Section 9.4: Plane Wave Scattering: Potential Energy Steps

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We begin our study of quantum-mechanical scattering by looking at a general case where we have two regions (Regions I and II) and a change in the potential energy at the interface between the two regions. Typically we choose V = 0 in Region I and V = V0 in Region II, and the change occurs at x = 0. This situation is shown in Step Up Animation 1 and Step Up Animation 2. The wave functions when E > V0 are

    Ψinc = A exp{i(k1x − ωt)}         Ψrefl = B exp{−i(k1x + ωt)}        Ψtrans = C exp{i(k2x − ωt)}, (9.9)

where k12 = 2mE/ħ2 and k22 = 2m(E V0)/ħ2, while the momenta are just: pinc = ħk1, prefl = −ħk1, and ptrans = ħk2. In addition, we note that the ω's are the same. Why?  The ω's are related to the total energy of the plane wave. The total energy  must be the same in each region (the entire wave function represents one state of the Schrödinger equation) even when the k's are different in each region. Therefore, since E = ħω, the ω's must be the same. In addition, if the ω's were different, the solution to the boundary-value problem we will shortly consider would not necessarily be valid for all times.

To construct transmission and reflection coefficients, we must first calculate Jx for the incident, reflected, and transmitted waves from the equation4

 Jx = (ħ/2mi){ψ* [∂ Ψ/∂x] − [∂ Ψ*/∂x} . (9.12)

We find that

 Jinc = (ħk1/m) |A|2        Jrefl = −(ħk1/m) |B|2        Jtrans = (ħk2/m) |C|2 . (9.13)

For E > V we may use the above parts of the wave function, but we must match the wave function parts at the change in potential energy (the boundary at x = 0) since the parts in Eq. (9.7) represent just one wave function.We require that ψI(0) = ψII(0)  and that ψI'(0) = ψI'(0).  We can use time-independent wave functions (since the time dependence for all wave function parts is identical, it cancels):

Ψinc = A exp(ik1x) Ψrefl = B exp(−ik1x)        Ψtrans = C exp(ik2x),

where ΨI = Ψinc + Ψrefl.  When we match ΨI(0) = ΨII(0), this gives A + B = C. When we match ΨI'(0) = ΨI'(0), we get k1A k1B = k2C. From these conditions we solve for the fractions |B/A|2 and |C/A|2

|B/A|2 = (1 − k2/k1)2/(1 + k2/k1)2    and      |C/A|2 = 4/ (1 + k2/k1)2,

since they occur in the the current densities,  Eq.(9.12), which are necessary in order to calculate the transmission and reflection coefficients from Eq.(9.10).  We therefore find that

R = (1 − k2/k1)2/(1 + k2/k1)2 (9.14)


T = (k2/k1) [4/(1 + k2/k1)2]. (9.15)

Recall k12 = 2mE/ħ2 and k22 = 2m(EV0)/ħ2. Therefore, k22/k12 = (EV0)/E = 1 − V0/E.  As E increases, k2 k1, and therefore T → 1, as expected. With a bit of algebra, you can also convince yourself that T + R = 1 by adding Eqs. (9.14) and (9.15) together.

If we had the opposite circumstance, as shown in Step Down Animation 1  and Step Down Animation 2, with an incident plane wave subject to a potential energy V0 and then at x > 0 experiencing no potential energy, this would amount to k2 → k1 and k1 → k2. Multiplying this result by the square of the ratio of the k's gets us back to the original T and R in Eqs. (9.14) and (9.15). Therefore the transmission and reflection coefficients for these two different problems are actually the same.

Look at both Step Up Animation 1 and Step Up Animation 2Step Up Animation 1 shows you the quantum-mechanical plane wave in both regions and the top panel shows the energy diagram. Change the energy of the incident plane wave by moving the slider and note the effect on the plane wave and also the transmission and reflection coefficients.  Step Up Animation 2 shows the same situation but with a graph of the transmission and reflection coefficients as a function of energy. Click-drag the plane wave and drag up to increase the energy or drag down to decrease the energy. If you do this slowly enough, you can create a very nice graph of the transmission and reflection coefficients versus E for a fixed V0.

For E < V0, we can use the time-independent wave functions

        Ψinc = A exp(ik1x)         Ψrefl = B exp( ik1x) Ψtrans = C exp( −κ2x),

where k12 = 2mE/ħ2 and κ22 = 2m(|EV0|)/ħ2.  We find for this case

        Jinc= (ħk1/m) |A|2        Jrefl =  −(ħk1/m) |B|2 Jtrans = 0.

Now we must again match the parts of the wave function. We require that ΨI(0) = ΨII(0)  and that ΨI'(0) = ΨI'(0).  When we match ΨI(0) = ΨII(0), this gives A + B = C. When we match ΨI'(0) = ΨI'(0), we get ik1A ik1B = − κ2C. We eventually arrive at the relationship

|B/A|2 = [(k1 + iκ2)/(k1 iκ2)] / [(k1 iκ2)(k1 + iκ2)] = 1.

Note in this case, R = |B/A|2 = 1. From R = 1 we also expect T = 0, which is also clear from the fact that Jtrans= 0. In addition, if E = V0, we find that T = 0. These results agree with the classical scattering case except that there is a finite probability of finding the particle in the classically-forbidden region.

4You may be wondering why we care about the probability current density. The conservation of probability in its integral form in one dimension is

dP/dt + ∫ (dJx/dx) dx = 0 [integral from −∞ to +∞] , 

and for a scattering process, the total probability of finding the particle over all space does not change with time, or that dP/dt = 0. Therefore: ∫ (dJx/dx) dx = 0 or that Jx(∞) = Jx(−∞).  We can identify Jx(∞) ≡ Jtrans and Jx(−∞) ≡ Jinc+ Jrefl.  We find that Jtrans =   Jinc+ Jrefl and therefore Jinc= JtransJrefl , which leads directly to

1 = Jtrans/JincJrefl /Jinc.

The sign in front of the Jrefl term is due to the fact that this expression is always negative, therefore −Jrefl  will be positive.  We incorporate this sign in the definition of the reflection and transmission coefficients as

T = Jtrans/Jinc ≡  |Jtrans/Jinc| (9.10)


R = −Jrefl /Jinc ≡ |Jrefl /Jinc|. (9.11)

5Also look for the E = V0 case in the animations. The solution in Region II should be a straight line as discussed.

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