the commutator
In everyday algebra it never matters whether you multiply 3 times 5 or 5 times 3. For operators it can matter enormously, and the commutator is the exact measure of how much. It answers a physical question: does the order in which you perform two measurements change the result? When the answer is yes, you are looking at the algebraic root of nearly everything strange about quantum mechanics.
The commutator of two operators is defined as [A, B] = AB - BA. If it vanishes, A and B commute: they can be diagonalized in a common eigenbasis and hence measured simultaneously to arbitrary precision (they are compatible). If it does not vanish, no such shared basis exists and the two observables obey a trade-off. The cornerstone of the whole theory is the canonical commutation relation [x, p] = i hbar, position and momentum failing to commute by exactly one unit of i hbar. Angular momentum components close among themselves, [L_x, L_y] = i hbar L_z, generating the rotation algebra.
Its reach is enormous. The general uncertainty relation reads sigma_A sigma_B >= (1/2)|<[A, B]>|, so a nonzero commutator directly forbids joint sharpness. In canonical quantization the classical Poisson bracket is promoted to a commutator, {A, B} -> (1/i hbar)[A, B], which is the formal recipe that turns a classical theory into a quantum one. And ladder-operator tricks (the harmonic oscillator, angular momentum) run entirely on knowing a few commutators.
Applying [x, p] = xp - px to a test wavefunction with p = -i hbar d/dx gives [x, p]psi = i hbar psi for every psi, so [x, p] = i hbar — the single relation from which the position-momentum uncertainty principle follows.
One line of algebra yields [x, p] = i hbar, the seed of quantum indeterminacy.
Commuting means simultaneously diagonalizable and jointly measurable; a nonzero commutator, via sigma_A sigma_B >= (1/2)|<[A,B]>|, forbids both observables being sharp at once.