I know that the uncertainty principle is: $$\Delta p\Delta q \ge \frac{\hbar}{2}.$$
But do the units on the left-hand side of the equation always have to equal $\text{Js}$, i.e. $\text{energy} \times \text{time}$ (the same as Planck's constant) or is it simply the numerical value which matters in the inequality.
Answer
The uncertainty principle may be stated more generally for two observables $A$ and $B$ as $$ \Delta A \Delta B \geq \dfrac{1}{2}\left|\langle\left[\hat{A},\hat{B}\right]\rangle\right|, $$
where $\langle \hat{C}\rangle$ is the expected value of the observable $C$ and $[\cdot\,,\cdot]$ is the commutator (see here for details). From this equation, we can see that the units of both sides are automatically the same (i.e., both sides have the units of $A$ multiplied by the units of $B$).
In the case of momentum $P$ and position $Q$ (using your notation), one can show that $\left[\hat{P},\hat{Q}\right]=-i\hbar$, which, substituted into the previous equation, gives the uncertainty principle given in the OP.
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