Answer to Question #164244 in Quantum Mechanics for Nidhin S

Question #164244

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.


1
Expert's answer
2021-02-16T15:56:35-0500

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}"


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