Particle in a box

The particle in a box model is one of the very few problems in quantum mechanics that can be solved analytically, without approximations. Due to its simplicity, the model allows insight into quantum effects without the need for complicated mathematics. It serves as a simple illustration of how energy quantizations (energy levels), which are found in more complicated quantum systems such as atoms and molecules, come about. 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.

One-dimensional solution

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. The walls of a one-dimensional box may be seen as regions of space with an infinitely large potential energy. Conversely, the interior of the box has a constant, zero potential energy. 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 energy in this model is given as $${\displaystyle V(x)=\{\begin{array}{ll}0,& x_c-\frac{L}{2}<x<x_c+\frac{L}{2},\\ \mathrm{\infty },& \text{otherwise,}\end{array},}$$ where L is the length of the box, x_c is the location of the center of the box and x is the position of the particle within the box. Simple cases include the centered box (x_c = 0) and the shifted box (x_c = L/2) (pictured).

Position wave function

In quantum mechanics, the wave function 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 wave function. The wave function ${\displaystyle \psi (x,t)}$ can be found by solving the Schrödinger equation for the system $${\displaystyle i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }t}\psi (x,t)=-\frac{{\hslash }^2}{2m}\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{x}^2}\psi (x,t)+V(x)\psi (x,t),}$$ where ${\displaystyle \hslash }$ is the reduced Planck constant, ${\displaystyle m}$ is the mass of the particle, ${\displaystyle i}$ is the imaginary unit and ${\displaystyle t}$ is time.

Inside the box, no forces act upon the particle, which means that the potential ${\displaystyle V(x)}$ vanishes and the part of the wave function inside the box oscillates through space and time with the same form as a free particle:

where ${\displaystyle A}$ and ${\displaystyle B}$ are arbitrary complex numbers. The frequency of the oscillations through space and time is given by the wave number ${\displaystyle k}$ and the angular frequency ${\displaystyle \omega }$ respectively. These are both related to the total energy of the particle by the expression

$${\displaystyle E=\hslash \omega =\frac{{\hslash }^2{k}^2}{2m},}$$ which is known as the dispersion relation for a free particle. However, since the particle is not entirely free but under the influence of a potential, the energy of the particle is $${\displaystyle E=T+V,}$$ where T is the kinetic and V the potential energy. Therefore, the energy of the particle given above is not the same thing as ${\displaystyle E=p^2/2m}$ (i.e. the momentum of the particle is not given by ${\displaystyle p=\hslash k}$). Thus the wave number k above actually describes the energy states of the particle and is not related to momentum like the "wave number" usually is. The rationale for calling k the wave number is that it enumerates the number of crests that the wave function has inside the box, and in this sense it is a wave number. This discrepancy can be seen more clearly below, when we find out that the energy spectrum of the particle is discrete (only discrete values of energy are allowed) but the momentum spectrum is continuous (momentum can vary continuously), i.e., ${\displaystyle E\ne p^2/2m}$.

The amplitude of the wave function at a given position is related to the probability of finding a particle there by ${\displaystyle P(x,t)=|\psi (x,t){|}^2}$. The wave function must therefore vanish everywhere beyond the edges of the box. Also, the amplitude of the wave function may not "jump" abruptly from one point to the next. These two conditions are only satisfied by wave functions with the form $${\displaystyle {\psi }_n(x,t)=\{\begin{array}{ll}A\sin \left(k_n\left(x-x_c+\frac{L}{2}\right)\right)e^{-i{\omega }_{n}t}& x_c-\frac{L}{2}<x<x_c+\frac{L}{2}\\ 0& \text{otherwise}\end{array},}$$ where $${\displaystyle k_n=\frac{n\pi }{L},}$$ and $${\displaystyle E_n=\hslash {\omega }_n=\frac{k_n^2{\hslash }^2}{2m}=\frac{n^2{\pi }^2{\hslash }^2}{2m{L}^2},}$$ for positive integers ${\displaystyle n\in {\mathbb{Z}}_{>0}}$. The simplest solutions, ${\displaystyle k_n=0}$ or ${\displaystyle A=0}$ both yield the trivial wave function ${\displaystyle \psi (x)=0}$, which describes a particle that does not exist anywhere in the system. Here one sees that only a discrete set of energy values and wave numbers k are allowed for the particle. Usually in quantum mechanics it is also demanded that the derivative of the wave function in addition to the wave function itself be continuous; here this demand would lead to the only solution being the constant zero function, which is not what we desire, so we give up this demand (as this system with infinite potential can be regarded as a nonphysical abstract limiting case, we can treat it as such and "bend the rules"). Note that giving up this demand means that the wave function is not a differentiable function at the boundary of the box, and thus it can be said that the wave function does not solve the Schrödinger equation at the boundary points ${\displaystyle x=0}$ and ${\displaystyle x=L}$ (but does solve it everywhere else).

Finally, the unknown constant ${\displaystyle A}$ may be found by normalizing the wave function. That is, it follows from $${\displaystyle {\int }_0^L{\left|\psi (x)\right|}^{2}dx=1,}$$ that any complex number ${\displaystyle A}$ whose absolute value is $${\displaystyle \left|A\right|=\sqrt{\frac{2}{L}},}$$ yields the same normalized state.

It is expected that the eigenvalues, i.e., the energy ${\displaystyle E_n}$ of the box should be the same regardless of its position in space, but ${\displaystyle {\psi }_n(x,t)}$ changes. Notice that ${\displaystyle x_c-\frac{L}{2}}$ represents a phase shift in the wave function. This phase shift has no effect when solving the Schrödinger equation, and therefore does not affect the eigenvalue.

If we set the origin of coordinates to the center of the box, we can rewrite the spatial part of the wave function succinctly as: $${\displaystyle {\psi }_n(x)=\{\begin{array}{l}\sqrt{\frac{2}{L}}\sin (k_{n}x)\text{for }n\text{ even}\\ \sqrt{\frac{2}{L}}\cos (k_{n}x)\text{for }n\text{ odd}.\end{array}}$$

Momentum wave function

The normalized momentum wave functions, ${\displaystyle {\varphi }_n(p,t)}$, are the Fourier transform of the normalized position wave functions, ${\displaystyle {\psi }_n(x,t)}$. With ${\displaystyle p\in \mathbb{R}}$, as the Fourier conjugate variable to the space variable, ${\displaystyle x\in \mathbb{R}}$, the momentum wave function is given by $${\displaystyle \begin{array}{cc}{\varphi }_n(p,t)={\hat{\psi }}_n(p,t)& =\frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\psi }_n(x,t)e^{-ipx/\hslash }dx,\\ & =\sqrt{\frac{L}{\pi \hslash }}\left(\frac{\hslash n\pi }{\hslash n\pi +pL}\right)\mathrm{sinc}\left(\frac{pL}{2\hslash }-\frac{n\pi }{2}\right)e^{-ip{x}_c/\hslash }{e}^{i(n-1)\frac{\pi }{2}}{e}^{-i{\omega }_{n}t},\\ & =i^{(n-1)}\sqrt{\frac{L}{\pi \hslash }}\left(\frac{\hslash n\pi }{\hslash n\pi +pL}\right)\frac{\sin \left(\frac{pL}{2\hslash }-\frac{n\pi }{2}\right)}{\left(\frac{pL}{2\hslash }-\frac{n\pi }{2}\right)}{e}^{-ip{x}_c/\hslash }{e}^{-i{\omega }_{n}t},\end{array}}$$ where the sinc function, ${\displaystyle \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(x)=\sin (x)/x}$. Note that the variable, ${\displaystyle k=p/\hslash \in \mathbb{R}}$, that is often used to describe momentum wave functions, as a function of the wave vector, ${\displaystyle k}$, is not the special k_n from the boundary conditions, which is linked to the energy eigenvalues. For the centered box, x_c = 0, at time, ${\displaystyle t=0}$, the solution is the product of a strictly real-valued function, and a unit modulus complex phase, ${\displaystyle i^{n-1}}$. The function, ${\displaystyle \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(p)=\sin (p)/p}$ , is an even function of p.

It can be seen that the allowed values of the momentum in this wave packet is continuous, and one may conclude that for the energy state described by the wave number k_n, the momentum can, when measured, attain other values beyond, ${\displaystyle p=\pm \hslash k_n}$.

Hence, it also appears that, since the energy is $E_n=\frac{{\hslash }^2{k}_n^2}{2m}$ for the nth eigenstate, the relation $E=\frac{p^2}{2m}$ does not strictly hold for the measured momentum p; the energy eigenstate ${\displaystyle {\psi }_n}$ is not a momentum eigenstate, and, in fact, not even a superposition of two momentum eigenstates, as one might be tempted to imagine from equation (1) above: peculiarly, it has no well-defined momentum before measurement!

Position and momentum probability distributions

In classic 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 wave function as ${\displaystyle P(x)=|\psi (x){|}^2.}$ 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 $${\displaystyle P_n(x,t)=\{\begin{array}{ll}\frac{2}{L}{\sin }^2\left(k_n\left(x-x_c+\frac{L}{2}\right)\right),& x_c-\frac{L}{2}<x<x_c+\frac{L}{2},\\ 0,& \text{otherwise,}\end{array}}$$Thus, for any value of n greater than one, there are points, ${\displaystyle x_j}$, within the box for which ${\displaystyle P(x_j)=0}$, indicating that spatial nodes exist at which the particle's probability amplitude is also zero, i.e. ${\displaystyle \psi (x_j)=0.}$. If relativistic wave equations are considered, however, the probability density does not go to zero at the nodes (apart from the trivial case ${\displaystyle n=0}$).

In quantum mechanics, the average, or expectation value of the position of a particle is given by $${\displaystyle ⟨x⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{P}_n(x)\mathrm{d}x.}$$For the steady state particle in a box, it can be shown that the average position is always, ${\displaystyle ⟨x⟩=x_c}$, regardless of the state of the particle. For a superposition of states, the expectation value of the position will change based on the cross term, which is proportional to ${\displaystyle \cos (\omega t)}$.

The variance in the position is a measure of the uncertainty in position of the particle: $${\displaystyle \mathrm{V}\mathrm{a}\mathrm{r}(x)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(x-⟨x⟩{)}^2{P}_n(x)dx={\int }_{x_c-\frac{L}{2}}^{x_c+\frac{L}{2}}(x-x_c{)}^2{P}_n(x)dx=\frac{L^2}{12}\left(1-\frac{6}{n^2{\pi }^2}\right).}$$The probability density for finding a particle with a given momentum is derived from the wave function as ${\displaystyle P(p)=|\varphi (p){|}^2}$. As with position, the probability density for finding the particle at a given momentum depends upon its state, and is given by $${\displaystyle P_n(p)=\frac{L}{\pi \hslash }{\left(\frac{\hslash n\pi }{\hslash n\pi +pL}\right)}^2{\text{sinc}}^2\left(\frac{pL}{2\hslash }-\frac{n\pi }{2}\right),}$$for ${\displaystyle p\in \mathbb{R}}$, and ${\displaystyle n\in \mathbb{N}}$. The expectation value for the momentum is then calculated to always be zero, because the integration is over the entire real line, and the momentum probability density is an even function of the momentum, ${\displaystyle p}$, while the momentum variable itself is odd (${\displaystyle p=p^1}$, has power 1). The variance in the momentum is calculated to be: $${\displaystyle \mathrm{V}\mathrm{a}\mathrm{r}(p)={\left(\frac{\hslash n\pi }{L}\right)}^2}$$The uncertainties in position and momentum (${\displaystyle \mathrm{\Delta }x}$ and ${\displaystyle \mathrm{\Delta }p}$) are defined as the square root of their respective variances, also known as their respective Standard deviations, so that: $${\displaystyle \mathrm{\Delta }x\mathrm{\Delta }p=\frac{\hslash }{2}\sqrt{\frac{n^2{\pi }^2}{3}-2}}$$This product increases with increasing n, having a minimum for n = 1. The value of this product for n = 1 is about equal to 0.568 ${\displaystyle \hslash }$, which obeys the Heisenberg uncertainty principle, which states that the product will be greater than or equal to ${\displaystyle \hslash /2}$.

Another measure of uncertainty in position is the information entropy of the probability distribution H_x: $${\displaystyle H_x={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{P}_n(x)\log (P_n(x)x_0)dx=\log \left(\frac{2L}{e{x}_0}\right)}$$ where x₀ is an arbitrary reference length.

Another measure of uncertainty in momentum is the information entropy of the probability distribution H_p: $${\displaystyle H_p(n)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{P}_n(p)\log (P_n(p)p_0)dp}$$ $${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{H}_p(n)=\log \left(\frac{4\pi \hslash e^{2(1-\gamma )}}{L{p}_0}\right)}$$ where γ is Euler's constant. The quantum mechanical entropic uncertainty principle states that for ${\displaystyle x_0{p}_0=\hslash }$ $${\displaystyle H_x+H_p(n)\ge \log (e\pi )\approx \mathrm{2.14473...}}$$ (nats)

For ${\displaystyle x_0{p}_0=\hslash }$, the sum of the position and momentum entropies yields: $${\displaystyle H_x+H_p(\mathrm{\infty })=\log \left(8\pi e^{1-2\gamma }\right)\approx \mathrm{3.06974...}}$$ where the unit is nat, and which satisfies the quantum entropic uncertainty principle.

Energy levels

The energies that correspond with each of the permitted wave numbers may be written as $${\displaystyle E_n=\frac{n^2{\hslash }^2{\pi }^2}{2m{L}^2}=\frac{n^2{h}^2}{8m{L}^2}.}$$ The energy levels increase with ${\displaystyle n^2}$, 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, which is given by $${\displaystyle E_1=\frac{{\hslash }^2{\pi }^2}{2m{L}^2}=\frac{h^2}{8m{L}^2}.}$$ The particle, therefore, always has a positive energy. This contrasts with classical systems, where the particle can have zero energy by resting motionlessly. 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 $${\displaystyle \mathrm{\Delta }x\mathrm{\Delta }p\ge \frac{\hslash }{2}}$$ It can be shown that the uncertainty in the position of the particle is proportional to the width of the box. Thus, the uncertainty in momentum is roughly inversely proportional to the width of the box. The kinetic energy of a particle is given by ${\displaystyle E=p^2/(2m)}$, 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.