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Section 9.6: Plane Wave Scattering: Finite Barriers and Wells




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We now consider a barrier that exists only from −a < x < a. This situation is shown in Barrier Animation 1 and Barrier Animation 2.  This problem proceeds much like the previous cases, except there are now three regions of interest. For E > V0, we consider the wave functions in each region:

ΨI inc = A exp{i(k1x − ω1t)} ,         ΨI refl = B exp{−i(k1x + ω1t)} ,

ΨII R = C exp{i(k2x − ω2t)} ,         ΨII L = D exp{−i(k2x + ω2t)} ,

and

ΨIII trans = F exp{i(k1x − ω1t)} ,

where k12 = 2mE/ħ2 and k22 = 2m(EV0)/ħ2. We calculate Jx for the incident, reflected, and transmitted waves:

Jinc= (ħk1/m) |A|2 ,       Jrefl= −(ħk1/m) |B|2 ,    and   Jtrans= (ħk2/m) |F|2 ,

and since k1 = k3, we find that the transmission and reflection coefficients reduce to just T = |F/A|2 and R = |B/A|2.

To find these ratios, we must again match the pieces of the wave function at the boundaries of the regions  (ΨII = ΨII R + ΨII L).  We require that ΨI(−a) = ΨII(−a), ΨI'(−a) = ΨI'(−a), ΨI(a) = ΨII(a), and ΨI'(a) = ΨI'(a). We get the following result after much algebra:

(A/F) = [cos(2k2a) − i{(k12 + k22)/(2k1k2)} sin(2k2a)]

and therefore for 1/T we have

1/T = |(A/F)|2 = [1 + {(k12 k22)/(2k1k2)}2 sin2(2k2a)] . (9.16)

We note that T → 1 when E is large enough (E >> V0 and therefore k2 →  k1) and also when the sine term is zero (when 2k2a = nπ or that k2 = nπ/2a = 2π/λ2). Now, since k12 = 2mE/ħ2 and k22 = 2m(EV0)/ħ2, we can rewrite 1/T in terms of E and V0 as

1/T = |(A/F)|2 = [ 1 + (V02/{4E(EV0)})sin2(2k2a)] . (9.17)

Now, for E < V0 we have a similar situation.6  We consider the barrier that exists again only from −a < x < a.  This problem proceeds much like the E > V0 case we have considered before except in the second region we make the substitution ik2 → κ2 or that k2 → −iκ2 and therefore

ΨII R = C exp{−κ2xiω2t}          ΨII L = D exp{κ2xiω2t}

Using the substitution k2 → −iκ2, the transmission coefficient becomes:

1/T = |(A/F)|2 = [1 + {(k12 + κ22)/(2k1κ2)}2 sinh2(2κ2a)] (9.18)

since sin(iz) = i sinh(z). We can rewrite 1/T in terms of E and V0 using k12 = 2mE/ħ2 and κ22 = 2m(V0 E)/ħ2 as

1/T = |(A/F)|2 = [1 + (V02/{4E(V0 E)}) sinh2(2κ2a)] (9.19)

Notice that this equation implies for E < V0, T < 1, but T is not necessarily zero.

Scattering from a well, for E > 0 but with V0 < 0 or V0 = −|V0|, we have a situation which is shown in Well Animation 1 and Well Animation 2.  We consider the well that exists only from −a < x < a.  We may use the previous answer for the barrier with E > V0:

1/T = |(A/F)|2 = [1+{(k12 − k22)/(2k1k2)}2 sin2(2k2a)] (9.20)

and in terms of E and V0

1/T = |(A/F)|2 = [ 1 + (V02/{4E(E + |V0|)}) sin2(2k2a)] (9.21)

where k22 = 2m(E + |V0|)/ħ2.  When does T → 1?  Look at the second term in Eq. (9.21).  When that term goes to zero, T → 1. This happens when E >> V0 and in one other set of occurrences. If we now consider the sine function, it is zero when nπ = 2k2a and therefore k2 = nπ/2a = 2π/λ2, where 2a is the width of the well. Now given the relationship k2 = nπ/2a that we can insert into E + |V0| = ħ2k22/2m, we get E + |V0| = ħ2n2π2/2m(2a)2n2E1, where E1 is the energy of an infinite square well of width 2a. This effect is called the Ramsauer effect. This effect describes the scattering of low-energy electrons from atoms.


6Again, there is one more possibility, E = V0.  Look for this in the animations.

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