I gained a lot of physical intuition about commutators by reading this topic. What is the physical meaning of commutators in quantum mechanics?
I have similar questions about the anti-commutators. What does it mean physically when two operators anti-commute ?
Sorry but the analysis of what commutators mean (in the given link) although very good, does not provide intuition and does not generalise to anti-commutators.
Commutators used for Bose particles make the Klein-Gordon equation have bounded energy (a necessary physical condition, which anti-commutators do not do).
On the other hand anti-commutators make the Dirac equation (for fermions) have bounded energy (unlike commutators), see spin-statistics connection theorem.
In this sense the anti-commutators is the exact analog of commutators for fermions (but what do actualy commutators mean?). nice and difficult question to answer intuitively.
In a sense commutators (between observables) measure the correlation of the observables. Thus is also a measure (away from) simultaneous diagonalisation of these observables.
Elaborating a little on this.
Lets say we have a state $\psi$ and two observables (operators) $A$, $B$. When these operators are simultaneously diagonalised in a given representation, they act on the state $\psi$ just by a mere multiplication with a real (c-number) number (either $a$, or $b$), an eigenvalue of each operator (i.e $A\psi=a\psi$, $B\psi=b\psi$).
We know that for real numbers $a,b$ this holds $ab-ba=0$ identicaly (or in operator form $(AB-BA)\psi=0$ or $\left[A,B\right]\psi=0$) so the expression $AB-BA=\left[A,B\right]$ (the commutator) becomes a measure away from simultaneous diagonalisation (when the observables commute the commutator is identicaly zero and not-zero in any other case).
Another way to see the commutator expression (which is related to previous paragraph), is as taking an (infinitesimal) path from point (state) $\psi$ to point $A \psi$ and then to point $BA \psi$ and then the path from $\psi$ to $B \psi$ to $AB \psi$. If the operators commute (are simultaneously diagonalisable) the two paths should land on the same final state (point). If not their difference is a measure of correlation (measure away from simultaneous diagonalisation).
When talking about fermions (pauli-exclusion principle, grassman variables $\theta_1 \theta_2 = - \theta_2 \theta_1$), the commutators have to be adjusted accordingly (change the minus sign), thus become anti-commutators (in order to measure the same quantity).
Continuing the previous line of thought, the expression used was based on the fact that for real numbers (and thus for boson operators) the expression $ab-ba$ is (identicaly) zero.
However fermion (grassman) variables have another algebra ($\theta_1 \theta_2 = - \theta_2 \theta_1 \implies \theta_1 \theta_2 + \theta_2 \theta_1=0$, identicaly).
So what was an identical zero relation for boson operators ($ab-ba$) needs to be adjusted for fermion operators to the identical zero relation $\theta_1 \theta_2 + \theta_2 \theta_1$, thus become an anti-commutator.
Un-correlated observables (either bosons or fermions) commute (or respectively anti-commute) thus are independent and can be measured (diagonalised) simultaneously with arbitrary precision. If not, the observables are correlated, thus the act of fixing one observable, alters the other observable making simultaneous (arbitrary) measurement/manipulation of both impossible.
PS. See how the previous analysis can be generalised to another arbitrary algebra (based on identicaly zero relations), in case in the future another type of particle having another algebra for its eigenvalues appears.
Commutators and anticommutators are ubiquitous in quantum mechanics, so one shoudl not really restrianing to the interpretation provdied in the OP. There's however one specific aspect of anti-commutators that may add a bit of clarity here: one often u-ses anti-commutators for correlation functions.
Indeed, the average value of a product of two quantum operators depends on the order of their multiplication. We can however always write: $$AB = \frac{1}{2}[A, B]+\frac{1}{2}\{A, B\},\\ BA = \frac{1}{2}[A, B]-\frac{1}{2}\{A, B\}.$$ In the classical limit the commutator vanishes, while the anticommutator simply become sidnependent on the order of the quantities in it. One therefore often defines quantum equivalents of correlation functions as: $$ K_{AB}=\left\langle \frac{1}{2}\{A, B\}\right\rangle.$$
As an example see the use of anti-commutator see [the quantum version of the fluctuation dissipation theorem][1], where $$ S_{x}(\omega)+S_{x}(-\omega)=\int dt e^{i\omega t}\left\langle \frac{1}{2}\{x(t), x(0)\}\right\rangle$$ (I am trying to adapt to the notation of the Wikipedia article, but there may be errors in the last equation.)
