Physics Book - User contributions [en]

Linear Momentum

2025-12-03T00:47:18Z

Joshykim:

claimed by joshua kim fall 2025

==Linear Momentum==

It is a vector quantity that describes an object's motion by combining its mass and velocity. Because mass is a scalar while velocity is a vector, momentum always points in the same directions as an object's motion. Momentum is typically represented by the symbol <math>\vec{p}</math>, and its SI unit is the kilogram meter per second (kg*m/s). Momentum is sometimes referred to simply as "momentum", and the plural form may appear as either "momenta" or "momentums".

==The Main Idea==

Momentum reflects how difficult it is to stop or change the motion of an object. Increasing either the mass or the velocity of an object increases its momentum. Momentum plays a central role in understanding collision, motion in systems of particles, and many other physical processes because it is often conserved even when forces are complicated or short-lived.

===A Mathematical Model===

For a single particle, linear momentum is defined by <math>\vec{p} = m\vec{v}</math>, where <math>m</math> is the mass and <math>\vec{v}</math> is the velocity. This formula works well for everyday speeds, but for motion approaching the speed of light, the definition must be updated using relativistic momentum.

For systems of multiple particles, the total momentum is the vector sum of the momenta of all particles: <math>\vec{p}_{\text{total}} = \sum m_i\vec{v}_i</math>. This allows complex systems - such as a rolling disk with many moving pieces - to be treated as if all their mass were concentrated at a single point.

====Single Particles====
The momentum of a particle is defined as follows:

<math>\vec{p} = m\vec{v}</math>

where <math>\vec{p}</math> is the particle's linear momentum, <math>m</math> is the particle's mass, and <math>\vec{v}</math> is the particle's velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.

====Multiple Particles====

The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:

<math>\vec{p}_{system} = \sum_i \vec{p}_i</math>

Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:

<math>\vec{p}_{system} = M_{tot}\vec{v}_{COM}</math>

This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small "mass elements," all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.

====Impulse====
Impulse measures the change in momentum during a time interval. It is defined as the integral of the force over time, and reduces to <math>\vec{J} = \vec{F}\Delta t</math>. This relationship is extremely useful in analyzing collisions, sports physics, and any situation where short-duration forces cause a measurable change in motion.

====In Relation to Other Physics Topics====

When no net external force acts on a particle, its momentum remains constant (Newton's First Law). When a force does act, the rate at which the momentum changes equals the net force (Newton's Second Law). In situations where an impulse is applied, the momentum changes by an amount equal to that impulse. For systems in which the total external force is zero, momentum is conserved even if internal forces act between particles.

===Momentum in Rocket Propulsion==

Rockets operate by expelling high-speed exhaust gases backward, which produces a forward momentum change in the rocket. Even in deep space, where there is "nothing to push against", momentum conservation ensures that ejecting mass at high velocity generates thrust. This highlights why rockets require large amounts of fuel: without a constant supply of propellant, the rocket cannot continue changing its momentum.

===A Common Misconception: Newton's Cradle===

Although Newton's cradle is often use to demonstrate conservation of momentum, the device actually relies on both momentum and energy conservation to produce its familiar motion. If momentum alone were conserved, the balls would not swing back with the same amplitude. The cradle highlights how both conservation laws work together in elastic collisions.

===A Computational Model===

In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in "time steps," or iterations of a loop representing a short time interval. In each time step, the particles' momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle's momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:

This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.

https://trinket.io/glowscript/ce43925647

For more information, see [[Iterative Prediction]].

==Examples==

===1. (Simple)===
Find the momentum of a ball that has a mass of 70kg and is moving at <1,2,3> m/s.

Steps:

1. Initialize Variables

m = 70 kg

v = <1,2,3> m/s

2. Momentum Principle

p = m * v

= 70 kg * <1,2,3> m/s

= <70,140,210> kg m/s

Therfore, the momentum of the ball is <70,140,210> kg m/s.

===2. (Middling)===
A car has 20,000 N of momentum.
How would the momentum of the car change if:

a) the car slowed to half of its speed?

b) the car completely stopped?

c) the car gained its original weight in luggage?

Steps:

1. Initialize variables:

p = 20,000 N
p_n = ? (new momentum)

a. Half speed:

p = m * v = 20,000 N If velocity was halved:

v = 1/2 v

p_n = m * 1/2 v

= 1/2 * m * v

= since m * v = 20,000 N, then

p_n = 10,000 N

In other words, if the velocity was halved, then momentum would
correspondingly be halved.

b. Car completely stopped:

If the car was completely stopped, then its velocity would be 0.

Since momentum is m * v, then the new momentum would be 0 as well.

c. Gained original weight in luggage:

Double weight = 2m

p_n = 2m * v

= 2 * m * v

= 40,000 N

The car's new momentum would be doubled if the mass was doubled.

===3. (Difficult)===
You and your friends are watching NBA highlights at home and want to practice your physics. You notice at the beginning of a clip a basketball ball is rolling down the court at 23.5 m/s to the right. At the end, it is rolling at 3.8 m/s in the same direction. The commentator tells you that the change in its momentum is 17.24 kg m/s to the left. Curious at how many basketballs you can carry, you want to find the mass of the ball.

Steps:

1. Initialize Variables:

v_i = 23.5 m/s v_f = 3.8 m/s

p = -17.24 kg m/s (negative because direction is left)

2. Plug in values and solve for m:

-17.24 = m(3.8-23.5)

m = -17.24/-19.7

= 0.8756 kg (approximately)

===4. (Difficult)===
A system is comprised of two particles. One particle has a mass of 6kg and is travelling in a direction 30 degrees east of north at a speed of 8m/s. The total momentum of the system is 3kg*m/s west. The second particle has a mass of 3kg. What is the second particle's velocity? Give your answer in terms of a north-south component and an east-west component.

Steps:

1. Initialize Variables:

m_1 = 6 kg v_1 = 8 m/s 30 degrees east of north

m_2 = 3 kg p_total = 3 kg m/s west

2. Break Down Velocities into Components for Particle 1:

North Component: v_1N = v_1 * cos(30) = 6.9282 m/s

Eastward Component: v_1E = v1 * sin(30) = 4 m/s

3. Calculate Momentum Components for Particle 1:

North Momentum: p_1N = m_1 * v_1N = 6 * 6.9282 = 41.5692 kg m/s

East Momentum: p_1E = m_1 * v_1E = 6 * 4 = 24 kg m/s

4. Express Total Momentum Components:

North total momentum:

p_total,N = 0 kg m/s (no north-south component in total momentum)

Eastward total momentum (negative because west is negative east):

p_total,E = -3 kg m/s

5. Conservation of Momentum:

p_total,E = p_1E + p_2E

Particle 2 North Momentum: 0 = 41.5692 + p_2N, p_2N = -41.5692 kg m/s

Particle 2 East Momentum: -3 = 24 + p_2E, p_2E = -27 kg m/s

6. Calculate Velocity Components:

North Velocity: v_2N = p_2N / m_2 = -13.8564 m/s

East Velocity: v_2E = p_2E / m_2 = -9 m/s

Therefore, the second particle's velocity components are:

13.86 m/s to the South

9.00 m/s to the West

==Connectedness==

===Scenario: runaway vehicle===

Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.

==History==

The oldest known attempt at quantifying motion using both an object's speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).

== See also ==

*[[Mass]]
*[[Velocity]]
*[[Vectors]]
*[[Newton's Second Law: the Momentum Principle]]
*[[Impulse and Momentum]]
*[[Conservation of Momentum]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.

===External links===

<ol>
<li>https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum</li>
</ol>

==References==

[[Category: Momentum]]
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html
*https://science.jrank.org/pages/4419/Momentum.html
*https://www.famousscientists.org/john-wallis/
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html

Linear Momentum

2025-12-02T22:55:50Z

Joshykim:

claimed by joshua kim fall 2025

This page defines the linear momentum of particles and systems.

==The Main Idea==

Linear momentum is a vector quantity describing an object's motion. It is defined as the product of an object's [[Mass]] (<math>m</math>) and [[Velocity]] (<math>\vec{v}</math>). Note that mass is a scalar while velocity is a vector, so an object's linear momentum is always in the same direction as its velocity. Linear momentum is a vector quantity. Linear momentum is represented by the letter <math>\vec{p}</math> and is often referred to as simply "momentum." The most commonly used metric unit for momentum is the kilogram*meter/second. The plural of momentum is momenta or momentums.

===A Mathematical Model===

====Single Particles====
The momentum of a particle is defined as follows:

<math>\vec{p} = m\vec{v}</math>

where <math>\vec{p}</math> is the particle's linear momentum, <math>m</math> is the particle's mass, and <math>\vec{v}</math> is the particle's velocity. This formula accurately describes the momentum of particles at everyday speeds, but for particles traveling near the speed of light, the formula for [[Relativistic Momentum]] must be used for concepts such as the momentum principle to remain true.

====Multiple Particles====

The total momentum of a system of particles is defined as the vector sum of the momenta of the particles that comprise the system:

<math>\vec{p}_{system} = \sum_i \vec{p}_i</math>

Although the proof does not appear on this page, it can be shown that the total momentum of a system of particles is equal to the total mass of the system times the velocity of its [[Center of Mass]]:

<math>\vec{p}_{system} = M_{tot}\vec{v}_{COM}</math>

This formula makes it significantly easier to calculate the momentum of certain objects. For example, consider a disk rolling along the ground. The disk is comprised of an infinite number of infinitely small "mass elements," all of which have different momenta; the ones at the bottom of the disk are hardly moving while the ones at the top are moving very quickly. Without the formula above, one would have to use an integral to add the momenta of all of the mass elements to find the total momentum of the disk. However, the formula above tells us that we can simply multiply the mass of the disk by the velocity of its center of mass, which is in this case the geometric center of the disk. This reveals that the fact that the disk is rotating does not affect its momentum.

====Impulse====
Impulse is the change in momentum. It is the integral of force over a time interval. Therefore, impulse equals to the force, F, multiplied by the change in time, t. The unit for impulse is newton*second. Since impulse is the change in momentum, an alternative formula of impulse is mass, m, multiplied by change in velocity, v. Both the formulae mentioned are important to remember to solve impulse-related questions.

====In Relation to Other Physics Topics====

In the absence of a net force, the momentum of a particle stays constant over time, as stated by Newton's first law.

When a force is applied to a particle, its momentum evolves over time according to Newton's second law. For more information, see [[Newton's Second Law: the Momentum Principle]].

When an impulse is applied to a particle, its momentum changes in a specific way: the change in momentum <math>\Delta\vec{p}</math> is equal to the impulse <math>\vec{J}</math>. This is a consequence of the Momentum Principle. For more information, see [[Impulse and Momentum]].

When the net external force on a system of particles is 0, the system's momentum is conserved (that is, constant over time) even if the particles within the system interact with each other (exert internal forces on each other). For more information, see [[Conservation of Momentum]].

===A Computational Model===

In computational simulations of particles using [[Iterative Prediction]], a momentum vector variable is assigned to each particle. Such simulations usually in "time steps," or iterations of a loop representing a short time interval. In each time step, the particles' momenta are updated based on the forces acting on it. Then their velocities are calculated by dividing each particle's momentum by its mass. Finally, the velocities are used to update the positions of the particles. Below is an example of such a simulation:

This vPython simulation shows a cart (represented by a rectangle) whose motion is affected by a gust of wind applying a constant force.

https://trinket.io/glowscript/ce43925647

For more information, see [[Iterative Prediction]].

==Examples==

===1. (Simple)===
Find the momentum of a ball that has a mass of 70kg and is moving at <1,2,3> m/s.

Steps:

1. Initialize Variables

m = 70 kg

v = <1,2,3> m/s

2. Momentum Principle

p = m * v

= 70 kg * <1,2,3> m/s

= <70,140,210> kg m/s

Therfore, the momentum of the ball is <70,140,210> kg m/s.

===2. (Middling)===
A car has 20,000 N of momentum.
How would the momentum of the car change if:

a) the car slowed to half of its speed?

b) the car completely stopped?

c) the car gained its original weight in luggage?

Steps:

1. Initialize variables:

p = 20,000 N
p_n = ? (new momentum)

a. Half speed:

p = m * v = 20,000 N If velocity was halved:

v = 1/2 v

p_n = m * 1/2 v

= 1/2 * m * v

= since m * v = 20,000 N, then

p_n = 10,000 N

In other words, if the velocity was halved, then momentum would
correspondingly be halved.

b. Car completely stopped:

If the car was completely stopped, then its velocity would be 0.

Since momentum is m * v, then the new momentum would be 0 as well.

c. Gained original weight in luggage:

Double weight = 2m

p_n = 2m * v

= 2 * m * v

= 40,000 N

The car's new momentum would be doubled if the mass was doubled.

===3. (Difficult)===
You and your friends are watching NBA highlights at home and want to practice your physics. You notice at the beginning of a clip a basketball ball is rolling down the court at 23.5 m/s to the right. At the end, it is rolling at 3.8 m/s in the same direction. The commentator tells you that the change in its momentum is 17.24 kg m/s to the left. Curious at how many basketballs you can carry, you want to find the mass of the ball.

Steps:

1. Initialize Variables:

v_i = 23.5 m/s v_f = 3.8 m/s

p = -17.24 kg m/s (negative because direction is left)

2. Plug in values and solve for m:

-17.24 = m(3.8-23.5)

m = -17.24/-19.7

= 0.8756 kg (approximately)

===4. (Difficult)===
A system is comprised of two particles. One particle has a mass of 6kg and is travelling in a direction 30 degrees east of north at a speed of 8m/s. The total momentum of the system is 3kg*m/s west. The second particle has a mass of 3kg. What is the second particle's velocity? Give your answer in terms of a north-south component and an east-west component.

Steps:

1. Initialize Variables:

m_1 = 6 kg v_1 = 8 m/s 30 degrees east of north

m_2 = 3 kg p_total = 3 kg m/s west

2. Break Down Velocities into Components for Particle 1:

North Component: v_1N = v_1 * cos(30) = 6.9282 m/s

Eastward Component: v_1E = v1 * sin(30) = 4 m/s

3. Calculate Momentum Components for Particle 1:

North Momentum: p_1N = m_1 * v_1N = 6 * 6.9282 = 41.5692 kg m/s

East Momentum: p_1E = m_1 * v_1E = 6 * 4 = 24 kg m/s

4. Express Total Momentum Components:

North total momentum:

p_total,N = 0 kg m/s (no north-south component in total momentum)

Eastward total momentum (negative because west is negative east):

p_total,E = -3 kg m/s

5. Conservation of Momentum:

p_total,E = p_1E + p_2E

Particle 2 North Momentum: 0 = 41.5692 + p_2N, p_2N = -41.5692 kg m/s

Particle 2 East Momentum: -3 = 24 + p_2E, p_2E = -27 kg m/s

6. Calculate Velocity Components:

North Velocity: v_2N = p_2N / m_2 = -13.8564 m/s

East Velocity: v_2E = p_2E / m_2 = -9 m/s

Therefore, the second particle's velocity components are:

13.86 m/s to the South

9.00 m/s to the West

==Connectedness==

===Scenario: runaway vehicle===

Imagine that you are standing at the bottom of a hill when a runaway vehicle comes careening down. If it is a bicycle, it would be much easier to stop than if it were a truck moving at the same speed. One explanation for this is that the truck would have a greater mass and therefore a greater momentum. In order to be brought to rest, the truck must therefore experience a large change in momentum, which means a large impulse must be exerted on it.

==History==

The oldest known attempt at quantifying motion using both an object's speed and mass was that of René Descartes (1596–1650) [https://science.jrank.org/pages/4419/Momentum.html (source)]. John Wallis (1616 – 1703) was the first to use the phrase momentum, and to define it as the product of mass and velocity (rather than speed). He recognized that this notion of momentum is conserved in closed systems [https://www.famousscientists.org/john-wallis/ (source)], a concept confirmed by experiments performed by Christian Huygens (1629-1695) [https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html (source)]. Finally, Isaac Newton (1643-1727) wrote his famous three laws relating momentum to force in his book Principia Mathematica in 1687. These offered a theoretical explanation for conservation of momentum. These foundations were so logically sound and experimentally observable that they would go unquestioned for centuries, until Albert Einstein (1879-1955) had to adjust the formula for momentum to maintain its accuracy for objects moving at relativistic speeds (see [[relativistic momentum]]).

== See also ==

*[[Mass]]
*[[Velocity]]
*[[Vectors]]
*[[Newton's Second Law: the Momentum Principle]]
*[[Impulse and Momentum]]
*[[Conservation of Momentum]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 1). Raleigh, North Carolina: Wiley.

===External links===

<ol>
<li>https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum</li>
</ol>

==References==

[[Category: Momentum]]
*http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html
*http://study.com/academy/lesson/linear-momentum-definition-equation-and-examples.html
*https://science.jrank.org/pages/4419/Momentum.html
*https://www.famousscientists.org/john-wallis/
*https://www2.stetson.edu/~efriedma/periodictable/html/Hg.html

Quantum Tunneling through Potential Barriers

2025-12-02T21:20:22Z

Joshykim: /* References */

'''Page Claimed By: Joshua Kim Fall '25'''

'''Quantum tunneling''' is a fundamental effect in quantum mechanics in which a particle has a nonzero probability of appearing on the far side of a potential barrier, even when its classical energy is insufficient to cross that barrier. This behavior arises from the wave-like nature of matter and is described mathematically through the Schrodinger equation and related quantum principles, such as the Heisenberg uncertainty principle.

==Overview==
In classical physics, a particle with energy lower than a barrier height cannot enter the barrier at all. Quantum mechanics, however, treats particles as wavefunctions rather than solid objects. These wavefunctions can extend into and beyond regions that would be forbidden in classical mechanics, giving rise to a small--but finite--chance that the particle will emerge on the opposite side.
A typical tunneling scenario involves a finite potential barrier of Height V_0 and width L. When a particle of energy E < V_0 approaches the barrier, part of its wavefunction decays inside the barrier yet continues far enough to "leak through", creating a transmitted wave.

===Classical Expectations===
Under classical mechanics, the criterion for overcoming a barrier is simple: the particle's energy must exceed the potential energy of the barrier. If E < V_0, the particle should reflect entirely. No motion is predicted within the barrier, and transmission probability is strictly zero.

===Quantum Theory===
====Wave Behavior of Particles====
Quantum mechanics models particles with wavefunctions that satisfy the time-independent Schrodinger equation. These wavefunctions can be decomposed into incident, reflected, and transmitted components. Because waves generally do not vanish abruptly at boundaries, the wavefunction must extend into the barrier. Within this region, the wave amplitude decays exponentially.
When the decaying wave reaches the far side of the barrier, a small fraction survives and becomes a traveling wave in the output region. This survival mechanism produces tunneling.

====Heisenberg Uncertainty Principle====
A qualitative way to appreciate tunneling is through the energy-time variant of the Heisenberg uncertainty principle. Over extremely short time intervals, a particle's energy cannot be known exactly. This uncertainty permits temporary fluctuations that may allow the particle to explore regions classically inaccessible to it. While not a precise calculation tool, this principle offers intuitive support for why penetration into forbidden regions is possible.

====The Schrödinger Equation====
The [https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation 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 Quantum Well | solution for a single particle in a semi-infinite well]] when <math>V_0 > 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>-\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>0 \le x \le L</math>. Given that, we can define the potential as follows:

<math> V(x) = \begin{cases}
0 & x < 0 \\
V_0 & 0\leq x\leq L \\
0 & x> L
\end{cases} </math>

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

* '''''Region I''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} = E \psi(x) </math>

* '''''Region II''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V_0 \psi (x) = E \psi(x) </math>

* '''''Region III''''' : <math> -\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> x = 0 </math> and <math> x = L </math>. The general solution for each region,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \psi(x)_{III} = Fe^{ikx} + Ge^{-ikx} </math>

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

There is only one direction in which the wave in region III is traveling (<math> +\infty </math>). Thus the wave functions are now,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \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> x = 0 </math>
* B: reflected wave at <math> x = 0 </math>
* C: incident wave at <math> x = L </math>
* D: reflected wave at <math> 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> |F|^2</math> and the incident wave intensity <math> |A|^2 </math>. To solve for these constants, we must apply boundary conditions.

At <math> x = 0 </math>, we know region I and II must agree and create a continuous wave function.

<math> Ae^{ikx} + Be^{-ikx} = Ce^{-\alpha x } + De^{\alpha x} </math>

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

<math> ik(Ae^{ikx} - Be^{-ikx}) = \alpha(-Ce^{-\alpha x } + De^{\alpha x}) </math>

The same applies at <math> x = L </math>.

<math> Ce^{-\alpha x } + De^{\alpha x} = Fe^{ikx} </math>

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

<math> \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>
\frac{F}{A} = \frac{e^{-ikL}}{cosh(\alpha L) +i(\gamma /2)sinh(\alpha L)} </math>[https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers]

where <math> \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> T \approx 16 \frac{E}{V_0} (1-\frac{E}{V_0}) e^{-2kL} </math>

where,
<math> 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.

<iframe src="https://phet.colorado.edu/sims/cheerpj/quantum-tunneling/latest/quantum-tunneling.html?simulation=quantum-tunneling"
width="800"
height="600"
allowfullscreen>
</iframe> [https://phet.colorado.edu/en/simulations/quantum-tunneling]

==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> E = 5.0 eV </math>
* <math> V_0 = 10.0 eV</math>
* <math> L = 1.00 nm = 1.00 \times 10^{-9} m </math>

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

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

<math> 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> 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===

==Applications==
'''''Nuclear Fusion in Stars'''''

Quantum tunneling enables protons to overcome their mutual electrostatic repulsion, allowing fusion reactions to occur at stellar core temperatures.

'''''Scanning Tunneling Microscopy (STM)'''''

Tunneling current between a sharp metallic tip and a surface allows imaging of materials with atomic resolution.

'''''Semiconductor Devices'''''

Tunnel diodes and modern transistors exploit controlled tunneling currents.

==Historical Background==
The notion of barrier penetration was introduced in the late 1920s by physicist Friedrich Hund, who noticed that quantum wavefunctions could leak between adjacent molecular wells. Soon after, George Gamow applied tunneling to explain alpha decay, firmly establishing it as a central quantum phenomenon.

==See Also==
[[Heisenberg Uncertainty Principle]]

[[Wave-Particle Duality]]

[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]

[[Application of Statistics in Physics]]

===Related Topics===
* [[Heisenberg Uncertainty Principle]]
* [[ Wave-Particle Duality]]
* [[ Schrodinger Equation]]
* [[Quantum Wells and Bound States]]

===External Resources===
* [https://phet.colorado.edu/en/simulations/quantum-tunneling]
* [https://www.youtube.com/watch?v=cTodS8hkSDg]
* [https://www.youtube.com/watch?v=RF7dDt3tVmI]

==References==
* https://openstax.org/details/books/university-physics-volume-3
* http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
* https://physicstoday.scitation.org/doi/10.1063/1.1510281
* https://phet.colorado.edu/en/simulations/quantum-tunneling

[[Category: Quantum Mechanics]]

Quantum Tunneling through Potential Barriers

2025-12-02T21:19:45Z

Joshykim: /* References */

'''Page Claimed By: Joshua Kim Fall '25'''

'''Quantum tunneling''' is a fundamental effect in quantum mechanics in which a particle has a nonzero probability of appearing on the far side of a potential barrier, even when its classical energy is insufficient to cross that barrier. This behavior arises from the wave-like nature of matter and is described mathematically through the Schrodinger equation and related quantum principles, such as the Heisenberg uncertainty principle.

==Overview==
In classical physics, a particle with energy lower than a barrier height cannot enter the barrier at all. Quantum mechanics, however, treats particles as wavefunctions rather than solid objects. These wavefunctions can extend into and beyond regions that would be forbidden in classical mechanics, giving rise to a small--but finite--chance that the particle will emerge on the opposite side.
A typical tunneling scenario involves a finite potential barrier of Height V_0 and width L. When a particle of energy E < V_0 approaches the barrier, part of its wavefunction decays inside the barrier yet continues far enough to "leak through", creating a transmitted wave.

===Classical Expectations===
Under classical mechanics, the criterion for overcoming a barrier is simple: the particle's energy must exceed the potential energy of the barrier. If E < V_0, the particle should reflect entirely. No motion is predicted within the barrier, and transmission probability is strictly zero.

===Quantum Theory===
====Wave Behavior of Particles====
Quantum mechanics models particles with wavefunctions that satisfy the time-independent Schrodinger equation. These wavefunctions can be decomposed into incident, reflected, and transmitted components. Because waves generally do not vanish abruptly at boundaries, the wavefunction must extend into the barrier. Within this region, the wave amplitude decays exponentially.
When the decaying wave reaches the far side of the barrier, a small fraction survives and becomes a traveling wave in the output region. This survival mechanism produces tunneling.

====Heisenberg Uncertainty Principle====
A qualitative way to appreciate tunneling is through the energy-time variant of the Heisenberg uncertainty principle. Over extremely short time intervals, a particle's energy cannot be known exactly. This uncertainty permits temporary fluctuations that may allow the particle to explore regions classically inaccessible to it. While not a precise calculation tool, this principle offers intuitive support for why penetration into forbidden regions is possible.

====The Schrödinger Equation====
The [https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation 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 Quantum Well | solution for a single particle in a semi-infinite well]] when <math>V_0 > 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>-\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>0 \le x \le L</math>. Given that, we can define the potential as follows:

<math> V(x) = \begin{cases}
0 & x < 0 \\
V_0 & 0\leq x\leq L \\
0 & x> L
\end{cases} </math>

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

* '''''Region I''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} = E \psi(x) </math>

* '''''Region II''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V_0 \psi (x) = E \psi(x) </math>

* '''''Region III''''' : <math> -\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> x = 0 </math> and <math> x = L </math>. The general solution for each region,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \psi(x)_{III} = Fe^{ikx} + Ge^{-ikx} </math>

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

There is only one direction in which the wave in region III is traveling (<math> +\infty </math>). Thus the wave functions are now,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \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> x = 0 </math>
* B: reflected wave at <math> x = 0 </math>
* C: incident wave at <math> x = L </math>
* D: reflected wave at <math> 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> |F|^2</math> and the incident wave intensity <math> |A|^2 </math>. To solve for these constants, we must apply boundary conditions.

At <math> x = 0 </math>, we know region I and II must agree and create a continuous wave function.

<math> Ae^{ikx} + Be^{-ikx} = Ce^{-\alpha x } + De^{\alpha x} </math>

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

<math> ik(Ae^{ikx} - Be^{-ikx}) = \alpha(-Ce^{-\alpha x } + De^{\alpha x}) </math>

The same applies at <math> x = L </math>.

<math> Ce^{-\alpha x } + De^{\alpha x} = Fe^{ikx} </math>

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

<math> \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>
\frac{F}{A} = \frac{e^{-ikL}}{cosh(\alpha L) +i(\gamma /2)sinh(\alpha L)} </math>[https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers]

where <math> \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> T \approx 16 \frac{E}{V_0} (1-\frac{E}{V_0}) e^{-2kL} </math>

where,
<math> 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.

<iframe src="https://phet.colorado.edu/sims/cheerpj/quantum-tunneling/latest/quantum-tunneling.html?simulation=quantum-tunneling"
width="800"
height="600"
allowfullscreen>
</iframe> [https://phet.colorado.edu/en/simulations/quantum-tunneling]

==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> E = 5.0 eV </math>
* <math> V_0 = 10.0 eV</math>
* <math> L = 1.00 nm = 1.00 \times 10^{-9} m </math>

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

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

<math> 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> 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===

==Applications==
'''''Nuclear Fusion in Stars'''''

Quantum tunneling enables protons to overcome their mutual electrostatic repulsion, allowing fusion reactions to occur at stellar core temperatures.

'''''Scanning Tunneling Microscopy (STM)'''''

Tunneling current between a sharp metallic tip and a surface allows imaging of materials with atomic resolution.

'''''Semiconductor Devices'''''

Tunnel diodes and modern transistors exploit controlled tunneling currents.

==Historical Background==
The notion of barrier penetration was introduced in the late 1920s by physicist Friedrich Hund, who noticed that quantum wavefunctions could leak between adjacent molecular wells. Soon after, George Gamow applied tunneling to explain alpha decay, firmly establishing it as a central quantum phenomenon.

==See Also==
[[Heisenberg Uncertainty Principle]]

[[Wave-Particle Duality]]

[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]

[[Application of Statistics in Physics]]

===Related Topics===
* [[Heisenberg Uncertainty Principle]]
* [[ Wave-Particle Duality]]
* [[ Schrodinger Equation]]
* [[Quantum Wells and Bound States]]

===References===
* [https://openstax.org/details/books/university-physics-volume-3]
* [http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html]
* [https://physicstoday.scitation.org/doi/10.1063/1.1510281]

==References==
* https://openstax.org/details/books/university-physics-volume-3
* http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
* https://physicstoday.scitation.org/doi/10.1063/1.1510281
* https://phet.colorado.edu/en/simulations/quantum-tunneling

[[Category: Quantum Mechanics]]

Quantum Tunneling through Potential Barriers

2025-12-02T21:19:10Z

Joshykim: /* External Links */

'''Page Claimed By: Joshua Kim Fall '25'''

'''Quantum tunneling''' is a fundamental effect in quantum mechanics in which a particle has a nonzero probability of appearing on the far side of a potential barrier, even when its classical energy is insufficient to cross that barrier. This behavior arises from the wave-like nature of matter and is described mathematically through the Schrodinger equation and related quantum principles, such as the Heisenberg uncertainty principle.

==Overview==
In classical physics, a particle with energy lower than a barrier height cannot enter the barrier at all. Quantum mechanics, however, treats particles as wavefunctions rather than solid objects. These wavefunctions can extend into and beyond regions that would be forbidden in classical mechanics, giving rise to a small--but finite--chance that the particle will emerge on the opposite side.
A typical tunneling scenario involves a finite potential barrier of Height V_0 and width L. When a particle of energy E < V_0 approaches the barrier, part of its wavefunction decays inside the barrier yet continues far enough to "leak through", creating a transmitted wave.

===Classical Expectations===
Under classical mechanics, the criterion for overcoming a barrier is simple: the particle's energy must exceed the potential energy of the barrier. If E < V_0, the particle should reflect entirely. No motion is predicted within the barrier, and transmission probability is strictly zero.

===Quantum Theory===
====Wave Behavior of Particles====
Quantum mechanics models particles with wavefunctions that satisfy the time-independent Schrodinger equation. These wavefunctions can be decomposed into incident, reflected, and transmitted components. Because waves generally do not vanish abruptly at boundaries, the wavefunction must extend into the barrier. Within this region, the wave amplitude decays exponentially.
When the decaying wave reaches the far side of the barrier, a small fraction survives and becomes a traveling wave in the output region. This survival mechanism produces tunneling.

====Heisenberg Uncertainty Principle====
A qualitative way to appreciate tunneling is through the energy-time variant of the Heisenberg uncertainty principle. Over extremely short time intervals, a particle's energy cannot be known exactly. This uncertainty permits temporary fluctuations that may allow the particle to explore regions classically inaccessible to it. While not a precise calculation tool, this principle offers intuitive support for why penetration into forbidden regions is possible.

====The Schrödinger Equation====
The [https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation 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 Quantum Well | solution for a single particle in a semi-infinite well]] when <math>V_0 > 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>-\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>0 \le x \le L</math>. Given that, we can define the potential as follows:

<math> V(x) = \begin{cases}
0 & x < 0 \\
V_0 & 0\leq x\leq L \\
0 & x> L
\end{cases} </math>

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

* '''''Region I''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} = E \psi(x) </math>

* '''''Region II''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V_0 \psi (x) = E \psi(x) </math>

* '''''Region III''''' : <math> -\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> x = 0 </math> and <math> x = L </math>. The general solution for each region,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \psi(x)_{III} = Fe^{ikx} + Ge^{-ikx} </math>

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

There is only one direction in which the wave in region III is traveling (<math> +\infty </math>). Thus the wave functions are now,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \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> x = 0 </math>
* B: reflected wave at <math> x = 0 </math>
* C: incident wave at <math> x = L </math>
* D: reflected wave at <math> 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> |F|^2</math> and the incident wave intensity <math> |A|^2 </math>. To solve for these constants, we must apply boundary conditions.

At <math> x = 0 </math>, we know region I and II must agree and create a continuous wave function.

<math> Ae^{ikx} + Be^{-ikx} = Ce^{-\alpha x } + De^{\alpha x} </math>

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

<math> ik(Ae^{ikx} - Be^{-ikx}) = \alpha(-Ce^{-\alpha x } + De^{\alpha x}) </math>

The same applies at <math> x = L </math>.

<math> Ce^{-\alpha x } + De^{\alpha x} = Fe^{ikx} </math>

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

<math> \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>
\frac{F}{A} = \frac{e^{-ikL}}{cosh(\alpha L) +i(\gamma /2)sinh(\alpha L)} </math>[https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers]

where <math> \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> T \approx 16 \frac{E}{V_0} (1-\frac{E}{V_0}) e^{-2kL} </math>

where,
<math> 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.

<iframe src="https://phet.colorado.edu/sims/cheerpj/quantum-tunneling/latest/quantum-tunneling.html?simulation=quantum-tunneling"
width="800"
height="600"
allowfullscreen>
</iframe> [https://phet.colorado.edu/en/simulations/quantum-tunneling]

==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> E = 5.0 eV </math>
* <math> V_0 = 10.0 eV</math>
* <math> L = 1.00 nm = 1.00 \times 10^{-9} m </math>

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

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

<math> 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> 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===

==Applications==
'''''Nuclear Fusion in Stars'''''

Quantum tunneling enables protons to overcome their mutual electrostatic repulsion, allowing fusion reactions to occur at stellar core temperatures.

'''''Scanning Tunneling Microscopy (STM)'''''

Tunneling current between a sharp metallic tip and a surface allows imaging of materials with atomic resolution.

'''''Semiconductor Devices'''''

Tunnel diodes and modern transistors exploit controlled tunneling currents.

==Historical Background==
The notion of barrier penetration was introduced in the late 1920s by physicist Friedrich Hund, who noticed that quantum wavefunctions could leak between adjacent molecular wells. Soon after, George Gamow applied tunneling to explain alpha decay, firmly establishing it as a central quantum phenomenon.

==See Also==
[[Heisenberg Uncertainty Principle]]

[[Wave-Particle Duality]]

[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]

[[Application of Statistics in Physics]]

===Related Topics===
* [[Heisenberg Uncertainty Principle]]
* [[ Wave-Particle Duality]]
* [[ Schrodinger Equation]]
* [[Quantum Wells and Bound States]]

===References===
* [https://openstax.org/details/books/university-physics-volume-3]
* [http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html]
* [https://physicstoday.scitation.org/doi/10.1063/1.1510281]

==References==
* https://slidetodoc.com/qm-review-and-shm-in-qm-review-and/
* https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers
* http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html
* https://phet.colorado.edu/en/simulations/quantum-tunneling
* https://physicstoday.scitation.org/doi/10.1063/1.1510281#:~:text=Friedrich%20Hund%20(1896%E2%80%931997),series%20of%20papers%20in%201927.

[[Category: Quantum Mechanics]]

Quantum Tunneling through Potential Barriers

2025-12-02T21:18:12Z

Joshykim: /* Further Reading */

'''Page Claimed By: Joshua Kim Fall '25'''

'''Quantum tunneling''' is a fundamental effect in quantum mechanics in which a particle has a nonzero probability of appearing on the far side of a potential barrier, even when its classical energy is insufficient to cross that barrier. This behavior arises from the wave-like nature of matter and is described mathematically through the Schrodinger equation and related quantum principles, such as the Heisenberg uncertainty principle.

==Overview==
In classical physics, a particle with energy lower than a barrier height cannot enter the barrier at all. Quantum mechanics, however, treats particles as wavefunctions rather than solid objects. These wavefunctions can extend into and beyond regions that would be forbidden in classical mechanics, giving rise to a small--but finite--chance that the particle will emerge on the opposite side.
A typical tunneling scenario involves a finite potential barrier of Height V_0 and width L. When a particle of energy E < V_0 approaches the barrier, part of its wavefunction decays inside the barrier yet continues far enough to "leak through", creating a transmitted wave.

===Classical Expectations===
Under classical mechanics, the criterion for overcoming a barrier is simple: the particle's energy must exceed the potential energy of the barrier. If E < V_0, the particle should reflect entirely. No motion is predicted within the barrier, and transmission probability is strictly zero.

===Quantum Theory===
====Wave Behavior of Particles====
Quantum mechanics models particles with wavefunctions that satisfy the time-independent Schrodinger equation. These wavefunctions can be decomposed into incident, reflected, and transmitted components. Because waves generally do not vanish abruptly at boundaries, the wavefunction must extend into the barrier. Within this region, the wave amplitude decays exponentially.
When the decaying wave reaches the far side of the barrier, a small fraction survives and becomes a traveling wave in the output region. This survival mechanism produces tunneling.

====Heisenberg Uncertainty Principle====
A qualitative way to appreciate tunneling is through the energy-time variant of the Heisenberg uncertainty principle. Over extremely short time intervals, a particle's energy cannot be known exactly. This uncertainty permits temporary fluctuations that may allow the particle to explore regions classically inaccessible to it. While not a precise calculation tool, this principle offers intuitive support for why penetration into forbidden regions is possible.

====The Schrödinger Equation====
The [https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation 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 Quantum Well | solution for a single particle in a semi-infinite well]] when <math>V_0 > 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>-\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>0 \le x \le L</math>. Given that, we can define the potential as follows:

<math> V(x) = \begin{cases}
0 & x < 0 \\
V_0 & 0\leq x\leq L \\
0 & x> L
\end{cases} </math>

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

* '''''Region I''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} = E \psi(x) </math>

* '''''Region II''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V_0 \psi (x) = E \psi(x) </math>

* '''''Region III''''' : <math> -\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> x = 0 </math> and <math> x = L </math>. The general solution for each region,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \psi(x)_{III} = Fe^{ikx} + Ge^{-ikx} </math>

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

There is only one direction in which the wave in region III is traveling (<math> +\infty </math>). Thus the wave functions are now,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \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> x = 0 </math>
* B: reflected wave at <math> x = 0 </math>
* C: incident wave at <math> x = L </math>
* D: reflected wave at <math> 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> |F|^2</math> and the incident wave intensity <math> |A|^2 </math>. To solve for these constants, we must apply boundary conditions.

At <math> x = 0 </math>, we know region I and II must agree and create a continuous wave function.

<math> Ae^{ikx} + Be^{-ikx} = Ce^{-\alpha x } + De^{\alpha x} </math>

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

<math> ik(Ae^{ikx} - Be^{-ikx}) = \alpha(-Ce^{-\alpha x } + De^{\alpha x}) </math>

The same applies at <math> x = L </math>.

<math> Ce^{-\alpha x } + De^{\alpha x} = Fe^{ikx} </math>

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

<math> \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>
\frac{F}{A} = \frac{e^{-ikL}}{cosh(\alpha L) +i(\gamma /2)sinh(\alpha L)} </math>[https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers]

where <math> \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> T \approx 16 \frac{E}{V_0} (1-\frac{E}{V_0}) e^{-2kL} </math>

where,
<math> 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.

<iframe src="https://phet.colorado.edu/sims/cheerpj/quantum-tunneling/latest/quantum-tunneling.html?simulation=quantum-tunneling"
width="800"
height="600"
allowfullscreen>
</iframe> [https://phet.colorado.edu/en/simulations/quantum-tunneling]

==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> E = 5.0 eV </math>
* <math> V_0 = 10.0 eV</math>
* <math> L = 1.00 nm = 1.00 \times 10^{-9} m </math>

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

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

<math> 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> 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===

==Applications==
'''''Nuclear Fusion in Stars'''''

Quantum tunneling enables protons to overcome their mutual electrostatic repulsion, allowing fusion reactions to occur at stellar core temperatures.

'''''Scanning Tunneling Microscopy (STM)'''''

Tunneling current between a sharp metallic tip and a surface allows imaging of materials with atomic resolution.

'''''Semiconductor Devices'''''

Tunnel diodes and modern transistors exploit controlled tunneling currents.

==Historical Background==
The notion of barrier penetration was introduced in the late 1920s by physicist Friedrich Hund, who noticed that quantum wavefunctions could leak between adjacent molecular wells. Soon after, George Gamow applied tunneling to explain alpha decay, firmly establishing it as a central quantum phenomenon.

==See Also==
[[Heisenberg Uncertainty Principle]]

[[Wave-Particle Duality]]

[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]

[[Application of Statistics in Physics]]

===Related Topics===
* [[Heisenberg Uncertainty Principle]]
* [[ Wave-Particle Duality]]
* [[ Schrodinger Equation]]
* [[Quantum Wells and Bound States]]

===External Links===
* [https://phet.colorado.edu/en/simulations/quantum-tunneling]
* [https://www.youtube.com/watch?v=cTodS8hkSDg]
* [https://www.youtube.com/watch?v=RF7dDt3tVmI]

==References==
* https://slidetodoc.com/qm-review-and-shm-in-qm-review-and/
* https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers
* http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html
* https://phet.colorado.edu/en/simulations/quantum-tunneling
* https://physicstoday.scitation.org/doi/10.1063/1.1510281#:~:text=Friedrich%20Hund%20(1896%E2%80%931997),series%20of%20papers%20in%201927.

[[Category: Quantum Mechanics]]

Quantum Tunneling through Potential Barriers

2025-12-02T21:17:10Z

Joshykim: /* History */

'''Page Claimed By: Joshua Kim Fall '25'''

'''Quantum tunneling''' is a fundamental effect in quantum mechanics in which a particle has a nonzero probability of appearing on the far side of a potential barrier, even when its classical energy is insufficient to cross that barrier. This behavior arises from the wave-like nature of matter and is described mathematically through the Schrodinger equation and related quantum principles, such as the Heisenberg uncertainty principle.

==Overview==
In classical physics, a particle with energy lower than a barrier height cannot enter the barrier at all. Quantum mechanics, however, treats particles as wavefunctions rather than solid objects. These wavefunctions can extend into and beyond regions that would be forbidden in classical mechanics, giving rise to a small--but finite--chance that the particle will emerge on the opposite side.
A typical tunneling scenario involves a finite potential barrier of Height V_0 and width L. When a particle of energy E < V_0 approaches the barrier, part of its wavefunction decays inside the barrier yet continues far enough to "leak through", creating a transmitted wave.

===Classical Expectations===
Under classical mechanics, the criterion for overcoming a barrier is simple: the particle's energy must exceed the potential energy of the barrier. If E < V_0, the particle should reflect entirely. No motion is predicted within the barrier, and transmission probability is strictly zero.

===Quantum Theory===
====Wave Behavior of Particles====
Quantum mechanics models particles with wavefunctions that satisfy the time-independent Schrodinger equation. These wavefunctions can be decomposed into incident, reflected, and transmitted components. Because waves generally do not vanish abruptly at boundaries, the wavefunction must extend into the barrier. Within this region, the wave amplitude decays exponentially.
When the decaying wave reaches the far side of the barrier, a small fraction survives and becomes a traveling wave in the output region. This survival mechanism produces tunneling.

====Heisenberg Uncertainty Principle====
A qualitative way to appreciate tunneling is through the energy-time variant of the Heisenberg uncertainty principle. Over extremely short time intervals, a particle's energy cannot be known exactly. This uncertainty permits temporary fluctuations that may allow the particle to explore regions classically inaccessible to it. While not a precise calculation tool, this principle offers intuitive support for why penetration into forbidden regions is possible.

====The Schrödinger Equation====
The [https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation 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 Quantum Well | solution for a single particle in a semi-infinite well]] when <math>V_0 > 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>-\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>0 \le x \le L</math>. Given that, we can define the potential as follows:

<math> V(x) = \begin{cases}
0 & x < 0 \\
V_0 & 0\leq x\leq L \\
0 & x> L
\end{cases} </math>

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

* '''''Region I''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} = E \psi(x) </math>

* '''''Region II''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V_0 \psi (x) = E \psi(x) </math>

* '''''Region III''''' : <math> -\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> x = 0 </math> and <math> x = L </math>. The general solution for each region,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \psi(x)_{III} = Fe^{ikx} + Ge^{-ikx} </math>

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

There is only one direction in which the wave in region III is traveling (<math> +\infty </math>). Thus the wave functions are now,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \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> x = 0 </math>
* B: reflected wave at <math> x = 0 </math>
* C: incident wave at <math> x = L </math>
* D: reflected wave at <math> 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> |F|^2</math> and the incident wave intensity <math> |A|^2 </math>. To solve for these constants, we must apply boundary conditions.

At <math> x = 0 </math>, we know region I and II must agree and create a continuous wave function.

<math> Ae^{ikx} + Be^{-ikx} = Ce^{-\alpha x } + De^{\alpha x} </math>

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

<math> ik(Ae^{ikx} - Be^{-ikx}) = \alpha(-Ce^{-\alpha x } + De^{\alpha x}) </math>

The same applies at <math> x = L </math>.

<math> Ce^{-\alpha x } + De^{\alpha x} = Fe^{ikx} </math>

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

<math> \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>
\frac{F}{A} = \frac{e^{-ikL}}{cosh(\alpha L) +i(\gamma /2)sinh(\alpha L)} </math>[https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers]

where <math> \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> T \approx 16 \frac{E}{V_0} (1-\frac{E}{V_0}) e^{-2kL} </math>

where,
<math> 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.

<iframe src="https://phet.colorado.edu/sims/cheerpj/quantum-tunneling/latest/quantum-tunneling.html?simulation=quantum-tunneling"
width="800"
height="600"
allowfullscreen>
</iframe> [https://phet.colorado.edu/en/simulations/quantum-tunneling]

==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> E = 5.0 eV </math>
* <math> V_0 = 10.0 eV</math>
* <math> L = 1.00 nm = 1.00 \times 10^{-9} m </math>

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

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

<math> 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> 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===

==Applications==
'''''Nuclear Fusion in Stars'''''

Quantum tunneling enables protons to overcome their mutual electrostatic repulsion, allowing fusion reactions to occur at stellar core temperatures.

'''''Scanning Tunneling Microscopy (STM)'''''

Tunneling current between a sharp metallic tip and a surface allows imaging of materials with atomic resolution.

'''''Semiconductor Devices'''''

Tunnel diodes and modern transistors exploit controlled tunneling currents.

==Historical Background==
The notion of barrier penetration was introduced in the late 1920s by physicist Friedrich Hund, who noticed that quantum wavefunctions could leak between adjacent molecular wells. Soon after, George Gamow applied tunneling to explain alpha decay, firmly establishing it as a central quantum phenomenon.

==See Also==
[[Heisenberg Uncertainty Principle]]

[[Wave-Particle Duality]]

[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]

[[Application of Statistics in Physics]]

===Further Reading===
*[https://www.quantamagazine.org/quantum-tunnel-shows-particles-can-break-the-speed-of-light-20201020/]
*[http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html]
*[https://openstax.org/books/university-physics-volume-3/pages/7-7-quantum-tunneling]

===External Links===
* [https://phet.colorado.edu/en/simulations/quantum-tunneling]
* [https://www.youtube.com/watch?v=cTodS8hkSDg]
* [https://www.youtube.com/watch?v=RF7dDt3tVmI]

==References==
* https://slidetodoc.com/qm-review-and-shm-in-qm-review-and/
* https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers
* http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html
* https://phet.colorado.edu/en/simulations/quantum-tunneling
* https://physicstoday.scitation.org/doi/10.1063/1.1510281#:~:text=Friedrich%20Hund%20(1896%E2%80%931997),series%20of%20papers%20in%201927.

[[Category: Quantum Mechanics]]

Quantum Tunneling through Potential Barriers

2025-12-02T21:15:50Z

Joshykim: /* Connectedness */

'''Page Claimed By: Joshua Kim Fall '25'''

'''Quantum tunneling''' is a fundamental effect in quantum mechanics in which a particle has a nonzero probability of appearing on the far side of a potential barrier, even when its classical energy is insufficient to cross that barrier. This behavior arises from the wave-like nature of matter and is described mathematically through the Schrodinger equation and related quantum principles, such as the Heisenberg uncertainty principle.

==Overview==
In classical physics, a particle with energy lower than a barrier height cannot enter the barrier at all. Quantum mechanics, however, treats particles as wavefunctions rather than solid objects. These wavefunctions can extend into and beyond regions that would be forbidden in classical mechanics, giving rise to a small--but finite--chance that the particle will emerge on the opposite side.
A typical tunneling scenario involves a finite potential barrier of Height V_0 and width L. When a particle of energy E < V_0 approaches the barrier, part of its wavefunction decays inside the barrier yet continues far enough to "leak through", creating a transmitted wave.

===Classical Expectations===
Under classical mechanics, the criterion for overcoming a barrier is simple: the particle's energy must exceed the potential energy of the barrier. If E < V_0, the particle should reflect entirely. No motion is predicted within the barrier, and transmission probability is strictly zero.

===Quantum Theory===
====Wave Behavior of Particles====
Quantum mechanics models particles with wavefunctions that satisfy the time-independent Schrodinger equation. These wavefunctions can be decomposed into incident, reflected, and transmitted components. Because waves generally do not vanish abruptly at boundaries, the wavefunction must extend into the barrier. Within this region, the wave amplitude decays exponentially.
When the decaying wave reaches the far side of the barrier, a small fraction survives and becomes a traveling wave in the output region. This survival mechanism produces tunneling.

====Heisenberg Uncertainty Principle====
A qualitative way to appreciate tunneling is through the energy-time variant of the Heisenberg uncertainty principle. Over extremely short time intervals, a particle's energy cannot be known exactly. This uncertainty permits temporary fluctuations that may allow the particle to explore regions classically inaccessible to it. While not a precise calculation tool, this principle offers intuitive support for why penetration into forbidden regions is possible.

====The Schrödinger Equation====
The [https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation 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 Quantum Well | solution for a single particle in a semi-infinite well]] when <math>V_0 > 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>-\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>0 \le x \le L</math>. Given that, we can define the potential as follows:

<math> V(x) = \begin{cases}
0 & x < 0 \\
V_0 & 0\leq x\leq L \\
0 & x> L
\end{cases} </math>

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

* '''''Region I''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} = E \psi(x) </math>

* '''''Region II''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V_0 \psi (x) = E \psi(x) </math>

* '''''Region III''''' : <math> -\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> x = 0 </math> and <math> x = L </math>. The general solution for each region,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \psi(x)_{III} = Fe^{ikx} + Ge^{-ikx} </math>

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

There is only one direction in which the wave in region III is traveling (<math> +\infty </math>). Thus the wave functions are now,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \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> x = 0 </math>
* B: reflected wave at <math> x = 0 </math>
* C: incident wave at <math> x = L </math>
* D: reflected wave at <math> 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> |F|^2</math> and the incident wave intensity <math> |A|^2 </math>. To solve for these constants, we must apply boundary conditions.

At <math> x = 0 </math>, we know region I and II must agree and create a continuous wave function.

<math> Ae^{ikx} + Be^{-ikx} = Ce^{-\alpha x } + De^{\alpha x} </math>

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

<math> ik(Ae^{ikx} - Be^{-ikx}) = \alpha(-Ce^{-\alpha x } + De^{\alpha x}) </math>

The same applies at <math> x = L </math>.

<math> Ce^{-\alpha x } + De^{\alpha x} = Fe^{ikx} </math>

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

<math> \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>
\frac{F}{A} = \frac{e^{-ikL}}{cosh(\alpha L) +i(\gamma /2)sinh(\alpha L)} </math>[https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers]

where <math> \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> T \approx 16 \frac{E}{V_0} (1-\frac{E}{V_0}) e^{-2kL} </math>

where,
<math> 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.

<iframe src="https://phet.colorado.edu/sims/cheerpj/quantum-tunneling/latest/quantum-tunneling.html?simulation=quantum-tunneling"
width="800"
height="600"
allowfullscreen>
</iframe> [https://phet.colorado.edu/en/simulations/quantum-tunneling]

==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> E = 5.0 eV </math>
* <math> V_0 = 10.0 eV</math>
* <math> L = 1.00 nm = 1.00 \times 10^{-9} m </math>

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

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

<math> 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> 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===

==Applications==
'''''Nuclear Fusion in Stars'''''

Quantum tunneling enables protons to overcome their mutual electrostatic repulsion, allowing fusion reactions to occur at stellar core temperatures.

'''''Scanning Tunneling Microscopy (STM)'''''

Tunneling current between a sharp metallic tip and a surface allows imaging of materials with atomic resolution.

'''''Semiconductor Devices'''''

Tunnel diodes and modern transistors exploit controlled tunneling currents.

==History==
In his discussion of molecular spectra theory in 1927, Friedrich was the first to utilize the quantum barrier penetration. When he submitted his papers on this theory in 1926, he was supported by household names Niels Bohr and Werner Heisenberg [https://physicstoday.scitation.org/doi/10.1063/1.1510281#:~:text=Friedrich%20Hund%20(1896%E2%80%931997),series%20of%20papers%20in%201927].
111

==See Also==
[[Heisenberg Uncertainty Principle]]

[[Wave-Particle Duality]]

[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]

[[Application of Statistics in Physics]]

===Further Reading===
*[https://www.quantamagazine.org/quantum-tunnel-shows-particles-can-break-the-speed-of-light-20201020/]
*[http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html]
*[https://openstax.org/books/university-physics-volume-3/pages/7-7-quantum-tunneling]

===External Links===
* [https://phet.colorado.edu/en/simulations/quantum-tunneling]
* [https://www.youtube.com/watch?v=cTodS8hkSDg]
* [https://www.youtube.com/watch?v=RF7dDt3tVmI]

==References==
* https://slidetodoc.com/qm-review-and-shm-in-qm-review-and/
* https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers
* http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html
* https://phet.colorado.edu/en/simulations/quantum-tunneling
* https://physicstoday.scitation.org/doi/10.1063/1.1510281#:~:text=Friedrich%20Hund%20(1896%E2%80%931997),series%20of%20papers%20in%201927.

[[Category: Quantum Mechanics]]

Quantum Tunneling through Potential Barriers

2025-12-02T19:25:41Z

Joshykim:

'''Page Claimed By: Joshua Kim Fall '25'''

'''Quantum tunneling''' is a fundamental effect in quantum mechanics in which a particle has a nonzero probability of appearing on the far side of a potential barrier, even when its classical energy is insufficient to cross that barrier. This behavior arises from the wave-like nature of matter and is described mathematically through the Schrodinger equation and related quantum principles, such as the Heisenberg uncertainty principle.

==Overview==
In classical physics, a particle with energy lower than a barrier height cannot enter the barrier at all. Quantum mechanics, however, treats particles as wavefunctions rather than solid objects. These wavefunctions can extend into and beyond regions that would be forbidden in classical mechanics, giving rise to a small--but finite--chance that the particle will emerge on the opposite side.
A typical tunneling scenario involves a finite potential barrier of Height V_0 and width L. When a particle of energy E < V_0 approaches the barrier, part of its wavefunction decays inside the barrier yet continues far enough to "leak through", creating a transmitted wave.

===Classical Expectations===
Under classical mechanics, the criterion for overcoming a barrier is simple: the particle's energy must exceed the potential energy of the barrier. If E < V_0, the particle should reflect entirely. No motion is predicted within the barrier, and transmission probability is strictly zero.

===Quantum Theory===
====Wave Behavior of Particles====
Quantum mechanics models particles with wavefunctions that satisfy the time-independent Schrodinger equation. These wavefunctions can be decomposed into incident, reflected, and transmitted components. Because waves generally do not vanish abruptly at boundaries, the wavefunction must extend into the barrier. Within this region, the wave amplitude decays exponentially.
When the decaying wave reaches the far side of the barrier, a small fraction survives and becomes a traveling wave in the output region. This survival mechanism produces tunneling.

====Heisenberg Uncertainty Principle====
A qualitative way to appreciate tunneling is through the energy-time variant of the Heisenberg uncertainty principle. Over extremely short time intervals, a particle's energy cannot be known exactly. This uncertainty permits temporary fluctuations that may allow the particle to explore regions classically inaccessible to it. While not a precise calculation tool, this principle offers intuitive support for why penetration into forbidden regions is possible.

====The Schrödinger Equation====
The [https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation 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 Quantum Well | solution for a single particle in a semi-infinite well]] when <math>V_0 > 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>-\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>0 \le x \le L</math>. Given that, we can define the potential as follows:

<math> V(x) = \begin{cases}
0 & x < 0 \\
V_0 & 0\leq x\leq L \\
0 & x> L
\end{cases} </math>

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

* '''''Region I''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} = E \psi(x) </math>

* '''''Region II''''' : <math> -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V_0 \psi (x) = E \psi(x) </math>

* '''''Region III''''' : <math> -\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> x = 0 </math> and <math> x = L </math>. The general solution for each region,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \psi(x)_{III} = Fe^{ikx} + Ge^{-ikx} </math>

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

There is only one direction in which the wave in region III is traveling (<math> +\infty </math>). Thus the wave functions are now,

* '''''Region I''''': <math> \psi(x)_I = Ae^{ikx} + Be^{-ikx} </math>

* '''''Region II''''': <math> \psi(x)_{II} = Ce^{-\alpha x} + De^{\alpha x } </math>

* '''''Region III''''': <math> \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> x = 0 </math>
* B: reflected wave at <math> x = 0 </math>
* C: incident wave at <math> x = L </math>
* D: reflected wave at <math> 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> |F|^2</math> and the incident wave intensity <math> |A|^2 </math>. To solve for these constants, we must apply boundary conditions.

At <math> x = 0 </math>, we know region I and II must agree and create a continuous wave function.

<math> Ae^{ikx} + Be^{-ikx} = Ce^{-\alpha x } + De^{\alpha x} </math>

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

<math> ik(Ae^{ikx} - Be^{-ikx}) = \alpha(-Ce^{-\alpha x } + De^{\alpha x}) </math>

The same applies at <math> x = L </math>.

<math> Ce^{-\alpha x } + De^{\alpha x} = Fe^{ikx} </math>

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

<math> \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>
\frac{F}{A} = \frac{e^{-ikL}}{cosh(\alpha L) +i(\gamma /2)sinh(\alpha L)} </math>[https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers]

where <math> \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> T \approx 16 \frac{E}{V_0} (1-\frac{E}{V_0}) e^{-2kL} </math>

where,
<math> 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.

<iframe src="https://phet.colorado.edu/sims/cheerpj/quantum-tunneling/latest/quantum-tunneling.html?simulation=quantum-tunneling"
width="800"
height="600"
allowfullscreen>
</iframe> [https://phet.colorado.edu/en/simulations/quantum-tunneling]

==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> E = 5.0 eV </math>
* <math> V_0 = 10.0 eV</math>
* <math> L = 1.00 nm = 1.00 \times 10^{-9} m </math>

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

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

<math> 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> 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===

==Connectedness==
'''''How is this topic connected to something that you are interested in?'''''

This topic is connected to quantum mechanics, a topic that I have been interested in for a while. It amazes me how a particle can tunnel through a higher potential barrier and be observed on the other side when we currently have no means of observing it in the barrier.

'''''How is it connected to your major?'''''

I am a physics major...

'''''Is there an interesting industrial application?'''''

Although not industrial, quantum tunneling can be attributed to the reason why stars shine. Also, the electron microscope uses quantum tunneling to create a detailed model of nanoscopic objects without coming in contact with them.

==History==
In his discussion of molecular spectra theory in 1927, Friedrich was the first to utilize the quantum barrier penetration. When he submitted his papers on this theory in 1926, he was supported by household names Niels Bohr and Werner Heisenberg [https://physicstoday.scitation.org/doi/10.1063/1.1510281#:~:text=Friedrich%20Hund%20(1896%E2%80%931997),series%20of%20papers%20in%201927].
111

==See Also==
[[Heisenberg Uncertainty Principle]]

[[Wave-Particle Duality]]

[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]

[[Application of Statistics in Physics]]

===Further Reading===
*[https://www.quantamagazine.org/quantum-tunnel-shows-particles-can-break-the-speed-of-light-20201020/]
*[http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html]
*[https://openstax.org/books/university-physics-volume-3/pages/7-7-quantum-tunneling]

===External Links===
* [https://phet.colorado.edu/en/simulations/quantum-tunneling]
* [https://www.youtube.com/watch?v=cTodS8hkSDg]
* [https://www.youtube.com/watch?v=RF7dDt3tVmI]

==References==
* https://slidetodoc.com/qm-review-and-shm-in-qm-review-and/
* https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers
* http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html
* https://phet.colorado.edu/en/simulations/quantum-tunneling
* https://physicstoday.scitation.org/doi/10.1063/1.1510281#:~:text=Friedrich%20Hund%20(1896%E2%80%931997),series%20of%20papers%20in%201927.

[[Category: Quantum Mechanics]]