Why Scatteing? B Lectue Notes Scatteing Theoy I Scatteing of paticles off taget has been one of the most impotant applications of quantum mechanics. It is pobably the most effective way to study the stuctue of matte at small distances. Due to the uncetainty pinciple x p h, ) in ode to pobe stuctue at the distance scale x d, we need momentum tansfe p > h/d and hence need a high-momentum scatteing expeiment. On the othe hand, to study the excitation spectum of a system, one can study the esonances and/o scatteing at elatively low enegies. Ruthefod scatteing expeiment, scatteing of α-paticles off gold foil, is the ealiest impotant quantum mechanical scatteing expeiment of the fist type, and evealed the fact that the positive chage in an atom is concentated at the cente athe than diffusely distibuted thoughout the atom, the plum-pudding model by J.J. Thomson. Fo non-elativistic poblems, time-independent fomalism is convenient, whee you study the stationay poblem as in the bound state poblems. It is somewhat confusing that you can study the scatteing pocess as a stationay poblem, but this becomes clea late once we discuss wave packets. Fo elativistic poblems, howeve, time-dependent fomalism is moe convenient because one can keep manifest Loentz covaiance in the fomulation. Lippmann Schwinge Equation We fist study time-independent fomalism fo scatteing. Imagine a paticle coming in and getting scatteed by a shot-anged potential V x) located aound the oigin x 0. The time-independent Schödinge equation is simply h m + V x) ψ x) Eψ x). )
Because we assumed that the potential is shot-anged, V x) 0 beyond a cetain distance x a whee a is the size of the scattee. Theefoe, the Schödinge equation Eq. ) educes to the fee equation at x a h ψ x) Eψ x), 3) m and hence the enegy eigenvalues and eigenfunctions ae given by h m ei k x h k m ei k x Ee i k x. 4) The question is how this eigenfunction is modified in the pesence of the potential tem in Eq. ). In ode to answe this question, we fist go to the ket notation x ψ ψ x) and ewite Eq. ) as H 0 + V ) ψ E ψ. 5) Hee, H 0 p /m is the fee-paticle Hamiltonian opeato. Because we ae inteested in the effect of the potential, we eode it as Naively, we can wite the solution to Eq. 6) as E H 0 ) ψ V ψ. 6) ψ E H 0 V ψ. 7) But this is not coect because E H 0 can be zeo and this fomal expession is ill-defined. Fist of all, we need to specify how we go aound the pole of /E H 0 ); this coesponds to specifying a bounday condition to the solution. Fo ou pupose, we choose /E H 0 +iɛ) whee we take the limit ɛ +0 at the end of the calculations. The meaning of this choice becomes clea when we discuss wave-packets in the next section. Second, because E H 0 can be zeo, thee ae solutions E H 0 ) φ 0, which ae nothing but the fee plane wave solutions. Theefoe, we can wite the solution to this equation as ψ φ + V ψ. 8) E H 0 + iɛ
One can easily check that Eq. 8) satisfies the equation Eq. 6). This equation is called Lippmann Schwinge equation. Because φ is a solution to the fee equation, we nomally take it a plane wave h k with E h k /m and momentum h k. To gain moe insight into the Lippmann-Schwinge equation, let us take the position epesentation of this equation, by taking the inne poduct of Eq. 8) with the ba x. We find ψ x) x ψ π h) 3/ ei k x + x E H 0 + iɛ V ψ π h) 3/ ei k x + d x x E H 0 + iɛ x x V ψ π h) 3/ ei k x + d x x E H 0 + iɛ x V x )ψ x ). 9) At the vey last step, we used the fact that the potential opeato is diagonal in the position epesentation x V x V x)δ x x ). Then the question is the exact fom of the Geen s function G x, x ) x E H 0 + iɛ x. 0) By inseting the complete set of states in the momentum epesentation, and using the fact that H 0 is diagonal in the momentum space, we find G x, x ) d p x p E p /m + iɛ p x ei x p/ h d p π h) 3/ E p /m + iɛ e i x p/ h π h) 3/ d p ei x x ) p/ h π h) 3 E p /m + iɛ. ) Thee ae many ways to do this integation. One way is to use pola coodinates fo p defining the pola angle elative to the diection of x x such that G ) 0 π π h) 3 π p cos θ/ h eip dp d cos θ dφ 0 π h) 3 E p /m + iɛ 0 p dp eip/ h e ip/ h ip/ h 3 E p /m + iɛ
eip/ h m pdp π h) i p me iɛ mi π h) pe ip/ h dp p me iɛ)p + me + iɛ). ) Because of the numeato e ip/ h, we can extend the integation contou to go along the eal axis and come back at the infinity on the uppe half plane. Then the contou integal picks up only the pole at p me +iɛ hk+iɛ, and we find G ) Going back to Eq. 9), we now obtain ψ x) mi πi hkeik π h) hk m h e ik 4π. 3) π h) 3/ ei k x m h d x e ik x x 4π x x V x )ψ x ). 4) This esult allows a simple intepetation. The fist tem plane wave) is the incident paticle with a fixed thee-momentum, while the second tem is a spheical wave oiginated fom the scatteing. If we had chosen the opposite bounday condition /E H 0 iɛ), we had obtained the second tem coming fom infinity and conveging at the oigin, which is pactically impossible to aange. This intepetation is futhe justified using wave-packets in the next section. The expeiment is done typically by placing the detecto fa away fom the scattee x a whee a is the size of the scattee. The integation ove x, on the othe hand, is limited within the size of the scattee because of the V x ) facto. Theefoe, we ae in the situation x x, and hence can use the appoximation x x x x x x x x. 5) x Unde this limit, the Lippmann Schwinge equation Eq. 4) becomes ψ x) π h) 3/ ei k x m e ik h d x 4π e i k x V x )ψ x ), 6) 4
whee x and k k x is the wave-vecto of the scatteed wave. Note that k k. It is customay to wite this equation in the fom with ψ x) ) e i k x + f k π h), k) eik, 7) 3/ f k, k) π h)3 4π m h h k V ψ, 8) which has a dimension of length. The advantage of using f k, k) is that it is diectly elated to the scatteing coss section as we will see late. dσ dω f k, k) 9) 3 Wave-Packets The time-independent solution we discussed in the pevious section is not easy to undestand intuitively because both the incident and scatteed waves appea and both waves ae spead out ove the entie space. What we nomally pictue as a scatteing pocess is that thee is a paticle coming in with a fixed momentum, and it gets scatteed to a diffeent diection, o it does not get scatteed and comes out with the same momentum. In fact, the time-dependent solution we discussed in the pevious section actually gives pecisely this pictue once we fom wave-packets out of it. We put in paticles and detect paticles fa away fom the scattee, and we can use the asymptotic fom Eq. 7). 3. Fee Wave Packets As an execise, let us fist study wave packets of plane waves. The plane wave e i k x has a definite momentum p h k p 0), but is spead out in the whole space x ) consistent with the uncetainty pinciple. We can fom a nomalized wave-packet which has a momentum appoximately 5
h k, but has a finite spatial extent d, by taking a linea combination of many plane waves. We take the so-called Gaussian wave packet ψ f x) d π 3 ) 3/4 d qe i q x e q k) d ) 3/4 e i k x e x /4d. 0) πd Fom the last expession, it is clea that the wave packet is concentated aound x 0 with the spead x d, while fom the middle expession, the momentum is distibuted aound h k with k q k /d, consistent with the uncetainty pinciple. Time evolution of couse moves the wave packet. At time t 0, the above wave function acquies a phase e iet/ h whee E h q /m fo the plane wave with momentum h q. Theefoe, ) d 3/4 ψ f x, t) d qe i q x e i h q /m)t/ h e q k) d π 3 ) d 3/4 ) 3/ π exp x 4i k xd 4 k d 4 π 3 d + i ht/m 4d + i ht/m) k d. ) This expession looks complicated, but it is easy to intepet the pobability distibution ψ x, t) : with ) 3/ ψ f x, t) e x h kt/m) /dt), ) πdt) dt) d4 + ht/m) ) /. 3) d The wave packet is somewhat lage at t 0 and is located at the position x h kt/m as expected. As long as the expeiment is done within the time t md / h, o in othe wods if the momentum is sufficiently well detemined, we can ignoe the change in the size of the wave packet. This coesponds to doing the integal ove the phase facto e iet/ h with E h q /m appoximated as E h k m + h k q k) + O q m k) 4) 6
and dopping the last coection. Then the wave packet is simply ) 3/4 ψ f x, t) e i k x e i h k /m)t/ h e x h kt/m) /4d. 5) πd 3. Wave Packet with Scatteing We now take the wave packet out of Eq. 7) with the same Gaussian weight, and show that it gives the incoming wave at t 0, while gives both the unscatteed outgoing wave with the same momentum and the scatteed outgoing wave emeging fom the cente at t 0. This would match ou intuition of a scatteing expeiment. The wave packet we conside is theefoe ψ x, t) d π 3 ) 3/4 d q e i q x + f q, q) eiq ) e i h q /m)t/ h e q k) d. 6) Of couse we assume that the tansvese size of the wave packet is lage enough d a so that entie scatteing egion is pobed by the wave packet. The injection and detection time is not too long that the size of the wave packet does not incease siginificantly t md / h, but lage enough that the entie packet exits the scatteing egion t md/ hk, which is possible as long as the wave packet has a well-defined momentum k /d k. The fist tem in the paenthesis gives pecisely the same wave packet as the fee case, which comes in and goes out at the position x h kt/m. We hence only need to discuss the second tem. Because of the Gaussian damping facto, q is highly concentated aound q k and we can eplace q by k almost eveywhee. Howeve, a cae must be taken in the phase factos because the integal vanishes if the integand oscillates vey apidly. The exponent in the phase facto is expanded as iq i h q m t h ik i h k t m h + i hk m t ) k k q k) + Oq k). 7) The phase facto is stationay only when hk t 0 but this is clealy m possible only when t > 0. Theefoe, this integal vanishes due to apidly oscillating phase facto fo t md/ hk, and only the plane wave piece emains: ) 3/4 ψ x, t) e i k x e i h k /m)t/ h e x h kt/m) /4d, t md ) πd hk 8) 7
which is the same as Eq. 5). This is nothing but the incoming wave packet befoe the scatteing takes place. On the othe hand, fo t md/ hk, the phase can be stationay 7), and the scatteed wave contibutes. q integation in the scatteed wave packet is d qf q, q) eik i h k /m)t/ h e i hk/m)t) k q k)/k e q k) d f k, k) eik i h k /m)t/ h π d ) 3/ e hk/m)t) /4d, 9) and hence ψ x, t) ) 3/4 e i h k /m)t/ h πd e i k x e x h kt/m) /4d + f k, k) eik /4d e hk/m)t) ). t md ) hk 30) This wave packet consists of two pieces: one that went though the scatteing egion unscatteed the fist tem) and the othe that emeged fom scatteing and expands as hk/m)t spheically. Howeve, note that these two waves intefee at x h k/m)t. If you detect the wave at a diection diffeent fom the oiginal plane wave, the fist tem simply does not contibute, and the only the second tem is detected. The pobability to detect a paticle along the diection θ, φ) in a solid angle dω d cos θdφ is given by P Ω)dΩ 0 0 πd ddω ψ x, t) ) 3/ f d k, πd k) /d e hk/m)t) dω f k, k) dω. 3) At the last step, we assumed that t is sufficiently lage so that the integation only fo > 0 can be appoximated by a full Gaussian integal. 3.3 Scatteing Coss Section Now we ae in the position to define the diffeential coss section. You wait fo a paticle to ente the detecto in the solid angle dω fo each paticle you 8
put in. The useful quantity is to detect the scatteed paticle fo the given pobability density of the paticle you injected pe unit aea. The incoming wave packet was given in Eq. 5), and had the density of dz ψ f x, t) dz ) 3/ e x h kt/m) /d +y )/d πd e x πd 3) whee we assumed that k is along the z-axis. At the location of the scattee x y 0, the density is /πd. Theefoe, the pobability of the scatteed paticle to be detected at a given solid angle dω fo a paticle pe aea is dσ dω πd P Ω) f k, k). 33) This is the definition of the diffeential coss section. It is usually phased as the numbe of scatteed paticles in the solid angle dω pe unit time pe unit luminosity, which is the numbe of paticles injected pe unit aea pe unit time. Obviously two definitions ae the same based on the pobabilistic intepetation of quantum mechanics. The total coss section is simply the diffeential coss section integated ove the entie solid angle σ dω dσ dω. 34) It is impotant to note, howeve, that the fomula fo the coss section does not take the unscatteed wave into account. Theefoe, the diffeential coss section at the fowad diection is not the actual numbe of paticles detected thee, but athe the numbe of scatteed paticles slightly off fom the fowad diection extapolated to the fowad diection. The actual numbe of paticles detected at the fowad diection should take both the unscatteed and scatteed waves, and impotantly thei intefeence must be taken into account. Because the total numbe of paticles detected must equal the total numbe of paticles injected, the scatteed wave at the fowad diection should satisfy a special equiement. This is the optical theoem discussed in the next section. 4 Optical Theoem Thee ae many ways to deive the optical theoem. Hee we continue using the wave packets and equie that the nomalization of the wave function 9
should always satisfy d x ψ x, t) fo any t, as guaanteed by the unitaity of time evolution opeato e iht/ h. This equiement leads to a special equiement on the scatteed wave, and hence f k, k). We stat with the wave packet Eq. 6). When t md/ hk befoe the scatteing), the scatteed wave vanishes as discussed in the pevious section, and hence the wave function is Eq. 8) which is popely nomalized. On the othe hand, when t md/ hk afte the scatteing), the wave packet is Eq. 30), and its nomalization is not obviously unity. We theefoe equie that ) 3/ d x πd ei k x e x h kt/m) /4d + f k, k) eik /4d e hk/m)t). 35) The absolute squae of the fist tem is also popely nomalized as it is the same as in the fee case. Theefoe, the coss tem and the second tem absolute squaed must cancel. The second tem absolute squaed is obtained easily as ) 3/ d x πd f k, k) eik /4d e hk/m)t) ) 3/ dω d f k πd, k) e hk/m)t) /d σ, 36) πd whee σ is the total coss section Eq. 34). The coss tem is slightly moe complicated but is staightfowad to calculate. Note that it is impotant only in the fowad egion whee both the unscatteed and scatteed waves coexist, and we can eplace f k, k) by f0) f k, k). It is: ) 3/ d x e i k x e x h k/m)t) /4d f0) eik /4d e hk/m)t) + c.c. πd ) 3/ d cos θdφ d cos θ πd f0)eik ik e hk cos θ/m)t+ hk/m) t )/4d e hk/m)t+ hk/m) t )/4d + c.c. ) 3/ π df0) πd hkt 0 ik 4md [ e 4 hk/m)t+ hk/m) t )/4d e + hk/m) t )/4d e ik] + c.c. 0 37)
The tem hkt/4md in the denominato is negligible compaed to k within ou assumption that the size of the wave packet does not gow significantly ht/md ) and can be dopped. The second tem in the squae backet is exponentially suppessed at lage time compaed to the fist tem whee the exponent vanishes at hkt/m. Theefoe, ) 3/ π d i f0)e hk/m)t) /d + c.c. πd 0 k 4π If0). 38) k πd Requiing that the second tem absolute squaed Eq. 36) and the coss tem Eq. 38) cancel exactly, we find σ 4π k If0). 39) This is what is called the optical theoem. The meaning of this theoem is clea. Because the scatteed wave takes the pobability away to diffeent diections, the total pobability fo the paticle to go to the fowad diection unscatteed) should decease. This decease is caused by the intefeence between the unscatteed and scatteed waves and hence is popotional to f0). On the othe hand, the amount of decease in the fowad diection should equal the total pobability at othe diections, which is popotional to the total coss section.