)..3) l= Using the addition theorem of the spherical harmonics P l cos θ) = 4π l + ) l l= m= l Y lm ˆk)Y lm ˆr), we find that the partial wave expansion of the general plane wave e ik r = e ikr cos θ with θ = k, r) is e ik r = 4π l l= m= l i l j l kr)y lm ˆk)Y lm ˆr)..3)

12 Chapter. Scattering by a central potential.4. Partial wave expansion of the scattering wave function For the case of a central potential V r) and an incident plane wave along the z-axis, the stationary scattering wave function must be independent of φ: Φr) = R l r)p l cos θ),.3) l= where R l r) is the radial function satisfying the radial equation [ d dr + ] d ll + ) r dr r Ur) + k R l r) =.33) with Ur) = µv r), k = µe. If Ur) and k are very small with respect to r when r, equation.33) becomes [ d dr + ] d ll + ) r dr r R l r) =.34) which has solutions of the form r λ with λ = l or λ = l + ). A general solution of.34) is a linear combination of r l and r l+) : R l r) a r lr l + b l r l+)..35) Since r l+) diverges at the origin for l, we must have b l =. For r, Ur) and.33) takes the form [ d dr + ] d ll + ) r dr r + k R l r) =..36) The solutions of this equation are j l kr) and n l kr). We can therefore write r R l r) kr A r lj l kr) + B l n l kr)..37) From the asymptotic forms of the spherical Bessel and Neumann functions, we have that [ ) R l r) A l sin B l cos kr lπ It is always possible to define c l and δ l such that kr lπ )]..38) A l = c l cos δ l, B l = c l sin δ l or We can then rewrite.38) as tan δ l = B l c l = A A l + B l. l R l r) r kr c l sin kr lπ ) + δ l..39) The real quantity δ l is called the phase shift for the partial wave l. It reflects the intensity of the interaction potential. If Ur) =, the solution of.33) is simply the regular spherical Bessel function

13 Chapter. Scattering by a central potential 3. δ Repulsive u r) -.. δ Attractive u r) r Figure.3: Reduced radial wave functions for a) repulsive and b) attractive finite-range potentials with l =. The dashed lines correspond to the reduced radial wave function for V r) =. j l and we have B l =, δ l =. For a repulsive potential, i.e. V r) > for all r, the radial wave function is pushed away from the origin see figure.3). Its nodes are displaced by δ l /k towards larger values of r, and so the phase shift is negative. For an attractive potential, the radial wave function is pulled in towards the origin. Its nodes are thus displaced towards smaller values of r, and the phase shift is positive. It is clear that the stronger the interaction potential Ur), the larger δ l. The phaseshift must be found by solving analytically or numerically the radial equation.33) whenever possible, otherwise it must be determined by approximation methods. Since the phaseshift depends on the collision energy E or momentum k, it can be denoted δ l E) or δ l k). Let us consider a potential with a finite range, i.e. For r a, the radial function is or V r) = for r a. R l r) = A l j l kr) + B l n l kr).4) R l r) = A l [j l kr) tan δ l n l kr)].4) with tan δ l = B l..4) A l Suppose that the log-derivative of the solution at r = a, γ l a) = R l a) R l a),

14 Chapter. Scattering by a central potential 4 has been determined by solving the problem in the region r a. Since we can write R l a) = A l [j l ka) tan δ l n l ka)] R l a) = A lk [ j l ka) tan δ ln l ka)], γ l a) = k j l ka) tan δ ln l ka) j l ka) tan δ l n l ka). The tangent of the phaseshift is therefore given in terms of γ l a) as tan δ l = kj l ka) γ lj l ka) kn l ka) γ ln l ka)..43).4.3 Partial wave expansion of the scattering amplitude We now impose on the scattering wave function.3) Φr, θ) = R l r)p l cos θ) l= the boundary conditions.5) Φr, θ) r eikz + fθ) r eikr. To do this, we first introduce the partial wave expansion.3) of the plane wave and develop fθ) as This yields Φr, θ) r l= fθ) = f l P l cos θ). l= [ ] l + )i l j l kr) + f l r eikr P l cos θ). We further replace the spherical Bessel function j l kr) by its asymptotic form.3) j l kr) r sinkr lπ/) = [e ikr lπ/) e ikr lπ/)] kr ikr to obtain the asymptotic form of the scattering wave function Φr, θ) r l= [ ) l + r ik + f l e ikr l+ l + + ) ik e ikr ] P l cos θ)..44) The quantity between the square brackets can obviously be identified with the asymptotic form of R l r). This is also given by.39), which can be rewritten as R l r) r r ik c l [ ] e ikr i l e iδ l e ikr i l e iδ l..45) By equating the coefficients of the incoming wave e ikr in equations.44) and.45), we find that r ik c l i l e iδ l = l + )l+ r ik.46)

15 Chapter. Scattering by a central potential 5 which implies c l = l + )i l e iδ l. By substituting c l into.45) and equating the coefficients of the outgoing wave e ikr in equations.44) and.45), we find that r ik c l i l e iδ l = ) l + r ik + f l.47) and hence f l = [ ik l + ) We thus obtain the important relation fθ) = ik l= = k l= ] e iδ l l + ) = l + e iδ l sin δ l. k [ ] e iδ l P l cos θ) l + )e iδ l sin δ l P l cos θ)..48) The quantity S l = e iδ l is complex and is known as the scattering matrix or S-matrix element for angular momentum l. A related quantity is the K-matrix element K l = tan δ l which is real. It is related to S l by since we have S l = + ik l ik l + i tan δ l i tan δ l = cos δ l + i sin δ l cos δ l i sin δ l = eiδl e iδ l = eiδ l. The asymptotic form.38) of R l r) can be written in term of the K-matrix element as [ R l r) r kr A l sin kr lπ ) + tan δ l cos kr lπ )].49) with A l = c l cos δ l = l + )i l e iδ l cos δ l. It is clear that the partial wave expansion method should be used only if a small number of partial waves l contribute to the scattering amplitudes. This is true at low collision energy, where for large values of l the projectile cannot penetrate through the centrifugal barrier ll + )/r into the region where the interaction potential V r) is important. More precisely, scattering only occurs for values of l for which the classical turning point lies within the range a of the potential. This turning point is defined as the distance where the kinetic energy equals the potential energy, i.e. when k = ll+)/r cl, so that r cl l/k. For r < r cl, the amplitude of the wave function decays rapidly see figure.4). If the turning point lies beyond a, the effect of the potential is negligible and the phase shift is vanishingly small. Scattering therefore only occurs when r c a, i.e. for l ka.

16 Chapter. Scattering by a central potential l = u r) l = 5 l =. -. l = r Figure.4: Reduced radial wave functions for a finite-range attractive potential with l =,,5 and. The dashed lines correspond to the reduced radial wave function for V r) =.

17 Chapter. Scattering by a central potential Total Cross section The differential cross section is given by.7) dσ θ, φ) = fθ, φ) dω = k ll l + )l + )e iδl δ l ) sin δ l sin δ l P l cos θ)p l cos θ). The total cross section is obtained by integrating the differential cross section dσ σ = θ, φ) dω. dω 4π The integration over dφ gives simply π, since fθ) is independent of φ. relation between the Legendre polynomials.8), we find Using the orthogonality This can also be written as σ = 4π k l + ) sin δ l..5) l= σ = σ l.5) l= where the cross section in the partial wave l is defined as The maximum possible value of σ l is σ l = 4π k l + ) sin δ l..5) σ max l = 4π l + ). k This value is reached for δ l = n + /)π with n an integer. On the other hand, if δ l = nπ, σ l =. From.48), we have Comparing with.5), we see that which is the optical theorem.8). Imfθ = ) = k σ = 4π k l + ) sin δ l. l= Imfθ = )

18 Chapter. Scattering by a central potential 8.5 Applications.5. Scattering by a hard sphere Consider a potential V r) such that Ur) = for r a Ur) = for r > a. U a r Figure.5: Hard sphere potential For r a, we must have R l r) =. The general solution for r a has the form R l r) = A l j l kr) + B l n l kr). Since the wave function must be continuous, R l a) = A l j l ka) + B l n l ka) =, and therefore tan δ l = B l A l = j lka) n l ka)..53) The behaviour of tan δ l at low energy can be determined by considering the behaviour of j l ρ) and n l ρ) for ρ. We find tan δ l ka) l+ k l + )!!l )!!. Thus tan δ l decreases rapidly as l increases for fixed values of k and a). At low energy, scattering is therefore dominated by the s wave l = ) and is thus isotropic. The phaseshift is determined from sin δ tan δ k ka. The differential cross section is thus given by and the total cross section is dσ dω σ k a. k 4πa.

19 Chapter. Scattering by a central potential 9 Similarly, the behaviour of tan δ l at high energy can be determined by considering the behaviour of j l ρ) and n l ρ) for ρ. We find tan δ l k ka + lπ for l ka. We note that the phase shifts do not vanish but tend to. The total cross section tends to σ k l 4π max ) lπ k l + ) sin ka with l max ka. Using the fact that sin θ ± π/) = cos θ, we have sin ka) + 3 sin ka π ) + 5 sin ka π) + 7 sin ka 3π = sin ka) + sin ka π ) + sin ka π 3 sin ka 3π ) + 4 sin ka 3π ) +... = sin ka) + cos ka) + sin ka π ) + cos ka π 4 sin ka 3π = ) +... l= ) +... ) + sin ka π) + 3 sin ka π) + We thus find that as k, the total cross section is approximately equal to σ 4π l max k l πa. l= ) + 3 sin ka π) + 3 cos ka π) +.5. Scattering by a square well potential Consider the square well potential where U is a positive constant. Ur) = { U, r < a, r > a In the internal region r < a the radial equation.33) becomes [ d dr + ] d ll + ) r dr r + κ R l r) = with κ = k + U ) /. This is of the form.36) and so the regular solution is R l r) = N l k)j l κr) where N l k) is an energy-dependent normalisation factor. In the external region r > a the effective potential Ur) is zero, and the radial wave function is given by.4). Continuity of the radial wave function and its derivative then gives tan δ l = kj l ka) γ lj l ka) kn l ka) γ ln l ka)

20 Chapter. Scattering by a central potential U internal region external region a r U Figure.6: Square well potential with γ l a) = κj l κa) j l κa). In particular, for l = and using the explicit forms.5) of the spherical bessel functions j ρ), n ρ), we have k tanκa) κ tanka) tan δ = κ + k tanκa) tanka), from which δ = ka + tan k κ tan κa ). The scattering length α is defined as tan δ α = lim. k k For a square well potential, we have )] kκ α = lim [ ka k k tan + tan tan κa = k [ )] k ka + tan κa κ with κ U and where we have used the relation tanɛ + θ) ɛ + tan θ. The scattering length is therefore given by [ α = tanλ ] a) a λ a where we have defined λ = U. For weak couplings λ a ) the phase shift δ tends to zero and the scattering length is negative. When λ a = π/, the phase shift tends to π/ as k, while the scattering length is infinite. The s-wave cross section in this case diverges as k. If λ a is just above π/, so that the potential supports an s-wave bound state, the phase shift tends to π as k. It can in fact be shown that if the potential can support n l bound states of angular momentum l, then lim δ lk) = n l π. k This is known as Levinson s theorem, and is true for more general potentials than the square well.

21 Chapter. Scattering by a central potential.6 Resonances In some cases, the phaseshift, and therefore the cross section, varies rapidly as a function of the collision energy E. The partial scattering amplitude is given by It is always true that if tan θ = y/x, then f l = [ ] ik l + ) e iδ l. e iθ = eiθ cos θ + i sin θ = e iθ cos θ i sin θ = + i tan θ i tan θ = tan θ i ix = y tan θ + i y + ix..54) Using expression.43) of tan δ l, we can therefore write e iδ l = kj l γ lj l ikn l γ ln l ) kj l γ lj l + ikn l γ ln l ) = kj l in l ) γ lj l in l ) kj l + in l ) γ lj l + in l ) = j l in l j l + in l j l k in l j l in l γ l k j l +in l j l +in l γ l..55) Remembering that the phaseshift ξ l for scattering by a hard sphere of radius a is given by.53) tan ξ l = j lka) n l ka), we can rewrite the first factor in.55) using.54) as j l in l j l + in l = e iξ l. If we define we can rewrite k j l + in l j l + in l = r l + is l, k j l in l j l in l γ l k j l +in l j l +in l γ l = r l γ l is l r l γ l + is l. The modulus of this quantity is and we can always rewrite it as e iρ l with We can therefore rewrite equation.55) as tan ρ l = s l γ l r l. e iδ l = e iξ l e iρ l and hence δ l = ξ l + ρ l. By definition, the phase ξ l does not depend on the interaction potential and is a smooth function of the scattering energy. For instance, for l =, we have ξ l = ka. In contrast, ρ l depends on the

22 Chapter. Scattering by a central potential δ l σ tot σ max 3π/4 π/ σ max / π/4 E r Γ/ E r E r + Γ/ E E r Γ/ E r E r + Γ/ E Figure.7: Behaviour of the phase shift and the total cross section in the vicinity of a narrow resonance at energy E r interaction potential through the log-derivative γ l. If at E = E r, the denominator γ l r l =, we have tan ρ l = and ρ l = π/. Let us suppose that in the region of E E r, we have s l γ l r l Γ E r E). If we suppose that the contribution of ξ l is negligible this is generally not true), we obtain the Breit-Wigner formula ρ l δl r = Γ tan E r E). In this case, the phaseshift varies quickly and jumps by π. The corresponding scattering amplitude is f l = l + k Γ/ E E r + i Γ. If the contribution from the other partial waves is negligible, we have fθ) = f l P l cos θ) and therefore dσ dω = fθ) l + ) Γ /4 k P E E r ) + Γ l cos θ) 4 Vr) Figure.8: Potential giving rise to a shape resonance

23 Chapter. Scattering by a central potential 3 and π σ = π fθ) 4πl + ) Γ /4 sin θdθ k. E E r ) + Γ 4 The cross section has the shape of a Lorentzian, with a width at half-height equal to Γ. At E = E r, the cross section is maximum and has the value σ = σ max = 4π l + ). k Physically, a resonance occurs when the projectile is temporarily captured during the collision, in a quasi-bound state of an attractive potential as shown in figure.8). The lifetime of the resonance is τ = /Γ..7 Scattering by a coulomb potential We consider a particle of masss µ moving in a Coulomb potential α/r a positive value of α corresponds to a repulsive potential, a negative value of α corresponds to an attractive potential). The Schrödinger equation describing the motion is µ + α ) r E ψr) =.56) where E is the kinetic energy of the particle..7. Exact solution Defining.56) may be written in the form We try a solution of the form k = µ E.57) ν = µα.58) k νk ) + k ψr) =..59) r ψ = expik r)fr).6) and substitute it into.59). This yields an equation for fr) + ik z νk ) fr) =.6) r where the polar axis has been defined in the direction of the vector k. We now introduce the parabolic coordinates ξ, η, φ), ξ = r z η = r + z φ = tan y/x)

24 Chapter. Scattering by a central potential 4 where x, y, z) are the Cartesian components of r. The Laplacian in these coordinates is [ 4 ξ ) + η ) ] + ξ + η ξ ξ η η ξη φ while z = η η + ξ η ξ ). ξ Since the system has an axial symmetry about the z-axis, the function fr) is independent of φ and.6) may be written as [ ξ ) + η ) + ik η ξ ξ η η η ξ ) ] νk f =..6) ξ This equation is separable, and so writing f = f ξ)f η) and substituting into.6) we obtain the equations for f and f [ ξ d ) ] ikξ d dξ dξ β f ξ) = d dξ [ d dη η d dη ) ] + ikη d dη β f η) = where β + β = νk. Defining ξ = ikξ, η = ikη these equations may be written in the form [ ] d ξ dξ + ξ ) d β f =.63) dξ ik [ ξ d dη + η ) d + β dη ik ] f =.64) which is the confluent hypergeometric equation. Since the wave equation must be finite and singlevalued everywhere, only the regular solutions of.63) and.64) are required. These are f ξ) = A F i β ) ; ; ikξ k f η) = A F i β ) ; ; ikη k where A, A are normalisation constants. In order to determine these, we must consider the behaviour of the wave function ψ. The boundary condition requires that asymptotically ψ should represent an incident wave and an outgoing scattered wave. Thus the scattered part must contain a term of the form expikr)/r. This requires that fr) contains a term of the form expikr z))/r in order to cancel the term expikz) occuring in.6). This in turn implies that f does not depend on η, and so choosing β =, β = νk the appropriate solution is ) ψ = A expik r) F iν; ; ikr ik r. From the asymptotic form of the confluent hypergeometric function we find that [ ] A expπν/) ψ I + Sfθ) r Γ + iν).65)

25 Chapter. Scattering by a central potential 5 where ) I = exp ik r + iν lnkr k r) S = ) ikr r exp iν lnkr) ν ) fθ) = k sin θ/) exp iδ iν ln sin θ/).66).67).68) and cos θ = ˆk ˆr, δ = arg Γ + iν). The incident wave is represented by.66), the scattered wave by.67) while the Coulomb scattering amplitude fθ) is given by.68). From ) we see that the Coulomb potential distorts the incident and scattered waves even at asymptotic separations it has infinite range). From.65) we see that choosing A = exp πν/)γ + iν) normalises the incident wave to unit amplitude. Combining these results we obtain the solution of.59) satisfying outgoing scattered wave conditions: ψ = exp πν/)γ + iν) expik r) F iν; ; ikr ik r)..69) Using.68), the differential cross section for scattering by a Coulomb potential is given by dσ dω θ) = fθ, φ).7) = ν 4k sin 4 θ/).7) which is the Rutherford scattering formula as derived from classical mechanics. Note that in the forward direction θ = ), the differential cross section is infinite. The total cross section is therefore infinite. This is a consequence of the infinite range of the Coulomb potential: all particles are deflected no matter how far they are from the scattering center..7. Partial wave expansion By analogy with the partial wave expansion.3) for the plane wave, we write ψr) = kr l + )i l u l r)p l cos θ)..7) l= The radial function u l r) satisfies the differential equation [ d ll + ) dρ ρ ν ] ρ + u l ρ) =.73) where we have introduced the variable ρ = kr. We then try a solution of the form u l ρ) = ρ l+ e iρ F l ρ). Substituting into.73) yields [ρ d dρ + l + + iρ) d ] + i l + + iν) F l ρ) =. dρ

26 Chapter. Scattering by a central potential 6 This can be written in the form of the confluent hypergeometric equation by the change of variable x = iρ. We thus obtain F l ρ) = C l F l + + iν; l + ; iρ). The asymptotic form of the radial function is therefore u l r) r C l l + )! e πν/ l Γl + + iν) sin kr lπ/ ν logkr) + σ l) where σ l = arg Γl + + iν) is the Coulomb phase shift in the partial wave l. We choose the normalisation constant C l so that the wave function behaves asymptotically as u l r) The regular radial Coulomb function is defined as F l ν, kr) = e iσ l u l r) r eiσ l sin kr lπ/ ν logkr) + σ l )..74) = e πν/ Γl + + iν) l + )! The partial wave expansion for ψr) is therefore ψr) = kr kr) l+ e ikr F l + + iν; l + ; iρ)..75) l + )i l e iσ l F l ν, kr)p l cos θ)..76) l=.8 R-matrix theory for potential scattering We consider the scattering of an electron by a short-range central potential Ur), in the partial wave s l = ). The Schrödinger equation for the reduced radial wave function ur) is ) d + Ur) + k dr ur) =.77) where Ur) =, r a. The boundary conditions.38) to be imposed on ur) are u) = ur) = A k sin kr + K cos kr) r a).78) where K = tan δ. We choose to develop the solution ur), in the region r a, in the complete basis of eigensolutions of the following equation ) d dr + Ur) + k λ u λ r) =.79) with the boundary conditions a u λ a) du λ dr u λ ) =.8) = b.8) r=a

27 Chapter. Scattering by a central potential r a ) Figure.9: R-matrix basis functions for Ur) = /r, l =, a = a. The first five states displayed have an energy of -.5, -.535, , and.65 Hartree respectively. Following conditions.8), their derivative at r = a is null. The wave functions full lines) are compared to those dashed lines) of the first 5 exact states of hydrogen. The inner region is large enough to encompass the s and s states, allowing these to be reproduced almost exactly. This is not true for the n 3 states, whose energies and orbitals differ substantially from those of hydrogen. where b is any constant number, usually chosen to be zero. In this case,.8) implies that the derivative of all basis functions u λ r) are null at the boundary r = a, as shown in figure.9. The functions u λ are called R-matrix basis functions. Since they are eigenfunctions of the same hermitian operator, they are orthonormal a u λ r)u λ r) dr = δ λλ..8) The lowest state has zero node, the first excited state has one node, etc. We expand the solution of the Schrödinger equation.77) on the basis of R-matrix functions ur) = a λ u λ r) r a)..83) λ=

28 Chapter. Scattering by a central potential 8 This series converges uniformly, except at r = a. This implies in particular that du du λ dr a λ r=a dr..84) r=a λ= The expansion coefficients a λ can be found by considering the integrals a ) d u λ r) + Ur) + k ur) dr =.85) dr a ) d ur) dr + Ur) + k λ u λ r) dr =..86) By substracting the second equation from the first one, we obtain a ) u λ r) d d ur) ur) dr dr u λr) dr = kλ k ) a u λ r)ur) dr = k λ k ) a λ.87) which is non-zero as the kinetic operator is not hermitian when the integral is defined over a finite range of r. The left-hand side can be evaluated by integrating by part and using the boundary conditions.78) and.8): a u λ r) d dr ur) dr = u λr) d dr ur) a a d dr u λr) d ur) dr dr a a d dr u λr) d ur) dr dr = u λ a) d dr r=a ur) ur) d dr u λr) dr = ua) d dr u a λr) d r=a dr u λr) d ur) dr.88) dr u λ a) d dr r=a ur) ua) d dr u λr) = kλ k ) a λ..89) r=a We therefore find d a λ = kλ u λa) k dr ur) b ) r=a a ua) = a kλ u λa) a d ) k dr r=a ur) bua)..9) We now define the element of the R-matrix as R = a k λ= λ u λ a)..9) k Substituting.9) into.83) and evaluating at r = a, we get ua) = a λ u λ a) = u λ a) a k λ= λ= λ a dur) ) k dr bua) r=a [ = R a dur) ] dr bua)..9) r=a which implies [ R = ua) a dur) dr bua)]..93) r=a

29 Chapter. Scattering by a central potential 9 If b =, R is the inverse of the log-derivative of the wave function at r = a. We substitute.78) into.93) to obtain sin ka + K cos ka R =.94) ak cos ka Kk sin ka) bsin ka + K cos ka) which can be inversed to yield the K-matrix element Rka cos ka b sin ka) KRka sin ka + b cos ka) = sin ka + K cos ka Kcos ka + R[ka sin ka + b cos ka]) = sin ka + Rka cos ka b sin ka) K = sin ka + Rka cos ka b sin ka) cos ka + Rka sin ka + b cos ka)..95) Once the eigenvalues of the inner region Hamiltonian.79) and the corresponding eigenvector surface amplitudes have been determined, the R-matrix and thus the K-matrix are readily obtained for different collision energies using relations.9) and.95). The collisional problem has therefore been reduced to a structure problem where the main computational effort consists in diagonalizing the hamiltonian of the collisional system. This task can be performed efficiently using standard linear algebra packages such as LAPACK.