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Quantum Mechanics II Scattering Basics; Scattering Phase-Shift; Scattering Phase-Shift Examples Tristan Hübsch Department of Physics and Astronomy, Howard University, Washington DC http://physics1.howard.edu/~thubsch/ Q M II Scattering Basics So far: quantum object (particle or wave) subject to (classical) agents represented by a surrounding potential Turn this around: the (classical) scattering target is localized the quantum object: (plane) wave or a beam of particles incoming from away, interacts with the target, scatters …and is detected away from the ⇥ target = ⇤ g—General Theory JS µ J I ds h̄ M ⇥ ⇤S~rY S Y =m ⇤ 1) ( Det. ⇥) = (2 r d2 W = sin(q )dq df solid angle = f q J S ⇤ ⇥ ⇤ ⇤ ⇥ ⇤ ⇤ z h̄ h̄ y ~ ~ r J I = M =meY r Y J = = m Y r Y (1) oI I S YI M S S ⇤~ ⇥ r h lT YI a q r diff. (2) cross-section m 2 e n d WGe= sin(q )dq df h̄ = q h i h i f d M 1 ds ds JS ds — = q g d )function of !: : 2 := 2 = in Determine J SI /JI as (a = q r e 2 2 dW rr dW d W W d JI r d W in att s c | {z } 2W= :S d := s (q, f) 2 Q M II Scattering Basics Linear momentum conservation: the target does move Calculate in CM (center-of-mass)-frame Compare with Lab-frame measurements Lab CM ~v1 ~v ~v 0 T ~v MM + MT q1 ~v 00 ~v2 Must be able to translate between the two v1 cos(q1 ) = v0 cos(q ) + V sin(q ) ) tan(q1 ) = 0 cos(q ) + vV0 v1 sin(q1 ) = v sin(q ) MT v 2 1 2 M M + MT M + MT v M + MT 2 ! 1 1 Mv = 2M 2 MT M + MT ! Mv MT M + MT = 0 = v 0 2 Mv0 3 + 1 2M v MT v00 00 2 | q {z } ⇣ ⌘ ~ ~v M+MMT =: V — velocity of the CM-frame in Lab in elastic collisions M are unchanged ) T speeds ⇢ 0 v = MM+TMvT ) v v00 = MM + MT Q M II Scattering ) tan(q1 ) = Basics sin(q ) cos(q ) + V v0 Relating the CM-calculation to the Lab-frame measurement: Lab CM ~v1 ~v | ~ V q1 T := ~v MM + MT ~v 00 ~v2 | ~v 0 {z q ~v M+MMT } in particular, for the “differential cross-section” s1 (q, f) sin(q1 )dq1 df1 = s(q, f) sin(q )dq df p ⇣ ∂(q, f) ⌘⌘ (1 + 2b cos(q ) + b2 )3 = s(q, f) s1 (q, f) = s(q, f) |1 + b cos(q )| ∂(q1 , f1 ) Lab b := CM V v0 There is no transformation in ". In fact, for the most part, the target and so the scattered wave will be "-independent. To first order at least. 4 Q M II Scattering Scattering Phase-Shift Scattering is a 2-body interaction b b= H ~R := h̄2 ~ h̄2 ~ r2 + W (~r1 2M1 r1 2Mr 2 M M1rr~1 + M M2~rr2 ~r := ~r1 ~r2 M1 + M2 CM-position b= H Thus ~r2 ) µ := MM11+MM22 b relative position 2 2 h̄ ~~ r 2( M1 + M2 ) R h̄ 2µ reduced mass + W (~r ) motion as a whole relative dynamics ⇥(free particle) ⇤ ~J I := Choose: h̄ µ ⇥ ⇤ ⇤ ⇤ ⇥ ⇥ ~ yI (~r ) ⇤ ~JS := =m yI (~r )r r yI (~r ) = A e i~k·~r 1 r h̄2 2µ ⇤ + W (~r ) y(~r ) = Ey(~r ) y(~r ) = yI (~r ) + yS (~r ) incident scattered h̄ µ ⇥ ⇤ ~ yS (~r ) =m yS (~r )r ⇤ ~J I = | A|2 h̄~k~k ⇠ | A|2~v uniform plane-wave µ yS (~r ) = A f (q, f) e I ⇥ i |~k||~r | ~JS = | A|2 | f (q, f)|2 h̄|~k| 12 êr µ rI 5 spherical wave S2 d2~s · ~JS = const. Q M II Scattering Scattering Phase-Shift Simple example: spherical potential p 2 ⇥ ~ y(~r ) + k2 r y(~r ) =  `m ⇥ ⇤ ⇤ U (r ) y(~r ) = 0 u ` (r ) m a` m Y` (q, f) r k := u00` (r ) + ⇥ ⇥ 2µE h̄ k2 2µ U (r ) := 2 W (r ) h̄ `(`+1) ⇤ U (r ) u ` (r ) r2 =0 Now: for very large r, ignore U(r) and the centrifugal barrier u00` (r ) + k2 u` (r ) ⇠ 0 for r ! • ) u` (r ) ⇠ e±ikr BTW, the Coulomb potential cannot be ignored for large r! ⇠ ) ⇠ ! Restrict to “short-range” potentials i~k·~r yI = e = Â(2`+1) i` j` (kr ) P` (cos (q )) Know: ` Try: y = y + y = (2`+1) i` A R (r ) P (cos (q )) satisfying I ⇥ 1 d2 r dr2 (rR` ) S ⇤  ⇥ ` 2 + k ` U 6 ` `(`+1) ⇤ R` r2 ` U = 0, = 0 where spherical Bessel eq. Q M II Scattering ⇥ ~ ⇤ ⇥ ⇤ y) = Â(2`+1⇠ ) i` A` R` (r ) P`! (cos (q )) ⇠ Scattering Phase-Shift ⇥ ⇤ ⇥ ⇤ yI = eik·~r = ` Â(2`+1) i` j` (kr) P` (cos (q )) ` As the solution of the spherical Bessel equation, a`2 + b`2 = 1 R` = a` j` (kr ) + b` n` (kr ) = cos(d` ) j` (kr ) + sin(d` ) n` (kr ) ⇠ ⇠ sin(kr 21 p `) cos(d` ) kr sin(kr 1 2 p` kr + d` ) + sin(d` ) cos(kr kr phase-shift 1 2 p `) for kr 1 at the detector! The phase-shifts are not multiples of #/2 and n$(kr) is not ruled out …because the origin (where the n$(kr) diverge) is excluded by the target ⇠ Equating y(~r ) = yI (~r ) + yS (~r ) = yI (~r ) + f (q, f) h ` ( 2 `+ 1 ) i P` (cos (q )) A`  ` sin(kr 1 2 p `+d` ) kr 7 eikr r sin(kr 12 p `) i eikr = f (q, f) r kr Q M II Scattering Scattering Phase-Shift  ` h h ` (2`+1)i P (cos (q )) A ` ` sin(kr 1 2 p `+d` ) kr i sin(kr 12 p `)i ikr = f (q, f) e r kr i Equating coefficients hof e –ikr (projecting w/e –ikr): ` ( 2 `+ 1 ) i P` (cos (q )) A` exp(kr  1 2 p `+d` ) exp(kr 1 2 ` Projecting with P$(cos(!)): A` = eid` Equating coefficients of e +ikr (projecting w/e +ikr): f (q, f) = 1 2ik ` ( 2 `+ 1 ) i e  ` Â(2`+1) e id` Thus: s(q, f) = f (q, f) 2 = 1 k ` s := Z ⇥⇥ p `/2 e2id` 4p k2 p `) = 0 ⇤⇤ 1 P` (cos (q )) sin(d` ) P` (cos (q )) d2 W s(q, f)) = = i This is indeed independent of " 2 ( 2 `+ 1 ) sin (d` )  ` So, calculating &(!,") and & reduces to calculating '$. 8 Q M II Scattering u00` (r ) + Phase-Shift Examples A spherically limited potential W (r ) = nU ( r ) 0 ⇥ k 2 U (r ) `(`+1) ⇤ u ` (r ) r2 =0 for r < a for r > a The inside solution R$(r) = u$(r)/r must be regular at r = 0, so u$(r) ~ r b, we must have b ≥ 1 @r ⇠ 0 ⇥ u00` (r ) + b 2 b ( b 1)r `(`+1) ⇤ u ` (r ) r2 b 2 `(`+1)r ⇠0 ⇠0 b= n `+1 ` regular singular As long asnr2 U(r) < ∞ for r ~ 0, one solution is regular, one singular Turns out, if r4 U(r) ~ ∞ for r ~ 0, both solutions are singular @ r ~ 0. Assume that r2 U(r) < ∞ for r ~ 0, solve the radial equation , compute (in) g` h d ln ru(in) (r ) i ` and match this to the logarithmic := dr r ! a derivative computed from the outside Matching u$(r) itself will match the amplitudes & the normalization 9 Q M II Scattering Phase-Shift Examples h i ⇥ 00 n `+1 2 ` (r ) + k bu= ` b ( b 1)r b U (r ) 2 `(`+1)r b `(`+1) ⇤ u ` (r ) r2 2 =0 The outside solution is R$(r)!= cos('$) j$(kr) – sin('$) n$(kr) hhd ln R(out) (r ) ii h (in) (out) ` h i : d ln ru ( r ) g` = (in) ` dr : r ! a g = dr r`! a ⇥ ⇤ ⇥ dr r!a 0 hh 0 0 i k cos(d` ) j` (z) sin (d` ) n` (z) (out) ! (in) ⇥ h i = g = h d ln R` (r ) ` (out) cos(d` ) j` (z)g sin:= (d` ) n` (z) z!ka ` dr r!a ⇥ ⇤i This is the equation that determines hthe phase-shifts, ' $ 0 0 k cos(in)(d` ) j` (z) sin(d` ) n` (z) 0 k j`= (ka) g` j` (ka) cos ( d ) j ( z ) sin ( d ) n ( z ) z!ka tan d` = ` ` ` ` (in) k n0` (ka) g` n` (ka) k j`0 (ka) g`(in) j` (ka) (in) Quick example: impenetrable inside solution = 0 tansphere. d` = The , g !• (in) 0 ` • ) k n` ``(ka n2s` (ka) z` sg ` ( 1 ) ( s +`) ! j` (ka) j` (z) = 2 z  s!(2s+2`+1)! z ⇡ (2`+1)!! tan d` = s =0 z⌧1 n` (ka) • `+ 1 s ( 1) ( 1) (s `)! (2` 1)!! n` (z) = 2` z`+1  s!(2s 2`)! z2s ⇡ z`+1 so tan d` ⇡ (2`+1) 2`+1 ( ka ) [(2`+1)!!]2 s =0 ka ⌧ 1 Lowest-ℓ terms contribute most. 10 Q M II Scattering tan d` ⇡ Phase-Shift Examples For very low energies, ka ≪ 1, s= ⇡ 2 Â(2`+1) sin 4p k2 2 ( 2 `+ 1 ) tan (d` )) = =  = (d` )) = ` ⇥ 2 = 4pa 1 + 4 1 ( ka ) 3 + tan2 (d` ) (2`+1) 1+tan2 (d ) ` 4p k2  4p k2 Â(2`+1) ` 1 405 ( ka ) ` 8 +... rapid convergence ⇤ 4 × the ⇥ geometric cross-section ⇤ For very high energies, j` (ka) ⇡ tan(d` ) = n` (ka) ka ⌧ 1 4p k2 ` (2`+1) 2`+1 ( ka ) 2 [(2`+1)!!] 1 ka sin( ka 1 ka cos( ka ⇣ (2`+1) 2`+1 ( ka ) [(2`+1)!!]2 `p/2) = tan ka ` p2 `p/2)) must sum over all ℓ Summation (Sakuri, 1982) produces & = 2$ a2 …still 2 × the geometric cross-section 11 ⌘2 Q M II Scattering ⇥ u00` (r ) + Phase-Shift Examples k 2 U (r ) `(`+1) ⇤ u ` (r ) r2 =0 Consider a finite-strength spherical potential barrier U(r) = V0 > 0 for 0 ≤ r ≤ a (inside), U(r) = 0 outside. K0 : = Focus on $ = 0 u(in) 0 (r ) = 0r a p 2MV0 h̄ k := p p { := pK02 k2 2ME h̄ n A sinh({ r ) u(out) 0 (r ) = sin( kr + d0 ) A0 sin({ r ) r a Equating at r = a the outside and the inside logarithmic derivative ! 1 k { tanh({ a ) ! ka cot(ka + d0 ) = { 0 a cot({ 0 a) E > V0 d0 = tan 1 {k0 tan({ 0 a) ⌧ 0 ka{⌧ ⇡ 1 k ⌧ K0 { ⇡ K0 However, for⌧low energies, ⌧ ⌧ 2`+1 ⇠ tan(d` ) ⇠ (ka) ⌧ 1 ka + d0 ⌧ 1 cot(ka+d0 ) ⇡ ka+1 d0 tanh(K0 a) K0 a ka 1 ) d0 ⇡ ka K0 a ka+d0 ⇡ tanh(K0 a) ka cot(ka + d0 ) = { a coth({ a) E < V0 12 d0 = tan s0 ⇡ 4pa2 tanh(K0 a) K0 a ka ka 1 2 Q M II Scattering u00` (r ) + Phase-Shift Examples ⇥ k 2 U (r ) Consider a spherical '-function bubble aZ+e Focus on $ = 0 00 0 0 `(`+1) ⇤ u ` (r ) 2 r ⇥ =0 dr u0 (r ) = u0 ( a+e) u0 (r e) = la u0 ( a) h h 0 i i 0 au0 ( a+e) au0 (r e) a e lim u (a) =l u ( a ) 0 i !e!0 h 0 The solution is n A sin(kr ) for r < a n u0 (r ) = sin(kr +d0 ) for r > a …for which the boundary condition is ⇥ ⇤ k cot(ka+d0 ) cot(ka) = l and produces: ⇥ ⇣ ⌘⇤ tan(ka) 1 l d0 = tan ka tan(ka) 1 + l ka while continuity yields: s tan2 (ka) + 1 A= tan(ka) 2 2 tan (ka) + 1 + l ka 13 just like a particle in a hard box En ⇡ n2 p 2 h̄2 2µa2 Resonances! tan(k n a) ⇡ 0 k n ⇡ n pa Quantum Mechanics II Now, go forth and calculate!! Tristan Hübsch Department of Physics and Astronomy, Howard University, Washington DC http://physics1.howard.edu/~thubsch/