(1) Introduction: a new basis set
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1 () Introduction: a new basis set In scattering, we are solving the S eq. for arbitrary VV in integral form We look for solutions to unbound states: certain boundary conditions (EE > 0, plane and spherical waves at infinity, as opposed to EE < 0 and zero at infinity for bound states) An important special case we already encountered for bound states: potentials of spherical symmetry. Our basic goal: solve the S eq. in integral form We reduced this to finding nn TT ii or nn VV ii At high energies applied the plane wave basis set because the scattered wave function can be written as ψψ + ii + cccccccccccccccccccc. We used properties of the scattered particle, but none of the scatterer. If VV is spherically-symmetric, then calculating αα VV ββ are easy in the spherical wave basis set. This is the idea of the PW method.
2 () Introduction: a new basis set HH 0 = pp, LL, LL zz are compatible operators. For instance, HH 0, LL zz = 0 from commutation relations. Also, if VV is spherically-symmetric VV, LL = 0 and VV, LL zz = 0 Then, HH, LL = 0, HH, LL zz = 0 where HH = HH 0 + VV HH, LL, LL zz are compatible we can use EE, ll, mm as variables (this basis set was already applied in undergraduate QM to solve for bound states with EE < 0) Apply a special basis set made of spherical wave states EEEEEE, just as we did for bound states in potentials of spherical symmetry. Note: EE > 0 for spherical wave states EEEEEE. This is a new basis set, different from the one applied in solving bound state problems Note: the bound-state solution of given EE, ll, mm is not a spherical wave and cannot be written as a sum over this basis set because it has EE < 0 This is elastic scattering only. Formally, TT, HH = 0 and SS, HH = 0. Also, TT nnnn commutes with LL and LL zz (from solution TT = VV + VV EE HH 0 VV + ). Therefore TT, LL = 0 and TT, LL zz = 0. Also, SS, LL = 0 and SS, LL zz = 0 because SS and TT differ in a pre-factor and the identity operator only The representation of SS and TT are diagonal in the llll basis set. We will see how, just as for bound states, the S eq. for each ll, mm is reducible to a D rr dependence eq. Note: this same basis set is used in classical EM, where it can describe elastic scattering from spherical potentials [more complex algebra because we have two vector fields instead of one scalar ψψ(xx), but it is basically the same idea of channels with given ll, mm (classical spherical harmonics)]
3 () EEEEEE states in the kk representation kk kk = δδ(kk kk ) is a complete and orthogonal basis set made of plane wave states We want a complete and orthogonal basis set EE ll mm EEEEEE = EE δδ llll δδ mmmm in the EE, ll, mm variables we introduced. The three variables kk xx,yy,zz are replaced by three variables EE, ll, mm In the kk representation kk EE, ll, mm = kkkkkk EEEEEE = δδ ħ kk EE YY mmmm ll mm ħ mm kk [units of RHS are = as on LHS] mmmm ħ kk kkkk The angular part comes from kkkkkk EEEEEE = kk EE θθθθ llll and θθθθ llll = YY mm ll (kk ) The energy part comes from the relation between kk and EE The pre-factor comes from assuring the correct units as dd 3 kk kk kk = and de EE EE = and kk EE kk EE kk 3 mmmm kk We can verify that this form is correct, by checking that the basis set is orthogonal EE ll mm EEEEEE = ddkk kk ddω EE ll mm kk kk EEEEEE = ddkk kk ddω mmkk δδ EE kk δδ llll δδ mmmm δδ EE EE ddkk kk mm k YY ll mm θθθθ YY ll mm θθθθ = k = δδ llll δδ mmmm δδ EE EE dddddd EE kk = δδ llll δδ mmmm δδ EE EE The basis set is also complete the spherical harmonics are complete in the θθ, φφ variables and every kk can be uniquely related to an energy with δδ ħ kk EE.
4 () EEEEEE states in the xx representation In the xx representation xx EE, ll, mm = iill jj ll kkkk are spherical Bessel functions Appendix: jj 0 = sin ρρ ρρ, jj = sin ρρ ρρ ħ cos ρρ, etc. ρρ mm ππ jj ll kkkk YY ll mm θθ, φφ Example: illustrate the simplest case of ll = 0, mm = 0 We have xx EE, 0,0 = mm ππ sin kkkk mmmm kk and kk EE, 0,0 = kkkk We can get one from the other with the relation of the previous slide xx EE, 0,0 = dd 3 kk xx kk kk EE, 0,0 = dd 3 kk xx kk = ddddddωkk ππ mmmm = dddd mm kk kk ddω = mmmm ππ = mmmm ππ = mmmm 3 3 ππ kk 3 ee iikk xx kk mmmm ππ ddωeeiikk xx 3 ee iikk xx kk mmmm ππ ππ sin θθ 0 sin kkkk kkkk iiiiii cos θθ dddd ee which is the expected result
5 () Plane waves as a sum over partial spherical waves We can represent a plane wave state in this basis set [each partial wave is a spherical wave] Note: here kk = pp (free plane wave) kk = dddd EEEEEE kkkkkk EEEEEE llll xx kk = dddd llll mmmm = dddd llll mmmm kk kk YY ll mm ii ll mm ππ jj ll kkkk (YY ll mm ) Integrate EE out and apply the addition theorem YY mm mm mm ll θθθθ YY ll (θθθθ) Then xx kk = ππ YY ll mm EEEEEE = ll+ PP ll(cos θθ) 3 ll ii ll ll + PP ll cos θθ jj ll (kkkk) [result will be useful in hard-sphere scattering] Note : aligning kk along the z-axis means that LL zz kk = 0 dddd llll LL zz EEEEEE = 0 mm = 0 (azimuthal symmetry) (no change in the plane wave on rotation about the z-axis) but LL kk 0 (many ll ss enter in the previous sum) Note : plane waves do not have spherical symmetry, which is why this result is relatively complicated Note 3: the PW method uses the normalization xx kk = eeiikk rr ππ 3/, instead of xx kk = eeiikk rr LL 3/. This makes both ff and TT measurable (Argand diagram) in contrast to the case of box normalization, where ff = mmll3 kk TT kk, and only ff is measurable (T is infinitesimal as LL ). ππ
6 (3) PW scattering phases VV representation is diagonal in the PW basis set TT is diagonal SS is diagonal independent ll channels αα = SS EEEEEE EE ll mm αα = SS ll EE EE δδ lllll δδ mmmmm (no change in EE, ll, mm) EE ll mm TT EEEEEE = TT ll EE EE δδ lllll δδ mmmmm EE ll mm SS EEEEEE = SS ll EE EE δδ lllll δδ mmmmm Therefore, channels of scattering defined by EE, ll, mm variables do not interchange norms Scattering channels are independent Scattering is interpreted as a sum over partial scatterings, each occurring in its own channel. Note: since mm = 0 it follows from the above that mm = 0 too. This index is always zero. The energy is also always the same in elastic scattering only the ll index varies. The conservation of probability condition SS + SS = becomes SS ll+ SS ll = for each channel ll EEEEEE SS ll+ SS ll EEEEEEEEEEEE = EE δδ llllll δδ mmmmmm ddddd EEEEEE SS ll+ EE ll mm EE ll mm SS ll EEEEEE = EE llllll δδ llllll δδ mmmmmm SS ll (EE)SS ll (EE) = SS ll kk = SS ll (kk) = ee iiδδ ll [note: SS ll (kk) has no units] The amplitude of the partial wave (or norm of the wave) will not be changed unitarity implies phase shifts only are possible Therefore, the usual scattering parameters (functions) for each channel TT ll EE or ff ll (θθ) are replaced by a special set of numbers δδ ll θθ this simplification came from the assumption that VV(rr) is sphericallysymmetric.
7 (4) Relating δδ ll to a partial wave amplitude ff ll By definition, n, i are eigenstates of H 0 (not of H ) EE nn,ii states are asymptotic free plane wave states, where kk = pp (pp is the eigenvalue of momentum operator pp ) [in general, kk and EE = kk are essentially the same variable, while kk and pp are not the same] In the EE basis set EE nn SS EE ii = δδ nnnn ππππ EE nn TT EE ii δδ EE nn EE ii In the kk basis set kk SS kk = δδ kk kk ππππ kkk TT kk δδ(ee kk EE kk ) EE ll mm TT EEEEEE = TT ll EE δδ(ee EE)δδ lllll δδ mmmmm kkkkkk TT kkkkkk = TT ll kk δδ kk kk δδ llll δδ mmmm δδ αααα = αα δδ xx δδ EEE EE = mm kk δδ(kk kk) Look at each channel separately Then EE nn SS ll EE ii = δδ nnnn ππππtt ll EE mm kk δδ(kk kk) SS ll (kk) = ππππtt ll kk [overall δδ(kk kk ) cancels] In general, the scattering amplitude is ff(kk, kk) TT kk, kk and we want a PW amplitude ff ll TT ll kk define ff ll kk ππtt ll EE kk Therefore, SS ll (kk) = + iiiiff ll kk Then ff ll = SS ll iiii = eeiiδδ ll sin δδ ll kk = kk cot δδ ll iiii [last expression useful for resonance]
8 (4) Relating δδ ll to usual scattering parameters EE ll mm TT EEEEEE = TT ll EE δδ(ee EE )δδ lllll δδ mmmmm We also have the representation of TT in the kk basis set as ff kk, kk = LL3 kk TT kk To find a relation between TT ll (EE) and ff, we have to make a change a basis insert identity, etc. mm ππ 3 Specifically, ff = kk TT kk = = TT ππ kk ll,mm ll EE This gives ff(θθ) = ππ (ll + )TT kk ll EE PP ll (cos θθ) ff θθ = ll=0 ll + ff ll kk PP ll (cos θθ) ff θθ = ll + kk eeiiδδ ll ll sin δδ ll PP ll (cos θθ) ll YY ll mm kk YY ll mm kk [mm = 0 everywhere] Derive σσ tttttt = kk (ll + ) sin δδ ll ll with ddωpp ll (cos θθ)pp ll (cos θθ) = ll+ δδ llll [from ddωyy ll mm θθ, φφ YY ll mm θθ, φφ = δδ llll δδ mmmm with mm = mm = 0] Note: σσ ll kk ll + EE Note: the optical theorem is satisfied as expected, because no approximations were made
9 (4) Example from proton proton scattering
10 Plane wave basis set Spherical wave basis set σσ = ff ddω dddd ddω = ff σσ = ll ll + kk sin δδ ll Differential cross-section not useful ff kk, kk = mmll3 ππ kk TT kk ff ll kk = ππtt ll EE kk SS nnnn has no units SS ll (kk) has no units TT nnnn units of energy TT ll (kk) has no units ff kk, kk ff ll (kk) has no units SS nnnn = δδ nnnn ππππππ EE nn EE ii TT nnnn SS ll kk = ππππtt ll (kk)
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