This post will be broken down into three parts: first, a few basics about commutators (for everyday use in physics), then the solutions to Problem Sheet 3, and finally an attempt at explaining the physical meaning of commutators (if you’re not a PHY251 student you might want to jump straight to that).

## All you need to know about commutators… when you’re a second year physics student

The commutator is an extremely powerful tool, able to capture essential properties of groups and algebras (most frequent use in physics) by making explicit the extent to which two objects, be they matrices, operators or group elements – and in fact they often turn out to be the same thingfail to commute:

This is crucial when ordering matters (as is the case in quantum mechanics and quantum field theory). Indeed, we know from our day-to-day experience of a world in three spatial dimensions with chirality that rotations, translations and flips give different results depending on the order in which we apply them to a given object; and so, the matrices we use to represent them mathematically usually do not commute. There are of course a few notable exceptions, and it is often as interesting to determine, when working out the properties of a new theory, which objects do and do not commute.

From the above definition, a few properties can be easily derived (by simply expanding out and matching):

If you really want to start thinking, walking and talking like a commutator, it’s a good exercise to prove the above relations. Last point worthy of mention: one can as easily define the anticommutator, which comes in especially handy when tackling fermionic quantities in QFT (while QM is inherently bosonic). It is defined as

and similar identities apply (can you derive them?).

## Solutions to Problem Sheet 3

(Again, the Problem Sheet is here)

1.a Evaluate the commutator $[\hat{A},\hat{A}^\dagger]$.

where we’ve made use of the canonical commutation relation (CCR) $\left[\hat{p},\hat{x}\right]=-i\hbar$.

1.b Show that you can express the Hamiltonian $\hat{H}$ in terms of $\hat{A}^\dagger$ and $\hat{A}$.

(Note that this is actually the reverse of the historical derivation – how can we factorise $\hat{H}$?) It should be fairly obvious that we should start by multiplying the two operators $\hat{A}^\dagger$ and $\hat{A}$. Depending on the ordering we will of course get two different answers, but they are equivalent up to $[\hat{A},\hat{A}^\dagger]$.

Almost magically (but you should know better), we’ve arrived at an expression very close to $\hat{H}$; we’re just off by a couple of constants. Let’s rectify that:

And as promised, there are two possible answers, equivalent through the commutation relation derived in question 1.a.

1.c Evaluate the commutator $[\hat{H},\hat{A}^\dagger]$.

(Check you get the same answer with the other definition of $\hat{H}$ from the previous question.)

1.d Evaluate $\hat{H}\hat{A}^\dagger\psi_n(x)$ where $\psi_n(x)$ is an eigenstate solution of the 1D quantum harmonic oscillator.

## Making sense of commutators in QM

### Compatible observables

One of the most frequent replies to the question “what the hell does $[\hat{p},\hat{x}]=-i\hbar$ mean?” is “position and momentum are not compatible observables”. What is meant by that is exactly that ordering matters, and that they fail to commute; in other words, measuring the position before the momentum (where by “measuring the position” we mean $\hat{x}\psi(x)=x\psi(x)$, and likewise for momentum) yields a different result from the measurement of momentum first, then of position. Different, but by how much? you ask – by $i\hbar$ precisely.

### Generating uncertainty

A similar explanation comes from a general theorem for non-commuting operators:

which clearly relates the uncertainty of measurement to the failure of the two observables to commute (for $\hat{A}=\hat{p}$ and $\hat{B}=\hat{x}$, this goes by the name of Heisenberg’s uncertainty principle, if you hadn’t guessed…).

### Time evolution

Now that’s a bit trickier, since we’ve been working with the time-independent Schrödinger equation so far. The full equation (complete with time-dependence) is

and so you see that $\hat{H}\sim\partial_t$. In QFT textbooks you’ll often read the statement that “the Hamiltonian generates the time evolution” of various quantities (operators, fields, etc.). This has obvious parallels with classical mechanics (at least in the Hamiltonian formulation), and should make sense when thinking about $\hat{H}$ as transforming kinetic energy into potential energy, back and forth. All that to say that commutators involving $\hat{H}$ take an extra, deeper meaning…

In fact, the Ehrenfest theorem shows that (in the time-dependent picture) if $\hat{\mathcal{O}}$ has no time dependence and commutes with $\hat{H}$, then it is a constant of motion:

### Meaning of $[\hat{A},\hat{A}^\dagger]$

By now you should know that we refer to $\hat{A}$ and $\hat{A}^\dagger$ as lowering and raising ladder operators, respectively. Indeed, they allow us to access $\psi_{n\pm 1}$ from $\psi_n$, up to a normalization constant. Acting with one then the other on a state should therefore bring it back to itself, right? Well, not exactly. In fact, you end up botching the normalization prefactor a bit (try it with the simple harmonic oscillator, you shouldn’t get back exactly $\psi_n$), hence the commutation relation $[\hat{A},\hat{A}^\dagger]=1$.

Another way to look at it is to consider the number operator $\hat{N}=\hat{A}^\dagger\hat{A}$, which returns the “number” of the energy eigenstate, e.g.:

and we can easily rewrite the Hamiltonian as $\hat{H}=\hbar\omega\left(\hat{N}+\frac{1}{2}\right)$.

The real reason these operators don’t commute, and have to take the representation described above (i.e. with different factors, ultimately generating the asymmetry) is that we require the energy spectrum to be bounded below: $\hat{A}\left\vert 0\right\rangle=0$. Consider the same operator action, but without the numerical factors (and call them $\hat{a}$ and $\hat{a}^\dagger$). Then:

where the last equality comes from the orthogonality of the eigenstates and the requirement $n\ge 0$. Therefore $\hat{a}\left\vert 0\right\rangle=0$. But now:

and obviously $[\hat{a},\hat{a}^\dagger] \cancel= 0$.

### Meaning of $[\hat{H},\hat{A}^\dagger]$

In the time-independent case, it doesn’t make sense to talk about “time evolution”; let’s interpret it in terms of energy levels instead. Finding out the value taken by the commutator is the same as asking, “how do the energy eigenvalues differ from one eigenstate to the other”. Indeed, expanding out the commutator, we see that it’s applying the operator $\hat{H}$ to $\hat{A}^\dagger\psi_n\sim\psi_{n+1}$, thus obtaining $E_{n+1}$ as eigenvalue, and subtracting $E_{n}$ (from $\hat{H}$ acting on $\psi_n$). As it doesn’t make sense to talk about energy differences of different eigenfunctions, we also need to raise $\psi_n$ to $\hat{A}^\dagger\psi_n\sim\psi_{n+1}$ in the second case to be able to take the difference.

In terms of measurements, this is asking, “does it matter if I measure the energy before or after changing the state of my system?” Of course, the answer should be yes! You can also see how this might be interpreted as some sort of evolution of the ladder operators – effectively generating the energy eigenspectrum.

Note: this commutator is in fact the definition of ladder operators. Were it to be zero, we’d have a degenerate spectrum, with many eigenfunctions corresponding to the same vacuum energy eigenstate. Using similar arguments as above, it should be straightforward to show that

### Bases

In question 1.d of the problem sheet, we used the fact that $\psi_n$ was an eigenstate of $\hat{H}$ with eigenvalue $E_n$; in this particular form, it is not necessarily an eigenstate of other operators. This is where the commutator comes into play: if we can find two operators $\hat{A}$ and $\hat{B}$ that commute, we can always find a basis for $\psi$. In fact, given eigenstates $a_n$ and $b_n$ of $\hat{A}$ and $\hat{B}$ respectively, we can write:

If we decide to measure $\hat{B}$, we’ll get the result $b_n$ with probability $\vert c_n\vert^2$ independently of whether or not we previously measured $\hat{A}$. Given an operator that commutes with $\hat{H}$, we have a powerful tool to diagonalize the Hamiltonian.

## A bit of Mathematica…

If you have access to Mathematica, try the following (for the simple harmonic oscillator in QM). First, set up your wavefunction and various operators:

a=Sqrt[m*omega/h]
psi[n_,x_]=1/Sqrt[2^n*n!]*(m*omega/(Pi*h))^(1/4)*Exp[-a^2/2*x^2]*HermiteH[n,a*x]
opa[x]=a/Sqrt[2]*(x*#+1/a^2*D[#,x])&
oph[x]=-h^2/(2*m)*D[#,{x,2}]+1/2*m*omega^2*x^2*#&


We can then compute the energy spectrum (using the Floor method to make sure Mathematica only picks integer values for $n$ – otherwise the wavefunction is no longer an eigenfunction of $\hat{H}$!):

Manipulate[Simplify[oph[x][psi[Floor@n,x]]/psi[Floor@n,x]],{n,0.,10.,1.}]


Now, a few questions:

1. Using the Manipulate method, can you show numerically (up to some $n$) that $[\hat{A},\hat{A}^\dagger]=1$?

2. What is $\hat{A}\psi_0(x)$?

3. What is $\frac{\hat{A}^\dagger\psi_0(x)}{\psi_1(x)}$?

4. What is $\frac{\hat{A}\psi_1(x)}{\psi_0(x)}$?

5. Do the results of the previous two questions generalize to higher $n$?

6. Setting any values for $\hbar$, $\omega$ and $m$, and using the Manipulate method, plot the first few eigenfunctions. What can be said about the parity of $\psi_n(x)$?

### Expanding on questions 2 and 5

Consider acting with the commutator of the ladder operators on some non-vacuum state:

But we know from questions 3 and 4 that for $n=0$ these coefficients are exactly $c_1=1$ and $c_2=0$. Thus:

(And you should have spotted that in general, $c_1$ and $c_2$ are given by the definition in the paragraph above about the number operator).