Question 2: finite square well in three dimensions 12 marks *Please note: in PHYS2111 we have...

@ A particle of mass m is subjected to the spherically symmetric square well potential defined as Enver S-voolrla Lornyata where is the distance is a very that ust sound from the origin of out â cocinorbrogub co-ordinate system of i notas rapwibon Noto Trog In the spherically symmetric potential V(r), the schrodinger equation is 30 Vy() +2m (6 - VODY() = 0 - now in spherical polas co-ordinate (r.O, o) vysti su (oray) + sino do lodo/ +1 24 arsino dor il (sino dy) where we have used the fact that the operator representation of L2 is given an. + [ šiuo Jo (sio So )stwo Opal nuus equation can be written as + (or ) + 2 [t-vo]y > LV - © In order to solve the above equation we have to use the method of separation of variables and write 417,0,0) = R(W) 0 (0) $(8) 2 Rim) (0,0) substituting equation & in equation @ we a YC039) (de) + Path [ 6-vm] em Y(0) - RW LY(0,0)

dividing by R(W) 7 (0,9) we get Root (or de 9+ z mo [e-ve] at your(0,0) = 2 where we set the terms equal to a constant & because the left hand side of the above equation depends only ons while the other term depends. only on o and a so Radial part of schrodinger equation is dR 2d R(x) + 2m (E-v(y) = f (l+1) h R(V) 20! 2me where a= 2 (2+1) , 820,1,2...

If we define new radial function by T U or (r) & R (8) R(8) 2 (r) then Above equation becomes dur2m idret tr al(4+1) t imre Tur)=0

Now for zero angular momentum a28 (2+1) 20 then it becomes dụw + 2m E-vu() = 0 Now since the potential is attractive so. E must be negative. Hence it becomes (dum + 2m (vo - 1 E 170 for okra- and dre te El dun - 2m 16 um 20 for orya Now if k= 2m (wo-1E1) and K2 2 2 m. 16) then these equations became du kiu=0 for olrla. Idret Idu - Kou=0 for pra the solution of these equations are u() = (Asinks + Blaskir) A, B, C, D are constants.

U() = c'exp(-K2Y) +. Desip' (kar) As so ur must tend to reso: This make B2o, are solution exp (K2) is not finite as ro Gence D=0 un solutions are Suu. Asinki for Orla! ura C exp(-K2Y) for a Applying the continuity conditions on u(r) and и о т о wе арн а у ) і

ST) 0 6 h I Beorp -8 1 A sink, a = exp(-K2 a.) or and A Kicos K, Q = -K2 ( exp(-K, a) Dividing one by other and multiplying throughout by a, we obtain to riservation of all foon Kacotk, a = - karist ol Now writing ki az and Kia z 8 h we have J mis is the transdental equation . we have put 8 2 2mvoa, which is the equation of a circle in ß8 plane with radius (2nvoa t rio of a To get the solution of p. corp against & is plotted along circles of radii jamvoar for different values of voa in the figure belows voitor 2 pwrp F AOZAJ ( Bora Botp=8 g; vol 2 0 fig.

For ñ ñ 3a un Bakia 2 pump Graphical solution of equation & and G. your values of voar.

© Two bound stakes exists is two intersection in the figure found. ruis happens if Radius 37 2mvoo 35 ✓ z 3 or 2mv.a a Vo 90th 8mar