# Weinberg's "Derivation" of Lie algebra commutation relations

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

$$U(T(\theta))=e^{it_a\theta^a}\tag{2.2.26}$$"

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.

• 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.
• 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 $(2') follows from the general 2.2.17 expansion. Commented Dec 16, 2019 at 22:18 • When you have multiple questions, it is advised to ask a separate question on physics.SE for each one. However, a hint for the second one: you have in this case$U(T(\theta))U(T(\bar\theta)=U(T(\theta+\bar\theta))\$.
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