Chapter 7: The infinite square well

In the previous chapter we considered the simplest possible scenario, that of a free particle where there are no forces acting whatsoever. Here we will begin our exploration of how quantum particles behave when there are forces acting. This is a much more natural and common situation to consider, and is the generic one we encounter in physics.

As already discussed in chapter, there are two distinct situations which we can consider. The first is where the forces confine the particle to a finite region of space. In this case, we often say that it is trapped inside a potential well. As we already stated, in quantum mechanics, it is in this situation where we end up with the famous quantisation of energy, the most notable example being the energy levels of atoms, which arise due to the fact that electrons are bound within the potential well of the nucleas (via the electromagnetic attraction).

In this chapter we will study the simplest example of a potential well, and see explicitly how the energy levels arise. This model isn’t very physical, but it is mathematically simple, and serves as an ideal starting point from which to study potential wells in quantum mechanics.

The second situation we can encounter — as we did in the previous chapter — is when the forces do not confine a particle to a finite region of space. We will return to more interesting examples — this time involving forces — in chapter x.

After introducing the infinite square well, we will show how to find the energy eigenfunctions. This will see that we need a new fact about energy eigenfuctions, relating to their continuity. Remarkably, it is in fact from this that the quantisation of energies will arise, as we will see in detail below.

We will then consider a simple example of the evolution in time of a particle inside the infinite square well, seeing that it is very different from what we saw in the previous chapter, for a free particle.

7.1The infinite square well potential¶

We are going to consider a situation where a particle is perfectly confined inside a box of width $L$. We will assume that there are walls at $x = 0$ and $x = L$, and want to model these are being perfectly elastic so that the particle definite bounces off of these walls. We can model this using the following potential energy function,

This potential well is depicted in Figure 7.1.

Why do we take the potential energy to be infinite outside of the box? Well, this means that the particle would have to have infinite energy to be outside. Thus, under the physical assumption that the particle can only have a finite energy then we see that this is a way of modelling a perfectly confined particle. This potential energy is itself unphysical, however we can take it as a good approximation to a deep but finite well. That is, one where the energy outside the well is $V_0$, which we take to be as large as we like.

7.2Energy eigenstates of the infinite square well¶

The first question we would like to ask about the infinite square well is what are the states of definite energy? We are interested in this question, as always, since it is the first step towards understanding the dynamics of a particle confined inside our box.

We are thus interested in finding the energy eigenstates of Hamiltonian, i.e. of finding solutions to (4.10). Since we have a potential energy which depends upon position, we will find the energy eigenstates by solving the TISE (4.12) instead. As we learnt in Chapter 4: Energy and the Hamiltonian, this is a completely equivalent way of finding the energy levels and energy eigenstates, and is the natural way to do so in this situation.

Because of the fact that the potential energy is specified in pieces, it means that the Hamiltonian operator $\op{H} = \op{K} + \op{V}$ will also be specified in pieces. What this means in practice is that we will need to consider the different regions separately, and then understand how to put them together afterwards. That is, we will need to consider the TISE in the region $x < 0$, the region $0 \leq x \leq L$, and the region $x > L$ separately. We will refer to these, respectively as regions I, II and III. If we denote a generic energy eigenfunction by $u_E(x)$, then we will specify this in pieces accordingly, via

7.2.1Energy eigenfunctions in Regions I and III¶

Region I, when $x < 0$, is the region outside and to the left of the box. In this region, $V(x) = \infty$ and hence also $\op{H} = \infty$. This is a strange situation! We however already implicitly stated the solution. The entire reason for considering the infinite square well is to have the particle confined within the region $0 \leq x \leq L$, inside the box. A different way of phrasing this mathematically is that the wavefunctions must necessarily always vanish outside of the well. Thus, we have

Region III, when $x > L$ is identical to region I, in that the potential energy is also infinite here. For the same reasons, we must also take any energy eigenfunction to vanish in this region, and so have

7.2.2Energy eigenfunctions in Region II¶

Region II is our primary region of interest, as this corresponds to $0\leq x\leq L$, inside the box. In this region $V(x) = 0$, and hence the Hamiltonian is simply

That is, inside the well, the potential energy vanishes and hence the particle is essentially free with no forces acting. We are thus in a very similar situation to Chapter 6: Free particle. We can use this to realise that we already what the energy eigenfunctions must be in region II: they must coincide with momentum eigenfunctions — i.e. they must be complex plane waves. We can check this directly. In particular, we can see that $e^{ikx}$ is an eigenfunction of $\op{H}$, for any value $k$,

and the corresponding eigenvalue is $E = \frac{\hbar^2 k^2}{2M}$.

However, as we also saw in Chapter 6: Free particle, we have degeneracy, and the complex plane wave with $k' = -k$ is an eigenfunction with the same energy eigenvalue

Since $e^{ikx}$ and $e^{-ikx}$ are both eigenfunctions with the same eigenvalue, as you will confirm in an exercise, any superposition of these two functions is also an eigenfunction with the same eigenvalue $E = \frac{\hbar^2 k^2}{2M}$. At this stage, we don’t yet know what is the correct superposition to consider, so we will keep things general, and say that the energy eigenfunction will be of the form

where $A$ and $B$ are arbitrary (complex) constants which we don’t yet know, and $E = \frac{\hbar^2 k^2}{2M}$.

7.2.3Continuity of energy eigenfunctions & energy quantisation¶

Where have we made it to? Well, summarising the above, we have analysed the TISE in each of the three regions separately, and so far can conclude that the energy eigenfunctions must have the form

In order to make progress, we now need to use a new fact about energy eigenfunctions that we haven’t needed to introduce until this point:

Energy eigenfunctions must be continuous

Recall that a continuous function is one which doesn’t have any jumps. That is, there cannot be two neighbouring points which differ by a finite amount. An example of a discontinuous function is shown in Figure 7.2.

We need to use this fact at the boundaries between the three regions at the walls of the box, at $x = 0$ and $x = L$. In particular, because of the fact that the eigenfunctions vanish for $x < 0$, continuity means that the wavefunction must also vanish at $x = 0$, at the wall. Substituting $x = 0$ into (7.9), we can therefore conclude that

That is, whereas before we didn’t know which superposition to take, continuity at $x = 0$ shows us that we can only take superpositions with $B = -A$, and hence in region II $u_E(x) = A(e^{ikx} - e^{-ikx})$. However, we can use Euler’s formula to express this as $u_E(x) = 2iA\sin kx$. We then see nicely, since $\sin0 = 0$, that the eigenfunction vanishes at the left-hand wall, as we require.

Turning now to the boundary between regions II and III, the right-hand wall, in order for the eigenfunction to vanish at $x = L$, we must take

One possible solution appears to be $A = 0$, however a little thought shows that this can’t be correct. If $A = 0$, then $u_E(x) = 0$, and hence we have a trivial mathematical solution, but not an interesting physical one. In particular, we need energy eigenfunctions — as valid quantum wavefunctions — to be normalised. A wavefunction that vanishes everywhere certainty isn’t normalised. Hence this isn’t what we are looking for.

The only other possibility is that $\sin kL = 0$. While this definitely isn’t true in general, there are special values of $k$ for which it will be true! In particular, we know that $\sin \theta = 0$ if $\theta = 0$, π, $2\pi$, $3\pi$, $\ldots$, that is, if $\theta = n\pi$ for $n$ an integer. Thus, we must choose $k$ such that $kL = n\pi$, i.e.

For these special values of $k$, we will have $\sin kL = 0$, so the eigenfunction vanishes at the right-hand wall.

Before going on, we can realise quickly that $n$ should be a positive integer. First, if $n = 0$, then $k = 0$ and so we have $\sin kx = \sin 0 = 0$. This is just the same as taking $A = 0$, which we already saw was mathematically correct, but not physically meaningful. We therefore are not interested in $n = 0$. For $n = -m$ (with $m$ a positive integer), we can use the fact that sine is an odd function, $\sin (-\theta) = -\sin \theta$. This means that $\sin n\pi x/L$ = $-\sin m\pi x/L$ and we have the same eigenfunction up to a minus sign. Since -1 is just a phase, and since we saw previously in Section 5.4 that an overall phase doesn’t change any of the physical properties of a quantum state, $n$ and $-n$ in fact lead to the same energy eigenfunction, hence we only need to keep one value of $n$. It is customary to keep the positive one.

Thus, putting everything together, the requirement of continuity has lead us to the conclusion that $k$ must be a positive multiple of $\pi/L$. Since the energy eigenvalues were nothing but $E = \hbar^2k^2/2M$, we therefore see that for a particle confined to an infinite square well, there are discrete energy levels

These are the only energies that the particle can have with certainty. It is simply impossible for the particle to have any other energy! The energy levels are depicted below in Figure 7.3

For the corresponding energy eigenfunctions $u_{E_n}(x)$, there is one final detail to take care of. We still appear to have one parameter left — $A$. This is however fixed by normalisation: we require $u_{E_n}(x)$ to be a normalised wavefunction. As an exercise, you show that the normalisation constant is $A = \frac{1}{2i}\sqrt{\frac{2}{L}}$ (independent of $n$). The energy eigenfunctions are thus given by

Note that here we have changed notation from $u_{E_n}(x)$ to $u_{n}(x)$. There are two reasons for doing this. The first is for brevity. The second is that it is much more common to label the energy levels by $n$, and leave it implicit that the energy is $E_n$ as given by (7.13). In fact $n$ is often referred to as a quantum number, and energy levels and energy eigenfunctions are labelled solely by the quantum number, in this case $E_n$ and $u_n(x)$.

We depict below in Figure 7.4 the first 6 energy eigenfunctions and in Figure 7.5 the associated probability density functions.

There is clearly some structure here, and also some surprises. The first surprising feature is that the minimum energy the particle can have is strictly bigger than the minimum of potential energy. That is, the lowest energy state — referred to as the ground state $\ket{E_1}$ — has energy $E_1 = \frac{\hbar^2\pi^2}{2ML^2} > 0 = V(x)$ inside the well. This is in stark constrast to classical physics, where a particle can be at rest at the minimum of a potential well, and hence have energy equal to the minimum potential energy. In quantum mechanics, in contrast, a particle has more energy than this. This additional energy is called the zero-point energy. It arises due to the HUP, and the fact that the kinetic and potential energy of a particle are incompatible properties. In particular, because the particle is confined within the well, it necessarily has a spread of momenta, and hence a spread of kinetic energy. So the confinement of the particle in fact demands that it will have kinetic energy! The ground state is in fact a balance between minimising the potential energy, and minimising the kinetic energy.

Maybe second most surprising is that when a particle has a definite energy, there are certain regions within the well where it is very unlikely to find the particle. For example, in the so-called first excited state $\ket{E_2}$ (with associated energy eigenfunction $u_2(x)$) it is very unlikely to find the particle close to the middle of the well!

7.3Evolution of a particle inside an infinite square well¶

Having obtained the energy eigenstates and energy eigenvalues, we now again have the ingredients we need in order to study the evolution of a particle inside the well. We will once again use the superposition principle to achieve this.

Consider a particle prepared initially at $t = 0$ with the wavefunction

We will start off assuming that we know the amplitudes $\alpha_n$, and come back to the question of how we determine these shortly below. (7.15) specifies the initial wavefunction as a superposition of energy eigenfunctions.

As we saw in the previous two chapters, given in this form, we can immediately write down the wavefunction $\psi(x,t)$ that it will evolve into: this is once again an application of the superposition principle: the evolution of the superpositon (of energy eigenfunctions) is just the superposition of the evolutions (of individual energy eigenfunctions). Each energy eigenfunction evolves in the usual simple way: by acquiring a phase $e^{-iE_n t/\hbar}$ in time. Thus the wavefunction at time $t$ will be

precisely as already given in (5.5) (except we updated our notation for energy eigenfunctions from $u_{E_n}(x)$ to $u_n(x)$, as explained above, and now we have infinitely many energy levels, rather than $d$).

Much like in the previously chapter, there is still a loose end that we should address, namely how to determine $\alpha_n$ if we are only given $\psi_\init(x)$?

In order to answer this, it is useful to go back to the full quantum state $\ket{\psi_\init}$. Written in terms of this, (7.15) is

and we recover (7.15) from this, by taking the scalar product on both sides with a position state $\bra{x}$, and recalling that $\inner{x}{\psi_\init} = \psi_\init(x)$ and $\inner{x}{E_n} = u_n(x)$.

If now however we take the scalar product on both sides with an energy eigenstate $\bra{E_m}$, recalling from (4.13) that the energy levels are orthogonal quantum states, then we see immediately that

On the other hand, we can use (1.4) to write

in terms of its wavefunction. Taking the scalar product on both sides of (7.19) with $\bra{E_n}$, we find

where in the second line we used the two facts that $\inner{x}{E_n} = u_n(x)$ by definition of the energy eigenfunctions, and that $\inner{E_n}{x} = \left(\inner{x}{E_n}\right)^*$. Combining (7.18) and (7.20), we finally arrive at

We can use (7.21) to write $\psi_\init(x)$ in the form assumed in (7.15). That is, if we aren’t given the amplitudes $\alpha_n$ initially, we can use (7.21) to calculate them.

Finally, just as we did for a free particle in (6.10), we can substitute (7.21) into (7.16), to obtain an expression for the wavefunction at time $t$ directly in terms of the initial wavefunction:

This is once again a formidible expression, but one which is in fact very similarly to (6.10). The difference is that whereas energy was continuous for a free particle, it is now discrete in the infinite square well, and we see this replaces one of the integrals with a summation. We also can realise that the factor $\frac{1}{2\pi \hbar} e^{ip(x-x')/\hbar}$ is really $v_p^*(x')v_p(x)$, where $v_p(x)$ is the momentum eigenfunction (2.5), which are the energy eigenfunctions of a free particle. Finally, the term $e^{-ip^2t/2M\hbar}$ is $e^{-iEt/\hbar}$, with $E = p^2/2M$ the energy eigenvalue of $v_p(x)$ for a free particle. Thus the form in both cases is in fact identical, up to the change from continuous to discrete energies. This is reassuring, as it shows that what we have done is the same, just in a different context here.

7.4Exercises¶