Why doesn't the anticommutator $\{x,p_x\}$ have an unique value? The commutator of position and momentum, $[x,p_x]$, has a unique value given by $i\hbar$. Why doesn't the anticommutator $\{x,p_x\}$ also have a definite value?
 A: If both the commutator and anticommutator have unique values, then by linearity, that would imply that $\hat{x}\hat{p}$ has a unique value. This would be inconsistent with the fact that the matrix element $\langle x |\hat{x}\hat{p}| p\rangle= xp\langle x | p\rangle$ depend on the bra $\langle x |$ and the ket $| p\rangle$.
A: A consistent (and physically relevant) quantum theory can either postulate the commutator of $\hat x$ and $\hat p$ or their anticommutator. Since $\{\hat x,\hat p\}=[\hat x,\hat p]+2\hat p\hat x$, if a quantum theory was to postulate both the commutator and the anticommutator to be certain constant values, it would imply a constant value for the operator $\hat p\hat x$ (and thus, for the operator $\hat x \hat p$) as well. 
This would be utterly problematic for the following reason: Let's say $|p\rangle$ is an eigenstate of $\hat p$. Then, $\hat x\hat p|p\rangle=p\hat x|p\rangle$. But now, if $\hat x\hat p$ has to have a constant value, it would mean that each $|p\rangle $ would have to be an eigenstate of $\hat x$ as well. This would further mean that the commutator $[\hat x,\hat p]$ actually vanishes. And since, furthermore, $\hat x\hat p$ has a constant value, $\hat x$ and $\hat p$ will share all the eigenstates. This stands in as much contradiction to the spirit of a theory that we (should) set out to construct. 
Thus, in order to construct a quantum theory of physically relevant operators, we postulate the commutator $[\hat x,\hat p]$ to have a constant value and let the expectation value of the anticommutator depend on the state.
A: If both the commutator $[x,p_x]$ and the anticommutator $\{x,p_x\}$ had constant values, then $x p_x$ and $p_x x$ would also have constant values. This would be bizarre since these values should depend on the detailed dynamics of the system. There is no reason to expect the momentum operator to be inversely related to the position operator for every system.
