Derivation of the Infinitesimal Translation Operator $$J(d \vec x') = \left(1 - \frac{i \vec p \cdot d \vec x'}{\hbar}\right)$$
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. 
 A: 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.
A: 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 $U^{-1}(x_0)$
