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On page 54 of Weinberg's QFT I, he says that an element $T(\theta)$ of a connected Lie group can be represented by a unitary operator $U(T(\theta))$ acting on the physical Hilbert space. Near the identity, he says that

$U(T(\theta)) = 1 + i\theta^a t_a + \frac{1}{2}\theta^a\theta^bt_{ab} + ...$

Weinberg then states that $t_a$, $t_{ab}$, ... are Hermitian. I can see why $t_a$ must be by expanding to order $\mathcal{O}(\theta)$ and invoking unitarity. However, expanding to $\mathcal{O}(\theta^2)$ gives

$t_at_b = \frac{1}{2}(t_{ab} + t^\dagger_{ab})$,

so it seems the same reasoning cannot be used to show that $t_{ab}$ is Hermitian. Why, then, is it?

Thanks!

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1 Answer 1

I) OP's last equation(v1) is not correct. If

$$\tag{2.2.17} U ~=~ {\bf 1} + i\theta^a t_a + \frac{1}{2}\theta^a\theta^b t_{ab} + {\cal O}(\theta^3), \qquad \theta^a \in \mathbb{R},\qquad t_{ab}~=~t_{ba}, \qquad $$

then the unitarity condition

$$U^{\dagger}U~=~{\bf 1}~=~UU^{\dagger}$$

yields to second order in $\theta$ that

$$ t^{\dagger}_a~=~t_a, $$

and

$$ t^{\dagger}_{ab}+t_{ab}+\{t_a,t_b\}~=~0. $$

II) However, in fact

$$U~=~e^{i\theta^a t_a}~=~{\bf 1} + i\theta^a t_a -\frac{1}{4}\theta^a\theta^b \{t_a,t_b\}+ {\cal O}(\theta^3),$$

so that the generator

$$ t_{ab} ~=~-\frac{1}{2}\{t_a,t_b\}~=~t^{\dagger}_{ab} $$

is indeed Hermitian as Weinberg claims.

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