Commutator Constant

Question

I have seen a lot of commutators in quantum mechanics having a constant factor $i\hbar$. I have read about Dirac supplanting Poisson Brackets with commutators having a constant $i\hbar$. I want to know why does this factor appear every now and then? Does every commutator somehow contain this factor $i\hbar$?

Answer 1

Does every commutator somehow contain this factor $i \hbar$?

No. Some commutators vanish, and some are real (see the counterexamples at the end). However, this form is kind of expected to happen in most commutators of interest. One of the reasons is canonical quantization (as mentioned in the comments to the question), but I'll focus on more intuitive reasons.

Why $\hbar$?

$\hbar$ yields the dimensions in which QM becomes relevant. It is the dimension in which classical physics is expected to be nearly useless in predicting the outcomes of experiments. Since commuting observables is a hallmark of classical physics ($px = xp$ in classical systems) and $\hbar$ is the hallmark of QM, it is expected that $\hbar$ measures, in some way, how two observables will fail to commute in quantum mechanics.

Why $i$?

This is a consequence of an exercise for the reader: let $A$ and $B$ be two Hermitian operators. Then $[A,B]^\dagger = - [A,B]$. This is the operator analogue of an imaginary number. The $i$ factor appears in the commutator of observables because the observables are Hermitian, and hence their commutator has to be skew-Hermitian. This means the eigenvalues of the commutator are all imaginary, and hence can be expressed as a real number times $i$.

Counterexamples

There are still some commuting operators in QM. For example, $p_x$ and $p_y$ (momenta in the $x$ and $y$ directions, respectively) do commute. Hence, their commutator is not proportional to $i \hbar$ (I mean, unless you consider $0$ to be proportional). If you consider operators that are not observables, such as the ladder operators, you'll also find their commutator is different: $[a, a^\dagger] = 1$.