PX156 - D5 - the finite potential barrier

in regions $1$ and $3$ , the corresponding TISEs are:

\begin{aligned} \frac{d^{2} ϕ_{1}}{d x^{2}} + \frac{2 m E}{ℏ^{2}} ϕ_{1} & = 0 \\ \frac{d^{2} ϕ_{3}}{d x^{2}} + \frac{2 m E}{ℏ^{2}} ϕ_{3} & = 0 \end{aligned}

solutions to these are:

\begin{aligned} ϕ_{1} & = A e^{i k x} + B e^{- i k x} \\ ϕ_{3} & = F e^{i k x} + G e^{- i k x} \\ k & = \frac{\sqrt{2 m E}}{ℏ} \end{aligned}

the incident wave is $ϕ_{1 +} = A e^{i k x}$
the reflected wave is $ϕ_{1 -} = A e^{- i k x}$
the transmitted wave is $ϕ_{3 +} = F e^{i k x}$
there is no left travelling wave in region $3$ , therefore $G = 0$
the transmission probability:

T = \frac{| ϕ_{3 +}^{2} |}{| ϕ_{1 +}^{2} |} = \frac{F F^{*}}{A A^{*}}

in region 2, the TISE is:

\frac{d^{2} ϕ_{2}}{d x^{2}} + \frac{2 m}{ℏ^{2}} (E - V) ϕ_{2} = 0

the solution is:

\begin{aligned} ϕ_{2} & = C e^{i k^{'} x} + D e^{- i k^{'} x} \\ k^{'} & = \frac{\sqrt{2 m (E - V)}}{ℏ} \end{aligned}

since $E < V$ , $k^{'}$ must be imaginary, therefore define another wavenumber:

k_{2} = - i k^{'} = \frac{\sqrt{2 m (V - E)}}{ℏ}

the solution:

ϕ_{2} = C e^{- k_{2} x} + D e^{k_{2} x}

boundary conditions at $x = 0 : ϕ_{1} = ϕ_{2}, \frac{d ϕ_{1}}{d x} = \frac{d ϕ_{2}}{d x}$
boundary conditions at $x = L : ϕ_{2} = ϕ_{3}, \frac{d ϕ_{2}}{d x} = \frac{d ϕ_{3}}{d x}$
hence:

\begin{aligned} A + B & = C + D \\ i k_{1} A - i k_{1} B & = - k_{2} C + k_{2} D \\ C e^{- k_{2} L} + D e^{k_{2} L} & = F e^{i k_{3} L} \\ - k_{2} C e^{- k_{2} L} + k_{2} D e^{k_{2} L} & = i k_{3} F e^{i k_{3} L} \end{aligned}

a general solution for $A / F$ ratio is possible, but assuming a large barrier (relative to the particle energy) and that the barrier is fairly wide, a good approximation is:

\frac{A}{F} = (\frac{1}{2} + \frac{i k_{2}}{4 k_{i}}) e^{(i k_{1} + k_{2}) L}

this leads to transmission probability:

T = \frac{F F^{*}}{A A^{*}} = \frac{16 e^{- 2 k_{2} L}}{4 + {(\frac{k_{2}}{k_{1}})}^{2}}

but:

{(\frac{k_{2}}{k_{1}})}^{2} = \frac{\frac{2 m (V - E)}{ℏ^{2}}}{\frac{2 m E}{ℏ^{2}}} = \frac{V}{E} - 1

therefore, the leading term in the transmission probability is given y the exponent and:

T \propto e^{- 2 k_{2} L} \propto \exp (- \frac{2 L \sqrt{2 m (V - E)}}{ℏ})