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# ph135cS2 - Ph135c Solution set#2 1 First consider the bound...

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Ph135c. Solution set #2, 4/15/10 1. First consider the bound state. As shown in the book, its energy is given by solving the following transcendental equation: K cot( Kr o ) = - k where: K = p 2 μ ( V o - E B ) k = p 2 μE B here μ is the reduced mass for the proton-neutron system, μ m p 2 . The scattering length a t is defined by: a t = - lim k 0 δ 0 k where δ 0 is the phase shift in the s -wave scattered off the potential. For a square well with depth V 0 and radius r 0 , δ 0 is given by solving: k tan( Kr o ) = K tan( kr 0 + δ 0 ) where now: K = p 2 μ ( V 0 + E ) This gives: δ 0 = tan - 1 ( k K tan( Kr 0 )) - kr 0 To obtain the scattering length, we need to extract the lowest order (in k ) term of this expression. Since the argument of the inverse tangent function is small for small k , we can approximate it by tan - 1 x x , and we find: δ 0 k ( tan( K 0 r 0 ) K 0 - r 0 ) where K 0 = 2 mV 0 . The scattering length is then: a t = r 0 - tan( K 0 r 0 ) K 0 1

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Plugging in the known values E B = 2 . 22 MeV and a t = 5 . 4 fm , these two equations can be solved numerically, and we find: r 0 = 2 . 1 fm V 0 = 34 . 2 MeV Next we need to extract the effective range of the scattering. To do this, we’ll need to determine δ 0 to the next order in the energy. This is tedious to do by hand, but using Mathemtica we find: δ 0 = k K 0 tan( K 0 r 0 ) - K 0 r 0 + k 3 6 K 0 3 3 K 0 r 0 - 3 tan( K 0 r 0 ) + 3 K 0 r 0 tan 2 ( K 0 r 0 ) - 2 tan 3 ( K 0 r 0 ) + ... Or, if we note from above that tan( K 0 r 0 ) = K 0 ( r 0 - a t ) and K 0 2 = 2 mV 0 , this can be rewritten as: = - a t k + k 3 a t 4 μV 0 + 1 6 r 0 3 - 1 2 a t 2 r 0 + 1 3 a t 3 We write this as δ 0 = - a t k + bk 3 , where numerically b 27 fm 3 . Then we have: k cot( δ 0 ) = k cot( - a t k + bk 3 ) = - 1 a t + k 2 a t 3 - b a t 2 + ... The expression in parentheses is then half the effective range, which we can compute to be: r e = 1 . 7 fm 2. a) Typically the ground state is associated with = 0, as a non-zero introduces an extra positive contribution to the potential which tends to raise the allowed energies. However,
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