Question

Obtain the smoothness conditions at the boundaries between regions for the $\mathit{E}\mathbf{<}{\mathit{U}}_{\mathbf{0}}$barrier (i.e., tunneling) case.

Short answer

The four conditions that are boxed above are the smoothness condition for the given condition.

$$\begin{gathered} \boxed{A+B=C+D} \\ \boxed{ik\left(A-B\right)=\alpha \left(C-D\right)} \end{gathered}$$

$$\begin{gathered} \boxed{C{e}^{\alpha L}+D{e}^{-\alpha L}=F{e}^{ikL}} \\ \boxed{\alpha \left(C{e}^{\alpha L}-D{e}^{-\alpha L}\right)=ikF{e}^{ikL}} \end{gathered}$$

Step-by-step solution

Concept involved

Tunneling is a quantum phenomenon that states that a particle can escape confinement even though it doesn’t have enough energy.

In the given case, the wave function inside the tunnel is real while wave vector and momentum are imaginary.

Determine the boundary conditions

Applying boundary conditions for$x<0$and $0<x<L$:

$$\begin{gathered} {\psi }_{x<0}\left(0\right)={\psi }_{0<x<L}\left(0\right) \\ A{e}^{ik0}+B{e}^{-ik0}=C{e}^{\alpha 0}+D{e}^{-\alpha 0} \\ \boxed{A+B=C+D} \end{gathered}$$

Also,

$$\begin{gathered} \frac{{\displaystyle d{\psi }_{x<0}}}{{\displaystyle dx}}\left|x=0=\frac{{\displaystyle d{\psi }_{0<x<L}}}{{\displaystyle dx}}\right|x=0 \\ ikA{e}^{ik0}-ikB{e}^{-ik0}=\alpha C{e}^{\alpha 0}-\alpha D{e}^{-\alpha 0} \\ \boxed{ik(A-B)=\alpha (C-D)} \end{gathered}$$

Now applying boundary conditions for$0<x<L$and$x>L$:

$$\begin{gathered} {\psi }_{0<x<L}\left(L\right)={\psi }_{x>0}\left(L\right) \\ \boxed{C{e}^{\alpha L}+D{e}^{-\alpha L}=F{e}^{ikL}} \end{gathered}$$

Also,

$$\begin{gathered} \frac{{\displaystyle d{\psi }_{0<x<L}}}{{\displaystyle dx}}\left|x=L=\frac{{\displaystyle d{\psi }_{x>0}}}{{\displaystyle dx}}\right|x=L \\ \alpha \left(C{e}^{\alpha L}-D{e}^{-\alpha L}\right)=ikF{e}^{ikL} \end{gathered}$$

The four conditions that are boxed above are the smoothness condition for the given condition.