Quantum tunneling

Quantum tunneling is the ability of a particle to overcome a potential barrier, even though it does not have sufficient energy to do so. This is conceptually similar to a Resurrection, changing from one physical status to another contrary to traditional laws of physics.

An example is an electron being fired at one side of a barrier and reappearing on the other side. The most common example is when the barrier is an insulator, and an electron on one side of the insulator moves to the other side. The term "tunneling" is a bit of a misnomer because the electron does not actually travel through the insulator (insulators do not conduct electricity). It reappears with the same energy it started with.

Quantum tunneling is important in radioactive decay, when an unstable nucleus emits a particle, it utilizes quantum tunneling to do so.

Quantum tunneling is forbidden under classical physics. The instantaneous nature of quantum tunneling seems contrary to the theory of relativity, but some argue that it is compatible.

Quantum tunneling is considered instantaneous in the Copenhagen interpretation, which is the orthodox view of quantum mechanics. In an experiment published in July 2020, rubidium atoms tunneled through a 1.3 micrometer barrier in 0.61 milliseconds.

Mathematics

To see a quantum tunneling effect, consider a rectangular potential barrier of width <math>a</math>, <math>U(x)</math>, defined as:

<math> U(x)=

\begin{cases}

      U_0 &\quad\text{if } 0 \le x \le a\\
      0 &\quad\text{otherwise} \\

\end{cases} </math>

and also consider a particle moving from left to right. We will consider a particle of energy <math>E</math> that is less than <math>U_0</math> as this is where quantum mechanics differs from classical mechanics. To find the wavefunction, <math>\psi(x)</math>, we must solve the time-independent Schrodinger equation. The wavefunction is defined in the three regions as:

<math> \psi(x)= \begin{cases}

      \psi_1 (x) &\quad\text{if } x<0\\
      \psi_2 (x) &\quad\text{if } 0 \le x \le a\\
      \psi_3 (x) &\quad\text{if } x>a\\

\end{cases} </math>

The Schrodinger equation can be rearranged to give:

<math> \frac{d^2 \psi}{dx^2} = -k^2_0 \psi </math>

outside the potential and:

<math> \frac{d^2 \psi}{dx^2} = -k^2_1 \psi </math>

inside it, where:

<math>k_0 = \frac{\sqrt{2mE}}{\hbar}</math>
<math>k_1 = \frac{\sqrt{2m(E-U_0)}}{\hbar}</math>

Hence the wavefunction can be written as:

<math> \psi(x)=

\begin{cases}

      \psi_1 (x) = A e^{ik_0 x} + B e^{-ik_0 x}\\
      \psi_2 (x) = C e^{ik_1 x} + D e^{-ik_1 x}\\
      \psi_3 (x) = E e^{ik_0 x} + F e^{-ik_0 x}\\

\end{cases} </math>

Where capital letters <math>A</math> to <math>F</math> are unknown constants. Since in <math>\psi_3 (x)</math>, the <math>F</math> term corresponds to a particle moving to the right, and we know this is not possible, <math>F=0</math>. Even so, it is clear that <math>\psi_1 (x)</math> and <math>\psi_3 (x)</math> correspond to sinusoidal wavefunctions, similar to that of free particles. Hence it is clear that although the particle starts on the left of the barrier and does not have sufficient energy to overcome it, there is some probability that it will be found on the other side. Solving for boundary conditions, it can be found that the transmission probability (probability that the particle tunnels through the barrier), <math>T</math> is equal to:

<math> T = |E/A|^2 = \frac{1}{1 - \frac{1}{4} \sin{k_1 a} (\frac{k_0}{k_1} - \frac{k_1}{k_0})^2} </math>

Since <math>k_1</math> is imaginary, using:

<math> k_2 = ik_1 = \frac{\sqrt{2m(U_0 - E)}}{\hbar} </math>

it can also be written in a form easier to compute as:

<math> T = \frac{1}{1 - \frac{1}{4} \sinh{k_2 a} (\frac{k_0}{k_2} + \frac{k_2}{k_0})^2} </math>

The probability the particle is reflected is <math>R=1-T</math>

Derivation of the Tunneling Probability

This section has advanced mathematical concepts such as complex numbers and hyperbolic trigonometric functions.

First, we start from the split wavefunction above, remembering that we have already shown <math>F=0</math>. Then the continuity conditions of a wavefunction (the wavefunction and its first derivative must be continuous) result in four simultaneous equations:

<math>\psi_1 (0) = \psi_2 (0)</math>
<math>\psi_2 (a) = \psi_3 (a)</math>
<math>\psi'_1 (0) = \psi'_2 (0)</math>
<math>\psi'_2 (a) = \psi'_3 (a)</math>

Since there are five variables, only four equations and solutions are linear, we can set <math>A</math> equal to 1. This means that we need to find <math>E</math> to determine the tunneling probability. This gives the equations (with some rearrangement):

<math>1+B=C+D</math>
<math>1-B = \frac{k_1}{k_0} (C-D)</math>
<math>C e^{ik_1 a} + D e^{-ik_1 a} = E e^{ik_0 a}</math>
<math>C e^{ik_1 a} - D e^{-1k_1 a} = E e^{ik_0 a} \frac{k_0}{k_1}</math>

We can eliminate <math>B</math> to get:

<math>2=C(1 + \frac{k_1}{k_0}) +D(1- \frac{k_1}{k_0} </math>

We can find <math>C</math> and <math>D</math> as:

<math>C = \frac{1}{2} E e^{ia(k_0 - k_1)} (1 + \frac{k_0}{k_1})</math>
<math>D = \frac{1}{2} E e^{ia(k_0 + k_1)} (1 - \frac{k_0}{k_1})</math>

Substituting these values for <math>C</math> and <math>D</math>, with some rearranging give:

<math> \frac{4k_0 k_1 e^{-ik_0 a}}{E} = 2k_0 k_1 (e^{ik_1 a} + e^{-ik_1 a}) + (k_0^2 + k_1^2)(e^{-ik_1 a} - e^{ik_1 a}) </math>

Using Euler's identity, the exponential terms can be rewritten in terms of trigonometric functions:

<math> \frac{4k_0 k_1 e^{-ik_0 a}}{E} = 2k_0 k_1 \cos{(k_1 a)} - 2i(k_0^2 + k_1^2) \sin{(k_1 a)} </math>

Hence:

<math> E = \frac{2k_0 k_1 e^{-ik_0 a}}{2k_0 k_1 \cos{(k_1 a)} - i(k_0^2 + k_1^2) \sin{(k_1 a)}} </math>

The transmission probability is given by <math>|E|^2 = EE^*</math>, so we must find the complex conjugate, <math>E^*</math>:

<math> E^* = \frac{2k_0 k_1 e^{-ik_0 a}}{2k_0 k_1 \cos{(k_1 a)} + i(k_0^2 + k_1^2) \sin{(k_1 a)}} </math>

<math> EE^* = \frac{1}{1 - \frac{1}{4} \sin{k_1 a} \frac{k^4_0 -2k^2_0 k^1_0 +k^4_1}{k^2_0 k^2_1}} </math>

<math> T = EE^* = \frac{1}{1 - \frac{1}{4} \sin{k_1 a} (\frac{k_0}{k_1} - \frac{k_1}{k_0})^2} </math>

To use hyperbolic functions instead, use the substitution:

<math> k_1 = \frac{\sqrt{2m(E-U_0)}}{\hbar} = ik_2 = \frac{\sqrt{2m(U_0 - E)}}{\hbar} </math>

and the fact <math>\sin{(ix)} = i \sinh{x}</math> to get:

<math> T = \frac{1}{1 - \frac{1}{4} \sinh{k_2 a} (\frac{k_0}{k_2} + \frac{k_2}{k_0})^2} </math>

where

<math>k_0 = \frac{\sqrt{2mE}}{\hbar}</math>
<math>k_1 = \frac{\sqrt{2m(E-U_0)}}{\hbar}</math>
<math>k_2 = \frac{\sqrt{2m(U_0-E)}}{\hbar}</math>

See also

• Quantum mechanics
• Radioactivity
• Biblical Scientific Foreknowledge
• Essay:Rebuttal to Biblical scientific foreknowledge