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 \begin{equation} U(T(\theta)))=1+i\theta^a t_a +\frac{1}{2} \theta_b\theta_c t_{bc} + \dots \end{equation} 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?


1 Answer 1


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?

  • $\begingroup$ 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. $\endgroup$
    – Philip
    Sep 4, 2020 at 14:15
  • $\begingroup$ Oh thanks! I think I am looking at an older version of the book. I will have a look at the newer version! $\endgroup$
    – Jiahao Fan
    Sep 4, 2020 at 16:53
  • $\begingroup$ @mike stone Hi! I want to know how do you get the reprinting version of the book? I have not found the digital version on the net.Thanks! $\endgroup$
    – Jiahao Fan
    Sep 6, 2020 at 12:28
  • $\begingroup$ I have a paperback copy. The printing number is on the back of the title page. I bought my three volumes new on Amazon. I imagine that all "new" copies will be up to date. It looks like there were quite a few careless errors in the early versions $\endgroup$
    – mike stone
    Sep 6, 2020 at 13:04

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