Def: given incident flux nv particles per unit area and unit time. (n is density and v speed of particles)

Transcription

1 8 Scatteing Theoy I 8.1 Kinematics Poblem: wave packet incident on fixed scatteing cente V () with finite ange. Goal: find pobability paticle is scatteed into angle θ, φ fa away fom scatteing cente. Solve S.-eqn. with bounday condition that at t = the wave function is a wave packet incident on scatteing ct. Decompose packet into component waves e ik, then eithe this wave is scatteed o it s not. If it s scatteed, at lage distances we expect a spheical wave. So at lage ou soln. should have fom of linea combs. of ψ k (, t) e ik + f k (θ, φ) eik e iωt, (1) whee hω = h 2 k 2 /2m is the enegy of the asymptotic plane wave befoe scatteing o afte it has scatteed (elastically). Intuitively 1st tem in (1) is unscatteed pat, 2nd tem is scatteed wave with angula dependence f k. f k called scatteing amplitude. 2nd tem u = f k e ik / (2) vaies as 1/, so intensity of scatteed wave falls off as u 2 1/ 2 as it must. To veify this, constuct pobability cuent j = i h 2m (ψ ψ ψ ψ ). (3) Fo scatteed flux density (eplace ψ by u above), find j scatt = i h 2m (u u u u ) = hk m 3 f(θ, φ) 2 (4) Diffeential scatteing coss section dσ dω Def: given incident flux nv paticles pe unit aea and unit time. (n is density and v speed of paticles) 1

2 Paticles collected in detecto of aea A at angles θ, φ fom incident diection. A theefoe subtends δω = A/ 2 steadians. Recall classically dn dt = (nv) dσ dω ate of detecting paticles in A = incident flux density Fom eq. (4) above, have dn dt = j scatta = hk m δω (5) diffeential scatteing coss sec. subtended solid angle (6) 3 f(θ, φ) 2 A = hk m dσ dω A (7) 2 so dσ dω (θ, φ) = f(θ, φ) 2 (8) Total scatteing coss-section σ Total coss-section defined by integating ove all angles: σ = dω dσ dω = dω f(θ, φ) 2 (9) Reminde classical analog & oigin of name coss-section : Conside n pt. paticles/vol. incident on sphee, adius a, flux nv. Paticle scattes if its impact paamete is less than a, so net scatteed flux is nv π 2, total σ = πa 2. 2

3 8.2 Optical theoem A consevation law elates diminished amplitude of unscatteed beam to flux scatteed out of beam. Incident flux density in Fouie component e ikz of incident beam is v classical = hk m (1) Role of classical density of paticles n played by ψ beam 2 = 1. Fom (9) net scatteing ate is nvσ = hk m dω fk 2 (11) so outgoing unscatteed beam must be diminished by this amount. Outgoing means e ikz plus fowad scatteing pat of scatteed wave, ψ = e ikz + f() eik (12) Fo angles close to fowad scatteing, so at lage z, 1 = z 2 + x 2 z + x2 2z (13) ψ e ikz 1 + f() kx 2 z ei 2z (14) Now want to compute pobability flux in fowad diection. Rapidly vaying pat of (14) is e ikz, so ψ ikψẑ (15) ψ ψ ikψ ψ = ik 1 + f () e ikx2 2z 1 + f() kx 2 z z ei 2z (16) ik z Re f()eikx2 2z (17) 1 Peebles calls this ψ f fo fowad, but I m too lazy, so you have to emembe I m always talking about small angles 3

4 So to lowest ode flux density in fowad diection is j = i h 2m (ψ ψ c.c.) (18) hk m z Re f()eikx2 2z (19) The fist tem is the same as the incident flux density, the second is the eduction of the outgoing flux by fowad scatteing. If we go out fa enough along z, even fo small angles the x can get lage, so we need to integate all the scatteed flux (2nd tem) in the fowad diection: 2 eduction of flux by fowad scatteing = v cl 2πxdx 2 z Re f()e ikx2 2z (21) see footnote = v cl π 2 z Imf() 2z k v clσ (22) since vσ is the definition of the total flux emoved fom the beam by scatteing! So Intepetation: σ = 4π k Imf() Optical theoem (23) Incident plane wave bings in pobability cuent density in z diection. Some of it gets scatteed into vaious diections. Must give ise to decease in cuent density behind taget (θ = ) due to destuctive intefeence of incident plane wave and scatteed wave in fowad diection. So total flux scatteed (vσ) elated to fowad scatteing amplitude f(). 2 This is a mathematically vey ill-defined pocedue, and you shouldn t believe it until you see we get the same esult by phase shift analysis late. But the integation goes as follows: 2xdx e ikx2 /(2z) = 2z i du e u = 2iz ik k eu i (2) We now want to ague that we can neglect the exponential at its uppe limit because it is apidly oscillating, i.e. we want to be able to set e i. Peebles agues physically: that a eal beam will consist of a finite spead of incident diections and cut off the incease in the ate of oscillations as z. That the oscillations at then aveage out to zeo is guaanteed by the Riemann-Lesbegue lemma 3. 4

5 8.3 Bon appoximation. Valid fo weak scatteing o fast paticles! Want to solve S. s eqn fo scatteing potential V () with bounday condition ψ e ik + f eik fo suff. lage. We e looking fist fo solutions with (24) E = h2 k 2 scatteing states (25) 2m i.e., not bound states whee the paticle is tapped by the taget. So S.-eqn. looks like 2 ψ + k 2 ψ = 2mV h 2 ψ ɛuψ (26) Now seek powe seies expansion of ψ (in powes of scatteing potential): plugging into (26) gives ψ() = e ik + ɛψ (1) + ɛ 2 ψ (27) ɛ( 2 ψ (1) + k 2 ψ (1) ) + ɛ 2 ( 2 ψ (2) + k 2 ψ (2) ) = as usual equate powes of ɛ: = ɛue ik + ɛ 2 Uψ (1) +... (28) 2 ψ (1) + k 2 ψ (1) = Ue ik (29) 2 ψ (2) + k 2 ψ (2) = Uψ (1), etc. (3). (31) See that we get sequence of coupled eqns., each coupled to pevious one on hs. This is called Bon s seies. If we want to solve to linea ode in ɛ, take only (29). Solve by Fouie tansfom; and define 5

6 and the invese ψ k = ψ (1) ()e ik d 3 (32) ψ (1) () = ψ k e ik d3 k (2π) 3 (33) So multiply (29) by e ik and integate lhs by pats 4 to get which yields immediately (k 2 k 2 )ψ k = U()e i(k k ) d 3 (35) ψ (1) () = = 1 (2π) 3 1 (2π) 3 d 3 k e ik k 2 k 2 d 3 U( )e i(k k ) (36) d 3 U( )e ik d 3 k eik ( ) k 2 k 2 }{{} I( ) (37) Now we spend some tedious but hopefully useful time showing how to calculate integals of this type. Fist evaluate I( ). Use pola coods., pola axis along y, θ is angle between y and k : So angula pat of I is so left with 2π k y = k y cos θ (38) d 3 k = k 2 dk sin θdθdφ. (39) π dφ sin θdθ e ik y cos θ Tick: note integand is even in k, wite I as k 2 dk sin k y I = 4π k 2 k 2 k y 1 = 2π d cos θe ik y cos θ 1 = 4π sin k y k y (4) (41) (42) I = 2π i k dk e ik y k 2 k 2 y because cos pat odd in k, theefoe doesn t contibute (43) 4 Explicitly, use ( 2 ψ())e ik d 3 = k 2 ψ k (34) 6

7 This is an impope integal due to the singulaity at k = ±k. Howeve we use physics to egulaize: so fa we haven t used condition that we want w. fctn. ψ at to look like scatteed wave. This coesponds in integal scatteing fomulation to specifying contou fo I in complex plane. One possibility is to handle singulaities by following contou shown below. Use Cauchy integal fomula, 5 complete the contou in uppe half-plane, shink the contou to the pole at +k, to get: I = 2π i e iky y 1 dk = 2π 2 eiky 2 k k }{{} y (45) 2πi (46) This pocedue may seem athe abitay at 1st sight, but a look at othe choices suggests it s ight thing to do. If we take contou above both poles, get zeo! If we take contou opposite to figue (above k and below k) o below both poles, get a contibution I e iky. Recall y = ; if is in asymptotic egion,, I e ik, which is ingoing, not outgoing scatteed wave. So choice of contou effectively implements bounday condition. Continuing, find ψ (1) () = 1 4π Finally, look at ψ in asymptotic egion, : use d 3 U( )e ik eik (47) + ˆ (48) e ik 1 eik e ikˆ (49) define k s kˆ (5) to wite ψ (1) () 1 ( 4π d 3 U( )e i(k ks) ) e ik (51) 5 If you don t like the integal fomula, note this choice is equivalent to eplacing (29) by i.e. shifting the zeos of the integand of I to k = ±(k + iɛ/2). 2 ψ (1) + (k 2 + iɛk)ψ (1) = Ue ik (44) 7

8 Now can compae diectly to Eq. (1), to identify (use U = 2mV/ h 2 ). f(θ, φ) = m 2π h 2 d 3 V ( )e i(k k s) (52) k along incident diection k s along scatteed diection This is Bon appoximation. Diff. scatt. coss sec. will be dσ/dω = f 2. Low-enegy limit We conside only elastic scatteing, E f = E i = h 2 k 2 /2m. 6 Note main contibution to integal in (53) is fom s inside ange of scatteing potential (meaning V ( ) small fo.) Thus if we conside low-enegy scatteing such that k 1, agument of exponential in (53) is 1, so find immediately Isotopic! dσ/dω m2 4π 2 h Sceened Coulomb scatteing ( d 3 V ( ) )2, k 1 (53) Scatteing by bae Coulomb potential vey singula, not obvious we can teat it popely with fomalism intoduced hee (V () was assumed to have finite ange). Howeve we ll teat a sceened Coulomb o Yukawa 7 potential, V () = e 2 e α (54) which has ange 1/α, and show that in limit α we indeed ecove Ruthefod diffeential coss section fo Coulomb scatteing. Bon appox. fo Yukawa potential. 6 So in scatteing pocess diection changes fom k to k s, but momentum magnitude hk doesn t change. 7 Yukawa wote it down as a phenomenological model fo the stong foce in nuclei to explain p p scatteing expeiments, and was led to popose the existence of the pion as a mediating paticle. The fom is also chaacteistic of the Coulomb inteaction between chages in a chaged medium capable of sceening, hence the title of this section. It s not a bad model fo atomic scatteing, since a fast electon penetating the electonic cloud suounding the nucleus sees moe and moe chage the close it gets. 8

9 Define q = k k s. Wok hee is only in calculating integal So diff. scatt. coss section is I d 3 e iq α (55) = d e α dωe iq (56) = d e α i2π(e iq e iq )/q (57) = 4π q Im 1 α iq = 4π α 2 + q 2 (58) dσ/dω m2 4π 2 h 4I = 4m 2 e 4 h 4 (59) (α 2 + q 2 ) 2 α = 4m 2 e 4 q 4 h 4 (6) e 4 4m 2 v 4 sin 4 (θ/2) (61) whee v = hk/m is classical velocity, and we used geomety to elate momentum tansfe q to incoming momentum k, q = 2k sin θ/2, as seen fom pictue. Eq. 62 is in fact famous Ruthefod fomula fo Coulomb scatteing so we did get ight answe afte all. Note geat iony hee: classical esult obtained coectly using leading-ode Bon appoximation, so h doesn t play majo ole! Yet measuement of Ruthefod coss-section poved existence of pointlike nucleus, and it was poblem of stable obits aound nucleus which led to Boh s quantum mechanics! 9