Quantum Tunnelling

Classical Mechanics Level 5

A one-dimensional "universe" consists of the linear space between x = 0 x = 0 and x = x f x = x_f (see below). There is a single particle in this universe, and its behavior is governed by a variant of the time-independent Schrodinger equation, expressed in terms of a non-unitary wave function Ψ N ( x ) \Psi_N (x) .

The quantities V V and E E are the potential energy and total energy, respectively.

− d 2 d x 2 Ψ N ( x ) + V ( x ) Ψ N ( x ) = E Ψ N ( x ) \large{-\frac{d^2}{d x^2} \, \Psi_N (x) + V(x) \, \Psi_N (x) = E \, \Psi_N (x)}

Let us consider a case in which E = 1 E = 1 . The particle is situated inside an infinite potential well, which contains a finite potential region (see graphic). The potential varies as follows:

V ( x ) = ∞ x < 0 V ( x ) = 0 0 ≤ x < x 1 V ( x ) = 1.95 x 1 ≤ x < x 2 V ( x ) = 0 x 2 ≤ x ≤ x f V ( x ) = ∞ x > x f x 1 = 15 4 π x 2 = 19 4 π x f ≈ 7.55 π (a zero crossing of Ψ N ( x ) ) V (x) = \infty \,\,\,\,\,\,\,\, x < 0 \\ V (x) = 0 \,\,\,\,\,\,\,\, 0 \leq x < x_1 \\ V (x) = 1.95 \,\,\,\,\,\,\,\, x_1 \leq x < x_2 \\ V (x) = 0 \,\,\,\,\,\,\,\, x_2 \leq x \leq x_f \\ V (x) = \infty \,\,\,\,\,\,\,\, x > x_f \\ x_1 = \frac{15}{4} \pi \\ x_2 = \frac{19}{4} \pi \\ x_f \approx 7.55 \pi \, \text{(a zero crossing of } \Psi_N (x))

The values of Ψ N ( x ) \Psi_N (x) and its first spatial derivative at x = 0 x = 0 are (see graphic):

Ψ N ( 0 ) = 0 ( d d x Ψ N ) ( 0 ) = 1 \Psi_N (0) = 0 \\ \Big ( \frac{d}{dx} \, \Psi_N \Big ) (0) = 1

Note that the potential energy V ( x ) V(x) exceeds E E inside the finite-potential region, and yet the particle still has a finite non-zero probability of being measured anywhere inside the infinite well (except for the extreme boundaries).

The probability of detecting the particle within the region a ≤ x ≤ b a \leq x \leq b is:

P ( a ≤ x ≤ b ) = ∫ a b ∣ Ψ N ( x ) ∣ 2 d x ∫ 0 x f ∣ Ψ N ( x ) ∣ 2 d x \large{P(a \leq x \leq b) = \frac{\int_a^b |\Psi_N (x)|^2 \, dx }{\int_0^{x_f} |\Psi_N (x)|^2 \, dx }}

What is the probability of detecting the particle in the region ( x 2 ≤ x ≤ x f ) ( x_2 \leq x \leq x_f ) ?

Note: This problem lends itself well to numerical solution


The answer is 0.05276.

Steven Chase by Steven Chase , 37, USA

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1 solution

Mark Hennings
Jan 19, 2019

The wave function is Ψ ( x ) = { 0 x < 0 sin ⁡ x 0 < x < x 1 A e κ ( x − x 1 ) + B e − κ ( x − x 1 ) x 1 < x < x 2 C sin ⁡ x + D cos ⁡ x x 2 < x < x f 0 x > x f \Psi(x) \; = \; \left\{ \begin{array}{lll}0 & \hspace{2cm} & x < 0 \\ \sin x & & 0 < x < x_1 \\ Ae^{\kappa(x-x_1)} + Be^{-\kappa(x-x_1)} & & x_1 < x < x^2 \\ C\sin x + D\cos x & & x_2 < x < x_f \\ 0 & & x > x_f \end{array}\right. where κ 2 = 0.95 \kappa^2 = 0.95 and A , B , C , D A,B,C,D are to be chosen to that Ψ \Psi and Ψ ′ \Psi' are continuous for 0 < x < x f 0 < x < x_f . Thus we need to solve the simultaneous equations sin ⁡ x 1 = A + B C sin ⁡ x 2 + D cos ⁡ x 2 = A e κ π + B e κ π cos ⁡ x 1 = κ A − κ B C cos ⁡ x 2 − D sin ⁡ x 2 = κ A e κ π − κ B e − κ π \begin{array}{rclcrcl} \sin x_1 &= & A + B & \hspace{2cm} & C\sin x_2 + D\cos x_2 & = & Ae^{\kappa \pi} + Be^{\kappa \pi} \\ \cos x_1 &=& \kappa A - \kappa B & & C\cos x_2 - D\sin x_2 & = & \kappa Ae^{\kappa \pi} - \kappa Be^{-\kappa \pi} \end{array} obtaining A = 0.00918473 B = − 0.716292 C = − 0.0432851 D = − 0.273480 A \; = \; 0.00918473 \hspace{1cm} B \; = \; -0.716292 \hspace{1cm} C \; = \; -0.0432851 \hspace{1cm} D \; = \; -0.273480 and hence the zero of Ψ \Psi near 7.55 π 7.55\pi is x f = 23.7189 x_f = 23.7189 .

It is now easy to calculate ∫ 0 x f ∣ Ψ ( x ) ∣ 2 d x = 6.73691 ∫ x 2 x f ∣ Ψ ( x ) ∣ 2 d x = 0.355416 \int_0^{x_f} |\Psi(x)|^2\,dx \; = \; 6.73691 \hspace{2cm} \int_{x_2}^{x_f} |\Psi(x)|^2\,dx \; = \; 0.355416 making the desired probability 0.0527565 \boxed{0.0527565} .

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