# Why multiply the infinitesimal generator for a rotation $R$ by $i$ when constructing $U(R)$?

I'm sure this is a silly question, but I can't figure out the answer. Current I'm reading chapter 4 in Weinberg's Lectures on Quantum mechanics. Earlier in the book, he asserts that unitary operators close to the identity look like

$$U \approx 1 + i T,$$ for some Hermitian matrix $T$. In this chapter, he deduces that generators close to the identity look like $$R \approx 1 + W,$$ where $W$ is real and skew-symmetric (and "infinitesimal") . So far, so good. This is just an intuitive way to think about the relevant Lie algebras.

He then says that the unitary transformation generated by an infinitesimal rotation $W$ is $$U(1 + W) \approx 1 + \frac{i}{2h} \sum w_{ij} J_{ij}.$$

I'm having trouble understanding this. Shouldn't the transformation generated by an infinitesimal rotation be some kind of rotation, and hence orthogonal? If so, why does the $i$ appear, bringing complex numbers into the picture?

A slightly different way to put this is, why does going from rotation to the associated infinitesimal rotation to the generated unitary transformation not give the original rotation back again?

• $U(1+W)$ is a unitary operator on a (possibly infinite dimensional) Hilbert space, while $1+W$ is a rotation in ordinay euclidean space . Therefore the operator $U$ is not itself a rotation, it is an operator representing a rotation. – mike stone May 4 '17 at 21:30