Answer to Question #196894 in Quantum Mechanics for Mathushika Murugia

Question

A particle strikes a potential barrier of height U and width L. Derive an expression for the approximate transmission probability, if the energy of the particle E < U

Answer 1

Let's solve the (stationnary) Schrodinger equation in three consecutive domains : before the barrier, in the barries, after the barrier :

1. $-\frac{\hbar^2}{2m} \psi'' = E \psi , \psi = A_i e^{ikx}+A_re^{-ikx}$ , $k = \sqrt{2mE}/\hbar$
2. $-\frac{\hbar^2}{2m} \psi'' = -(U-E)\psi, \psi = B_i e^{\chi x} + B_r e^{-\chi x}$ , $\chi = \sqrt{2m(U-E)}/\hbar$
3. $-\frac{\hbar^2}{2m} \psi'' = E\psi, \psi = C_i e^{ikx} + C_r e^{-ikx}$

Now by using the continuity of $\psi$ and $\psi'$, and taking $A_i=1$ (as the particle is incident on the barrier) and $C_r=0$ (as no particle comes from the other side) we get a set of equations :

$\begin{cases} 1 + A_r = B_i + B_r \\ ik-A_r ik = \chi B_i - \chi B_r \\B_i e^{\chi a}+B_r e^{-\chi a} = C_i e^{ika} \\ B_i \chi e^{\chi a}-\chi B_r e^{-\chi a} = ikC_i e^{ika} \end{cases}$

From the first pair of equations we get $2ik = (\chi +ik)B_i - (\chi-ik)B_r$ and thus $B_i = \frac{2ik+(\chi-ik)B_r}{\chi + ik}$. We get a system :

$\begin{cases} \frac{2ik}{\chi+ik} e^{\chi a} + B_r(e^{-\chi a}+\frac{\chi - ik}{\chi+ik}e^{\chi a})=C_i e^{ika} \\ \frac{2ik \chi}{\chi+ik} e^{\chi a} - \chi(e^{-\chi a}-\frac{\chi - ik}{\chi+ik}e^{\chi a})B_r=ikC_i e^{ika} \end{cases}$

From this we find $\chi C_i e^{ika}(e^{-\chi a}-\frac{\chi - ik}{\chi+ik}e^{\chi a})+ik C_i e^{ika}(e^{-\chi a}+\frac{\chi - ik}{\chi+ik}e^{\chi a}) = 2\frac{2ik\chi}{\chi+ik}$ and thus, finally

$C_i = \frac{4ik\chi e^{-ika}}{(\chi+ik)(\chi e^{-\chi a}-\frac{(\chi-ik)^2}{\chi+ik}e^{\chi a}+ike^{-\chi a})}$ , $C_i = \frac{4ik\chi e^{-ika}}{(\chi+ik)^2e^{-\chi a} - (\chi-ik)^2 e^{\chi a}}=\frac{2ik\chi e^{-ika}}{-(\chi^2-k^2)\sinh(\chi a)+2ik\chi \cosh(\chi a)}$ , $C_i = \frac{e^{-ika}}{\cosh(\chi a) +i\frac{\chi^2-k^2}{2k\chi}\sinh(\chi a)}$

Thus $T = |C_i|^2 = \frac{1}{\cosh^2(\chi a)+\frac{(\chi^2-k^2)^2}{4k^2\chi^2}\sinh^2(\chi a)} = \frac{1}{1+\frac{(\chi^2+k^2)^2}{4k^2\chi^2}\sinh^2(\chi a)}$ , $T=\frac{1}{1+ \frac{((U-E)+E)^2}{4E(U-E)}\sinh^2(\chi a)} = \frac{1}{1+\frac{U^2}{4E(U-E)}\sinh^2(\chi a)}$

When $\frac{U^2}{E(U-E)}\sinh^2(\chi a)\gg 1$ (wide, tall barrier) we get $T\sim 16\frac{E}{U}(1-\frac{E}{U})e^{-2\chi a}$