Particle In A Box (Physics): Equation, Derivation & Examples

The difference between classical mechanics and quantum mechanics is huge. While in classical mechanics particles and objects have clearly-defined positions, in quantum mechanics (prior to a measurement) a particle can only be said to have a range of possible positions, which are described in terms of probabilities by the wave function.

The Schrodinger equation defines the wave function of quantum mechanical systems, and learning how to use and interpret it is an important part of any course in quantum mechanics. One of the simplest examples of a solution to this equation is for a particle in a box.

In quantum mechanics, a particle is represented by a ​wave function​. This is usually denoted by the Greek letter psi (​Ψ​) and it depends on both position and time, and it contains everything that can be known about the particle.

The modulus of this function squared tells you the probability that the particle will be found at position ​x​ at time ​t​, provided the function is "normalized." This just means adjusted so that it is certain to be found at ​some​ position ​x​ at the time ​t​ when the results at every location are summed over, i.e. the normalization condition says that:

\(\int_{-\infty}^\infty
\vertΨ\vert^2 = 1\)

You can use the wave function to calculate the expectation value for the position of a particle at time ​t​, where the expectation value just means the average value you'd get for ​x​ if you repeated the measurement a large number of times. Of course, this doesn't mean it will be the result you'd get for any given measurement – that is ​effectively​ random, although some locations are usually substantially more likely than others.

There are many other quantities you can calculate expectation values for, such as momentum and energy values, as well as many other "observables."

The Schrodinger equation is a differential equation that is used to find the value for the wave function and the eigenstates for the energy of the particle. The equation can be derived from the conservation of energy and the expressions for the kinetic and potential energy of a particle. The simplest way to write it is:

\(H(Ψ) =iℏ\frac{\partialΨ}{\partial t}\)

But here ​H​ represents the ​Hamiltonian operator​, which in itself is a fairly long expression:

\(H = \frac{−ℏ}{2m} \frac{\partial^2}{\partial x^2} + V(x)\)

Here, ​m​ is the mass, ℏ is Planck's constant divided by 2π, and ​V​ (​x​) is a general function for the potential energy of the system. The Hamiltonian has two distinct parts – the first term is the kinetic energy of the system and the second term is the potential energy.

Every observable value in quantum mechanics is associated with an operator, and in the time-independent version of the Schrodinger equation, the Hamiltonian is the energy operator. However, in the time-dependent version shown above, the Hamiltonian generates the time evolution of the wave function too.

Combining all of the information contained in the equation, you can describe the evolution of the particle in space and time and predict the possible energy values for it too.

The time-dependent part of the equation can be removed – to describe a situation that doesn't notably evolve with time – by separating the wave function into space and time parts: ​Ψ​(​x​, ​t​) = ​Ψ​(​x​) ​f​(​t​). The time-dependent parts can then be cancelled out of the equation, which leaves the time-independent version of the Schrodinger equation:

\(H Ψ(x) = E (Ψ (x))\)

​E​ is the energy of the system. This has the exact form of an eigenvalue equation, with ​Ψ​(​x​) being the eigenfunction, and ​E​ being the eigenvalue, which is why the time-independent equation is often called the eigenvalue equation for the energy of a quantum mechanical system. The time function is simply given by:

\(f(t) = e^{-iEt/ℏ}\)

The time-independent equation is useful because it simplifies the calculations for many situations where time evolution isn't particularly crucial. This is the most useful form for "particle in a box" problems and even for determining the energy levels for electrons around an atom.

One of the simplest solutions to the time-independent Schrodinger equation is for a particle in an infinitely deep square well (i.e. an infinite potential well), or a one-dimensional box of base length ​L​. Of course, these are theoretical idealizations, but it gives a basic idea of how you solve the Schrodinger equation without accounting for many of the complications that exist in nature.

With the potential energy set to 0 outside the well where probability density is also 0, the Schrodinger equation for this situation becomes:

\(\frac{−ℏ^2}{2m} \frac{d^2Ψ(x)}{dx^2} = E Ψ(x)\)

And the general solution for an equation of this form is:

\(Ψ(x) = A \sin (kx) + B \cos (kx)\)

However, looking at the boundary conditions can help narrow this down. For ​x​ = 0 and ​x​ = L, i.e. the sides of the box or the walls of the well, the wave function has to go to zero. The cosine function has a value of 1 when the argument is 0, so for the boundary conditions to be satisfied, the constant ​B​ must equal zero. This leaves:

\(Ψ(x) = A \sin (kx)\)

You can also use the boundary conditions to set a value for ​k​. Since the sin function goes to zero at values ​n​π, where quantum number ​n​ = 0, 1, 2, 3... and so on, this means when ​x​ = ​L​, the equation will only work if ​k​ = ​n​π / ​L​. Finally, you can use the fact that the wave function has to be normalized to find the value of ​A​ (integrate across all possible ​x​ values, i.e. from 0 to ​L​, and then set the result equal to 1 and re-arrange), to arrive at the final expression:

\(Ψ(x) = \sqrt{\frac{2}{L}} \sin \bigg(\frac{nπ}{L}x\bigg)\)

Using the original equation and this result, you can then solve for ​E​, which yields:

\(E = \frac{n^2ℎ^2}{8mL^2}\)

Note that the fact that ​n​ is in this expression means that the energy levels are ​quantized​, so they can't take ​any​ value, but only a discrete set of specific energy level values depending on the mass of the particle and the length of the box.

The same problem gets a little more complicated if the potential well has a finite wall height. For example, if the potential ​V​ (​x​) takes the value ​V​₀ outside the potential well and 0 inside it, the wave function can be determined in the three main regions covered by the problem. This is a more involved process, though, so here you'll only be able to see the results rather than run through the whole process.

If the well is at ​x​ = 0 to ​x​ = ​L​ again, for the region where ​x​ < 0 the solution is:

\(Ψ(x) = Be^{kx}\)

For the region ​x​ > ​L​, it is:

\(Ψ(x) = Ae^{-kx}\)

Where

\(k = \sqrt{\frac{2me}{ℏ^2}}\)

For the region inside the well, where 0 < ​x​ < ​L​, the general solution is:

\(Ψ(x) = C \sin(wx) + D\cos(wx)\)

Where

\(w = \sqrt{\frac{-2m(E+V_0)}{ℏ^2}}\)

You can then use the boundary conditions to determine the values of the constants ​A​, ​B​, ​C​ and ​D​, noting that as well as having defined values at the walls of the well, the wave function and its first derivative has to be continuous everywhere, and the wave function has to be finite everywhere.

In other cases, such as shallow boxes, narrow boxes and many other specific situations, there are approximations and different solutions you can find.