Quantum tunneling

Quantum tunnelling is ability of a particle to overcome a potential barrier, even though it does not have sufficient energy to do so. 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 "tunnelling" 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 tunnelling is important in radioactive decay, when an unstable nucleus emits a particle, it utilizes quantum tunnelling to do so.

Quantum tunnelling is forbidden under classical physics. The instantaneous nature of quantum tunnelling appears to defy the theory of relativity, but it is in fact compatible.

Mathematics

To see a quantum tunnelling effect, consider a rectangular potential barrier of width Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U(x)} , defined as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U(x)= \begin{cases} U_0 &\quad\text{if } 0 \le x \le a\\ 0 &\quad\text{otherwise} \\ \end{cases} }

and also consider a particle moving from left to right. We will consider a particle of energy Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} that is less than Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_0} as this is where quantum mechanics differs from classical mechanics. To find the wavefunction, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)} , we must solve the time-independent Schrodinger equation. The wavefunction is defined in the three regions as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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} }

The Schrodinger equation can be rearranged to give:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2 \psi}{dx^2} = -k^2_0 \psi }

outside the potential and:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2 \psi}{dx^2} = -k^2_1 \psi }

inside it, where:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_0 = \frac{\sqrt{2mE}}{\hbar}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_1 = \frac{\sqrt{2m(E-U_0)}}{\hbar}}

Hence the wavefunction can be written as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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} }

Where capital letters Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} are unknown constants. Since in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_3 (x)} , the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} term corresponds to a particle moving to the right, and we know this is not possible, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F=0} . Even so, it is clear that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_1 (x)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_3 (x)} 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), Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T} is equal to:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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} }

Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_1} is imaginary, using:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_2 = ik_1 = \frac{\sqrt{2m(U_0 - E)}}{\hbar} }

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = \frac{1}{1 - \frac{1}{4} \sinh{k_2 a} (\frac{k_0}{k_2} + \frac{k_2}{k_0})^2} }

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F=0} . Then the continuity conditions of a wavefunction (the wavefunction and its first derivative must be continuous) to find four simultaneous equations:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_1 (0) = \psi_2 (0)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_2 (a) = \psi_3 (a)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi'_1 (0) = \psi'_2 (0)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi'_2 (a) = \psi'_3 (a)}

Since there are five variables, only four equations and solutions are linear, we can set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} equal to 1. This means that we need to find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} to determine the tunnelling probability. This gives the equations (with some rearrangement):

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1+B=C+D}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1-B = \frac{k_1}{k_0} (C-D)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C e^{ik_1 a} + D e^{-ik_1 a} = E e^{ik_0 a}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C e^{ik_1 a} - D e^{-1k_1 a} = E e^{ik_0 a} \frac{k_0}{k_1}}

We can eliminate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B} to get:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2=C(1 + \frac{k_1}{k_0}) +D(1- \frac{k_1}{k_0} }

We can find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D} as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C = \frac{1}{2} E e^{ia(k_0 - k_1)} (1 + \frac{k_0}{k_1})}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D = \frac{1}{2} E e^{ia(k_0 + k_1)} (1 - \frac{k_0}{k_1})}

Substituting these values for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D} , with some rearranging give:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}) }

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)} }

Hence:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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)}} }

The transmission probability is given by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |E|^2 = EE^*} , so we must find the complex conjugate, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^*} :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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)}} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}} }

$T=EE^{*}={\frac {1}{1-{\frac {1}{4}}\sin {k_{1}a}({\frac {k_{0}}{k_{1}}}-{\frac {k_{1}}{k_{0}}})^{2}}}$

To use hyperbolic functions instead, use the substitution:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_1 = \frac{\sqrt{2m(E-U_0)}}{\hbar} = ik_2 = \frac{\sqrt{2m(U_0 - E)}}{\hbar} }

and the fact Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin{(ix)} = i \sinh{x}} to get:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = \frac{1}{1 - \frac{1}{4} \sinh{k_2 a} (\frac{k_0}{k_2} + \frac{k_2}{k_0})^2} }

where

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_0 = \frac{\sqrt{2mE}}{\hbar}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_1 = \frac{\sqrt{2m(E-U_0)}}{\hbar}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_2 = \frac{\sqrt{2m(U_0-E)}}{\hbar}}

See also

• Biblical Scientific Foreknowledge
• Quantum mechanics
• Radioactivity