I already attached the solution to those question but I would like to know how to get the same solution The solutions ar

Post by **answerhappygod** » Sun Apr 10, 2022 8:50 am

I already attached the solution to those question but I would
like to know how to get the same solution

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We will explore this situation for the soft-sphere potential: V(r) {o" -V if r <a 0 ifra = = (a) (/10) Find the general solution (for all 1) to the radial Schrodinger equa- tion inside the sphere (for r < a) for which the energy of the particle is E = h²K2/(2m), where K? k2 + kä, and ko V2mVo/h2. Call your solution 4,?! (r) - make sure to ensure that you solution does not diverge at r = 0. Do the same outside the sphere (for r > a), for which the energy of the particle is E = ħ²k2/(2m) and call your solution 4, 2) (r). (b) (/20) The phase shift of the l-th partial wave, di, can be found in terms of the spherical Bessel functions ji(kr) and ni(kr) as ji(ka) – nij(ka) tan di (1) n(ka) – nını(ka) =
where j/(ka) denotes the derivative dji(kr)/dr evaluated at r = a (likewise for ní(ka)) and dų{1}(r) 1 i n= 4,{"}(r) dr r=a Calculate tan 81=0 (s-wave) and tan dizi (p-wave) in the low-energy limit ka < 1. Express your answers in terms of 70 and 71. NOTE: the following spherical Bessel functions may be useful: jo(x) sin (2)/x , ji(2) sin (2)/(x2) – cos (x)/x , no(x) = - cos (2)/2 , and n1(x) = -cos ()/(x2) - sin(x)/x. =-
(b) We need to calculate each of the terms in kj/(ka) -(ka) tan di kní(ka) – vini(ka) (1) for 1 0 and 1 = = 1. For 1 no(x) = -cos (x)/x so that 0, we have that jo(ka) = sin (ka)/ka and k ka cos(ka)-sin(ka) k ka sin(ka)+cos(ka) ka2 tan(do) sin(ka) 70 ka cos(ka) + yo ka ka2 where yo K cot(Ka) - a In the limit ka + 0, we get that yoka? tan(80) = (2) 1+ya Likewise, for 1 = 1, we have that ji(x) = sin(x)/(x2) – cos (x)/x and
ni(2) cos (x)/(x2) – sin (x)/x. The expression for tan(81) is long, we will separate the answer into two parts: the numerator is given by ka cos(ka) – 2 sin(ka) ka sin(ka) + cos(ka) ) k2a3 ka2 (ka)2 ka psin(ka) cos(ka)] - + - 71 and the denominator is given by ka sin(ka) – 2 cos(ka) k2a3 * ka cos(ka) – sin(ka) + 71 ka2 cos(ka) + sin(ka)] + (ka)2 so that ka cos(ka)-2 sin(ka) k223 ka sin(ka)+cos(ka) ka2 71 cos(ka) ka tan(81) sin(ka) (ka)2 cos(ka) (ka)2 ka sin(ka) -2 cos(ka) k2q3 ka cos(ka)-sin(ka) ka2 +71 + sin(ka) ka where 71 = Ka 1-K a cot(Ka) à. In the limit ka → 0, we get that a (ka): 1 – Ya tan(81) = - 3 2 + 71a (3 (3)