Step potentials

Here we shall make a simplification and assume that is constant over a range of $x$ and it varies in steps as shown in Figure 20.2 instead of varying smoothly as shown in Figure 20.1

Figure 20.2:

We calculate the wavefunction of a particle of energy $E$ inside a region of constant potential $V$. We have

$$\frac{d^2 X}{d x^2} = - \frac{2m}{\hbar^2} (E- V) X$$ (20.45)

There are two possibilities. The first possibility, shown in Figure 20.3, is where $E>V$. The particle's momentum inside the potential is

$$p' = \sqrt{2m(E-V)}\,.$$ (20.46)

Writing equation (20.45) in terms of this we have

$$\frac{d^2X}{dx^2} = \frac{-{p'}^2}{\hbar^2} X$$ (20.47)

which has a solution

(20.48)

Figure 20.3:

The wave function inside the potential is

$$\psi(x,t) = e^{-iE t/\hbar} \left[ A_1 e^{i p' x/\hbar} + A_2 e^{-i p' x /\hbar} \right]$$ (20.49)

and for the same particle outside where $V=0$ we have

$$\psi(x,t) = e^{-iE t/\hbar} \left[ A_1 e^{i p x/\hbar} + A_2 e^{-i p x /\hbar} \right]$$ (20.50)

where $p=\sqrt{2 m E}$. We see that the wavefunctions frequency is the same both inside and outside the potential, whereas the wave number is different inside the potential (Figure 20.3) and outside. The potential is like a change in the refractive index, the wavelength changes. We shall discuss matching of boundary conditions later.

The second possibility, shown in Figure 20.4 is where $E<V$.

Figure 20.4:

In classical mechanics the particle is never found in this region as the momentum is imaginary which is meaningless. Defining

(20.51)

we write equation (20.45) as

$$\frac{d^2X}{dx^2} = \frac{q^2 }{\hbar^2} X$$ (20.52)

which has solutions

$$X(x) = A_1 e^{-q x/\hbar}+A_2 e^{ q x/\hbar}$$ (20.53)

The second solution blows up as $x \to \infty$ and if the region to the right extends to infinity then $A_2 = 0$ and we have the solution

$$X(x) = A_1e^{-q x/\hbar}$$ (20.54)

in the region inside the potential. The corresponding wavefunction is

$$\psi(x,t)= A_1 e^{-i E t/\hbar} e^{- q x/\hbar}$$ (20.55)

The wave function decays exponentially inside the potential. There is finite probability of finding the particle in a region $E<V$ The probability decays exponentially inside the region where $E<V$. The decay rate increases with $V$.