Chapter 8: The quantum harmonic oscillator

We now turn our attention to arguably the most important system in all of quantum mechanics — the quantum harmonic oscillator. Here we will study the quantum mechanics of a particle whose force is of the form $F(x) = - kx$. That is, a restorative force proportional to the displacement from the origin. We know that in classical mechanics systems with such forces lead to simple harmonic motion. Our goal now is to understand the quantum mechanics of such systems.

Why is this one of the most important systems to study? The first reason is that it provides an approximation to the behaviour of almost all potential wells, as we outline briefly as an aside. Second, you will see later in your degree that the quantum harmonic oscillator reappears in a huge range of situations, from describing light in quantum field theory, to describing phonons in condensed matter physics and far beyond.

The method we will use to find the energy eigenstates of the quantum harmonic oscillator is also a very deep and powerful method. It is often called the algebraic approach, and provides a fascinating way of turning one energy eigenstate into another. The alternative approach is called the analytical approach, and relies on solving the TISE directly. Most textbooks choose one approach or the other (or present them both). Here, we will focus on the former, as it is also the approach which is widely used in other contexts.

8.1The harmonic oscillator potential¶

For the classical harmonic oscillator, the relation between the spring-constant $k$, the mass $M$ and angular frequency ω is $\omega = \sqrt{k/M}$. For this reason, it is customary to write $k = M \omega^2$ and take the definition of the harmonic oscillator potential to be

This is depicted below in Figure 8.1.

8.2Generating energy eigenstates using the creation and annihilation operators¶

Our first goal, as always, is to determine the energy eigenstates of the quantum harmonic oscillator. In this chapter instead of going about this directly — by solving the TISE with the potential (8.4) (which is in fact not very straightforward to do at all) — we are going to take a different approach. While at first sight this might seem strange, we will ultimately realise that this is a very powerful technique for finding the energy eigenstates.

The approach we will take is to find a special pair of operators called the creation and annihilation operators. These remarkable operators transform one energy eigenstate into another. In fact, we will see that the creation operator generates the next energy level up, while the annihilation operator generates the next one down. In this way, once we have found just one energy eigenstate (by whatever means!) we can then use the creation and annihilation operators to find them all!

What are these remarkable creation and annihilation operators? To define them, we first need to define a characteristic length scale and a characteristic momentum scale for the harmonic oscillator. As you will show in Exercise 8.1, the physical quantities of the harmonic oscillator — the mass $M$, the angular frequency ω (which specifies the scale of the force through the spring constant), along with $\hbar$ (the key quantum constant which has dimensions) — can be uniquely combined to specify a length $\ell$ and a momentum $q = \hbar/\ell$, given by

With these defined, we can now define the annihilation and creation operators, which we denote by $\hat{a}$ and $\hat{a}^\dagger$ respectively, as

That is, the creation operator is a dimensionless combination of the position and momentum operators, with the momentum operator multipled by the imaginary number $i$ (and an overall normalisation factor of $1/\sqrt{2}$).

Why did we call the creation operator $\hat{a}^\dagger$? Because it is indeed the Hermitian conjugate operator to $\hat{a}$: Recall that since position and momentum are observable quantities the operators $\hat{X}$ and $\hat{P}$ are Hermitian, satisfying $\hat{X}^\dagger = \hat{X}$ and $\hat{P}^\dagger = \hat{P}$. Thus, when we take the Hermitian conjugate of $\hat{a}$, the only change is to the sign of the imaginary number $i$, which is precisely the definition of $\hat{a}^\dagger$.

So, how do we see that these operators do as stated? The claim is that if we act with the creation operator on an energy eigenstate $\ket{E_n}$ (of the quantum harmonic oscillator), then we will obtain (up to normalisation) the energy eigenstate $\ket{E_{n+1}}$. To be an energy eigenstate means that $\hat{H} \ket{E_n} = E_n \ket{E_n}$. Thus, the claim above, expressed mathematically, is that

since if this equation is satisfied, it means that $\left(\hat{a}^\dagger\ket{E_n}\right)$ is an energy eigenstate of $\hat{H}$, with eigenvalue $E_{n+1}$, i.e. it is the next energy level.

In order to show (8.7) holds, it is easiest to return to notion of the commutator that we introduced previously in Section 3.4. In particular, if we calculate $[\hat{H},\hat{a}^\dagger]$, and find that it is equal to the operator $\hat{b}$ (which we will come to below), then this would mean that $\hat{H}\hat{a}^\dagger - \hat{a}^\dagger \hat{H} = \hat{b}$, and so by re-arranging, $\hat{H}\hat{a}^\dagger = \hat{a}^\dagger \hat{H} + \hat{b}$. We can then substitute this into the left-hand side of (8.7) to obtain

At first sight this might not look too helpful. However, at this stage it is useful to determine $\hat{b}$. In Exercise 8.2 below, by using the explicit form of $\hat{a}^\dagger$ from (8.6), as well as the explicit form of the Hamiltonian operator, with $\hat{V} = \frac{1}{2} M \omega^2 \hat{X}^2$ from (8.4), and the two commutators

both of which you prove as problems in problem classes, you will show that

(i.e. $\hat{b} = \hbar \omega \hat{a}^\dagger$). Substituting this into (8.8), we therefore see that

This means that $(\hat{a}^\dagger \ket{E_n})$ is an energy eigenstate of energy $(E_n + \hbar \omega)$ (up to normalisation). This is in fact the energy of the next energy level $E_{n+1}$. We will not prove this here, but state it as a fact (which has been shown to be true). One crucial point to note is that the spacing between the energy levels of the harmonic oscillator is constant, and equal to $\hbar \omega$. This is a special feature of the harmonic oscillator and for other potential wells, the spacing is not constant (e.g. as we saw for the infinite square well in (7.13), the spacing grows).

In a similar fashion, in Exercise 8.3 you will similarly show the commutation relation

between the Hamiltonian and the annihilation operator. From this, you will also show that

This shows that $(\hat{a}\ket{E_n})$ is an energy eigenstate of $\hat{H}$ with energy $(E_n - \hbar \omega) = E_{n-1}$. That is, the annihilation operator transforms an energy level into the next lowest energy level (up to normalisation).

We thus have a procedure, starting from any given energy level $\ket{E_n}$, to create any other energy level, by either successively applying the creation operator (to ‘climb up’ the energy levels) or by applying $\hat{a}$ (to ‘climb down’ them). All we need is a way to find one eigenstate, and then we have everything we need. We will now see how we can obtain the ground state of the oscillator.

8.3The ground state of the harmonic oscillator¶

You might have noticed something strange in the above — we appear to be able to apply the annihilation operator as many times as we like, and continually create new energy levels of progressively lower energy. However, we know that at some point we must hit the state of lowest energy — the ground state. We seem to thus have a problem with this approach.

A little thought shows that there is a way out of this problem: we pointed out a couple of times above that $\hat{a}\ket{E_n}$ isn’t neccessarily normalised, and in fact we will see below that it isn’t, and this is the key to escaping our problem. In particular, let us imagine that we have indeed found the ground state, which we will denote by $\ket{E_0}$. According to the above, $\hat{a}\ket{E_0}$ will be proportional to a state of energy $E_0 - \hbar \omega$. However, if the proportionality constant vanishes, then our sequence terminates. That is, if $\hat{a}\ket{E_0} = 0$, then we don’t obtain a new state, and hence we can no longer climb down!

Remarkably, this insight also allows us to find the ground state $\ket{E_0}$. More precisely, we can use it to find the ground state eigenfunction $u_0(x)$. How do we do this? Well, we know that eigenfunctions satisfy the same equation as eigenstates, except we have to use operators on wavefunctions instead. Therefore, $\hat{a}\ket{E_0} = 0$ is equivalent to

where $\op{a} = \frac{1}{\sqrt{2}}(\op{X}/\ell + i\ell \op{P}/\hbar)$, as in (8.6). Substituting in $\op{P} = -i\hbar \frac{\partial}{\partial x}$ and $\op{X} = x$, and re-arranging, we arrive at a simple differential equation that must be satisfied by $u_0(x)$:

It isn’t too difficult to spot the solution to this: it is nothing but a Gaussian:

where $N_0$ is a normalisation constant, that we will fix below. This shows that the ground state of the harmonic oscillator is a gaussian wavefunction, with standard deviation $\Delta x = \ell/\sqrt{2}$ (i.e. confirming that, indeed, $\ell$ is really a characteristic length scale of the harmonic oscillator). We can thus use (3.1) to fix the normalisation, and so the ground state wavefunction is

As an exercise, by evaluating $\op{H}u_0(x)$, you will show that the energy of the ground state is

Before determining the rest of the energy eigenstates and eigenvalues, we will first introduce the number operator as this will allow us to take care of the normalisation of energy eigenstates.

8.4The number operator¶

Having found the energy of the ground state, and given that we saw above that the energy spacing is $\hbar \omega$ (as we saw in (8.11)), this means we now in fact know the energy of all of the energy levels, which is given by

Note, that whereas for the energy levels of the infinite square well we took $n$ to start at $n = 1$, for the harmonic oscillator we take $n$ to start at $n = 0$. This is standard and somewhat unfortunate, but has reason (which we will explain below).

We can use (8.19) to learn something very interesting about the product operator $\hat{a}^\dagger \hat{a}$. Using the definitions (8.6), we see that

Using the definition of $\ell$ (8.5), and the canonical commutation relation (3.10), this can be further simplified to give

The first two terms in the bracket are nothing but the kinetic energy operator and potential energy operator of the harmonic oscillator, and so together give the Hamiltonian $\hat{H}$. Thus, by re-arranging, we arrive at a special equation,

which shows how we can express the Hamiltonian in terms of the creation and annihilation operators! Finally, comparing to (8.19), we see that we can give $\hat{a}^\dagger \hat{a}$ a very nice interpretation. Since $\hat{H}\ket{E_n} = E_n \ket{E_n}$ by definition, we must have

We therefore call $\hat{a}^\dagger \hat{a}$ the number operator,

such that $\ket{E_n}$ are eigenstates, with eigenvalue $n$ — the quantum number of the energy level.

8.5Energy eigenstates of the harmonic oscillator¶

We can use the number operator to figure out how the creation and annihilation operators change the norm of a state. Consider that we have an energy eigenstate that is normalised, satisfying $\| \ket{E_n} \| = \sqrt{\inner{E_n}{E_n}} = 1$. Then, the norm of $\hat{a}\ket{E_n}$ will be

where to obtain the second line we used (8.23), and pulled $n$ out the front. Thus, combining this with (8.13), we can conclude that the action of $\hat{a}$ on an energy eigenstate is

To find out how $\hat{a}^\dagger$ changes the norm of a state, we can apply $\hat{a}^\dagger$ to $(\hat{a}\ket{E_{n+1}})$. We see on the one hand that

On the other hand, from (8.26), $\hat{a}\ket{E_{n+1}} = \sqrt{n+1}\ket{E_n}$, and so

Equating the right-hand side of (8.27) and (8.28), and dividing by $\sqrt{n+1}$, we therefore see that

We finally have everything we need in order to determine the energy eigenstates of the harmonic oscillator! We can generate the $n^\mathrm{th}$ energy eigenstate by applying $\hat{a}^\dagger$ $n$ times to the ground state $\ket{E_0}$ (which we have already found), and dividing $\sqrt{n!}$ (where $n! = 1\times 2 \times \cdots \times n$),

This way, we ‘climb up’ the energy eigenstates, and remain normalised as we go, successively generating all of them.

What is more interesting and useful however is to obtain the energy eigenfunctions which we can also do in a similar manner: In fact, we have

For example,

Repeating this procedure, what we find in general is that all of the energy eigenfunctions have the form

where $N_n$ is a normalisation factor, and $h_n(y)$ are functions known as Hermite polynomials. These are polynomials of degree $n$ (and this is finally why we start counting at $n = 0$ and not $n = 1$), which alternate between being even and odd (with $n$). The first few Hermite polynomials (as well as the associated normalised energy eigenfunction) are given in the following table:

The energy eigenfunctions and corresponding probability densities of the first few energy eigenstates are plotted in Figure 8.2 and Figure 8.3 respectively.

It is insightful to compare to corresponding plots for the eigenfunctions and probability densities of the infinite square well potential from the previous chapter, as given in Figure 7.4 and Figure 7.5. Even though the functional form is very different, graphically they are remarkably similar! This is a type of universality: low energy states generically have similar energy eigenfunctions (even though the energies in general will be very different).

8.6Tunnelling and barrier penetration¶

There is one new and very important aspect of quantum mechanics that arises for the harmonic oscillator, which isn’t present in the infinite square well potential, known as tunnelling or alternatively barrier penetration.

Looking back at the potential energy $V(x)$ in Figure 8.1, it is evident that as the particle moves away from th origin, its potential energy increases quadratically. We have now found the energy eigenstates, and their corresponding energy eigenfunctions, as listed in Table 8.1 and depicted in Figure 7.4 and Figure 7.5.

Now, while the gaussian function decays rather rapidly, it is in fact non-zero for all values of x, and hence when a particle has a definite energy in the harmonic oscillator, there is a non-zero probability to find it in any finite region, no matter how far from the origin.

This is however very strange, as above we just noted that the potential energy increases as we move away from the origin. This means that, no matter what finite energy the particle has (and all the energy levels have finite energy), it has a chance of being found far away in a region where the potential energy is larger than the total energy $E_n$ of the particle!

This is known as tunnelling or barrier penetration: a quantum particle manages to enter a region in which classically it does not have enough energy to enter!

Remarkably, this isn’t even a small effect! In Figure 7.4 and Figure 7.5 we have added vertical lines on the axis to denote the position where the potential energy $V(x) = E_n$. Beyond these points, classically the particle should not be found. As you can see, there is a significant probability to be found in the classically forbidden region, even in the ground state.

Although we won’t study it in this course, tunnelling is interesting and important. Consider for example that we have two potential wells, separated by an energy barrier. Classically a particle can be trapped in one of them (for example, a planet can be bound in orbit around one star, even if there is a second star that it could equally be bound in orbit around). In quantum mechanics, because a particle tunnels, it is able to move between the two wells. This effect is extremely important, for example explaining α decay in particle physics, and being utilised in semiconductor devices.

In this unit we won’t go further into this fascinating effect, but just highlight its existence, and leave it for future courses to elaborate on the physical consequences of it.

8.7Exercises¶