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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.

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

$\frac{{鈭�}^2\mathrm{蠄}}{鈭�{\mathrm{x}}^2}+\frac{{鈭�}^2\mathrm{蠄}}{鈭�{\mathrm{y}}^2}+\frac{{鈭�}^2\mathrm{蠄}}{鈭�{\mathrm{z}}^2}+\frac{8{\mathrm{蟺}}^2\mathrm{m}}{{\mathrm{h}}^2}(\mathrm{E}鈭�{\mathrm{U}}_0)\mathrm{蠄}=0$

And

$\mathrm{蠄}(\mathrm{x},\mathrm{t})=\mathrm{A}\left[\cos (\mathrm{kx}鈭�\mathrm{蝇迟})\text{鈥�}+\text{鈥�}\mathrm{i}\text{鈥�}\sin (\mathrm{kx}鈭�\mathrm{蝇迟})\right]$

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

The solution of schrodiger equation will be given by,

$\mathrm{蠄}(\mathrm{x},\mathrm{t})=\mathrm{A}\left[\cos (\mathrm{kx}鈭�\mathrm{蝇迟})\text{鈥�}+\text{鈥�}\mathrm{i}\text{鈥�}\sin (\mathrm{kx}鈭�\mathrm{蝇迟})\right]$

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,

$\frac{{鈭�}^2\mathrm{蠄}}{鈭�{\mathrm{x}}^2}+\frac{8{\mathrm{蟺}}^2\mathrm{m}}{{\mathrm{h}}^2}(\mathrm{E}鈭�{\mathrm{U}}_0)\mathrm{蠄}=0$

Becomes,

$\frac{{鈭�}^2\mathrm{蠄}}{鈭�{\mathrm{x}}^2}=0$

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

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

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

Continuity of derivative gives$\mathbf{ik}(\mathbf{A}鈭�\mathbf{B})=\mathbf{C}鈰�鈰�鈰�鈰�鈰�\left(2\right)$

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

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

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

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

$\mathrm{A}+\mathrm{B}=(1鈭�\mathrm{ikL}){\mathrm{Fe}}^{\mathrm{ikL}}鈰�鈰�鈰�鈰�鈰�鈰�\left(6\right)$

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

$\mathrm{A}鈭�\mathrm{B}={\mathrm{Fe}}^{\mathrm{ikL}}鈰�鈰�鈰�鈰�鈰�鈰�\left(7\right)$

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

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

Hence,

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

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

As, $\mathrm{L}鈫�\mathrm{鈭�},\text{鈥夆赌�}\mathrm{T}鈫�0$which can be seen from the Figure given below. The tunneling probability becomes zero.

Fig. 鈥� Reflection and Transmission probabilities for a potential step

Hence Figure 6.4 makes sense.