Scattering in Three Dimensions

Transcription

1 Scatteing in Thee Dimensions Scatteing expeiments ae an impotant souce of infomation about quantum systems, anging in enegy fom vey low enegy chemical eactions to the highest possible enegies at the LHC. This set of notes gives a bief teatment of some of the main ideas of scatteing of a single paticle in thee dimensions fom a static potential. Patial Wave Expansion A patial wave expansion is useful fo scatteing fom a spheically symmetic potential at low enegy. The basic move is to expand the wave function into states of definite obital angula momentum. Expeiments neve input waves of definite angula momentum, athe the input is a beam of paticles, along a cetain diection, almost always taken to be the z axis. Theefoe, one of the fist esults needed is the expansion of a plane wave along the z axis into patial waves. (The wod patial wave is synonymous with a state of definite angula momentum.) Setting z = cos θ, the expansion eads e ik cos θ = l (i) l j l (k)(2l + )P l (cos θ). The P l (cos θ) ae the Legende polynomials. The j l (k) ae the so-called spheical Bessel functions. They ae actually simple functions, involving sin k and cos k and invese powes of k. The main popeties of the j l ae the following: and j l (k) sin(k lπ 2 ) k as k, j l (k) C l (k) l as 0. The facto i l in the expansion is fo convenience. With this facto pulled out, the j l ae eal. The j l satisfy the fee adial Schödinge equation, ( ) ( 2 l(l + ) 2 ) + k 2 j 2 l (k) = 0 Tuning to the case when a potential is pesent, we can epesent the full continuum wave function by the fomula Ψ k ( x) = e ik cos θ + Ψ sc ( x), whee Ψ sc is the pat of the wave function diectly coming fom the potential. It should be noted that fo the notion of scatteing to make sense, the potential must vanish sufficiently apidly as x. We may expand the full wave function in patial waves in a simila manne to the expansion of a plane wave; Ψ k ( x) = (i) l R l ()(2l + )P l (cos θ), l

2 whee R l () is the adial wave function and satisfies the following adial Schödinge equation, ( ) ( 2 l(l + ) 2 ) U() + k 2 R 2 l = 0, whee U() = 2mV ()/ h 2. As will be seen below, the adial wave function R l is not eal in geneal. Howeve, it is useful in dealing the the adial wave function to solve it using a eal function, R l. The function R l satisfies the same equation as R l, ( ) ( 2 l(l + ) 2 ) U() + k 2 R 2 l = 0. As, R l appoaches a eal solution of the fee Schödinge equation. This is witten as follows, R l () sin(k lπ + δ 2 l) as, k whee δ l is an angle called the phase shift. In a low enegy expeiment, usually only a few patial waves ae needed, and the goal of the expeiment will be to detemine the phase shifts. Now since R l and R l satisfy the same Schödinge equation, they must be elated by a constant facto. This facto can be detemined by compaing the ingoing waves of each function. (An ingoing wave has adial dependence exp( ik), while an outgoing wave has adial dependence exp(ik).) Using the fomula fo sin k in tems of exponentials, we have fo the ingoing wave of R l, R l 2ik exp( i(k lπ 2 + δ l)) as. Meanwhile fo R l the ingoing wave must be exactly the ingoing wave of the plane wave, since the scatteing pocess only poduces outgoing waves. We have fo the ingoing wave of R l : R l 2ik exp( i(k lπ 2 )) Fom this we see that the two functions ae elated by a facto involving the phase shift, R l = e iδ l Rl To summaize, using the infomation that the functions must be elated by a constant since they both satisfy the same diffeential equation, we have gone to a egion whee both functions take a simple fom (lage ) and found the facto by compaing the ingoing wave in each function. Having found the needed facto, it will be useful in what follows to note that the outgoing wave of R l is fo lage given by: Outgoing wave of i l R l : i l R l 2ik exp(i(k + 2δ l))

3 Example fo l = 0 An example whee many details can be woked out is the δ shell potential, discussed in much moe detail in the ealie edition of Gottfied s book. We have U() = λδ( 0 ) Hee we will wok out the poblem fo l = 0 only. The adial Schödinge equation becomes simple if we define a modified adial wave function, The function u 0 satisfies R 0 = u 0 ( 2 U() + k 2 )u 0 () = 0. To keep R 0 finite at = 0, u 0 must vanish at = 0. Now the potential is only active at = 0, so u 0 satisfies the fee Schödinge equation fo < a and > a. We have u 0 () = A sin(k) < 0 u 0 () = B sin(k + δ 0 ) > 0 Demanding that u 0 be continuous and its deivative have the jump equied by a δ function potential gives and A sin(k 0 ) = B sin(k 0 + δ 0 ), kb cos(k 0 + δ 0 ) ka cos(k 0 ) = λa sin(k 0 ) Taking the atio of these two equations gives tan(k 0 + δ 0 ) = Using standad tigonometic identities, we have tan(k 0 ) + λ k tan(k 0). tan(δ 0 ) k = λ k 2 sin 2 (k 0 ) + λ k sin(k 0) cos(k 0 ). () We note that tan δ 0 k as k 0. This behavio genealizes to highe values of l as follows: tan δ l k 2l+ as k 0 The oigin of this small k behavio is that fo a fixed l as k 0, to keep the angula momentum fixed, the egion whee the wave function is appeciable must move to lage and lage. This can be easily seen by making the classical estimate hl hk m, whee m is the position of the wave function maximum. Clealy as k 0, m must get lage and lage.

4 Scatteing Length Fo l = 0 scatteing at vey low enegy, the scatteing is dominated by a quantity called the scatteing length. The scatteing length a (o sometimes a s ) is defined by tan δ 0 lim = a. k 0 k (An equivalent fomula fo /a is often witten using the cot δ 0 instead of tan δ 0.) In the example of the delta shell potential, we have fom Eq.() Fom this fomula, we have a = λ 0 + λ 0 0 (2) a > 0 λ > 0 a < 0 λ < 0, + λ 0 > 0 a > 0 λ < 0, + λ 0 < 0 We see that if the delta shell potential is epulsive (λ > 0), then a > 0. Howeve, if the delta shell potential is attactive, a can have eithe sign. Roughly speaking, weak attaction leads to a < 0, and stong attaction leads to a > 0. It is possible to gain insight into these esults by consideing the equation fo u 0 () fo k 2 = 0. This is 2 u 0 () = U()u 0 () Fo the delta shell potential we can solve this equation vey easily. We have u 0 () = A, < 0 u 0 () = B( a), > 0 (3) It is not obvious at this point that the paamete a in Eq.(3) is the scatteing length as given in Eq.(2). This will be justified shotly. Demanding continuity and the coect delta function jump condition at = 0 gives A 0 = B( 0 a) A( + λ 0 ) = B. Taking the atio of these equations leads to Eq.(2) fo a, which establishes that the a in Eq.(3) is indeed the scatteing length. Now conside a geneic shot ange potential, which is eithe zeo o vanishes vey apidly fo 0, whee 0 is a paamete chaacteizing the ange of the potential. (The delta shell is cetainly an example, but now we want to discuss an abitay shot ange potential.) Fo > 0, the k = 0 wave function u 0 () will cetainly appoach a staight line of the fom B( a), and it can be established that this a is indeed the scatteing length. At this point, we see that the scatteing length is defined by taking the lage fom of the k = 0 wave function u 0 (), and asking whee does this fom each 0? The value of this intecept is the scatteing length.

5 Finally, we can undestand the sign of a, by looking at the k = 0 equation fo < 0, 2 u 0 () = U()u 0 (). We must have u 0 (0) = 0, and we can cetainly stat off fom = 0 with u 0 () > 0. Then fo an attactive (U() < 0), we have 2 u 0 () < 0, i.e. the potential is tying to cuve u 0 () towad the u 0 = 0 axis. Fo a stongly attactive potential, by the time > 0, u 0 () will have developed a negative slope and the staight line it matches onto must coss the axis at > 0. Fo a weakly attactive potential, the slope of u 0 () will still be positive at = 0, and the scatteing length will be negative. The bounday between weak and stong attaction comes when the potential becomes stong enough to have a bound state. A stiking example of this is in the scatteing of a neuton and a poton. The scatteing length is lage by nuclea standads, a +5.2fm, and thee is a bound state in the system, the deuteon. It is easy to show using fomulas fom the next section that at vey low values of k, the total cossection is given in tems of the scatteing length by σ tot = a 2. Resonant Scatteing Resonant scatteing is an impotant phenomenon in low enegy scatteing. It occus when the system in question has an almost bound state. The system has an attactive potential, and may have bound states, but a esonance occus in the actual scatteing egion. A tue esonance coesponds to a paticle with a finite lifetime. Resonances abound in paticle physics, nuclea physics, and atomic physics. They ae had to model with simple potentials, so only a basic desciption is given hee. A typical example is a esonance in an l = patial wave. If the scatteing enegy is E, the esonance enegy is E and the so-called width of the esonance is Γ/2. then the fomula fo exp(2iδ ) would be o equivalently, e 2iδ = E E i Γ 2 E E + i Γ 2 e iδ sin δ = E E + i Γ 2 Fom these two fomulas, it is easy to see that a esonance coesponds to a phase shift that is positive fo E < E R, ises apidly and eaches π/2 at the esonance, and then continues on. A shap esonance coesponds to a a elatively small value of Γ. In many cases, even though seveal patial waves may be active, a esonance in one of them can lead to a shap peak in the total cossection. This is the way many esonances wee discoveed. Anothe hallmak of a esonance is the location of the pole in the complex enegy plane. We see that this occus at E = E iγ/2. Consideations of causality equie that the pole occu in the lowe half of the complex enegy plane. Γ 2

6 Patial waves and the full scatteing amplitude As, the scatteing pat of the wave function becomes popotional to the scatteing amplitude, Ψ sc f(θ, ϕ) e ik as Fom ou fomula fo the patial wave expansion of the full scatteing wave function along with the outgoing wave tem in R l, we have fo the outgoing wave of the full scatteing wave function, Ψ out k 2ik eik l e 2iδ l (2l + )P l (cos θ) as. Similaly, the outgoing wave pat of the plane wave is, fom ou pevious fomulas, (e ikz ) out 2ik eik l (2l + )P l (cos θ) as. Now the scatteing outgoing wave comes fom the diffeence between the outgoing wave of the full wave function and the plane wave, Using the two fomulas above, we have f(θ, ϕ) e ik = Ψ out k (e ikz ) out f(θ, ϕ) e ik = eik l 2ik (e2iδ l )(2l + )P l (cos θ), which finally gives fo the scatteing amplitude, f(θ, ϕ) = l e iδ sin δ l l k (2l + )P l(cos θ). (Fo a spheically symmetic potential, the scatteing amplitude depends only on the pola angle θ.) It is of inteest to see how the optical theoem is satisfied by the patial wave expansion. The optical theoem is k Im(f(θ = 0)) = σ T = f(θ, ϕ) 2 dω Fom ou fomula fo the patial wave expansion of f, and P l (cos θ = ) =, we have fo the left hand side of the optical theoem, k Im(f(θ = 0) = k l sin 2 δ l (2l + ). k

7 Fo the integal on the ight hand side, we have f(θ, ϕ) 2 dω = dω (e iδ sin δ) k 2 l (e iδ sin δ) l l,l (2l + )(2l + )P l (cos θ)p l (cos θ)) Using the othogonality popeties of the Legende polynomials, we have dωp l (cos θ)p l (cos θ) = 2l +. This educes the double sum on l, l to a single sum and we have σ T = sin 2 δ k 2 l (2l + ), l which is the same as the left hand side we calculated above. It is impotant to note that the optical theoem is satisfied patial wave by patial wave. This makes sense, since the optical theoem is ultimately a statement of pobability consevation, and says that fo a given patial wave, the outgoing wave must have the same stength as the incoming wave, and theefoe can only diffe by a phase facto, which in tems of δ l is exp(2iδ l ). The full scatteing amplitude NOTE In the equations of this section x and x ae thee dimensional vectos. The Schödinge equation fo the full scatteing wave function is ( 2 + k 2 )Ψ k (x) = U(x)Ψ k (x). In scatteing theoy, it is useful to tun this into an integal equation, Ψ k (x) = Ψ 0 k(x) + G k (x x )U(x )Ψ k (x )d 3 x The function Ψ 0 k is the fee plane wave input, and satisfies ( 2 + k 2 )Ψ 0 k(x) = 0. The function G k (x x ) is the scatteing Geen function, and must satisfy ( 2 + k 2 )G k (x x ) = δ 3 (x x ). This is necessay so that Ψ k satisfies the full Schödinge equation. The scatteing Geen function can be found by Fouie tansfom, G k (x x ) = Applying ( 2 + k 2 ) to G k (x x ) gives e k (x x ) Gk (k ) d3 k (2π) 3. (k 2 (k ) 2 ) G k (k ) =

8 The coect fom of G k (x x ) is obtained fom the Fouie tansfom by adding iϵ to k 2 in the denominato of the integal. (This insues that G k (x x ) involves only outgoing waves.) We have G k (x x ) = e k (x x ) k 2 + iϵ (k ) 2 d 3 k (2π) 2. The esult of computing the Fouie tansfom can be obtained by contou integation and is G k (x x ) = x x eik x x The scatteing integal equation now takes the fom, Ψ k (x) = Ψ 0 k(x) e ik x x x x U(x )Ψ k (x )d 3 x. Fo eading off the scatteing amplitude, we want to take the coodinates vey lage compaed to the ange of U(). We have x x as whee = x, and x x = x 2 2x x + (x ) 2 ˆ x as. Using these fomulas to get the lage distance fom of Ψ sc, we have Ψ k (x) e ikz eik e ikˆ x U(x )Ψ k (x ) By definition, the scatteing amplitude is the coefficient of exp(ik)/, so the scatteing amplitude is then f(θ, ϕ) = e ikˆ x U(x)Ψ k (x) This is an exact fomula, and equies the knowledge of the exact scatteing wave function. It is a useful fomula nevetheless, since it only equies that the exact wave function be known inside the ange of U(). It is also useful fo fomulating appoximations to f(θ, ϕ). Bon appoximation The Bon appoximation is extemely useful fo scatteing at high enegy, in paticula fo U() << k 2. In using the Bon appoximation, it is convenient to think of the scatteing as going fom an initial wave vecto to a final wave vecto, whee these ae defined by k f = kˆ, k i = kẑ.

9 The Bon appoximation involves eplacing the exact wave function in the integal fo the scatting amplitude by the plane wave, so we have f B (k f, k i ) = e ikˆ x U(x)e ikz d 3 x To see how a calculation of f B goes, take the potential U() = γ e β. This so-called Yukawa potential can also be used to model a Coulomb potential fo β vey small. The Bon appoximation is f B (k f, k i ) = e i(k i k f ) x γ e β d3 x It is seen that the Bon appoximation only depends on the diffeence of initial and final wave vectos. Setting q = k i k f, and doing the elementay solid angle integal, we have f B = γ sin q q e β This adial integal is also elementay and we obtain 2 d. f B (k f, k i ) = γ β 2 + q 2 Note that if we take an attactive potential, γ < 0, the Bon amplitude is positive. The diffeential coss section in Bon appoximation is given by dσ B dω = f B 2 = γ 2 β 2 + q 2 2 Witing out q 2 in tems of the scatteing angle, we have q 2 = k 2 f 2k f k i + k 2 i = 2k 2 ( cos θ) = 4k 2 sin 2 θ 2, so the diffeential coss section in Bon appoximation is given by dσ B dω = f B 2 = γ 2 β 2 + 4k 2 sin 2 θ 2 2 Fo lage k 2, this fomula shows a shap peak as θ 0. The scatteing at lage angles is dominated by the 4k 2 sin 2 θ facto in the denominato, and is the way Ruthefod 2 demonstated that the nucleus of an atom looks like a point chage to α paticle scatteing at the enegy he was using.