# Quantum Tunneling

**Quantum tunneling** refers to the nonzero probability that a particle in quantum mechanics can be measured to be in a state that is forbidden in classical mechanics. Quantum tunneling occurs because there exists a nontrivial solution to the Schrödinger equation in a classically forbidden region, which corresponds to the exponential decay of the magnitude of the wavefunction.

To illustrate the concept of tunneling, consider trying to confine an electron in a box. One could try to pin down the location of the particle by shrinking the walls of the box, which will result in the electron wavefunction acquiring greater momentum uncertainty by the Heisenberg uncertainty principle. As the box gets smaller and smaller, the probability of measuring the location of the electron to be outside the box increases towards one, despite the fact that classically the electron is confined inside the box.

The easiest solvable example of quantum tunneling is in one dimension. However, tunneling is responsible for a wide range of physical phenomena in three dimensions such as radioactive decay, the behavior of semiconductors and superconductors, and scanning tunneling microscopy.

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## Scattering from a Potential Barrier in One Dimension

Scattering particles off of a potential barrier in one dimension look like this:

Suppose that the height of the potential barrier is $V_0$ and the width is $L$ and that scattering particles have energy $E <V_0$. Then the picture can be divided into three regions:

**Region 1**$(-\infty < x<0)$: $E > V(x)$**Region 2**$(0 \le x \le L)$: $E <V(x)$**Region 3**$(L < x< \infty)$: $E > V(x),$ where $V(x)=0; x<0$, $V(x)=V_0;x=0, 0<x<L$ and $V(x)=0, x>L.$

In **Region 1**, the potential is zero. A moving wave thus has energy greater than the potential. This is also true in **Region 3**. However, in **Region 2**, the energy of the wave is less than the potential. Therefore, the Schrödinger equation yields two different differential equations depending on the region:

**Region 1**and**Region 3**$\frac{{d}^{2}\psi}{d{x}^{2}}={k}^{2}\psi, \quad k=\sqrt{\frac{-2mE}{{\hbar}^{2}}}$**Region 2**$\frac{{d}^{2}\psi}{d{x}^{2}}={\kappa}^{2}\psi, \quad \kappa=\sqrt{\frac{2m(V-E)}{{\hbar}^{2}}}.$

The general solutions can be written as linear combinations of oscillatory terms in Regions 1 and 3, and as linear combinations of growing and decaying exponentials in Region 2:

$\psi (x)=\begin{cases} { Ae }^{ ikx }+{ Be }^{ -ikx } \quad &\text{: Region 1} \\ { Ce }^{ \kappa x }+{ De }^{ -\kappa x } \quad &\text{: Region 2} \\ { Fe }^{ ikx } &\text{: Region 3}. \end{cases}$

Note that plane-waves that travel to the right are of the form ${e}^{ikx}$ and plane-waves that travel to the left are of the form ${e}^{-ikx}$. In this experiment, a particle (plane-wave) enters from the left and will partially transmit and partially reflect. However, no particle enters from the right heading towards the left; therefore, there is no $Ge^{-ikx}$ term above in Region 3.

The coefficients above are fixed by the continuity of the wavefunction and its derivative at each point where the potential changes. One obtains two conditions from continuity at $x=0$ and $x=L$:

- 1) $A+B=C+D$
- 2) ${Ce}^{\kappa L}+{De}^{-\kappa L}= {Fe}^{ikL}$

and two conditions from continuity of the derivative at $x=0$ and $x=L$:

- 3) $Aik-Bik=C\kappa-D\kappa$
- 4) ${C\kappa e}^{\kappa L}-{D\kappa e}^{-\kappa L}= {Fike}^{ikL}.$

Dividing 3) by $ik$ and adding to 1) obtains

- 5) $2A=\left(1+\frac{\kappa}{ik}\right)C+\left(1-\frac{\kappa}{ik}\right)D.$

Similarly, dividing 4) by $\kappa$ and adding or subtracting from 2) obtains

- 6) $2C{e}^{\kappa L}=\left(1+\frac{ik}{\kappa}\right)F{e}^{ikL}$
- 7) $2D{e}^{-\kappa L}=\left(1-\frac{ik}{\kappa}\right)F{e}^{ikL}.$

Combining 5), 6), and 7) yields an equation for $A$ in terms of $F:$

$2A=\left(1+\frac{\kappa}{ik}\right)\left(1+\frac{ik}{\kappa}\right)\frac{F{e}^{ikL}{e}^{-\kappa L}}{2}+\left(1-\frac{\kappa}{ik}\right)\left(1-\frac{ik}{\kappa}\right)\frac{F{e}^{ikL}{e}^{\kappa L}}{2},$

which can be rearranged to

$\frac{A{e}^{-ikL}}{F}=\cosh{(\kappa L)}+i\left(\frac{{\kappa}^{2}-{k}^{2}}{2k\kappa}\right)\sinh{(\kappa L)}.$

Now the probability of a wave to tunnel through the barrier is equal to the probability of the wavefunction in **Region 3** divided by the probability of the wavefunction in **Region 1**. Multiplying the above equation by its conjugate and taking the inverse, the probability of transmission is therefore quantified by

$T=\frac{|F|^2}{|A|^2}={\left[{{\text{cosh}}^{2}(\kappa L)+\left(\frac{{\kappa}^{2}-{k}^{2}}{2k\kappa}\right)}^{2}{\text{sinh}}^{2}(\kappa L)\right]}^{-1}.$

Since ${\text{cosh}}^{2}(x)-{\text{sinh}}^{2}(x)=1$,

$T={\left[1+\left(\frac{{k}^{2}+{\kappa}^{2}}{2k\kappa}\right){\text{sinh}}^{2}(\kappa L)\right]}^{-1}.$

Defining $\beta=\left(\frac{{k}^{2}+{\kappa}^{2}}{2k\kappa}\right)$ makes the solution more compact:

$T=\frac{1}{1+{\beta}^{2}{\text{sinh}}^{2}(\kappa L)}.$

This can also be rewritten in terms of the energies:

$T = \left(1+ \frac{V_0^2}{4E(V_0 -E)} \sinh^2 \left(\frac{L}{\hbar} \sqrt{2m(V_0 - E)}\right)\right)^{-1} .$

Naturally, the probability of reflection is $1-T$; hence,

$R=\frac{{\beta}^{2}{\text{sinh}}^{2}(\kappa L)}{1+{\beta}^{2}{\text{sinh}}^{2}(\kappa L)}.$

Macroscopically, objects colliding against a wall will be deflected. This is analogous to the reflection probability being 100% and transmission probability being 0%. The above example shows that it is possible for matter waves to "go through walls" with some probability, given that a matter wave has sufficient energy or the barrier being sufficiently narrow $($small $L).$ Note that, for a very wide or tall barrier $(L$ very large$)$ or $V_0 \gg E,$ the $\sinh$ term in the expression for $T$ goes to $\infty$, yielding $T \approx 0:$ for a very wide or tall barrier, there is almost no transmission.

The below animation shows a localized wavefunction tunneling through the one-dimensional barrier by evolving the time-dependent Schrödinger equation:

Challenge problem: derive the transmission coefficient for the rectangular potential

well,$V(x) = \begin{cases} -V_0 \quad & 0<x<L \\ 0 \quad & \text{ otherwise}. \end{cases}$

We have

$T = \left(1+ \frac{V_0^2}{4E(V_0 +E)} \sin^2 \left(\frac{L}{\hbar} \sqrt{2m(V_0 + E)}\right)\right)^{-1} .$

## Gamow Model of Radioactive Decay

One of the first applications of quantum tunneling was to explain alpha decay, the radioactive decay of a nucleus leading to emission of an alpha particle (helium nucleus). The relevant model is called the **Gamow model** after its creator George Gamow [4]. Gamow modeled the potential experienced by an alpha particle in the nucleus as a finite square well in the nuclear region and Coulombic repulsion outside the nucleus, as displayed in the diagram:

The corresponding potential can be written formally as below:

$V = \begin{cases} -V_0 \quad &r < r_0\\\\ \frac{1}{4\pi \epsilon_0} \frac{2Z e^2}{r} \quad & r>r_0. \end{cases}$

A fact in advanced quantum mechanics is that the transmission probability $T$ is well approximated by

$T = e^{-2\gamma},$

where $\gamma$ is defined via

$\gamma = \frac{1}{\hbar} \int_{r_0}^{r_1} \sqrt{2m \big(V(r) - E\big)}\, dr,$

and $r_0, r_1$ are the points where the energy $E$ intersects the potential $V(r)$.

Performing the integration for the Gamow model under the assumption that $r_1 \gg r_0$, one obtains the result

$\gamma = \frac{\sqrt{2mE}}{\hbar} \left(\frac{\pi}{2} r_1 - 2\sqrt{r_0 r_1} \right).$

Although the radii $r_0$ and $r_1$ are not known *a priori*, the important part is the dependence of the logarithm of the lifetime on $E^{-1/2}$, which can be confirmed experimentally. The energy of emitted alpha particles can be computed using

$E = (m_p - m_d - m_{\alpha}) c^2,$

where $m_p$ is the mass of the nucleus before decay, $m_d$ is the mass of the nucleus of the decay product, and $m_{\alpha}$ is the alpha particle mass.

$\gamma = \dfrac{1}{\hbar} \int_{r_0}^{r_1} \sqrt{2m(V-E)}\, dr$

Suppose that the charge of a nucleus and the energy $E$ of particles in the nucleus change in such a way that the equation above is shifted to $\gamma - \frac{\ln 2}{2}$, where $r_0$ and $r_1$ are the points where the energy $E$ intersects the potential $V$.

By what factor does the probability of alpha decay in a given amount of time (transmission coefficient) change?

## More Applications of Quantum Tunneling

Quantum tunneling is responsible for many physical phenomena that baffled scientists in the early $20^\text{th}$ century. One of the first was radioactivity, both via Gamow's model of alpha decay discussed above as well as via electrons tunneling into the nucleus to be captured by protons. Another wide area of applicability of quantum tunneling has been to the dynamics of electrons in materials, such as microscopy, semiconductors, and superconductors.

**Scanning Tunneling Microscopy**

A scanning tunneling microscope is an incredibly sensitive device used to map the topography of materials at the atomic level. It works by running an extremely sharp tip of only a single-atom-thick over the surface of the material, with the tip at a higher voltage than the material. This voltage allows a non-negligible **tunneling current** to flow from electrons that tunnel from the surface of the material, through the potential barrier represented by the air, to the tip of the microscope, completing a circuit. By measuring the amount of current that flows at a given distance, the microscope can resolve where the atoms are on the surface of the material.

**Tunnel Diodes**

In a tunnel diode, two p-type and n-type semiconductors are separated by a thin insulating region called the depletion region. Recall that a p-type semiconductor is one that has been doped with impurity atoms that carry one less valence electron, while an n-type semiconductor has been doped with impurities carrying one more valence electron; both allow conduction to occur more easily due to the extra electrons or "holes" provided by the dopant. In the depletion region, there are no conduction electrons; the electrons have been depleted to other regions. The main effect of a tunnel diode is that an applied voltage can make electrons from the n-type semiconductor tunnel through the depletion region, causing a unidirectional current towards the p-type semiconductor at low voltages. As voltage increases, the current drops as the depletion region widens and then increases again at high voltages to function as a normal diode. The ability of tunnel diodes to direct current at low voltages due to tunneling allows them to operate at very high AC frequencies.

**Josephson Junctions**

Some semiconducting materials are superconductors, meaning that in certain temperature ranges a current can flow indefinitely without resistive heating occurring. In Josephson junctions, two superconducting semiconductors are separated by a thin insulating barrier. In the **Josephson effect**, superconducting pairs of electrons (Cooper pairs) can tunnel through this barrier to carry the superconducting current through the junction.

## References

[1] By The original uploader was Jean-Christophe BENOIST at French Wikipedia - Transferred from fr.wikipedia to Commons., CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=653747.

[2] Image from https://upload.wikimedia.org/wikipedia/commons/1/1d/TunnelEffektKling1.png under Creative Commons licensing for reuse and modification.

[3] By Yuvalr (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commons

[4] Griffiths, David J. *Introduction to Quantum Mechanics*. Second Edition. Pearson: Upper Saddle River, NJ, 2006.

[5] Illustration by Kristian Molhave for the Opensource Handbook of Nanoscience and Nanotechnology.

**Cite as:**Quantum Tunneling.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/quantum-tunneling/