0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

This is a difference of convention between mathematicians and physicists. Think of an antisymmetric real matrix. If you multiply it by i, you make it Hermitian instead of antiHermitian. Think of the J as having an i in it already because of physics conventions vs math conventions.

$\endgroup$
  • $\begingroup$ And most importantly: a hermitian matrix is an observable. Hence J is the observable angular momentum, instead of -ihJ $\endgroup$ – user12029 May 4 '17 at 21:01
  • $\begingroup$ $U(1+W)$ 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. $\endgroup$ – mike stone May 4 '17 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.