# What is the significance of taking the commutator?

I'm an undergrad studying QM, and sometimes we use the method of taking the commutator of two operators to make some statement about the group. I always get lost because I don't know WHY were taking the commutator. Of course [A,B]=0 means they commute, but what kinds of non-zero commutator are significant and what do they mean? What is the general reason for taking a commutator? What are you trying to accomplish in this method? If the composition of two operators is a multiple of another, what does this suggest about the commutator?

• Possible duplicates: physics.stackexchange.com/q/130800/2451 , physics.stackexchange.com/q/9194/2451 and links therein. – Qmechanic Aug 1 '17 at 15:43
• Have you studied classical mechanics yet? Unfortunately, the way UG physics is structured nowadays, students are introduced to QM before CM, and hence never see the Poisson bracket, which is actually fundamental to mechanics. The commutator is the Hilbert-space analogue of the Poisson bracket, roughly speaking, hence the form of Heisenberg's equations! – Dr. Ikjyot Singh Kohli Aug 1 '17 at 15:47

Uncertainty and Measurements

Suppose the operators $A$ and $B$ are Hermitian and represent observables. Then if we make respective measurements in an experiment, then the standard deviation in $A$ denoted $\sigma_A$ as well as in $B$, $\sigma_B$, must satisfy the inequality,

$$\sigma_A \sigma_B \geq \frac12 |\langle [A,B]\rangle|.$$

Thus, two operators not commuting has an implication on the limitations of measurement, that is, if we make some measurement with $\sigma_A$, there is a limit to how well we can measure $B$.

Constants of Motion

A further application of the commutator is to determine conserved quantities. If $Q$ is to be conserved, it must commute with the generator of time translations which is the Hamiltonian, and thus we require $[Q,H] = 0$. This follows from Ehrenfest's theorem,

$$\frac{d}{dt}\langle Q \rangle = \langle \partial_t A\rangle - \frac{i}{\hbar}\langle [Q,H]\rangle.$$

Canonical Quantisation

As noted in a comment, the analogue of the commutator in classical mechanics is the Poisson bracket, which has certain implications, and an interpretation in terms of a symplectic manifold known as the phase space of a system.

In quantum mechanics, as well as relativistic quantum mechanics, one can pass from a classical theory - at least in theory - to a quantum theory through canonical quantisation by changing the Poisson bracket to a commutator, as a rule of thumb:

$$\{A,B\} \to -\frac{i}{\hbar}[A,B].$$

There are many subtleties - too many to list here - as to how this can go wrong and need further refinements to quantise a theory. However, a common one is ordering ambiguities. Recall that in a classical theory we do not deal with operators and do not have to worry about ordering, which may differ by a commutator. In quantum mechanics, we do and require an ordering prescription.

When operators don't commute, their mutual eigenstates don't span the Hilbert space, so non-commuting observables are sometimes not simultaneously well-defined. In some cases (e.g. coordinates and their conjugate momenta), the commutator is a nonzero multiple of the identify, and no simultaneous eigenstates exist. Taking the trace of such a commutator proves the Hilbert space is infinite-dimensional. Anything commuting with the Hamiltonian is conserved.