Repulsive Square Well
المؤلف:
Sidney B. Cahn, Gerald D. Mahan And Boris E. Nadgorny
المصدر:
A GUIDE TO PHYSICS PROBLEMS
الجزء والصفحة:
part 2 , p 75
21-8-2016
1491
Repulsive Square Well
Consider in three dimensions a repulsive (V0 > 0) square well at the origin of width a. The potential is
(1)
Figure 1.1
A particle of energy E = h2k2/2m < V0 is incident upon the square well (see Figure 1.1).
a) Derive the phase shift for s-waves.
b) How does the phase shift behave as V0 → ∞?
c) Derive the total cross section in the limit of zero energy.
SOLUTION
a) If the radial part of the wave function is R(r) then define χ(r) = rR(r). Since R is well behaved at r → 0, χ = 0 in this limit. The function χ(r) obeys the following equation for s-waves:
(1)
where
(2)
and the theta function Θ(a – r) is 1 if a > r and 0 if a < r. For r > a the solutions are in the form of sin kr or cos kr. Instead, write it as sin(kr + δ) where the phase shift is δ(k). For r < a define a constant α according to α2 = k20 – k2 > 0. Then the eigenfunction is
(3)
For r < a the constraint that χ(0) = 0 forces the choice of the hyberbolic sine function. Matching the eigenfunction and slope at r = a gives
(4)
(5)
Dividing these equations eliminates the constants A and B. The remaining equation defines the phase shift.
(6)
(7)
b) In the limit that V0 → ∞, the argument of the arctangent vanishes, since the hyperbolic tangent goes to unity, and δ = -ka.
c) In the limit of zero energy, we can define
(8)
(9)
To find the s-wave part of the cross section at low energy, we start with
(10)
where the total cross section is σ.
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