Commuting operators

You might have heard of commuting and non commuting operators. What do they mean?

If $$\begin{eqnarray*} {\hat {A}} ({\hat {B}} \psi) & \neq & {\hat {B}} ({\hat {A}} \psi) \end{eqnarray*}$$, then we say that the operators  and B do not commute. The corresponding physical observables A and B are then said to be complementary observables.

The quantity $$\begin{eqnarray*} {\hat {A}} {\hat {B}} & - & {\hat {B}} {\hat {A}} \end{eqnarray*}$$ is expressed as [Â,B] and called the commutator of the operators  and B.

So when two operators do not commute

$$\begin{eqnarray*}[{\hat {A}},{\hat {B}}]& \neq & 0 \end{eqnarray*}$$

and when two operators do commute

$$\begin{eqnarray*}[{\hat {A}},{\hat {B}}]& = & 0 \end{eqnarray*}$$

In particular, quantum mechanical operators are postulated to satisfy the following commutation relations:

$$\begin{eqnarray*}[{\hat {q}},{\hat {p_{q^{/}}}}]& = & i \hbar \delta_{qq^{/}} \end{eqnarray*}$$

$$\begin{eqnarray*}[{\hat {q}},{\hat {q^{/}}}]& = & 0 \end{eqnarray*}$$

$$\begin{eqnarray*}[{\hat {p_{q}}}, {\hat {p_{q^{/}}}}]& = & 0 \end{eqnarray*}$$

The q's are the cartesian coordinates and the p_q's are the corresponding conjugate momenta. Needless to mention, they are a set of complementary observables and as we shall see soon, obey Heisenberg's uncertainty principle.