Why do momentum and position have the same direction in space in the Heisenberg Uncertainty Principle? It is known that $\Delta p  \Delta x \geq \frac{h}{4\pi}$. I read that the two uncertainties must be along same axes. Why is that so?
 A: On the one hand, the uncertainty principle is in fact a postulate of quantum mechanics, so the question why it is postulated the way it is does not make a lot of sense. On the other hand, few thing can be said with respect to the question.
First, the exist a more general form of the uncertainty principle usually refered to as the Robertson–Schrödinger uncertainty relation, which states that for any pair of observables $A$ and $B$ represented by self-adjoint operators $\hat{A}$ and $\hat{B}$ respectively,
$$
\sigma_{A}\sigma_{B} \geq \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right|,
$$
where $[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}$ is the commutator between the operators.
Secondly, for the specific case of position $\hat{q}$ and momentum in the same direction $\hat{p}_q$ operators, the commutator is known to be 
$$
    [\hat q,\hat p_q] = i\hbar \, ,
$$
where $q$ is $x$, $y$, or $z$. This is also a postulate, known as the canonical commutation relation, so the question why is again not quite appropriate here. However, one must note specifically that the canonical commutation relation is the relation between the so-called conjugate variables, the variables such that one is the Fourier transform of another. Only position and momentum in the same direction are such variables, while any other combination is not. In fact, commutators for any other pair ofobservables are all zero, so that in summary we have,
$$
    [\hat q,\hat p_{q'}] = i\hbar \delta_{q q'} \, , \quad [\hat q,\hat q'] = [\hat p_q,\hat p_{q'}] = 0 \, ,
$$
where $\delta_{q q'} = 1$ only if $q=q'$ and is zero otherwise. Physically it means that only position and momentum in the same direction could not be measured simultaneously, while all other pairs of observables $q$ and $p_q$ could.
