Physlets run in a Java-enabled browser, except Chrome, on the latest Windows & Mac operating systems. If Physlets do not run, click here for help on updating Java and setting Java security.

# Section 9.6: Plane Wave Scattering: Finite Barriers and Wells

Show Real and Imaginary Parts | Show Phase as Color

Please wait for the animation to completely load.

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* > *V*_{0}, we consider the wave functions in each region:

Ψ_{I inc }= *A* exp{*i*(*k*_{1}*x* − ω_{1}*t*)} , Ψ_{I refl }= B exp{−i(*k*_{1}*x* + ω_{1}*t*)} ,

Ψ_{II R }= C exp{i(*k*_{2}*x* − ω_{2}*t*)} , Ψ_{II L }= D exp{−i(*k*_{2}*x* + ω_{2}*t*)} ,

and

Ψ_{III trans }= F exp{*i*(*k*_{1}*x* − ω_{1}*t*)} ,

where *k*_{1}^{2 }= 2*mE*/*ħ*^{2} and *k*_{2}^{2 }= 2*m*(*E* − *V*_{0})/*ħ*^{2}. We calculate *J*_{x} for the incident, reflected, and transmitted waves:

*J*_{inc}= (*ħk*_{1}/*m*) |*A*|^{2} , *J*_{refl}= −(*ħk*_{1}/*m*) |*B*|^{2} , and *J*_{trans}= (*ħk*_{2}/*m*) |*F*|^{2} ,

and since *k*_{1 }= *k*_{3}, 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(2*k*_{2}*a*) −* i*{(*k*_{1}^{2 } +* k*_{2}^{2})/(2*k*_{1}*k*_{2})} sin(2*k*_{2}*a*)]

and therefore for 1/*T* we have

1/*T *= |(*A*/*F*)|^{2 }= [1 + {(*k*_{1}^{2 } − *k*_{2}^{2})/(2*k*_{1}*k*_{2})}^{2} sin^{2}(2*k*_{2}*a*)] . (9.16)

We note that *T *→ 1 when *E* is large enough (*E* >> *V*_{0} and therefore *k*_{2 }→ *k*_{1}) and also when the sine term is zero (when 2*k*_{2}*a* = *n*π or that *k*_{2 }= *n*π/2*a* = 2π/λ_{2}). Now, since *k*_{1}^{2} = 2*mE*/*ħ*^{2} and *k*_{2}^{2 }= 2*m*(*E* − *V*_{0})/*ħ*^{2}, we can rewrite 1/*T* in terms of *E* and *V*_{0} as

1/*T* = |(*A*/*F*)|^{2 }= [ 1 + (*V*_{0}^{2}/{4*E*(*E* − *V*_{0})})sin^{2}(2*k*_{2}*a*)] . (9.17)

Now, for *E* < *V*_{0} 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 *> *V*_{0} case we have considered before except in the second region we make the substitution *ik*_{2 }→ κ_{2} or that *k*_{2 }→ −*i*κ_{2} and therefore

Ψ_{II R }= *C* exp{−κ_{2}*x* − *i*ω_{2}*t*} Ψ_{II L }= *D* exp{κ_{2}*x* − *i*ω_{2}*t*}

Using the substitution *k*_{2 }→ −*i*κ_{2}, the transmission coefficient becomes:

1/*T* = |(*A*/*F*)|^{2 }= [1 + {(*k*_{1}^{2 }+ κ_{2}^{2})/(2*k*_{1}κ_{2})}^{2} sinh^{2}(2κ_{2}*a*)] (9.18)

since sin(*iz*) = *i* sinh(*z*). We can rewrite 1/*T* in terms of *E* and *V*_{0} using *k*_{1}^{2 }= 2m*E*/*ħ*^{2} and κ_{2}^{2 }= 2*m*(*V*_{0} −* E*)/*ħ*^{2} as

1/*T* = |(*A*/*F*)|^{2 }= [1 + (*V*_{0}^{2}/{4*E*(*V*_{0} −* E*)}) sinh^{2}(2κ_{2}*a*)] (9.19)

Notice that this equation implies for E < *V*_{0}, *T* < 1, but *T* is not necessarily zero.

Scattering from a well, for *E* > 0 but with *V*_{0 }< 0 or *V*_{0 }= −|*V*_{0}|, 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* > *V*_{0}:

1/*T* = |(*A*/*F*)|^{2 }= [1+{(k_{1}^{2 } − k_{2}^{2})/(2k_{1}k_{2})}^{2} sin^{2}(2*k*_{2}*a*)] (9.20)

and in terms of *E* and *V*_{0}

1/*T* = |(*A*/*F*)|^{2 }= [ 1 + (V_{0}^{2}/{4E(E + |*V*_{0}|)}) sin^{2}(2*k*_{2}*a*)] (9.21)

where *k*_{2}^{2 }= 2*m*(*E *+ |*V*_{0}|)/*ħ*^{2}. 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 *>> V_{0} and in one other set of occurrences. If we now consider the sine function, it is zero when *n*π = 2*k*_{2}*a* and therefore *k*_{2} = *n*π/2*a* = 2π/λ_{2}, where 2*a* is the width of the well. Now given the relationship *k*_{2 }= *n*π/2*a* that we can insert into *E *+ |*V*_{0}| = *ħ*^{2}*k*_{2}^{2}/2*m*, we get *E *+ |*V*_{0}| = *ħ*^{2}*n*^{2}π^{2}/2*m*(2*a*)^{2} = *n*^{2}E_{1}, where E_{1} is the energy of an infinite square well of width 2*a*. This effect is called the Ramsauer effect. This effect describes the scattering of low-energy electrons from atoms.

^{6}Again, there is one more possibility, *E* = *V*_{0}. Look for this in the animations.

« previous

next »