1.2 Partial Wave Analysis

Transcription

1 February, 205 Lecture X.2 Partia Wave Anaysis We have described scattering in terms of an incoming pane wave, a momentum eigenet, and and outgoing spherica wave, aso with definite momentum. We now consider the basis of free partice states with definite energy and anguar momentum (rather than inear momentum) that oo ie E,, m. These are eigenets of of H 0, L 2, and L z. We woud ie to expand our pane wave in terms of these spherica waves ie so,m E,, m E,, m (.) Then we can write the scattering ampitude f(, ) 2 T 2 de,m de,m E,, m E,, m T E,, m E,, m If the scattering potentia is sphericay symmetric, T is a scaar operator, and by WE,, m m, and E,, m T E,, m is independent of m. Then f(, ) 2 dede E,, m E,, m T E,, m E,, m,m 2 dede E,, m T E,, m (.2),m (.) The spherica scattering ampitude conserves anguar momentum. Now et s figure out E,, m. Consider the state ẑ. ẑ L z E,, m 0 (m 0) ẑ E,, m 0 (m 0) 2+ Aso ẑ E,, m 0 is independent of θ, φ,so, ẑ E,, m 0 g (). We can transform the z-direction momentum et into an arbitrary direction by a rotation. Then E,, m ẑ D E,, m D(α φ, β θ, 0) ẑ (.4) ẑ E,, m 0 E,, m 0 D E,, m

2 2 + g ()D 0,m 2 + g () 2 + Y m (θ, φ) g ()Y m (θ, φ) One more thing. H 0 E E,, m E,, m ( 2 2 E) 0 E,, m δ( 2 2 E) g () N δ( 2 2 E) To determine N et s try to normaize. E,, m E,, m d E,, m E,, m From which we get.2. ρ 0 ππ g 2 d dωn N Y m Y m δ( 2 2 E )δ( 2 2 E) 2 d N 2 δ( 2 2 E )δ( 2 2 E) m 2 de N 2 δ( 2 2 E )δ( 2 2 E) m 2 N 2 δ(e E )δ δ mm m δ(e 2 2 ) E, m, m Y m (θ, φ)δ(e 2 2 ) (.5) The ρ meson is spin and it decays to two spin 0 pions. Suppose that the ρ is in the, m state, where there is some z-axis defined by something. The fina state has the same anguar momentum quantum numbers and the ampitude to find a π with momentum in the ˆ direction is E,, m Y (ˆ) sin θ The anguar distribution of the π is Y 2 sin 2 θ 2

3 If we imagine producing ρ in e + e coisions where eectrons and positrons are poarized so that j z + aong the z-axis defined by the direction of the positron beam, then dσ dω (θ) sin2 θ.2.2 Bac to partia wave expansion Substituting into Equation.29 we have f(, ) 2 de de E,, m E,, m T E,, m E,, m,m,m 2 de de E,, m T E,, m,m 2 de de m Y m ( )δ(e 2 2 )T Y m ()δ(e 2 2,m m ) 2,m 2,m Y m 2 m Y m ( )Y m () T ( )Y m () T Let ẑ so that θ 0, φ? and then Y m () Y 0( ) 2+ δ m0. So ony m 0 contributes. Then 2+ P (cos θ) where θ is the ange between and. The scattering ampitude becomes f(, ) P (θ)t π (2 + )P (cos θ)t (.6) Define f () πt (E) and f(, ) (2 + )P (cos θ)f () (.7) f () is ampitude to scatter an incident partice with anguar momentum or impact parameter b such that b. Remember that the outgoing soution to the SE far outside the range of the potentia is ψ + [e (2π) ) iz + f(θ) eir /2 r

4 .2. Expansion of pane wave as spherica waves The radia part of the free partice Schrodinger equation is 2 d 2 u [V dr 2 + ( + ) 2 (r) + r 2 u Eu The soution to the free partice Schrodinger equation in spherica coordinates is x E,, m c j (r)y m (ˆr). Next expand the pane wave as a inear combination of incoming and outgoing spherica waves. e i x x (2π) de x E,, m E,, m /2,m dec j (r)y m (ˆr) δ(e 2 2,m m )Y m (ˆ) where we use the addition theorem Turns out that c i π m so that (2 + ) Y m P (ˆ ˆr) c j (r) m (ˆr)Y m 2 + ˆ) P (ˆ ˆr) e i x (2 + )i j (r)p (ˆ ˆr) As r, e i x (2 + ) ei(r)) e i(r (π)) P (cos θ) (.8).2.4 Partia wave expansion Now ψ + ψ + Remember that for arge r, [e (2π) ) + f(θ) eir /2 r (.9) i (2 + )A (r)p (cos θ) (.0) i (2 + )(c h (r) + c 2 h 2 (r))p (cos θ) (.) h ei(r (π/2), h 2 e i(r (π/2) ir ir (.2) 4

5 becomes, using j (r) (arge r) ei(r (π/2)) e i(r (π/2)) Anyway, we can write, the genera soution to the Schrodinger equation in the partia wave basis, far from the scattering potentia as Then since ψ eastic ψ + ψ + ψ eastic i (2 + ) η e i(r (π/2)) c e i(r (π/2)) (2 + ) η e i(r)) c e i(r (π)) eix f(θ)eir (2π) /2 (2π) /2 r we now that c so that the ingoing wave is the same. Therefore ψ eastic ψ + i (2 + )(η e i(r π/2) e i(r π/2) ) Probabity conservation requres η <. If η, the scattering is pure eastic and each partia wave gets some phase shift. If η 0 the scattering for that partia wave is purey ineastic. And we now that η 2i f ψ + [ (2 + )P (2π) /2 (cos θ) ) [ (2 + )P (2π) /2 (cos θ) ) (2π) /2 ) ( e ir e i(r π) ) + f(θ) eir r ( e ir e i(r π) ) + [ ( e ir (2 + )P (cos θ)( + 2if (θ)) ) + (2 + )P (cos θ)f () eir r (2 + )P (cos θ) e i(r π) Unitarity requires that fux is conserved for each anguar momentum state. Outgoing fux is no more than incoming. Therefore + 2if (θ) η (.) and equa to one for eastic scattering. For eastic scattering we define a phase shift The eastic partia wave ampitude + 2if (θ) e 2iδ (.4) f (θ) e2iδ 2i 5 eiδ sin δ

6 and The tota cross section is σ t f(θ) (2 + )e iδ sin δ P (cos θ) (.5) dω (2 + ) 2 f 2 P 2 (cos θ) (.6) (2 + ) f 2 (.7) π 2 (2 + ) η 2 π 2 (2 + )( η 2 + 2R(η )) (.8) If the scattering is eastic then the tota cross section is σ ea 2 dω (2 + ) 2 sin 2 δ P 2 (cos θ) 2 (2 + ) sin 2 δ (.9) where η e 2iδ. Suppose that there is an ineastic component, so that the magnitude of the outgoing wave at momentum in ψ scat is ess than the magnitude in the incoming pane wave. Then η <. The ineastic cross section is the piece ost from the outgoing, namey η 2. Therefore σ ineastic (2 + ) ( η 2 ) 2i 2 π 2 (2 + )( η 2 ) (.20) And the optica theorem? f(θ) (2 + ) η 2i P (cos θ) Imf(0) ( ) η (2 + )R 2 σ tot 2π 2 (2 + )R(η ) The ineastic cross section is the difference of the tota and the eastic σ ine σ tot σ eas 2π 2 (2 + )R(η ) π 2 (2 + )( η 2 + 2R(η )) π 2 (2 + )( η 2 ) 6

7 .2.5 Genera soution Far from the scattering center where V 0, the soution to the free partice Schrodinger equation is a inear combination of spherica besse functions. ψ + Remember that for arge r, i (2 + )A (2π) /2 (r)p (cos θ) (.2) i (2 + )(c (2π) /2 h (r) + c 2 h 2 (r))p (cos θ) (.22) h ei(r (π/2), h 2 e i(r (π/2) ir ir Meanwhie, in terms of the scattering ampitude and phase shift we write ψ + [ e 2iδ e ir (2 + )P (2π) /2 e i(r π) (.2) (.24) Comparison with Equation.49 gives c 2 e2iδ and c 2 2. Then A (r) c h (r) + c 2 h 2 (r) 2 e2iδ (j + in ) + 2 (j in ) 2 eiδ ((cos δ + i sin δ )(j + in ) + (cos δ i sin δ )(j in )) e iδ (cos δ j sin δ n ) 7

8 .2.6 Hard Sphere scattering The probabiity of finding the partice inside the radius R of the sphere is zero. It is zero at the boundary. So ψ + (R) j (R) cos δ n (R) sin δ 0 (.25) j (R) cos δ n (R) sin δ tan δ j (R) n (R) (.26) As we have determined a of the phase shifts, we can compute the exact differentia cross section. Let s consider the imits. Suppose that the energy of the incoming partice is very ow. So ow that the anguar momentum R. Then ony the 0 partia wave wi contribute. Partia waves with > 0 have zero ampitude of appearing at R. Aso consider the sma x imit of j (x), n (x). im j (x) x 0 2! (2 + )! x im n (x) (2)! x 0 2! im R 0 tan δ x + (2!) 2 (2 + )(2)! x(2+) Ceary the 0 partia wave is the most important component of the wave function when R. The 0 phase shift is tan δ 0 The differentia cross section for s-wave scattering is sin R/R cos R/KR tan R δ 0 R (.27) dσ dω f(θ) 2 sin R 2 R 2 (.28) No anguar dependence of course. The tota cross section is σ t R 2, four times the geometric cross section. In the imit of high energy, where R, we can use the asymptotic form for j and n. Then a partia waves with R wi contribute. Then im j (R) sin(r π 2 ) R R im n (R) cos(r π 2 ) R R (.29) (.0) (.) im R tan δ tan(r π 2 ) (.2) sin 2 δ sin 2 (R π 2 ) (.) dω f(θ) 2 2 (2 + )sin 2 (R π 2 ) (.4) 8

9 So et s sum to R. dω f(θ) 2 2 R 0 (2 + )sin 2 (R π 2 ) (.5) 2 (R(R + )) 2 (.6) 2πR 2 (.7) Now the tota cross section is ony twice the area of the sphere. What goes on? Let s consider the scattering ampitude again f (θ) e2iδ 2i f,scat + f,0 (.8) where f,scat is the part that is scattered off the sphere, and f,0 is the piece of the outgoing wave that was there in the first pace. We have that f scat 2 dω 2 (2 + )f,scat P (θ) dω 0 2 (2 + )(2 + ) (2 + ) (2 + ) f,scaty 0 (θ)f,scaty 0(θ) dω 0 0 R (2 + ) f,scat 2i 2 0 π(r)2 2 πr 2 9