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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:

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

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

and

Ψ_{III trans} = F exp{i(k₁x − ω₁t)} ,

where k₁² = 2mE/ħ² and k₂² = 2m(E − V₀)/ħ². We calculate J_x for the incident, reflected, and transmitted waves:

J_inc= (ħk₁/m) |A|² ,       J_refl= −(ħk₁/m) |B|² ,    and   J_trans= (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  (Ψ_II = Ψ_{II R} + Ψ_{II L}).  We require that Ψ_I(−a) = Ψ_II(−a), Ψ_I'(−a) = Ψ_I'(−a), Ψ_I(a) = Ψ_II(a), and Ψ_I'(a) = Ψ_I'(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(E − V₀)/ħ², we can rewrite 1/T in terms of E and V₀ as

1/T = |(A/F)|² = [ 1 + (V₀²/{4E(E − V₀)})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

Ψ_{II R} = C exp{−κ₂x − iω₂t}          Ψ_{II L} = D exp{κ₂x − iω₂t}

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)² =  n²E₁, 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.

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