Solving for constants A, B in the Schrodinger Equation

We now consider a simple set up such as shown in the sketch where we have a beam of electrons of energy kinetic energy W incident on a plane where there is a potential step such that Q is less than W. (E.g. the energy of the step is at energy Q, less than W). Since there is a sharp change in potential there must be also a sharp change in the electron wavelength, so the wave ought to behave like a light wave incident on a slab of glass where we expect partial transmission and partial reflection.

The amounts transmitted and reflected can be calculated as follows:

If we take the potential energy V(x) to be:

a) V - Q, x > 0

b) V = 0, (x less than 0)


We suppose the wave function for the respective cases to be:


a) U = [exp(2πi(Kx ) + A exp(-2πiKx] exp (-2πift)


b) U = exp{2πi(K' x - 2πift)} where:

K^2 = 2mW/ h^2 and we understand m = m(e) the mass of the electron.


Then K = 1/L where L denotes the wavelength as before, and K'^2 = 2m(W-Q)/ h^2.

The wave function here represents one electron per unit volume in the incident wave, therefore, ‖A‖^2 particles per unit volume occur i the reflected wave and ‖B‖^2 in the transmitted wave.

Thus, v particles cross unit area per unit time in the incident wave and v ‖B‖^2 in the transmitted wave. Since: v'/v = K'/K then the proportion of the beam reflected is: ‖A‖^2 and the proportion transmitted is: (K'/K)‖B‖^2.

The problem then is to calculate the constants A and B. Then we need to know the boundary conditions satisfied at x = 0. These are basically that U and dU/dx should be continuous. Integrating both sides of the Schrodinger equation from the previous blog:


dU/dx = - 8π^2 m/h^2 INT (0 to x) (W - V)U dx

so even if V is discontinuous, its integral must be continuous. So dU/dx is continuous, and hence U is continuous. Inserting these b.c.'s: , since U is continuous at x = 0 it follows that:

1 + A = B and since dU/dx is continuous at x = 0 it follows that:
K(1 - A) = K'B

Solving for A and B:

A = (K - K')/ (K + K')

B = 2K/ (K + K')


A most interesting thing to be deduced from the above is that the SUM of the proportions of the reflected and transmitted waves comes out to unity.

This is so if: ‖A‖^2 + K' ‖B‖^2 / K = 1

If it wasn't so there'd be something wrong with the wave equation, as it would predict creation or disappearance of the particles at the step. Consider now the case where: Q > W

Here we expect all particles to undergo reflection. For x less than 0 we have the wave function:

U = [exp(2πi(Kx ) + A exp(-2πiKx] exp (-2πift)

For x > 0:

U = [B exp(-2πc x ) + C exp(2πcx)] exp (-2πift)

c^2 = 2m(Q - W)/h^2

For similar reasons that applied in the previous case, to describe total reflection we take the solution that decreases with increasing x and thus set C = 0.

Then we obtain:

U = B exp (-2πc x - 2πift), x > 0

Putting in the b.c.'s as before we have:

1 + A = B

iK(1 - A) = - cB

Eliminating B, i.e. by setting: iK(1 - A) = -c (1 + A)

we obtain:

A = (iK + c)/ (iK - c)

The important thing to note here is that:

‖A‖^2 = [(iK + c)/ (iK - c)]^2 = 1

so the reflected wave has the same amplitude as the incident one, which should be obvious to any reader familiar with the blogs already done on the complex numbers!