91Ó°ÊÓ

Boundary conditions for E < U₀ are as follows.

1. Due to the requirement that the wave function value for the left side of the barrier must equal the wave function value for the right side of the barrier, ${\mathrm{ψ}}_{X<0}\left(x\right)={\mathrm{ψ}}_{X>0}\left(x\right)$
2. The first derivative on the left side must match the derivative on the right side since the wave function should be smooth.
   role="math" localid="1660028357372" ${\overline{)\frac{d{\mathrm{ψ}}_{x<0}}{dx}}}_{x-0}={\overline{)\frac{d{\mathrm{ψ}}_{x>0}}{dx}}}_{x-0}$
   

Consider the wave function which is given in equation 6.15

$\begin{array}{rcl}{\mathrm{ψ}}_{x<0}\left(x\right)& =& A{e}^{+ikx}+B{e}^{-ikx}\\ {\mathrm{ψ}}_{0<x-<L}\left(x\right)& =& C{e}^{+\mathrm{Î\pm }x}+D{e}^{-x}\\ {\mathrm{ψ}}_{x>L}& =& F{e}^{+ikx}\end{array}$

Apply condition (i),

$\begin{array}{rcl}{\mathrm{ψ}}_{x<0}\left(x\right)& =& {\mathrm{ψ}}_{x>0}\left(x\right)\\ A{e}^{+ikx}+{\overline{)B{e}^{-ikx}}}_{x=0}& =& C{e}^{+\mathrm{Î\pm }x}+{\overline{)D{e}^{-\mathrm{Î\pm }x}}}_{x=0}\\ A+B& =& C+D\end{array}$

Apply condition (ii),

$\begin{array}{rcl}{\overline{)\frac{d{\mathrm{ψ}}_{x<0}}{dx}}}_{x-0}& =& {\overline{)\frac{d{\mathrm{ψ}}_{x>0}}{dx}}}_{x-0}\\ ikA{e}^{+ikx}-{\overline{)ikB{e}^{-ikx}}}_{x-0}& =& {\overline{)\mathrm{Î\pm }C{e}^{+\mathrm{Î\pm }x}-\mathrm{Î\pm }D{e}^{-\mathrm{Î\pm }x}}}_{x-0}\\ ik\left(A-B\right)& =& \mathrm{Î\pm }\left(C-D\right)\end{array}$

Now, apply condition for x = L,

$\begin{array}{rcl}{\mathrm{ψ}}_{L<x}\left(L\right)& =& {\mathrm{ψ}}_{x>L}\left(L\right)\\ C{e}^{+\mathrm{Î\pm }L}+D{e}^{-\mathrm{Î\pm }L}& =& F{e}^{+ikL}\\ {\overline{)\frac{d{\mathrm{ψ}}_{L<x}\left(x\right)}{dx}}}_{x=L}& =& {\overline{)\frac{d{\mathrm{ψ}}_{L>x}\left(x\right)}{dx}}}_{x=L}\\ \mathrm{Î\pm }C{e}^{+\mathrm{Î\pm }L}-\mathrm{Î\pm }D{e}^{-L}& =& ikF{e}^{+ikL}\end{array}$

Write the expression for reflection coefficient of wave.

$R=\frac{B^{*}B}{A^{*}A}......\left(1\right)$

Write the relation between B and A.

role="math" localid="1660029338206" $B=A\frac{\sin h\left(\mathrm{Î\pm }L\right)}{\sqrt{\sin h^2\left(\mathrm{Î\pm }L\right)+4{\mathrm{Î\pm }}^2{k}^2/{\left(k^2+{\mathrm{Î\pm }}^2\right)}^2}}{e}^{-i\mathrm{β}}$

Determine $\frac{B}{A}$and $\frac{B^{*}}{A^{*}}$from above equation.

$\begin{array}{rcl}B& =& A\frac{\sin h\left(\mathrm{Î\pm }L\right)}{\sqrt{\sin h^2\left(\mathrm{Î\pm }L\right)+4{\mathrm{Î\pm }}^2{k}^2/{\left(k^2+{\mathrm{Î\pm }}^2\right)}^2}}{e}^{-i\mathrm{β}}\\ \left(\frac{B^{*}}{A}\right)& =& \frac{\sin h\left(\mathrm{Î\pm }L\right)}{\sqrt{\sin h^2\left(\mathrm{Î\pm }L\right)+4{\mathrm{Î\pm }}^2{k}^2/{\left(k^2+{\mathrm{Î\pm }}^2\right)}^2}}{e}^{-i\mathrm{β}}\end{array}$

Now, substitute the all values in equation (1)

$\begin{array}{rcl}R& =& \left(\frac{\sin h\left(\mathrm{Î\pm }L\right)}{\sqrt{\sin h^2\left(\mathrm{Î\pm }L\right)+4{\mathrm{Î\pm }}^2{k}^2/{(k^2+{\mathrm{Î\pm }}^2)}^2}}{e}^{-i\mathrm{β}}\right)\left(\frac{\sin h\left(\mathrm{Î\pm }L\right)}{\sqrt{\sin h^2\left(\mathrm{Î\pm }L\right)+4{\mathrm{Î\pm }}^2{k}^2/{(k^2+{\mathrm{Î\pm }}^2)}^2}}{e}^{-i\mathrm{β}}\right)\\ & =& \frac{\sin h^2\left({\displaystyle \frac{\sqrt{2m\left(U_0-E\right)}}{h}}L\right)}{\sin h^2\left(\frac{\sqrt{2m\left(U_0-E\right)}}{h}L\right)+4\left(\frac{\sqrt{2m\left(U_0-E\right)}}{h}L\right)k^2/{\left(k^2+{\left(\frac{\sqrt{2m\left(U_0-E\right)}}{h}L\right)}^2\right)}^2}\\ & =& \frac{\sin h^2\left({\displaystyle \frac{\sqrt{2m\left(U_0-E\right)}}{h}L}\right)}{\sin h^2\left(\frac{\sqrt{2m\left(U_0-E\right)}}{h}L\right)+4\left(\frac{2m\left(U_0-E\right)}{h}L\right)k^2/{\left(k^2+{\left(\frac{2m\left(U_0-E\right)}{h}L\right)}^2\right)}^2}\end{array}$

$k=\frac{\sqrt{2mE}}{h}and{\mathrm{Î\pm }}^2+k^2=\frac{2m{U}_0}{h^2}givesRinequation(6-16)$

Hence, it is verified that he reflection probability R given in equation (6.16) follows from this result.