# Particle in a box

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A classical particle, such as a ball, bouncing between two high walls. The particle moves forwards and backwards, and can not go out.
A particle in a box is free to move in a space surrounded by impenetrable barriers). When the barriers lie very close together, quantum effects are observed. For example, the particle is more likely to be found at certain positions than others and it may only occupy specific energy levels.

In quantum mechanics, the particle in a box model (also known as the infinite potential well or the infinite square well) describes a particle that can move freely between two walls. This is used to show the differences between classical and quantum systems. In classical systems, for example a ball trapped inside a heavy box, the particle can move at any speed within the box and it is no more likely to be found at one position than another. However, when the box becomes very narrow (on the scale of a few nanometers), quantum effects become important. The particle may only have certain energy levels. Likewise, it can never have zero energy, meaning that the particle can never "sit still". It is more likely to be found at certain places than at others, depending on its energy level. The particle may never be found at certain places, known as spatial nodes.

The particle in a box model provides one of the very few problems in quantum mechanics which can be solved without approximations. This means that the energy and position of the particle are related to the mass of the particle and the width of the well by simple mathematical expressions. Due to its simplicity, the model allows insight into quantum effects without the need for complicated mathematics. It is one of the first quantum mechanics problems taught in undergraduate physics courses, and it is commonly used as an approximation for more complicated quantum systems. See also: the history of quantum mechanics.

## One-dimensional solution

The barriers outside a one-dimensional box have infinitely large potential, while the interior of the box has a constant, zero potential.

The simplest form of the particle in a box model considers a one-dimensional system. Here, the particle may only move backwards and forwards along a straight line with impenetrable barriers at either end.[1] The walls of a one-dimensional box may be visualised as regions of space with an infinitely large potential. Conversely, the interior of the box has a constant, zero potential.[2] This means that no forces act upon the particle inside the box and it can move freely in that region. However, infinitely large forces repel the particle if it touches the walls of the box, preventing it from escaping. The potential in this model is given as

$V(x) = \begin{cases} 0, & 0 < x < L,\\ \infty, & \text{otherwise,} \end{cases},$

where $L$ is the length of the box and $x$ is the position of the particle within the box.

### Wavefunctions

In quantum mechanics, the wavefunction gives the most fundamental description of the behavior of a particle; the measurable properties of the particle (such as its position, momentum and energy) may all be derived from the wavefunction.[3] The wavefunction $\psi(x,t)$ can be found by solving the Schrödinger equation for the system

$\mathrm{i}\hbar\frac{\partial}{\partial t}\psi(x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t) +V(x)\psi(x,t)$

where $\hbar$ is the reduced Planck constant, $m$ is the mass of the particle, $\mathrm{i}$ is the imaginary unit and $t$ is time.

To solve for $\Psi(x,t)\;$ the first thing we do is assume that $\Psi(x,t)\;$ is the product of two functions, $\Psi(x,t)=\psi(x)\phi(t)\;$

Substituting into $\mathrm{i}\hbar\frac{\partial}{\partial t}\Psi(x,t)=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\Psi(x,t)+V(x)\Psi(x,t)$ gives us:

$\mathrm{i}\hbar\frac{\partial}{\partial t}\psi(x)\phi(t)=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\psi(x)\phi(t)+V(x)\psi(x)\phi(t)$

Since the $\phi(t)$ is a constant relative to x, and $\psi(x)$ is constant relative to t, we can pull out the 'constants' out of the partial derivatives:

$\mathrm{i}\hbar\psi(x)\frac{\partial}{\partial t}\phi(t)=-\frac{\hbar^{2}}{2m}\phi(t)\frac{\partial^{2}}{\partial x^{2}}\psi(x)+V(x)\psi(x)\phi(t)$

We can then divide the entire equation by $\psi(x)\phi(t)$

$\mathrm{i}\hbar\frac{\frac{\partial}{\partial t}\phi(t)}{\phi(t)}=-\frac{\hbar^{2}}{2m}\frac{\frac{\partial^{2}}{\partial x^{2}}\psi(x)}{\psi(x)}+V(x)$

Now, the entire equation on the left is dependent on t, and the entire equation on the right is dependent on x. However, the only way this could be true would be if both sides are a constant. We can call this constant C. So now we have two equations:

$\mathrm{i}\hbar\frac{\frac{\partial}{\partial t}\phi(t)}{\phi(t)}=C$

and

$-\frac{\hbar^{2}}{2m}\frac{\frac{\partial^{2}}{\partial x^{2}}\psi(x)}{\psi(x)}+V(x)=C$

#### Solving the Time-Dependent Portion

Now we'll focus on solving for $\phi(t)$ the time dependent part of the equation:

$\mathrm{i}\hbar\frac{\frac{\partial}{\partial t}\phi(t)}{\phi(t)}=C$

Some algebraic manipulation gives us:

$\frac{\partial\phi(t)}{\phi(t)}=\frac{-\mathrm{i}C}{\hbar}\partial t$ (because $(\frac{1}{i}=-i)$)

Solving this differential equation yields:

$\ln[\phi(t)]=\frac{-\mathrm{i}C}{\hbar}t$

We then raise e to the powers of both sides to isolate $\phi(t)$:

$\phi(t)=e^{\frac{-\mathrm{i}C}{\hbar}t}$

We can then expand $\phi(t)$ with Eulers equation, and we get:

$\phi(t)=\cos\left(\frac{C}{\hbar}t\right) - i\sin\left(\frac{C}{\hbar}t\right)$

So we see that it's a wave with a frequency $\omega=\frac{C}{\hbar}$ . Therefore, $C=\hbar\omega$

However, we can take it one step further, and say that in this case, $E=\hbar\omega$

Substituting back into $\phi(t)=e^{\frac{-\mathrm{i}C}{\hbar}t}$ gives us our final solution for the time dependent part of the function:

$\phi(t)=e^{-\mathrm{i(}\frac{E}{\hbar})t}$

#### Solving the Spatial Portion

Continuing from before, we have:

$-\frac{\hbar^{2}}{2m}\frac{\frac{\partial^{2}}{\partial x^{2}}\psi(x)}{\psi(x)}+V(x)=E$

As mentioned earlier, the definition of the 'particle in a box' scenario is that there is an infinite potential outside of the box, and no potential within the box. Therefore:

$V(x) = \begin{cases} 0, & 0 < x < L,\\ \infty, & \text{otherwise,} \end{cases},$

Substituting $V(x)=0\;$ within the box gives us:

$-\frac{\hbar^{2}}{2m}\frac{\frac{\partial^{2}}{\partial x^{2}}\psi(x)}{\psi(x)}=E$

Doing some algebra gives us:

$\frac{\partial^{2}}{\partial x^{2}}\psi(x)+\frac{2mE}{\hbar^{2}}\psi(x)=0$

The solutions for a differential equation of the form $ay''+by'+cy=0\;$ where $b^{2}-4ac<0\;$ are given to be

$y(t)=c_{1}e^{\lambda t}\cos(\mu t)+c_{2}e^{\lambda t}\sin(\mu t)\;$, where $\lambda=\frac{-b}{2a}\;$, and $\mu=\frac{\sqrt{b^{2}-4ac}}{2a*i}\;$

In our case, $b=0\;$, and $c = \frac{2m}{\hbar^{2}}E\;$, so $\lambda=0\;$ and $\mu=\sqrt{\frac{2m}{\hbar^{2}}E}\;$. For ease in writing, we'll set $k=\mu=\sqrt{\frac{2m}{\hbar^{2}}E}\;$, so our solution is $y(t)=c_{1}\cos(kt)+c_{2}\sin(kt)\;$

Therefore, we can say the form of $\psi(x)\;$ is: $\psi(x) = A\sin(kx) + B\cos(kx)\;$

The wave function needs to be continuous, and because it is 0 outside of the walls of the well, it needs to be 0 at the point $x = 0\;$. Therefore $B=0\;$, and $\psi(x) = A\sin(kx)\;$

On the same note, the wave function needs to be continuous, and as it is 0 outside of the walls of the well, it needs to be 0 at the point $x = L\;$. In order for that to be true, the sin must be 0. $\sin\;$ is 0 at integer values of $\pi\;$. Therefore $kL=n\pi\;$, and $k=\frac{n\pi}{L}\;$where $n\;$ is an integer.

Therefore, $\psi(x) = A\sin(\frac{n\pi x}{L})\;$

One of the conditions of the wave function is that the probability of finding it somewhere the enclosed well must be a total of 1 (100%). The size (or amplitude) of the wavefunction at a given position is related to the probability of finding a particle there by $P(x) = |\psi(x)|^2\;$. Therefore, $\int_{0}^{L}|\psi(x)|^{2}=1\;$, or

$\int_{0}^{L}A^{2}\sin^{2}(kx)dx=1\;$.
$A^{2}\int_{0}^{L}\frac{1-\cos(2kx)}{2}dx=1\;$
$\frac{A^{2}}{2}\int_{0}^{L}dx-\frac{A^{2}}{2}\int_{0}^{L}\cos(2kx)dx=1\;$

However, integrating the cos portion gives us:$\int_{0}^{L}\cos(2kx)=\frac{L\sin(2kL)}{2k}\;$. Substituting in $k=\frac{n\pi}{L}\;$ gives us $\frac{L\sin(2\frac{n\pi}{L}L)}{2\frac{n\pi}{L}}=\frac{L\sin(2n\pi)}{2\frac{n\pi x}{L}}=0\;$, because $\sin(2n\pi)\;$, where n is an integer is always 0.

Therefore, the only part remaining is the $\frac{A^{2}}{2}\int_{0}^{L}dx\;$, which gives us $\frac{A^{2}L}{2}=1\;$, and after some algebraic manipulation, we reach our result of:

$A=\sqrt{\frac{2}{L}}\;$

Now we have our solution for the spatial component: $\psi(x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})\;$, where n is an integer.

Initial wavefunctions for the first four states in a one-dimensional particle in a box

#### Putting them together

The first thing we did was assume that $\Psi(x,t)\;$ is the product of two functions, $\Psi(x,t) = \psi(x)\phi(t)\;$

$\phi(t) = e^{-\mathrm{i}\left(\frac{E}{\hbar}\right)t}$
$\psi(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)\;$, where n is an integer.

Therefore, our result for the time dependent Schrödinger equation is:

$\Psi(x,t) = \psi(x)\phi(t) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)e^{-\mathrm{i}\left(\frac{E}{\hbar}\right)t}\;$

### Energy levels

The energy of a particle in a box (black circles) and a free particle (grey line) both depend upon wavenumber in the same way. However, the particle in a box may only have certain, discrete energy levels.

Because $k=\sqrt{\frac{2m}{\hbar^{2}}E}\;$ and $kL=n\pi\;$, we can combine the two: $\sqrt{\frac{2m}{\hbar^{2}}E}L=n\pi\;$

After doing some algebra, we can solve for E (Energy) to be:

$E=\frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}\;$

We can get an alternate form by substituting $\hbar=\frac{h}{2\pi}\;$

$E_n = \frac{n^2 h^2}{8mL^2}$.

The energy levels increase with $n^2$, meaning that high energy levels are separated from each other by a greater amount than low energy levels are. The lowest possible energy for the particle (its zero-point energy) is found in state 1 (when $n=1$), which is given by[4]

$E_1 = \frac{\hbar^2\pi^2}{2mL^2}.$

The particle, therefore, always has a positive energy. This contrasts with classical systems, where the particle can have zero energy by resting motionless at the bottom of the box. This can be explained in terms of the uncertainty principle, which states that the product of the uncertainties in the position and momentum of a particle is limited by

$\Delta x\Delta p \geq \frac{\hbar}{2}$

It can be shown that the uncertainty in the position of the particle is proportional to the width of the box.[5] Thus, the uncertainty in momentum is roughly inversely proportional to the width of the box.[4] The kinetic energy of a particle is given by $E=p^2/(2m)$, and hence the minimum kinetic energy of the particle in a box is inversely proportional to the mass and the square of the well width, in qualitative agreement with the calculation above.[4]

### Spatial location

In classical physics, the particle can be detected anywhere in the box with equal probability. In quantum mechanics, however, the probability density for finding a particle at a given position is derived from the wavefunction as $P(x) = |\psi(x)|^2.$ For the particle in a box, the probability density for finding the particle at a given position depends upon its state, and is given by

$P_n(x) = \begin{cases} \frac{2 }{L}\sin^2\left(\frac{n\pi x}{L}\right); & 0 < x < L \\ 0; & \text{otherwise}. \end{cases}$

Thus, for any value of n greater than one, there are regions within the box for which $P(x)=0$, indicating that spatial nodes exist at which the particle cannot be found.

In quantum mechanics, the average, or expectation value of the position of a particle is given by

$\langle x \rangle = \int_{-\infty}^{\infty} \psi^*(x) x \psi(x)\,\mathrm{d}x.$

For the particle in a box, it can be shown that the average position is always $\langle x \rangle = L/2$, regardless of the state of the particle. In other words, the average position at which a particle in a box may be detected is exactly in the center of the quantum well; in agreement with a classical system.

## Higher-dimensional boxes

The wavefunction of a 2D well with nx=4 and ny=4

If a particle is trapped in a two-dimensional box, it may freely move in the $x$ and $y$-directions, between barriers separated by lengths $L_x$ and $L_y$ respectively. Using a similar approach to that of the one-dimensional box, it can be shown that the wavefunctions and energies are given respectively by

$\psi_{n_x,n_y} = \sqrt{\frac{4}{L_x L_y}} \sin \left( k_{n_x} x \right) \sin \left( k_{n_y} y\right)$,
$E_{n_x,n_y} = \frac{\hbar^2 k_{n_x,n_y}^2}{2m}$,

where the two-dimensional wavevector is given by

$\mathbf{k_{n_x,n_y}} = k_{n_x}\mathbf{\hat{x}} + k_{n_y}\mathbf{\hat{y}} = \frac{n_x \pi }{L_x} \mathbf{\hat{x}} + \frac{n_y \pi }{L_y} \mathbf{\hat{y}}$.

For a three dimensional box, the solutions are

$\psi_{n_x,n_y,n_z} = \sqrt{\frac{8}{L_x L_y L_z}} \sin \left( k_{n_x} x \right) \sin \left( k_{n_y} y \right) \sin \left( k_{n_z} z \right)$,
$E_{n_x,n_y,n_z} = \frac{\hbar^2 k_{n_x,n_y,n_z}^2}{2m}$,

where the three-dimensional wavevector is given by

$\mathbf{k_{n_x,n_y,n_z}} = k_{n_x}\mathbf{\hat{x}} + k_{n_y}\mathbf{\hat{y}} + k_{n_z}\mathbf{\hat{z}} = \frac{n_x \pi }{L_x} \mathbf{\hat{x}} + \frac{n_y \pi }{L_y} \mathbf{\hat{y}} + \frac{n_z \pi }{L_z} \mathbf{\hat{z}}$.

An interesting feature of the above solutions is that when two or more of the lengths are the same (e.g. $L_x = L_y$), there are multiple wavefunctions corresponding to the same total energy. For example the wavefunction with $n_x = 2, n_y = 1$ has the same energy as the wavefunction with $n_x = 1, n_y = 2$. This situation is called degeneracy and for the case where exactly two degenerate wavefunctions have the same energy that energy level is said to be doubly degenerate. Degeneracy results from symmetry in the system. For the above case two of the lengths are equal so the system is symmetric with respect to a 90° rotation.

## Applications

Because of its mathematical simplicity, the particle in a box model is used to find approximate solutions for more complex physical systems in which a particle is trapped in a narrow region of low electric potential between two high potential barriers. These quantum well systems are particularly important in optoelectronics, and are used in devices such as the quantum well laser, the quantum well infrared photodetector and the quantum-confined Stark effect modulator.

## References

1. Davies, p.4
2. Actually, any constant, finite potential $V_0$ can be specified within the box. This merely shifts the energies of the states by $V_0$.
3. Davies, p. 1
4. Bransden and Joachain, p. 159
5. Davies, p. 15

## Bibliography

• Bransden, B. H.; Joachain, C. J. (2000). Quantum mechanics (2nd ed.). Pearson Education, Essex. ISBN 0-582-35691-1.
• Davies, John H. (2006). The Physics of Low-Dimensional Semiconductors: An Introduction (6th reprint ed.). Cambridge University Press. ISBN 0-521-48491-X.
• Griffiths, David J. (2004). Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. ISBN 0-13-111892-7.