Question 6 of 7: Quantum Tunneling Through a Finite Square Barrier

Question 6: Quantum Tunneling Through a Finite Square Barrier (20 marks: (a) 5, (b) 5, (c) 5, (d) 5)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $V(x)=0$ for $x < 0$ and $x > a$, $V(x)=V_0$ for $0 < x < a$; a particle of mass $m$ and energy $E < V_0$ incident from the left.

Find. (a) TISE and its solution in each region; (b) the four boundary-matching equations and the method to get $R,T$; (c) numeric $T$; (d) the large-$\kappa a$ approximation.

Approach. (a)–(b) set up and describe the standard matching-conditions method for a 1-D step/barrier scattering problem; (c) substitutes the given numbers into the supplied closed-form $T$; (d) takes the $\kappa a\gg1$ limit of that same formula.

1. Part (a) — TISE and solutions in each region. The TISE is $-\dfrac{\hbar^2}{2m}\psi''+V(x)\psi=E\psi$ everywhere; only $V(x)$ changes by region.
   (i) $x < 0$ ($V=0$): $\psi''=-k^2\psi$, $k=\sqrt{2mE}/\hbar$, so $\psi_I(x)=Ae^{ikx}+Be^{-ikx}$ (incident $+$ reflected wave).
   (ii) $0 < x < a$ ($V=V_0 > E$): $\psi''=\kappa^2\psi$, $\kappa=\sqrt{2m(V_0-E)}/\hbar$ (real, since $E < V_0$), so $\psi_{II}(x)=Ce^{\kappa x}+De^{-\kappa x}$ (growing $+$ decaying, not oscillatory — classically forbidden region).
   (iii) $x > a$ ($V=0$): $\psi_{III}(x)=Fe^{ikx}$ (transmitted wave only; no wave is incident from $+\infty$).
2. Part (b) — the four matching equations. Continuity of $\psi$ and $\psi'$ at $x=0$ and $x=a$: $$ A+B=C+D,\qquad ik(A-B)=\kappa(C-D), $$ $$ Ce^{\kappa a}+De^{-\kappa a}=Fe^{ika},\qquad \kappa\!\left(Ce^{\kappa a}-De^{-\kappa a}\right)=ikFe^{ika}. $$ Method: use the first pair to write $C,D$ in terms of $A,B$; substitute into the second pair to relate $F$ and $B$ to $A$; solving the resulting linear system for the ratios $B/A$ and $F/A$ gives the reflection coefficient $R=|B/A|^2$ and transmission (tunneling) coefficient $T=|F/A|^2$, with $R+T=1$ guaranteed by conservation of probability current (no absorption anywhere in the potential).
3. Part (c) — evaluate $\kappa a$. Writing $\kappa$ with $mc^2$ to avoid unit conversion of $m$: $\kappa=\sqrt{2mc^2(V_0-E)}\,/(\hbar c)$. With $mc^2=511\times10^3$ eV, $V_0-E=5$ eV, $\hbar c=(6.58\times10^{-16}\,\text{eV}\cdot\text{s})(3\times10^8\,\text{m/s})=1.974\times10^{-7}$ eV·m: $$ \kappa = \frac{\sqrt{2(511\times10^3)(5)}}{1.974\times10^{-7}\,\text{eV}\cdot\text{m}}\,\text{eV} = 1.145\times10^{10}\ \text{m}^{-1}. $$ $$ \kappa a = (1.145\times10^{10}\,\text{m}^{-1})(0.53\times10^{-10}\,\text{m}) \approx 0.607. $$
4. Substitute into $T$. With $E=5$, $V_0=10$, $V_0-E=5$ (all eV): $$ \sinh(0.607)\approx0.645,\qquad \sinh^2(\kappa a)\approx0.416, $$ $$ T = \left[1+\frac{(10)^2(0.416)}{4(5)(5)}\right]^{-1} = \left[1+0.416\right]^{-1} $$ $$ \boxed{T\approx0.706\ \ (70.6\%)} $$ — a large tunneling probability because $\kappa a\sim0.6$ is not large; the barrier is thin/low relative to the electron's de Broglie wavelength scale.
5. Part (d) — large-$\kappa a$ limit. For $\kappa a\gg1$, $\sinh(\kappa a)=\frac{e^{\kappa a}-e^{-\kappa a}}{2}\approx\frac{e^{\kappa a}}{2}$, so $\sinh^2(\kappa a)\approx e^{2\kappa a}/4$. The bracket in $T$ becomes $$ 1+\frac{V_0^2}{4E(V_0-E)}\cdot\frac{e^{2\kappa a}}{4} = 1+\frac{V_0^2\,e^{2\kappa a}}{16E(V_0-E)}. $$ Since $e^{2\kappa a}\gg1$, the leading "$1+$" is negligible next to the exponential term, so $$ T\approx\left[\frac{V_0^2e^{2\kappa a}}{16E(V_0-E)}\right]^{-1} $$ $$ \boxed{T\approx\frac{16E(V_0-E)}{V_0^2}\,e^{-2\kappa a}} $$ — the familiar exponential-suppression law for a thick/high tunneling barrier.