What are commutators?

# What are commutators?

TL;DR – Commutators, like Poisson bracket, tell us how a quantity changes under a transformation generated by another.

Commutators play a fundamental role in quantum mechanics. Mathematically, they tell us how multiplication between operators behaves but this does not provide any physical insight. What is more interesting physically is that they are intimately related to infinitesimal transformations, which gives a better understanding of why they are related to Poisson brackets and why commutation relationships between position, momentum and spin must be what they are.

1. Conjugate quantities

In classical mechanics, conjugate quantities are pair of variables that form an independent degree of freedom. If you integrate over a pair, you get a quantity that no longer depends on the unit and coordinate used for those quantities.

In quantum mechanics, things work a bit differently mathematically. In quantum mechanics you have operators and parameters. If a quantity is an operator (like spin) it means you can write the state as a wave-function over that quantity. If a quantity is a parameter (like angle) it means you can make and infinitesimal transformation along it. The conjugate relationship is between operators and parameters: for each operator $A$ you have one parameter $\alpha$ and we say that $A$ generates the infinitesimal transformation $(1 + \frac{A d\alpha}{\imath \hbar})$, which is a unitary transformation.

For example, spin component $S_z$ is the operator while the angle $\theta_{xy}$ is the parameter, and $S_z$ generates the rotation $(1 + \frac{S_z d\theta_{xy}}{\imath \hbar})$. The confusing part is that some quantities, like position and momentum, can both be operators and parameters. So we have the $P$ operator and the $x$ parameter and $P$ generates the translation $(1 + \frac{P dx}{\imath \hbar})$, but also $X$ operator and $p$ parameter and $X$ generates the change in momentum $(1 + \frac{X dp}{\imath \hbar})$. Note how the upper case denotes an Hermitian operator while the lower case denotes a real number.

2. Commutators

Now we ask: suppose we have two operators $A$ and $B$. Suppose you perform the infinitesimal transformation generated by the second, over the parameter $\beta$. How does the operator $A$ change?

The infinitesimal transformation corresponds to the unitary operator $U=(1 + \frac{B d\beta}{\imath \hbar})$. An operator under a unitary transformation changes as $\hat{A} = U^\dagger A U$. Therefore we have:

\begin{aligned}
\hat{A}&=U^\dagger A U = (1 – \frac{B d\beta}{\imath \hbar}) A (1 + \frac{B d\beta}{\imath \hbar}) \\
&=A + A \frac{B d\beta}{\imath \hbar} – \frac{B d\beta}{\imath \hbar} A – \frac{B d\beta}{\imath \hbar} A \frac{B d\beta}{\imath \hbar} \\
&=A + \frac{(AB – BA)}{\imath \hbar}d\beta + \frac{BAB}{\hbar^2}d\beta^2
\end{aligned}

The first order for the change of $A$ is:

\begin{aligned}
\frac{dA}{d\beta}&=\frac{(AB – BA)}{\imath \hbar} \\
&= \frac{[A,B]}{\imath \hbar}
\end{aligned}

Therefore the commutator between $A$ and $B$ tells us how $A$ changes under a transformation generated by $B$. In a previous post we saw that the Poisson brackets in classical Hamiltonian mechanics had the same role. That is why we can do the formal substitution: they describe the same physical relationship in a different mathematical framework.

We can also understand why commutators between notable quantities must be what they are. For example, the infinitesimal transformation generated by the operator momentum $P_j$ is the translation over the position parameter $x_j$. So how will the operator position $X_i$ change under the transformation generated by $P_j$?

\begin{aligned}
\frac{[X_i,P_j]}{\imath \hbar} &= \frac{dX_i}{dx_j}=I \delta_{ij}
\end{aligned}

If $X_i$ and $P_j$ are along the same direction, then position operator will change of the same amount of the position parameter: they are physically the same thing. If they are along different directions there is no change.

Now, consider the operators $S_x$, $S_y$ and $S_z$ for the spin components. How will they change under the transformation generated by $S_z$, which is the rotation over $\theta_{xy}$?

\begin{aligned}
\frac{[S_x,S_z]}{\imath \hbar} &= \frac{dS_x}{d\theta_{xy}} = S_y \\
\frac{[S_y,S_z]}{\imath \hbar} &= \frac{dS_x}{d\theta_{xy}} = – S_x \\
\frac{[S_z,S_z]}{\imath \hbar} &= \frac{dS_z}{d\theta_{xy}} = 0
\end{aligned}

The last one is the easiest: the $z$ component of spin does not change under rotation over the $xy$ plane. For the other two, just remember that the $x$ component is rotated in the $y$ direction and the $y$ direction is rotated in the $-x$ direction.

These are indeed the commutation relationships one has in quantum mechanics. Given the physical meaning of the operators and their transformations, the commutation relationship can’t be anything else.

3. Conclusion

We have seen that commutators have a well defined physical/geometrical meaning, which is the same one for Poisson brackets in classical Hamiltonian mechanics. I personally not only find it insightful, but it helps me remember the correct sign for the commutation relationships.