Quantum Tunneling through Potential Barriers

Page Claimed By: Shreenithi Katta Spring 22

Quantum tunneling is a phenomenon of quantum mechanics in which a wave function can travel through a potential barrier and be observed on the other side. It utilizes the Heisenberg uncertainty principle and the Schrödinger equation to explain the classically forbidden phenomenon. Using these principles, the probability that a particle can tunnel through a potential barrier can be calculated.

Main Idea

A potential barrier can be defined as a bounded region of significantly high potential than the space it is surrounded by. For the purpose of quantum tunneling, the potential barrier is finite with a specified height, width, and some unknown potential defined in terms of the particle that is incident upon it ([math]\displaystyle{ V_0 \gt E }[/math]).

Classical Theory

According to classical mechanics, a particle of energy [math]\displaystyle{ E }[/math] does not have sufficient energy to pass through the barrier of energy [math]\displaystyle{ V_0 \gt E }[/math] - the particle is classically forbidden to enter the region. However, due to the Wave-Particle Duality, quantum mechanics predicts that the wavefunction of the particle has a chance of entering the classically forbidden region.

Quantum Theory

Wave Mechanics

Wave mechanics tells us that when a wave travels through a medium are partially absorbed, transmitted, and reflected. With a finite potential barrier of certain dimensions, we can create three regions the wave function travels through i) the initial region before the potential barrier ii) the finite potential barrier iii) the region after the potential barrier. Using the wave properties, we know a portion of the incident wave function is reflected, absorbed, and transmitted into region B where the same happens on the border of regions II and III.

Heisenberg Uncertainty Principle

The Heisenberg uncertainty principle can be used to explain quantum tunneling a bit more intuitively. The second uncertainty principle outlined by Heisenberg is concerned with energy and time. Given a short amount of time to examine a particle, the uncertainty in the energy it contains vastly increases. Applying that to this phenomenon is simple. If the uncertainty in the energy of the particle is great in a short amount of time, there is a possibility that the particle contains enough energy to enter the classically forbidden region. Whether it is reflected halfway through or transmits to the other side, the uncertainty in the energy essentially allows for a classical particle to exist in a potential barrier.

The Schrödinger Equation

The Schrödinger equation tells us that a wave function must be continuous at each boundary it encounters, and the derivative of the wave function must also be continuous except when the boundary height is infinite. Applying this to a finite potential barrier, we can find the probability a particle can tunnel through a potential barrier. Notably, the wave functions and calculations most resemble the solution for a single particle in a semi-infinite well when [math]\displaystyle{ V_0 \gt E }[/math] in regions II and III.

Mathematical Model

Beyond the conceptual understanding of how quantum tunneling works, calculations can be made to determine the probability a wave function can tunnel through a specific finite potential barrier. This is known as transmission probability and to understand this equation, we must look at the wave equations associated with the potential barrier.

Wave Equations

The time independent Schrodinger equation is given by the following equation.

[math]\displaystyle{ -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V(x)\psi (x) = E \psi(x) }[/math]

As mentioned before, the finite potential barrier creates three different regions with their own unique wave equations. Let's assume that the potential barrier is from [math]\displaystyle{ 0 \le x \le L }[/math]. Given that, we can define the potential as follows:

[math]\displaystyle{ V(x) = \begin{cases} 0 & x \lt 0 \\ V_0 & 0\leq x\leq L \\ 0 & x\gt L \end{cases} }[/math]

Thus, we can write the Schrodinger equations for the three regions.

Region I : [math]\displaystyle{ -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} = E \psi(x) }[/math]

Region II : [math]\displaystyle{ -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V_0 \psi (x) = E \psi(x) }[/math]

Region III : [math]\displaystyle{ -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} = E \psi(x) }[/math]

Boundary conditions require that the wave functions and their derivatives are continuous on each boundary i.e [math]\displaystyle{ x = 0 }[/math] and [math]\displaystyle{ x = L }[/math]. The general solution for each region,

Region I: [math]\displaystyle{ \psi(x)_I = Ae^{ikx} + Be^{-ikx} }[/math]

Region II: [math]\displaystyle{ \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } }[/math]

Region III: [math]\displaystyle{ \psi(x)_{III} = Fe^{ikx} + Ge^{-ikx} }[/math]

where [math]\displaystyle{ \alpha = \sqrt {\frac{2m}{\hbar^2} (V_0 - E)} }[/math] and [math]\displaystyle{ k = \sqrt {\frac{2m}{\hbar^2} E} }[/math].

There is only one direction in which the wave in region III is traveling ([math]\displaystyle{ +\infty }[/math]). Thus the wave functions are now,

Region I: [math]\displaystyle{ \psi(x)_I = Ae^{ikx} + Be^{-ikx} }[/math]

Region II: [math]\displaystyle{ \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } }[/math]

Region III: [math]\displaystyle{ \psi(x)_{III} = Fe^{ikx} }[/math]

It is important to note that the constants A, B, C, D, and F refer to the amplitude of the five waves in this problem:

A: incident wave at [math]\displaystyle{ x = 0 }[/math]
B: reflected wave at [math]\displaystyle{ x = 0 }[/math]
C: incident wave at [math]\displaystyle{ x = L }[/math]
D: reflected wave at [math]\displaystyle{ x = L }[/math]
F: transmitted wave

Tunneling Probability

The probability that a particle can tunnel through a potential barrier utilizes the wave functions discussed in the previous section. Specifically, tunneling probability is the ratio of the transmitted wave intensity [math]\displaystyle{ |F|^2 }[/math] and the incident wave intensity [math]\displaystyle{ |A|^2 }[/math]. To solve for these constants, we must apply boundary conditions.

At [math]\displaystyle{ x = 0 }[/math], we know region I and II must agree and create a continuous wave function.

[math]\displaystyle{ Ae^{ikx} + Be^{-ikx} = Ce^{-\alpha x } + De^{\alpha x} }[/math]

[math]\displaystyle{ \frac{d}{dx}(Ae^{ikx} + Be^{-ikx} = Ce^{-\alpha x } + De^{\alpha x}) }[/math]

[math]\displaystyle{ ik(Ae^{ikx} - Be^{-ikx}) = \alpha(-Ce^{-\alpha x } + De^{\alpha x}) }[/math]

The same applies at [math]\displaystyle{ x = L }[/math].

[math]\displaystyle{ Ce^{-\alpha x } + De^{\alpha x} = Fe^{ikx} }[/math]

[math]\displaystyle{ \frac{d}{dx}(Ce^{-\alpha x } + De^{\alpha x} = Fe^{ikx}) }[/math]

[math]\displaystyle{ \alpha(-Ce^{-\alpha x } + De^{\alpha x}) = ikFe^{ikx} }[/math]

After some lengthy algebra and calculations, we know that the ratio of the transmitted wave amplitude and incident wave amplitude is this messy expression.

[math]\displaystyle{ \frac{F}{A} = \frac{e^{-ikL}}{cosh(\alpha L) +i(\gamma /2)sinh(\alpha L)} }[/math][1]

where [math]\displaystyle{ \gamma = \frac{\alpha}{k} - \frac{k}{\alpha} }[/math].

We know that tunneling probability is a ratio of intensity. So when we multiply the ratio of amplitudes by the conjugate, we get the following equation for tunneling probability.

[math]\displaystyle{ T \approx 16 \frac{E}{V_0} (1-\frac{E}{V_0}) e^{-2kL} }[/math]

where, [math]\displaystyle{ k = \sqrt {\frac{2m(V_0-E)}{\hbar^2}} }[/math] and L is the width of the barrier.

Computational Model

The following simulation illustrates a wave function for various potentials in both plane wave and wave packet form.

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Examples

Simple

A particle has 5.0 eV of energy. What is the probability that it will tunnel through a potential barrier of 10.0 eV with a width/thickness of 1.00 nm?

Solution

First, let's define the given variables and equation we must use

[math]\displaystyle{ E = 5.0 eV }[/math]
[math]\displaystyle{ V_0 = 10.0 eV }[/math]
[math]\displaystyle{ L = 1.00 nm = 1.00 \times 10^{-9} m }[/math]

[math]\displaystyle{ T \approx 16\frac {E}{V_0} (1-\frac{E}{V_0})e^{-2kL} }[/math]
[math]\displaystyle{ k = \sqrt{\frac {2m(V_0 -E)}{\hbar^2}} }[/math]

Next, plug in our given values into the equation and solve.

[math]\displaystyle{ k = \sqrt {\frac {2(9.11 \times 10^{-31} kg)(10.0-5.0 eV)(1.602 \times 10^{-19} J)}{(1.055 \times 10^{-34})^2}} = 1.145 \times 10^{10} m^{-1} }[/math]

[math]\displaystyle{ T \approx 16\frac {5.0}{10.0} (1-\frac{5.0}{10.0})e^{-2(1.145 \times 10^{10} m^{-1})(1.00 \times 10^{-9})} = 4.52 \times 10^{-10} }[/math]

Middling

Difficult

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