# Hermitian operators in the expansion of symmetry operators in Weinberg's QFT

This is related to Taylor series for unitary operator in Weinberg and Weinberg derivation of Lie Algebra.

$$\textbf{The first question}$$

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 $$U(T(\theta)) = 1 + i\theta^a t_a + \frac{1}{2}\theta^a\theta^bt_{ab} + \ldots. \tag{2.2.17}$$ 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 $$t_at_b = -\frac{1}{2}(t_{ab} + t^\dagger_{ab})\tag{2},$$ so it seems the same reasoning cannot be used to show that $$t_{ab}$$ is Hermitian. Why, then, is it?

$$\textbf{The second question}$$

In the derivation of the Lie algebra in the first volume of Quantum Theory of Fileds by Weinberg, it is assumed that the operator $$U(T(\theta)))$$ in equation (2.2.17) is unitary, and the rhs of the expansion $$$$U(T(\theta)))=1+i\theta^a t_a +\frac{1}{2} \theta_b\theta_c t_{bc} + \dots$$$$ requires $$t_{bc}=-\frac{1}{2}[t_b,t_c]_+.$$ If this is the case there is a redundancy somewhere. In fact, by symmetry $$U(T(\theta))=1+i\theta_at_a+\frac{1}{2}\theta_a\theta_bt_{ab}+\dots\equiv 1+i\theta_at_a-\frac{1}{2}\theta_a\theta_bt_at_b+\dots$$ and it coincides with the second order expansion of $$\exp\left(i\theta_at_a\right)$$; the same argument would then hold at any order, obtaining $$U(T(\theta))=\exp\left(i\theta_at_a\right)$$ automatically. However, according to eq. (2.2.26) of Weinberg's book, the expansion $$U(T(\theta))=\exp\left(i\theta_at_a\right)$$ holds only (if the group is just connected) for abelian groups. This seems very sloppy and I think that Lie algebras relations could be obtained in a rigorous, self-consistent way only recurring to Differential Geometry methods.

There have been some answers or speculations for these two questions but I do not think they are solved. I think the crucial point for these two questions is that the $$t_{ab}$$ operator is $$\textbf{not}$$ hermitian unless the $$\{t_a\}$$ operators commute with each other. Here is why:

From the unitarity of $$U(t(\theta))$$ we have $$t_at_b = -\frac{1}{2}(t_{ab} + t^\dagger_{ab})\tag{2},$$ And from the expansion of $$f(\theta_a,\theta_b)$$ we have $$t_{ab} = t_a t_b - if^c_{ab} t_c.$$ So $$t_{ab}$$ is hermitian iff $$f^c_{ab}$$ is zero, which mean that $$\{t_a\}$$ group algebra is abelian.

I think that solves the problem. Any other opinions?

I looked at page 54 and Weinberg does not say that the $$t_{ab}$$ are Hermitian, only that the $$t_a$$ are Hermitian. I have the 7th reprinting of the paperback edition. Maybe it was wrong in earlier editions?
• The version I have says: "where $t_a, t_{bc} = t_{cb},$ etc. are Hermitian operators independent of the $\theta$s." I think the one I have is a very old edition, from around 1996, so perhaps it's been corrected. Sep 4, 2020 at 14:15