1.2 Differential cross section

Transcription

1 .2. DIFFERENTIAL CROSS SECTION Febuay 9, 205 Lectue VIII.2 Diffeential coss section We found that the solution to the Schodinge equation has the fom e ik x ψ 2π 3/2 fk, k + e ik x and that fk, k = 2 m d 3 x k x V x x ψ. Not eally much good since we need the solution to do the calculation, but we do lean something about the fom. We have been sloppy about nomalization. Multiply by V 2 whee V is volume of space. Then if the plane wave epesents the incoming flux, we have incident flux v/v = /mv. The flux scatteed adially outwad is v fk, k 2 V 2. Let dṅ be the numbe of paticles scatteed outwad pe unit time into the solid angle dω. The diffeential coss section dṅ = v V fk, k dω.2 dσ dω = dṅ Inc Flux = fk, k Pobability cuent The scatteed paticle pobability flux is j = m Imψ ψ m Im 8π 3 f e ik ik eik f eik 2 f + eik f At lage, all tems fall off faste than / 2 except the fist. Note that f will involve angula deivatives that all have / and then deivatives with espect to θ and φ. So vey fa away, j m Im f ik 8π 3 e ik eik f k f 2 m 8π 3 2

2 .2. DIFFERENTIAL CROSS SECTION The flux into the detecto with aea 2 dω will be F det = j 2 dω. The total incoming flux is j inc = k. Then the ate of scatteing into solid angle dω is 2π 3 m R = F inc dσ = j scat 2 dω dσ dω = j2 2 j inc = f 2 That s all well and good. But, pobability is conseved. So j fo the entie wave function ψ integated ove the entie sphee must be zeo. 2

3 .3. BORN APPROXIMATION.3 Bon appoximation We have an integal equation fo the scatteing amplitude but it is of limited valued since it includes the solution to Schodinge s equation. The fist ode Bon appoximation is petty simple. We assume that the potential is vey weak and that the exact solution is not vey diffeent fom the fee paticle state. Then we get f k, k = 2 m = 2 m d 3 x k x V x x k d 3 x eik k x 2π 3 V x.4 Hee, k, k ae in the diection of the incoming plane wave and the scatteed wave espectively. Define q = k k = 2k sin θ/2 whee θ is the scatteing angle. Then we can pefom the angula integal if we assume that V is spheically symmetic. Yukawa potential Conside the Yukawa potential f k, k = 2 m = 2 m = 2 m d 3 x eiq x 2π 3 V x.5 2 d2π. cos θ cos θ eiq 2π 3 V.6 2 d2π. cos θ cos θ eiq 2π 3 V.7 = 2 m 2 iq d2π eiq e iq 2π 3 V.8 = m iq de iq e iq V.9 = 2m q d sinqv.0 V = V 0e µ µ which educes to the Coulomb potential with µ 0 with V 0 /µ fixed. Substitution and integation gives f 2mV0 θ = µ q 2 + µ 2 2mV0 = µ 4k 2 sin 2 θ/2 + µ 2 Note that fo the fist ode Bon appoximation, the scatteing coss section is always independent of the sign of V, and the scatteing amplitude is always eal 3

4 .3. BORN APPROXIMATION.3. Highe Ode Bon Appoximation and tansition opeato T We would like to have an opeato that effects a tansition fom a plane wave intial state to a plane wave final state. Let s evisit the Schodinge equation fo the plane wave. whee E H ψ ± = 0. Solve Equation?? fo ψ ± = + and Then the tansition opeato H 0 φ = E φ H V φ = E φ E H φ = V φ. φ = E H ± iɛ V φ + ψ±.2 V ψ ± = T = E H ± iɛ V φ E H ± iɛ V φ E H ± iɛ V φ T φk = V ψ.3 whee φk is a plane wave with momentum k, and ψ is a solution to Schodinge s equation. The diffeential coss section fk, k = 2 m.3.2 Bon appoximation again Multiplying Lippmann-Schwinge Equation by V gives T φ = V φk T φk E H 0 T φ + V φ Assuming that latte is tue fo a complete set of base states, it must be a legitimate opeato equation. T = V E H 0 + iɛ T + V.4 On iteation we get something like T = E H 0 + iɛ E H 0 + iɛ T E H 0 + iɛ E H 0 + iɛ E H 0 + iɛ T = E H 0 + iɛ E H 0 + iɛ V E H 0 + iɛ V

5 .3. BORN APPROXIMATION and so on. The scatteing amplitude becomes fk, k = 2 m k V ψ.5 fk, k = 2 m k T k.6 The momentum eigneket k is scatteed to definite momentum plane wave state k. Then f k, k = 2 m 2π 3 d 3 x e ik x V x e ix k Next ode f 2 k, k = 2 m 2π 3 d 3 x d 3 x e ik x V x 2m Gx, x V x e ix k.7.8 5

6 .4. CURRENTS AND OPTICAL THEOREM.4 Cuents and optical theoem ψ 0 epesents the incoming fee paticle and is a solution to ψ = ψ 0 + ψ s.9 H 0 ψ 0 = i t ψ 0 whee H 0 = p2 2m. Then Hψ = i t ψ whee H = H0 + V and The flux of scatteed paticles into aea element da is j s ˆnda = m Imψ s ψ s ˆnda The flux of incoming paticles is = k f 2 8π 3 m 2 2 dω The diffeential coss section is j inc = m Imψ 0 ψ 0 = Then along with the divegence theoem, σ t = 8π3 m j s ˆnda = 8π3 m k 8π 3 m.20 dσ dω = j scat ˆnda = f 2.2 j inc j s dv By the continuity equation j s dv = t ψ s 2 dv Altogethe we find that σ t = m t ψ s 2 dv Substituting Equation. we have σ t = 8π3 m dv ψ ψ 2 2Rψ t 0ψ = 8π3 m ψ dv 0 t ψ 0 + ψ0 ψ 0 t + ψ t = 2 8π3 m i R = 2 8π3 m R dv H 0ψ 0ψ + ψ0h 0 + V ψ i dvψ 0V ψ 6 ψ ψ + ψ t 2R ψ 0 t ψ + ψ ψ 0 t

7 .4. CURRENTS AND OPTICAL THEOREM = Im 8π3 m = Im 8π3 m = Im k f0 2 k V ψ 2 2 m f0 The total coss section is popotional to the imaginay pat of the fowad scatteing amplitude. The flux of scatteed paticles is balanced by the imaginay pat of the fowad amplitude, the shadow. Thee is anothe way: The total wave function The flux density is ψ inc + ψ scat = e ikz + fθ eik.22 j = m Imψ ψ.23 The total flux density in the adial diection is j = m Im e ikz + f e ik ik cos θe ik cos θ + ikf 2 eik ˆ Since we ae inteested in, only the fist and two tems in the second backets will emain. Then j = m Im e ikz + f e ik ik cos θe ik cos θ + ikf eik ˆ The intefeence tem is j int = m Imik e ikcos θ f + f e ik cos θ cos θ ˆ] Next integate j int fowad cone ove a tiny cone in the fowad diection to show j int 2 dω = m fowad cone j int 2 dω = 2π k β m 2π k m 0 β 0 Imf0.24 k Imie ikcos θ f + cos θf e ik cos θ dcos θ Imie ik e ik cos θ f + f e ik e ik cos θ dcos θ = 2π k eik Imi m ik e ik cos θ f + f e ik ik eik cos θ β 0 7

8 .4. CURRENTS AND OPTICAL THEOREM = 2π k eik Imi m ik e ik cos β f + f e ik ik eik cos β = 2π k m Im k e ikcos β e ik f + f k eikcos β e ik As long as θ 0, the aveage of j int ove any small solid angle is zeo because. Assume fθ is a smooth function. In the limit β 0, and as, we use the aveage value fo e ±ik, namely zeo, we get fowad cone j int 2 dω = 2π m Im k f + f k = 2π m k Imf f = m k Imf0 = m k Imf0] In evaluating the uppe limit in the θ integation, assume that the limit of a function that oscillates as its agument appoaches infinity is equal to its aveage value. The total pobability cuent in the egion behind the taget poduces a depletion of paticles. It must be that the poduct of the incident flux and the total coss section is equivalent to hwat is depleted in the fowad diection. Theefoe j int 2 dω = m k Imf0 = m σ t fowad.4. Optical Theoem σ t = k Imf0 We begin with the basic Lippmann/Schwinge equation The scatteing amplitude is ψ ± = φ + E H 0 ± iɛ V ψ±.25 fk, k = 2 m φk T φk.26 whee T φ = V ψ ± by definition. Then fk, k = ψ ± ψ ± V V ψ ± E H 0 ± iɛ = ψ ± V ψ ± ψ ± V E H 0 ± iɛ V ψ± 8

9 .4. CURRENTS AND OPTICAL THEOREM The imaginay pat of the fowad scatteing amplitude is Imfk, k = 8π3 Imfk, k = Im 8π3 Imfk, k = Im 8π3 = Im 8π3 = Im 8π3 Im 8π3 2m k T k 2 2m φk T φ k φ k E H 0 ± iɛ φ k φ k T φk d3 k d 3 k d 3 k φk T φ k k 2 k 2 ± iɛ φ k T φk dk k 2 dω φk T φk k 2 k 2 ± iɛ φ k T φk dk k 2 dω φk T φk Imiπ 8π 3 2π 2 Im ik k σ t 8π 3 k ± iɛ + k k k ± iɛ k 2 k 2 dk k 2 dω φk T φk k ± iɛ + k k ± iɛ k φ k T φk k 2 dω φk T φk 2k φ k T φk dωf k, k fk, k φ k T φk Shadowing We wite the solution to Schodinge s equation as ψ + as the sum of an incoming plane wave that extends ove all space, and an outgoing spheical wave with angula distibution epesented as fθ. Conside scatteing fom a had sphee. The fowad diection along the axis of the incoming wave is shadowed. In that egion the pobability density ψ + 2 = 0. Thee must be destuctive intefeence between the incoming plane wave and the scatteing amplitude in the fowad diection. So the scatteing amplitude in the fowad diection cannot be zeo. Moe geneally we wite σ tot = k Imf elastic0.27 Only states scatteed elastically in the fowad diection will have the same enegy as the incident state, which is equied if thee is to be destuctive intefeence. Also the depletion of the fowad flux must account fo all scatteed states elastic o inelastic. 9