$$J(d \vec x') = \left(1 - \frac{i \vec p \cdot d \vec x'}{\hbar}\right)$$
This is the infinitesimal translation operator, as defined on p. 46 of Modern Quantum Mechanics by Sakurai & Napolitano.
How is this equation derived?
In particular, I am wondering why it is necessary for this operator to have an imaginary number.
I have derived it in a not so rigorous manner and attached a picture of it. The derivation is different from Sakurai but in the same spirit.
If it has some mistakes, please do rectify.
The proof starts with the observation (in 1d) that \begin{align} e^{-i x_0\hat p}\hat x e^{i x_0\hat p}&= \left(\hat 1 -i x_0\hat p +\ldots \right)\hat x \left(\hat 1 +i x_0\hat p +\ldots \right)\, ,\\ &=\hat x+ -ix_0\hat p\hat x + i x_0\hat x\hat p+\ldots\, ,\\ &=\hat x-i x_0[\hat p,\hat x]+\ldots \, ,\\ &=\hat x -i x_0 (-i\hbar)=\hat x-x_0 \end{align} which is just a translation by $x_0$. Thus, $$ e^{-i x_0\hat p}\psi(x)=\langle x\vert e^{-i x_0\hat p}\vert \psi\rangle = \langle x+x_0\vert\psi\rangle =\psi(x+x_0) $$ The 3d generalization is immediate.
Note that $U(x_0)=e^{-i x_0\hat p}$ must be unitary, meaning it must satisfy $U^{-1}(x_0)= U^\dagger(x_0)$. Since $\hat p$ is hermitian, $\hat p^\dagger= \hat p$; since $x_0$ is real, $x_0^*=x_0$. Thus, to guarantee unitarity we must have this form so that $$ U^\dagger(x_0)=\left(e^{-i x_0\hat p}\right)^\dagger=e^{+i x_0\hat p} $$ which is obviously $U^{-1}(x_0)$
