Question

Solving the potential barrier smoothness conditions for relationships among the coefficients $\mathbf{A}\mathbf{,}\mathbf{B}\mathbf{\text{\hspace{0.17em}and\hspace{0.17em}}}\mathbf{F}$giving the reflection and transmission probabilities, usually involves rather messy algebra. However, there is a special case than can be done fairly easily, through requiring a slight departure from the standard solutions used in the chapter. Suppose the incident particles’ energy$\mathbf{E}$is precisely${\mathbf{U}}_{\mathbf{0}}$.

(a) Write down solutions to the Schrodinger Equation in the three regions. Be especially carefull in the region$\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{L}$. It should have two arbitrary constants and it isn’t difficult – just different.

(b) Obtain the smoothness conditions, and from these find$\mathbf{R}\mathbf{}\mathbf{}\mathbf{\text{\hspace{0.17em}and\hspace{0.17em}}}\mathbf{T}$.

(c) Do the results make sense in the limit$\mathbf{L}\mathbf{\to }\mathbf{\infty }$?

Short answer

1. In the region$\mathrm{x}<0\mathrm{and}\mathrm{x}>\mathrm{L}$ , the solution of Schrodinger Equation is a straight line.
2. $\mathrm{R}=\frac{{\mathrm{k}}^2{\mathrm{L}}^2}{4+{\mathrm{k}}^2{\mathrm{L}}^2}$ and$\mathrm{T}=\frac{4}{4+{\mathrm{k}}^2{\mathrm{L}}^2}$
3. yes

Step-by-step solution

Concepts involved

Schrodinger wave equationis a mathematical expression which determines the position of an electron in space and time taking into account the matter wave property of the electron.

Tunneling is a phenomenon when a particle propagates through a potential barrier when the potential energy of the barrier is higher than the kinetic energy of the particle.

Formula used

The Schrodinger wave equation and its general solution is given by,

$\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{y}}^2}+\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{z}}^2}+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{{\mathrm{h}}^2}(\mathrm{E}-{\mathrm{U}}_0)\mathrm{\psi }=0$

And

$\mathrm{\psi }(\mathrm{x},\mathrm{t})=\mathrm{A}\left[\cos (\mathrm{kx}-\mathrm{\omega t})+\mathrm{i}\sin (\mathrm{kx}-\mathrm{\omega t})\right]$

Step 3(a): Solutions of Schrodinger Equation

Nothing special is observed special in the regions$\mathrm{x}<0\mathrm{and}\mathrm{x}>\mathrm{L}$

The solution of schrodiger equation will be given by,

$\mathrm{\psi }(\mathrm{x},\mathrm{t})=\mathrm{A}\left[\cos (\mathrm{kx}-\mathrm{\omega t})+\mathrm{i}\sin (\mathrm{kx}-\mathrm{\omega t})\right]$

But in the region $0<\mathrm{x}<\mathrm{L}$and you know that, $\mathrm{E}={\mathrm{U}}_0$,

Hence, the Schrodinger equation which is given by,

$\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{x}}^2}+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{{\mathrm{h}}^2}(\mathrm{E}-{\mathrm{U}}_0)\mathrm{\psi }=0$

Becomes,

$\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{x}}^2}=0$

Which results in a straight line ($\mathrm{\psi }\left(\mathrm{x}\right)=\mathrm{mx}+\mathrm{c}$).

Hence, in the region $\mathrm{x}<0\mathrm{and}\mathrm{x}>\mathrm{L}$, the solution of Schrodinger Equation is a straight line.

Step 4(b): Obtain the Smoothness Conditions and find R,T

Continuity at$\mathrm{x}=0$gives $\mathbf{A}+\mathbf{B}=\mathbf{D}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Continuity of derivative gives$\mathbf{ik}(\mathbf{A}-\mathbf{B})=\mathbf{C}\cdot \cdot \cdot \cdot \cdot \left(2\right)$

At, $\mathrm{x}=\mathrm{L}$continuity $\mathrm{CL}+\mathrm{D}={\mathrm{Fe}}^{\mathrm{ikL}}\cdot \cdot \cdot \cdot \cdot \left(3\right)$

Continuity of derivative $\mathrm{C}={\mathrm{ikFe}}^{\mathrm{ikL}}\cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)$.

Eliminating C by using Equations (3) and (4), we get,$\mathrm{D}=(1-\mathrm{ikL}){\mathrm{Fe}}^{\mathrm{ikL}}\cdot \cdot \cdot \cdot \cdot \cdot \left(5\right)$

By using equation (5) in equation(1), we get,

$\mathrm{A}+\mathrm{B}=(1-\mathrm{ikL}){\mathrm{Fe}}^{\mathrm{ikL}}\cdot \cdot \cdot \cdot \cdot \cdot \left(6\right)$

By using equation (4) in equation(2), we get,

$\mathrm{A}-\mathrm{B}={\mathrm{Fe}}^{\mathrm{ikL}}\cdot \cdot \cdot \cdot \cdot \cdot \left(7\right)$

By adding equation (6) and (7), we get,

$2\mathrm{A}=(2-\mathrm{ikL}){\mathrm{Fe}}^{\mathrm{ikL}}$

Hence,

$\begin{array}{l}\mathrm{T}=\frac{{\mathrm{F}}^{*}\mathrm{F}}{{\mathrm{A}}^{*}\mathrm{A}}=\frac{2{\mathrm{e}}^{\mathrm{ikL}}}{2+\mathrm{ikL}}\frac{2{\mathrm{e}}^{-\mathrm{ikL}}}{2-\mathrm{ikl}}=\frac{4}{4+{\mathrm{k}}^2{\mathrm{L}}^2}\\ \mathrm{R}=1-\mathrm{T}=\frac{{\mathrm{k}}^2{\mathrm{L}}^2}{4+{\mathrm{k}}^2{\mathrm{L}}^2}\end{array}$

Hence, $\mathrm{R}=\frac{{\mathrm{k}}^2{\mathrm{L}}^2}{4+{\mathrm{k}}^2{\mathrm{L}}^2}$ and$\mathrm{T}=\frac{4}{4+{\mathrm{k}}^2{\mathrm{L}}^2}$

Step 5(c): Tunneling Probability

As, $\mathrm{L}\to \mathrm{\infty },\mathrm{T}\to 0$which can be seen from the Figure given below. The tunneling probability becomes zero.

Fig. – Reflection and Transmission probabilities for a potential step

Hence Figure 6.4 makes sense.