Hydrogen Ground State — Probability Outside the Classically Allowed Region: 17-Phys-A4 Quantum Mechanics, May 2018, Q5

Question 5: Hydrogen Ground State — Probability Outside the Classically Allowed Region (20 marks)

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. Ground-state hydrogen wavefunction $\psi_{100}=\dfrac{1}{\sqrt{\pi a^3}}e^{-r/a}$ ($a$ = Bohr radius), ground-state energy $E_1=-e^2/8\pi\epsilon_0 a$, Coulomb potential $V(r)=-e^2/4\pi\epsilon_0 r$.

Find. The probability $P(r > r_c)$ that a position measurement finds the electron beyond the classical turning point $r_c$ (where $E_1=V(r_c)$).

Fig. 2 — Coulomb potential $V(r)$ (blue) versus the constant ground-state energy $E_1$ (red, dashed); the classically forbidden region $r > 2a$ (shaded) is where $V(r) > E_1$, i.e. where the classical kinetic energy $E_1-V(r)$ would be negative.

Approach. Find the classical turning point $r_c$ from $E_1=V(r_c)$, then integrate the quantum-mechanical radial probability density $|\psi_{100}|^2\,4\pi r^2\,dr$ from $r_c$ to infinity.

Locate the classical turning point. Classically the particle is confined to $E_1\ge V(r)$ (non-negative kinetic energy). Using $E_1=-e^2/8\pi\epsilon_0 a$ and $V(r)=-e^2/4\pi\epsilon_0 r$, the turning point solves $E_1=V(r_c)$: $$ \frac{1}{8a}=\frac{1}{4r_c}\quad\Longrightarrow\quad \boxed{r_c=2a}. $$ For $r < 2a$, $V(r)$ is more negative than $E_1$ (kinetic energy positive, classically allowed); for $r > 2a$, $V(r) > E_1$ (classically forbidden).
Radial probability density. Since $\psi_{100}$ has no angular dependence, integrating over solid angle gives a factor $4\pi$: $$ P(r)\,dr = |\psi_{100}|^2\,4\pi r^2\,dr = \frac{4r^2}{a^3}e^{-2r/a}\,dr. $$
Set up the integral and substitute $x=2r/a$. $$ P(r>2a)=\int_{2a}^{\infty}\frac{4r^2}{a^3}e^{-2r/a}\,dr \;\xrightarrow{x=2r/a}\; \frac12\int_{4}^{\infty}x^2e^{-x}\,dx. $$
Evaluate with the reduction formula. Using the exam's supplied integral $\int x^ne^{ax}dx=\frac{x^ne^{ax}}{a}-\frac{n}{a}\int x^{n-1}e^{ax}dx$ (with $a=-1$) twice reduces $\int_4^\infty x^2e^{-x}dx$ to $(4^2+2\cdot4+2)e^{-4}=26e^{-4}$: $$ P(r>2a)=\frac12(26e^{-4}) = 13e^{-4}. $$ $$ \boxed{P(r>2a) = 13e^{-4} \approx 0.238} $$

Final results
Quantity	Result
Classical turning point	$r_c=2a$
Probability outside classically allowed region	$13e^{-4}\approx0.238$ (23.8%)

Topic: Quantum tunneling into a classically forbidden region for a bound Coulomb state.

Key relations: Classical turning point $E=V(r_c)$; radial probability $P(r)dr=|R(r)|^2r^2dr$ (times $4\pi$ for an $s$-state); Gamma-function-type integral $\int_0^\infty x^ne^{-x}dx=n!$.

Why this works: Unlike a classical particle, a quantum particle in a bound state has a wavefunction that extends smoothly past the classical turning point (it must decay, not vanish, to remain differentiable) — a roughly 24% chance of finding the ground-state electron "outside" the classically allowed radius is a direct, quantitative signature of this purely quantum leakage, closely related to barrier tunneling in Q6.

Common pitfall: Omitting the $4\pi r^2$ radial weighting (using $|\psi|^2$ alone instead of the radial probability density), or mis-locating $r_c$ by using the wrong sign convention for $E_1$.

Source: Griffiths-QM §4.2 (Hydrogen ground state) and Problem 4.53 (classically forbidden region), §2.4 for the general turning-point argument.

Question 5 of 7: Hydrogen Ground State — Probability Outside the Classically Allowed Region

Question 5: Hydrogen Ground State — Probability Outside the Classically Allowed Region (20 marks)