0
$\begingroup$

I am studying group theory by myself and while i was reading "Physics from symmetry", Jakob Schwichtenberg's book, he said it was conventional in physics to write the generators of $SO(3)$ with an extra $i$, that is, multiplying the group generator matrix by $i$, but i am not understanding is why the generators have to be written like that? with the $i$, why this is conventional? what is the advantage?

In physics it’s conventional to define the generators of $SO ( 3)$ with an extra $i$. Concretely this means that instead of $e^{φ̃J}$ , we write $e^{iφJ}$ with with φ = − φ̃.

This is the quote that i am referring to in my question, right below equation 3.70, page 44.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ I hope you read from the 2nd edition, the 1st is a disaster. That $i$ in the exponent is to ensure that $J$ is self-adjoint for $\exp\left({i\varphi J}\right)$ being unitary (by Stone's theorem). $\endgroup$
    – DanielC
    Commented Oct 31, 2021 at 21:14
  • $\begingroup$ Yes, i am reading from 2nd edition, i appreciate the answer. I will take a look at this theorem. $\endgroup$
    – Gusklin
    Commented Oct 31, 2021 at 21:17
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/321230/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Nov 1, 2021 at 1:49

1 Answer 1

Reset to default
1
$\begingroup$

I'm not familiar with that book, but this is what you must have come across several times in college physics.

The generators of SO(N) may be represented by real antisymmetric matrices A, so the Lie algebra involves real quantities. Check its transposition invariance. Exponentiating real antisymmetric matrices yields real orthogonal matrices, the rotations of your mechanics. The advantage of this is working with real quantities throughout.

If you incorporated an i into these matrices, you'd get Hermitian matrices for the generators, but the algebra involves an extra i on the right hand side. Check its invariance under hermitian conjugation. Now you may make a unitary transformation of basis on these imaginary hermitian matrices and preserve hermiticity, but the resulting hermitian matrices are now complex, like the standard QM spin matrices. Their exponentials, group transformations, are not unitary, unless you put an i in front of them in the exponent: thus, unless you make the exponents antiHermitian! In QM, or spinor representation constructions, you want your transformations to be unitary, a cornerstone of probability preservation in the QM theory.

Improve this answer
$\endgroup$

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