Solution Set of Homework # 2. Friday, September 09, 2017

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1 Temple University Department of Physics Quantum Mechanics II Physics 57 Fall Semester 17 Z. Meziani Quantum Mechanics Textboo Volume II Solution Set of Homewor # Friday, September 9, 17 Problem # 1 In order to describe the interaction between two nucleons, Yuawa introduced the following potential which contain spherical symmetry: V (r) = A ( r exp r ) (1) r 1. Find the differential cross section for scattering, in the Born approximation. To find the differential cross section we first calculate the scattering amplitude which in the Born approximation is expressed by: = 1 d 3 r e i q r U(r ) () 4π = µ π d 3 r e i q r V (r ) (3) where q = i f and U(r) = µ V (r) (4) We need to expand the volume integration into its components. Therefore we find d 3 r e i q r V (r ) = = π = 4π q π r dr V (r ) (θ) = µ q After replacing V (r) by its expression we obtain (θ) = µa q r dr V (r ) π e iqr cos θ sin θ dθ dφ (5) +1 1 e iqr x dx (6) r V (r ) sin (qr )dr (7) 1 r V (r ) sin (qr )dr (8) e r /r sin (qr )dr (9)

2 leading to the differential cross section expression dω = 4µ A 4 q e r /r sin (qr )dr where leading to the e r /r sin (qr )dr = 1 e 1[1/r iqr dr 1 e 1[1/r +iqr dr (11) i i = 1 [ 1 i 1/r iq 1 (1) 1/r + iq q = 1/r + (13) q (1) dω = 4µ A 1 4 q ( 1/r + q ) (14) and since q = i f = i + f i f cos θ = (1 cos θ) = sin θ/ (15) with = i = f in the case of elastic scattering. We obtain finally dω = 4µ A 1 4 q ( 1/r + 4 sin θ/ ) (16). How does the total cross section behave at high energies, where the use of the Born approximation justified? To obtain the total cross section we need to integrate over the solid angle σ = sin θdθdφ = π sin θdθ (17) dω dω = π 4µ A r 4 π sin θdθ 4 ( r sin (θ/) ) (18) the change of variable x = sin (θ/) implies sin θdθ = xdx/( r ) and leads to σ = 8πµ A r r = 16πµ A r r xdx (1 + x ) (19) () = 16πµ A r µEr / (1) where we used = µe/. From the energy dependence of the total cross section above we see that at high energies we have E then σ 16πµ A r µEr () / Therefore, the Born approximation is better at higher energies as the higher order contributions become smaller.

3 3. Compare the result to the use of the following potential: V (r) = V for r < r, V (r) = for r > r. (3) For the constant potential described above we can evaluate (θ) = µ q where V (r ) = V for r < r. The expression of course gives zero for r > r We will use integration by parts, using (θ) = µv q r V (r ) sin (qr )dr (4) udv = uv r sin (qr )dr (5) vdu, (6) to evaluate the above integral. We will set u = r and dv = sin (qr )dr thus du = dr and v = 1/q cos (qr ) Leading to the scattering amplitude and to the differential cross section r sin (qr )dr = r q cos (qr ) r + r (θ) = µv q dω (θ) = 1 q cos (qr )dr (7) = r q cos (qr ) + 1 q sin (qr ) (8) ( r cos (qr ) sin (qr ) ) q (9) ( µv r 3 ) [sin (qr ) qr cos (qr ) (qr ) 6 (3) In this case the cross section fall faster with energy compared to the case of the Yuawa potential. Problem # Exercice 3a, page 96, Complement C VIII Scattering of the p Wave by a Hard Sphere. α. We first write the radial equation of u,1 (r) for r > r where the potential V (r) =. The general radial equation is: [ d dr + l(l + 1) r u,1 (r) = (31) With the condition u,1 () = (3) 3

4 The general solution is given by a linear combination of independent spherical Bessel j l (ρ) and spherical Neumann n l (ρ) functions. Here ρ = For the case where l = 1 we have sin () j 1 () = () note also that R,l (r) = u,l() () cos () = A l j l () + B l n l () (33) ; n 1 () = cos () () ) ( cos cos + B 1 + sin () ) + sin ( sin u,1 (r) = A 1 (35) Setting the constants A 1 and B 1 to A 1 = C and B 1 = Ca we obtain the general expression of the solution of the radial equation for r > r, namely [ sin u,l (r) = C ( ) cos cos + a + sin β. By definition δ 1 () is the p wave phase shift and is by definition obtained by comparing the behavior of the solution to the radial differential equation between the incoming and outgoing spherical wave at large r. In our case we need to express the solution u,1 (r) at large r in a form proportional to sin ( π/ + δ 1 ). (34) (36) u,1 (r) sin ( π/ + δ 1 ) = (sin ( π/) cos δ 1 + cos ( π/) sin δ 1 ) (37) Now at large r our solution reads = (cos cos δ 1 sin sin δ 1 ) (38) lim u,1(r) C ( cos + a sin ) (39) r comparing the two above expressions we find that cos δ 1 = C sin δ 1 = Ca = a = tan δ 1 () (4) γ. We can determine a using the condition that at r = r since the potential is infinite u,1 (r ) =. This condition reads [ sin u,l (r ) = C sin cos + a a = ( cos cos + a ) + sin = (41) ) = (4) ( cos + sin sin cos cos + sin (43) (44) a = sin cos cos + sin (45) 4

5 δ. As approaches zero we can we can expand sin and cos in terms of powers of ( ) in the expression of a and eep the dominant term. sin = cos = n= n= ( 1) n n + 1)! x( ) n+1 ( ) ( ) 3 6 ( 1) n n)! x( ) n 1 ( ) (46) (47) (48) Replacing these expansions into the expression of a we obtain a ( ) ( ) 3 ( ) (1 ( ) ) 6 1 ( ) + ( ) (( ) ( ) 3 ) (49) 6 ( ) 3 (5) 3 Since δ () = with 1, the low energy behavior of δ 1 () ( ) 3 is negligible compared to δ (). Problem # 3: Square Spherical Well :Bound States and Scattering Resonances. Exercice 3b, page 96, Complement C VIII α. i) For any central potential we have, φ lm ( r) = u l (r)/ryl m (θ, φ) and a radial equation [ d l(l + 1) + µ dr µr + V (r) u l (r) = µ u l(r) (51) and the condition at the origin u l (r = ) =. If we restrict ourselves to the s wave(l = ) the radial equation becomes: [ d dr + u (r) = for r > r (5) and [ d u, (r) = µ dr V µ u (r) for r < r. (53) The general solution to equation (5) is given by: u (r) = A + e +i + A e i for r > r (54) 5

6 where Since E is negative we define = ρ = i = µe (55) µe (56) The boundary condition, for E <, of lim r u (r) leads to A + =. We relabel A to be A and obtain: u (r) = Ae ρr for r > r (57) The general solution to equation (53) is where u (r) = B + e +ikr + B e ikr for r < r (58) K = µv + µe (59) The condition at the origin, u () = implies that B + = B = B, so U (r) for r < r has the form u (r) = B sin Kr for r < r (6) (ii) Now we can write the matching condition at r = r and deduce the only possible values of ρ which satisfy tan Kr = K ρ (61) Since the above state for r < r is a bound state at r = r we require continuity for the wave function u (r) and it s first derivative du (r)/dr. Thus we have Dividing the two equations above we obtain B sin Kr = Ae ρr (6) BK cos Kr = ρae ρr (63) tan Kr = K ρ (64) (iii)the number of s bound states is the number of values of ρ (hence E) which satisfy the condition tan Kr = K/ρ. To find this number we need to plot tan Kr and K/ρ on the same graph and loo for intersections between the plots of these quantities. K is restricted to be between and since V < E <. The scale of the LHS, tan Kr, is fixed by r. The scale of the R.H.S., K/ρ = 1/ ( /K ) 1. The two graphs in Fig. 1 and do not intersect for < π r (65) Define ξ, n via = (n + 1)π r + ξ where ξ < π r (66) 6

7 tan Kr π r π r 3π r 4π r 5π r K Figure 1: Graphical solution of tan (Kr ) as a function of K There will b n solutions, or n bound states, for this relation between V and r. It s worth showing that there is no solution for E < V (the condition E > is ruled out by the problem.). E < V K is imaginary. L.H.S = tan Kr = tan ρ r = i tanh ρ r (67) R.H.S = K ρ = ρ ρ = i ρ ρ (68) But ρ tanh ρ r is positive, while ρ is negative. q.e.d. (69) β. Scattering Resonances (E>) -K/ρρ K Figure : Graphical solution; bottom plot of f K/ρ and bottom plot of 7

8 (i) [ d dr + u, (r) = for r > r (7) [ d dr + K u, (r) = for r < r (71) The solutions are u (r) = the condition u, () = B + = B B/. { A+ e +i + A e i for r > r B + e +ik r + B e ik r for r < r (7) Since this is a scattering resonance (rather than a bound state) we do not have the condition u, ( ) =. Consider instead the condition at the origin, u, () =, and the fact that r >. This is equivalent to a one dimensional problem with V (r) = for r <, in which case A is the amplitude of the incident plane wave and A + is the amplitude of a reflected plane at large r. Since there is no transmission, we require ( ) A + = A u, (r) = A e i e iφ + + e i e iφ (73) which we can write which is a sine wave with a phase factor { A sin ( + δ ) for r > r u (r) = B sin (K (74) r) for r < r (ii) Again here we require continuity of u, (r) and du, (r)/dr at r = r { sin ( + δ ) = B sin (K r ) cos ( + δ ) = BK cos (K r ) (75) Square both equations Add and rearrange: sin ( + δ ) = B sin (K r ) cos ( + δ ) = B K cos (K r ) (76) [ 1 = B sin (K r ) + K cos (K r ) (77) B 1 = (78) [1 cos (K r ) + K cos (K r ) B = + cos (K r ) (79) 8

9 Now divide both equations (75) we then compare to giving the definition of α() tan ( + δ ) = K tan (K r ) (8) tan (α() = K tan (K r ) (81) to obtain the or α() = + δ (8) δ = + +α() (83) (iii) The minima occur at cos (K r ) = 1. The maxima occur at cos (K r ) = solving for that give a maximum B we find ((n + 1)π K (n + 1)π r = r (84) + (n + 1)π = r (85) = r ) where n is integer (86) At these values of, tan (K r ) blows up, as does tan (α()). α() = (m + 1)π r where m is integer (87) 1 B 1 1+( /) Figure 3: Graph representing B as a function of 9

10 In terms of the phase shifts, the total cross section is For 1, at resonance σ = 4π (l + 1) sin δ l (88) l= sin δ = sin ( + α()) (89) ( ) = sin (m + 1)π + (9) r = [( 1) m cos ( ) (91) = cos ( ) (no approximations yet) (9) = 1 ( ) + O [ ( ) 4 (93) Therefore sin δ is practically maximal (i.e., close to 1) at 1, and the contribution to σ from the s wave is sin δ γ. (i) This is apparent from the graphs in figures 1 and when combined which I have plotted in Fig. 4. If the dotted line is a little to the right of one of the dashed lines, then the two curves intersect at a value of K slightly less than, between the dashed and dotted line.thus there will be a bound state corresponding to K = ρ, while K <. Now let s show it mathematically, since we have guessed it from the graph Fig. 4. Guess: ρ ɛ for = (n + 1)π/ + ɛ with ɛ positive and ɛ 1. LHS: RHS: tan Kr = tan ρ r (94) [ ( ) (n + 1)π = tan (1 ɛ) 1/ + ɛ (95) [ ( + 1)π = tan + ɛ + (n + 1)π [ (n + 1)π ɛ = cot ɛ + + O(ɛ3 ) = 1 ɛ + ɛ [ ( + 1)π ɛ + O(ɛ3 ) (96) (97) O(ɛ ) (98) K ρ = ρ ρ = 1 [ 1 1 ɛ ɛ + O(ɛ 4 ) (99) (1) = 1 ɛ + ɛ + O(ɛ3 ) (11) We see that the L.H.S. = R.H.S. to first order in ɛ, note that if ɛ is not positive, we may not have ρ ɛ 1

11 (ii) The condition for resonance is plug in for : ( ) (n + 1)π = (1) r ( ) (n + 1)π = 1 [ (n + 1)π r r + ɛ (13) ( ) (n + 1)π = 1 ( ) (n + 1)π r r ( ) (n + 1)π ɛ r r (14) = ɛ [ (n + 1)π + ɛ (15) r r = ɛ r + O(ɛ ) (16) (iii) We deduce that the bound state disappears as in (i) when ɛ goes negative, and a resonance appears as in (ii) at about the same energy. tan Kr π r π r 3π r 4π r 5π r K Figure 4: Combined graphs of Figs.1 and 11