Determine the transmission coefficient for a rectangular barrier (same as Equation 2.145, only with $\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{+}{\mathbf{V}}_{\mathbf{0}}\mathbf{>}\mathbf{0}$ in the region$\mathbf{-}\mathbf{a}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{a}$ ). Treat separately the three cases$\mathbf{E}\mathbf{<}{\mathbf{V}}_{\mathbf{0}}\mathbf{,}\mathbf{E}\mathbf{=}{\mathbf{V}}_{\mathbf{0}}$ , and$\mathbf{E}\mathbf{>}{\mathbf{V}}_{\mathbf{0}}$ (note that the wave function inside the barrier is different in the three cases).

A differential equation describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

_{$V\left(x\right)=0$} When _{localid="1656057126624" $x\le -a$}

_{$V\left(x\right)=-V_0$} When _{$-\mathrm{a}<\mathrm{x}<\mathrm{a}$}

_{$V\left(x\right)=0$} When _{$x\ge -a$}

Schrodinger equation

$-\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\mathrm{\Psi }\left(\mathrm{x}\right)+\mathrm{V}\left(\mathrm{x}\right)\mathrm{\Psi }\left(\mathrm{x}\right)=\mathrm{E\Psi }\left(\mathrm{x}\right)$

We shall split into 3 regions

$\mathrm{\Psi }\left(\mathrm{x}\right)={\mathrm{Ae}}^{\mathrm{ikx}}+{\mathrm{Be}}^{-\mathrm{ikx}}$ $\mathrm{x}<-\mathrm{a}$

$\mathrm{\Psi }\left(\mathrm{x}\right)={\mathrm{Fe}}^{\mathrm{ikx}}$$\mathrm{x}>\mathrm{a}$

${\mathrm{Ae}}^{-\mathrm{ika}}+{\mathrm{Be}}^{\mathrm{ika}}={\mathrm{Ce}}^{-\mathrm{\mu a}}+{\mathrm{De}}^{\mathrm{\mu a}}$

$\mathrm{ik}({{\mathrm{Ae}}^{-}}^{\mathrm{ika}}-{\mathrm{Be}}^{\mathrm{ika}})=\mathrm{\mu }({\mathrm{Ce}}^{-\mathrm{\mu a}}-{\mathrm{De}}^{\mathrm{\mu a}})$

We have

_{${\mathrm{Ce}}^{\mathrm{\mu a}}+{\mathrm{De}}^{\mathrm{\mu a}}={\mathrm{Fe}}^{\mathrm{ika}}$}

_{${\mathrm{ikFe}}^{\mathrm{ika}}=\mathrm{\mu }({\mathrm{Ce}}^{-\mathrm{\mu a}}-{\mathrm{De}}^{\mathrm{\mu a}})$}

$\mathrm{B}=\frac{{\mathrm{e}}^{-2\mathrm{ika}}({\mathrm{k}}^2+{\mathrm{\mu }}^2)\sinh \left(2\mathrm{\mu a}\right)}{2\mathrm{ik\mu cosh}\left(2\mathrm{\mu a}\right)+({\mathrm{k}}^2-{\mathrm{\mu }}^2)\sinh \left(2\mathrm{\mu a}\right)}\mathrm{A}$

$\mathrm{C}=\frac{{{\mathrm{e}}^{-\mathrm{a\mu }}}^{-\mathrm{ika}}(-{\mathrm{k}}^2+\mathrm{k\mu i})}{2\mathrm{ik\mu cosh}\left(2\mathrm{\mu a}\right)+({\mathrm{k}}^2-{\mathrm{\mu }}^2)\sinh \left(2\mathrm{\mu a}\right)}\mathrm{A}$

$\mathrm{D}=\frac{{{\mathrm{e}}^{-\mathrm{a\mu }}}^{-\mathrm{ika}}(-{\mathrm{k}}^2+\mathrm{k\mu i})}{2\mathrm{ik\mu cosh}\left(2\mathrm{\mu a}\right)+({\mathrm{k}}^2-{\mathrm{\mu }}^2)\sinh \left(2\mathrm{\mu a}\right)}\mathrm{A}$

$\mathrm{F}=\frac{2{\mathrm{e}}^{-2\mathrm{ika}}\mathrm{k\mu i}}{2\mathrm{ik\mu cosh}\left(2\mathrm{\mu a}\right)+({\mathrm{k}}^2-{\mathrm{\mu }}^2)\sinh \left(2\mathrm{\mu a}\right)}\mathrm{A}$

$T={\left(\frac{F}{A}\right)}^2$

$\mathrm{T}={\left[\frac{2{\mathrm{e}}^{-2\mathrm{ika}}\mathrm{k\mu i}}{2\mathrm{ik\mu cosh}\left(2\mathrm{\mu a}\right)+({\mathrm{k}}^2-{\mathrm{\mu }}^2)\sinh \left(2\mathrm{\mu a}\right)}\right]}^2$

$\mathrm{T}=\frac{4{\mathrm{\mu }}^2{\mathrm{k}}^2}{({\mathrm{\mu }}^4+2{\mathrm{\mu }}^2{\mathrm{k}}^2+{\mathrm{k}}^4)\left[{\sinh }^2\right(2\mathrm{\mu a})+4{\mathrm{\mu }}^2{\mathrm{k}}^2{\cosh }^2(2\mathrm{\mu a}\left)\right]}$

$\mathrm{Put}{\cosh }^2\left(2\mathrm{\mu a}\right)=1+{\sinh }^2\left(2\mathrm{\mu a}\right)$

$\mathrm{T}=\frac{1}{{({\mathrm{\mu }}^2+{\mathrm{k}}^2)}^2{\sinh }^2\left(2\mathrm{\mu a}\right)/4{\mathrm{\mu }}^2{\mathrm{k}}^2+1}$

${\mathrm{T}}^{-1}=1+\frac{({\mathrm{\mu }}^2+{\mathrm{k}}^2){\sinh }^2\left(2\mathrm{\mu a}\right)}{4{\mathrm{\mu }}^2{\mathrm{k}}^2}$

Put the value of constant$\mathrm{\mu }$and$\mathrm{k}$we get,

${\mathrm{T}}^{-1}=1+\frac{{{\mathrm{V}}_0}^2}{4\mathrm{E}({\mathrm{V}}_0-\mathrm{E})}{\sinh }^2\left(\frac{2\mathrm{a}}{\mathrm{h}}\sqrt{2\mathrm{m}({\mathrm{V}}_0-\mathrm{E})}\right)$

When energy is equal to the potential but the barrier region we have

${\mathrm{\Psi }}^{\mathrm{n}}=0$

$\mathrm{\Psi }\left(\mathrm{x}\right)=\mathrm{Cx}+\mathrm{D}$

${\mathrm{Ae}}^{-\mathrm{ika}}+{\mathrm{Be}}^{\mathrm{ika}}=-\mathrm{Ca}+\mathrm{D}$

$\mathrm{ik}({\mathrm{Ae}}^{-\mathrm{ika}}-{\mathrm{Be}}^{\mathrm{ika}})=\mathrm{C}$

At, $\mathrm{x}=\mathrm{a}$

$\mathrm{Ca}+\mathrm{D}={\mathrm{Fe}}^{\mathrm{ika}}$

$\mathrm{C}={\mathrm{ikFe}}^{\mathrm{ika}}$

$\mathrm{B}=\left(\frac{{\mathrm{kae}}^{-2\mathrm{ika}}}{\mathrm{ka}+\mathrm{i}}\right)\mathrm{A}$

$\mathrm{C}=\left(\frac{{\mathrm{ke}}^{-\mathrm{ika}}}{-\mathrm{ka}-\mathrm{i}}\right)\mathrm{A}$

$\mathrm{D}={\mathrm{e}}^{-\mathrm{ika}}\mathrm{A}$

$\mathrm{F}=\left(\frac{{\mathrm{e}}^{-2\mathrm{ika}}}{1-\mathrm{ika}}\right)\mathrm{A}$

$\mathrm{T}={\left(\frac{\mathrm{F}}{\mathrm{A}}\right)}^2$

$\mathrm{T}=\frac{1}{1+\frac{2{\mathrm{mEa}}^2}{{\mathrm{h}}^2}}$

${\mathrm{T}}^{-1}=1+\frac{2{\mathrm{mEa}}^2}{{\mathrm{h}}^2}$

$-\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\mathrm{\Psi }\left(\mathrm{x}\right)+\mathrm{V}\left(\mathrm{x}\right)\mathrm{\Psi }\left(\mathrm{x}\right)=\mathrm{E\Psi }\left(\mathrm{x}\right)$

$-\frac{{\mathrm{h}}^2}{2\mathrm{m}}{\mathrm{\Psi }}^{\mathrm{n}}=(\mathrm{E}-{\mathrm{V}}_0)\mathrm{\Psi }$

${\mathrm{\Psi }}^{\mathrm{n}}=-\frac{2\mathrm{m}(\mathrm{E}-{\mathrm{V}}_0)}{{\mathrm{h}}^2}$

${\mathrm{\Psi }}^{\mathrm{n}}=-{\mathrm{\lambda }}^2\mathrm{\Psi }$

$\mathrm{\lambda }=\sqrt{\frac{2\mathrm{m}(\mathrm{E}-{\mathrm{V}}_0)}{{\mathrm{h}}^2}}$

${\mathrm{Ae}}^{-\mathrm{ika}}+{\mathrm{Be}}^{\mathrm{ika}}={\mathrm{Ce}}^{-\mathrm{i\lambda a}}+{\mathrm{De}}^{\mathrm{i\lambda a}}$

$\mathrm{ik}({{\mathrm{Ae}}^{-}}^{\mathrm{ika}}-{\mathrm{Be}}^{\mathrm{ika}})=\mathrm{i\lambda }({\mathrm{Ce}}^{-\mathrm{i\lambda a}}-{\mathrm{De}}^{\mathrm{i\lambda a}})$

We have

${\mathrm{Ce}}^{\mathrm{i\lambda a}}+{{\mathrm{De}}^{-}}^{\mathrm{i\lambda a}}={\mathrm{Fe}}^{\mathrm{ika}}$

${\mathrm{ikFe}}^{\mathrm{ika}}=\mathrm{i\lambda }({\mathrm{Ce}}^{-\mathrm{i\lambda a}}-{\mathrm{De}}^{\mathrm{i\lambda a}})$

$\mathrm{B}=\frac{{\mathrm{e}}^{-2\mathrm{ika}}({\mathrm{k}}^2-{\mathrm{\lambda }}^2)\sin \left(2\mathrm{\lambda a}\right)}{2\mathrm{k\lambda cos}\left(2\mathrm{\lambda a}\right)+({\mathrm{k}}^2+{\mathrm{\lambda }}^2)\sin \left(2\mathrm{\lambda a}\right)}\mathrm{A}$

$\mathrm{C}=\frac{{\mathrm{e}}^{-\mathrm{ia}(\mathrm{\lambda }+\mathrm{k})}(\mathrm{\lambda }+\mathrm{K})\mathrm{K}}{2\mathrm{k\lambda cos}\left(2\mathrm{\lambda a}\right)-\mathrm{i}({\mathrm{k}}^2+{\mathrm{\lambda }}^2)\sin \left(2\mathrm{\lambda a}\right)}\mathrm{A}$

$\mathrm{D}=\frac{{\mathrm{e}}^{-\mathrm{ia}(\mathrm{k}-\mathrm{\lambda })}(\mathrm{k}-\mathrm{\lambda })\mathrm{K}}{-2\mathrm{k\lambda cos}\left(2\mathrm{\lambda a}\right)+\mathrm{i}({\mathrm{k}}^2+{\mathrm{\lambda }}^2)\sin \left(2\mathrm{\lambda a}\right)}\mathrm{A}$

$\mathrm{F}=\frac{2{\mathrm{k\lambda e}}^{-2\mathrm{ika}}}{2\mathrm{k\lambda cos}\left(2\mathrm{\lambda a}\right)-\mathrm{i}({\mathrm{k}}^2+{\mathrm{\lambda }}^2)\sin \left(2\mathrm{\lambda a}\right)}\mathrm{A}$

$\mathrm{T}={\left(\frac{\mathrm{F}}{\mathrm{A}}\right)}^2={\left(\frac{2{\mathrm{k\lambda e}}^{-2\mathrm{ika}}}{2\mathrm{k\lambda cos}\left(2\mathrm{\lambda a}\right)-\mathrm{i}({\mathrm{k}}^2+{\mathrm{\lambda }}^2)\sin \left(2\mathrm{\lambda a}\right)}\right)}^2$

$\mathrm{T}=\frac{4{\mathrm{\lambda }}^2{\mathrm{k}}^2}{({\mathrm{\lambda }}^4-2{\mathrm{\lambda }}^2{\mathrm{k}}^2+{\mathrm{k}}^4){\sin }^2\left(2\mathrm{\lambda a}\right)+4{\mathrm{\lambda }}^2{\mathrm{k}}^2}$

$\mathrm{T}=\frac{1}{1+({\mathrm{\lambda }}^2-{\mathrm{k}}^2){\sin }^2\left(2\mathrm{\lambda a}\right)/4{\mathrm{\lambda }}^2{\mathrm{k}}^2}$

${\mathrm{T}}^{-1}=1+\frac{({\mathrm{\lambda }}^2-{\mathrm{k}}^2){\sin }^2\left(2\mathrm{\lambda a}\right)}{4{\mathrm{\lambda }}^2{\mathrm{k}}^2}$

${\mathrm{T}}^{-1}=1+\frac{{{\mathrm{V}}_0}^2}{4\mathrm{E}({\mathrm{V}}_0-\mathrm{E})}{\sin }^2\left(\frac{2\mathrm{a}}{\mathrm{h}}\sqrt{2\mathrm{m}({\mathrm{V}}_0-\mathrm{E})}\right)$