Electron Scattering of a Step Potential

The following post was initially one of my assignments for an independent study in modern physics in my penultimate year as an undergraduate. While studying this problem the text that I used to verify my answer was:

R. Eisberg and R. Resnick, Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles. John Wiley & Sons. 1985. 6. 

One of the hallmarks of quantum theory is the Schrödinger equation. There are two forms: the time-dependent and the time-independent equation. The former can be turned into the latter by way of assuming stationary states in which case there is no time evolution (i.e. the time derivative \partial_{t} \Psi(x,t)=0. \partial_{t}\Psi(x,t) denotes the derivative of the wavefunction \Psi(x,t) with respect to time.)

The wavefunction describes the state of a system, and it is found by solving the Schrödinger equation. In this post, I’ll be considering a step potential in which

V(x) = V_{0},  x < 0,  (1.1)

V(x) = 0, x > 0.  (1.2)

After solving for the wavefunction, I will calculate the reflection and transmission coefficients for the case where the energy of the electron is less than that of the step potential E < V_{0}.

First, we assume that we are dealing with stationary states, by doing so we assume that there is no time-dependence. The wavefunction \Psi(x,t) becomes an eigenfunction (eigen– is German for “characteristic” e.g. characteristic function, characteristic value (eigenvalue), and so on), \psi(x). There are requirements for this eigenfunction in the context of quantum mechanics: eigenfunction \psi(x) and its first order spatial derivative \frac{d\psi(x)}{dx} must be finite, single-valued, and continuous. Using the wavefunction, we write the time-independent Schrödinger equation as

-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x) = E\psi(x), (2)

where \hbar is the reduced Planck’s constant \hbar \equiv h(2\pi)^{-1}, m is the mass of the particle, V(x) represents the potential, and E is the energy.

Now, in electron scattering there are two cases regarding a step potential: the case for which E < V_{0} and the case for which E>V_{0} The focus of this post is the former case. In such a case, the potential is given mathematically by Eqs. (1.1) and (1.2) and can be depicted by the image below:


Image Credit:  http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/potential/barrier/STUDY-GUIDE.htm

The first part of this problem is to solve for \psi(x) when x<0. Then Eq.(2) becomes

\frac{d^{2}\psi_{I}(x)}{dx^{2}}=-\kappa_{I}^{2}\psi_{I}(x), (3)

where \kappa_{I}^{2}\equiv \frac{-2mE}{\hbar^{2}}. Eq.(3) is a second order linear homogeneous ordinary differential equation with constant coefficients and can be solved using a characteristic equation. We assume that the solution is of the general form


which upon taking the first and second order spatial derivatives and substituting into Eq.(3) yields


We can factor out the exponential and recalling that the graph of e^{x} \neq 0 (except in the limit x\rightarrow -\infty)we can then conclude that


Hence, r = \pm i\kappa_{I}. Therefore, we can write the solution Schrödinger equation in the region x < 0 as

\psi_{I}(x) = A\exp{+i\kappa_{I}x}+B\exp{-i\kappa_{I}x}. (4)

This is the eigenfunction for the first region. Coefficients A and B will be determined later.

Now we can use the same logic for the Schrödinger equation in the region x > 0 where V(x)=V_{0}:

\frac{d^{2}\psi_{II}(x)}{dx^{2}}= -\kappa_{II}^{2}\psi_{II}(x), (5)

where \kappa_{II}\equiv \frac{\sqrt[]{2m(E-V_{0})}}{\hbar}. The general solution for this region is

\psi_{II}(x) = C\exp{+i\kappa_{II}x}+D\exp{-i\kappa_{II}x}. (6)

Now the next step is taken using two different approaches: the first using a mathematical argument and the other from a conceptual interpretation of the problem at hand.  The former is this: Suppose we let x\rightarrow \infty. What results is that the first term on the right hand side of Eq.(6) diverges (i.e. becomes arbitrarily large). The second term on the right hand side ends up going to zero (it converges). Therefore, D remains finite, hence D\neq 0. However, in order to suppress the divergence of the first term, we let C=0. Thus we arrive at the eigenfunction for x>0

\psi_{II}(x) = D\exp{-i\kappa_{II}x}. (7)

The latter argument is this: In the region x < 0, the first term of the solution represents a wave propagating in the positive x-direction, while the second denotes a wave traveling in the negative x-direction. Similarly, in the region where x > 0, the first term corresponds to a wave traveling in the positive x-direction. However, this cannot be because the energy of the wave is not sufficient enough to overcome the potential. Therefore, the only term that is relevant here for this region is the second term, for it is a wave propagating in the negative x-direction.

We now determine the coefficients A and B. Recall that the eigenfunction must satisfy the following continuity requirements

\psi_{I}(x)=\psi_{II}(x), (8.1)

\psi^{\prime}_{I}(x)=\psi^{\prime}_{II}(x), (8.2)

evaluated when x=0. Doing so in Eqs. (4) and (7), and equating them we arrive at the continuity condition for \psi(x)

A+B = D. (9)

Taking the derivative of \psi_{I}(x) and \psi_{II}(x) and evaluating them for when x=0 we arrive at

A-B = \frac{i\kappa_{II}}{i\kappa_{I}}D. (10)

If we add Eqs. (9) and (10) we get the value for the coefficient A in terms of the arbitrary constant D

A =\frac{D}{2}(1+\frac{i\kappa_{II}}{i\kappa_{I}}. (11)

Conversely, if we subtract (9) and (10) we get the value for B in terms of D

B = \frac{D}{2}(1-\frac{i\kappa_{II}}{i\kappa_{I}}. (12)


Now the reflection coefficient defined as

R = \frac{B^{*}B}{A^{*}A}, (13)

which is the probability that an incident electron (wave) will be reflected. On the other hand, the transmission coefficient is the probability that an electron will be transmitted through the potential (e.g. barrier potential).  These two also must satisfy the relation

R+T =1. (14)

This means that the probability that the electron will be reflected or transmitted is 100%. Therefore, to evaluate R (the reason why I don’t calculate T will become apparent shortly), we take the complex conjugate of A and B and using them in Eq.(13) we get

R = \frac{\frac{D}{2}(1+\frac{\kappa_{II}}{\kappa_{I}})\frac{D}{2}(1-\frac{\kappa_{II}}{\kappa_{I}})} {\frac{D}{2}(1-\frac{\kappa_{II}}{\kappa_{I}}) \frac{D}{2}(1+\frac{\kappa_{II}}{\kappa_{I}})} = 1. (15)

What this conceptually means is that the probability that the electron is reflected is 100%. This implies that it is impossible for an electron to be transmitted through the potential for this system.


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