I have a question regarding the evolution of Lie algebra conditions in Weinberg's The Quantum Theory of Fields vol. 1: Foundations, chapter 2.

I will reproduce the text here and state my two questions throughout.

(Page 53-55)

"[...] a connected Lie group [...is a...] group of transformations $T(\theta)$ that are described by a finite set of real continuous parameters, say $\theta^a$, with each element of the group connected to the identity by a path within the group. The group multiplication law then takes the form

$ T(\bar{\theta})T(\theta) = T(f(\bar{\theta},\theta))\tag{2.2.15} $

With $f^a(\bar{\theta},\theta)$ a function of the $\bar{\theta}$'s and $\theta$'s. Taking $\theta^a =0$ as the coordinates of the identity, we must have

$ f^a(\bar{\theta},0) =f^a(0,\theta)=\theta^a\tag{2.2.16} $ "

He reminds the reader that transformations of such groups must be, in Hilbert space, represented by unitary operators $U(T(\theta))$.

We can take our representation of the unitary operator $UU^\dagger =1 \rightarrow UU^\dagger = e^{iT}e^{-iT} =1$, ie $U \equiv e^{iT}$ to motivate a power series expansion in the neighborhood of the identity:

"$ U(T(\theta)) = 1+i\theta^at_a+\frac{1}{2}\theta^b\theta^ct_{bc}+\ldots\tag{2.2.17} $

where $t_a$, $t_{bc} =t_{cb}$ are operators independent of the $\theta$'s, with $t_a$ hermitian.

Suppose that the $U(T(\theta))$ form an ordinary representation of this group [...] ie,

$ U(T(\bar{\theta}))U(T(\theta)) =U(T(f(\bar{\theta},\theta)))\tag{2.2.18} $ "

We next expand $f^a(\bar{\theta},\theta)$ to second order "according to Eq. (2.2.16)", getting

$ f^a(\bar{\theta},\theta) = \theta^a +\bar{\theta^a} +f^a_{bc}\bar{\theta^b}\theta^c \tag{2.2.19} $

where $f^a_{bc}$ are real coefficients. We can now expand Eq (2.2.18) as

$ \left[1+i\bar{\theta^a}t_a+\frac{1}{2}\bar{\theta^b}\bar{\theta^c}t_{bc}+\ldots \right]\times\left[1+i\theta^at_a+\frac{1}{2}\theta^b\theta^ct_{bc}+\ldots \right] = 1 +i(\theta^a+\bar{\theta^a}+f^a_{bc}\bar{\theta^b}\theta^c+\ldots )t_a+\frac{1}{2}(\theta^b+\bar{\theta^b}+\ldots)(\theta^c+\bar{\theta^c}+\ldots)t_{bc}+\ldots\tag{2.2.20} $

This is where my first question arises. Weinberg now says that "the terms of order $1$, $\theta$,$\bar{\theta}$,$\theta^2$,and $\bar{\theta^2}$ automatically match on both sides, but from the $\bar{\theta}\theta$ terms we obtain a non-trivial condition

$ t_{bc}=-t_bt_c-if^a_{bc}t_a\tag{2.2.21} $

But if I group terms which have only two mixed $\bar{\theta}$ or $\theta$, ie: $\bar{\theta}\theta$ or $\theta\bar{\theta}$ only, I get

$-\bar{\theta^a}\theta^at_at_a$ on the LHS and

$if^a_{bc}\bar{\theta^b}\theta^ct_a + \frac{1}{2}(\theta^b\bar{\theta^c}+\bar{\theta^b}\theta^c)t_{bc}$ on the RHS.

Now from here I don't see how we can cleanly get to Eq(2.2.21). This is my first point of departure. The $\theta$ and $\bar{\theta}$ are real, continuous variables, so if order and barring is unimportant, I can see how the RHS would become something like:

$ \left[if^a_{bc}t_a + t_{bc}\right](\bar{\theta^b}\theta^c) $

But a.) $\bar{\theta^b}\theta^c$ and $\bar{\theta^a}\theta^a$ are composed of different continuous variables, and I therefore do not see how they can be equated. And b.) even if they were, we would get:

$ -t_at_a=if^a_{bc}t_a + t_{bc} $

Which is pointedly not the same as Eq. (2.2.21).

If $-t_at_a \rightarrow-t_bt_c$, then this would work. But how can we be so care-free with the indexing of continuous variables? If they are interchangeable, why label them at all?

Eq. (2.2.21) is then used to establish the requirement that

$ [t_b,t_c]=iC^a_{bc}t_a\tag{2.2.22} $

Under the consistency condition that the operator $t_bc$ must be symmetric in b and c (because it is the second derivative of $U(T(\theta))$ with respect to $\theta^b$ and $\theta^c$.

The last point of confusion for me comes on page 55, where we consider the specific case where the group structure is given by $f^a(\theta,\bar{\theta})=\theta^a+\bar{\theta^a}\tag{2.2.24}$

Weinberg then says "In this case, it is easy to calculate $U(T(\theta))$ for all $\theta^a$. From Eqs. (2.2.18) and (2.2.24) we have for any integer N

$$U(T(\theta))=\left[U \left(T\left(\frac{\theta}{N}\right)\right)\right]^N\tag{1'} $$

Letting $N\rightarrow\infty$, and keeping only the first-order term in $U(T(\theta/N))$, we have then

$$ U(T(\theta))=\lim_{N\to\infty} \left[1+\frac{i}{N}\theta^at_a\right]^N\tag{2'} $$

And hence


I do not understand how Eq. (2.2.18) and (2.2.24) entail Eq. (1') and (2'), and in this case I'm not even sure where to start. Any help would be greatly appreciated.

  • 1
    $\begingroup$ Be careful, $(\bar\theta^at_a) \times (\theta^at_a)$ is ambiguous with Einstein notation. What it really means is $(\sum\bar\theta^at_a) \times (\sum\theta^at_a)=(\sum\bar\theta^at_a) \times (\sum\theta^bt_b)=\bar\theta^at_a\theta^bt_b=\bar\theta^a\theta^b t_at_b$. When indices are repeted, you can change their name, they are just dummy indices. When you have a product of sums, be careful and change the variables names to avoid this kind of mistakes. $\endgroup$
    – Adam
    Commented Dec 16, 2019 at 22:11
  • $\begingroup$ For the last part, 2.2.18 says that the product of two operators, both of them with argument $\theta/2$ is the operator with argument $f^a(\theta/2, \theta/2) = \theta/2 + \theta/2 = \theta$. Now cut it in N pieces (or use induction) to get $(1')$, then $(2') follows from the general 2.2.17 expansion. $\endgroup$
    – NickD
    Commented Dec 16, 2019 at 22:18
  • $\begingroup$ When you have multiple questions, it is advised to ask a separate question on physics.SE for each one. However, a hint for the second one: you have in this case $U(T(\theta))U(T(\bar\theta)=U(T(\theta+\bar\theta))$. $\endgroup$
    – Adam
    Commented Dec 16, 2019 at 22:18

1 Answer 1


In (2.2.20) you used the same indices in both brackets in lhs. But the sums in those brackets should be independent. So you should write them using different indices, e.g. $\bar{a},\bar{b},\bar{c}$ in the first bracket. Then instead of $$\bar{\theta}^a\theta^a t_a t_a$$ we actually have $$\bar{\theta}^{\bar{a}}\theta^a t_{\bar{a}} t_a$$ where we can rename indices $\bar{a}\mapsto b, a\mapsto c$ and get Weinberg's formula

When (2.2.24) is true we can substitute $\bar{\theta}\mapsto \theta/N$,$\theta\mapsto\theta/N$ into (2.2.18). Then you'll get $$U(T(2\theta/N))=\Big[U(T(\theta/N))\Big]^2$$ Then we can multiply it by $U(T(\theta/N))$ to get, $$U(T(3\theta/N))=\Big[U(T(\theta/N))\Big]^3$$ Repeating the process we get (1'). Then if $N$ is large you neglect terms starting with $1/N^2$ and get (2') which is then you use Euler's formula for the exponential function to get the answer


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