Power Series Expansion of Unitary Operators in Weinberg

Question

For a Lie group $T(\theta)$ depending on a finite set of real parameters $\theta^a$,Weinberg (in his QFT book - Equations 2.2.17, 2.2.18, 2.2.19, p54) expands the unitary representations of $T$ in the neighbourhood of the identity as $$U(T(\theta)) = 1 + \theta^a t_a + \frac{1}{2}\theta^b\theta^c t_{bc}+ ... \tag{2.2.17}$$ and for the product of two representations $$U(T(\bar{\theta}) U(T(\theta)) = U(T(f(\bar{\theta},\theta)))\tag{2.2.18}$$ does the expansion $$f^a(\bar{\theta},\theta) = \theta^a + \bar{\theta}^a + f^a_{bc}\bar{\theta}^b\theta^c.\tag{2.2.19}$$

The main thing I don't understand is how the coefficients and indices are being determined in these expansions. The expansion of $U(T(0))$ is presumably a taylor series expansion around 0, but then I don't understand why only the parameter $\theta^a$ appears in the second term and then is replaced by $\theta^b\theta^c$ in the third term, or why $t_a$ changes to $t_{bc}$ and isn't squared in the third term. I have similar confusions about the change in parameters in the third term of the product expansion.

If someone could tell me precisely what formula Weinberg is using I would appreciate it.

Answer 1

It's just the multivariable Taylor expansion. For functions of a single variable, $$f(x) = f(0) + x f'(0) + \frac{1}{2} x^2 f''(0) + \ldots$$ For functions of two variables, $$f(x,y) = f(\mathbf 0) + x f_x(\mathbf 0) + yf_y (\mathbf 0) + \frac{1}{2}\big(x^2 f_{xx}(\mathbf 0) + xyf_{xy}(\mathbf 0) + yx f_{yx}(\mathbf 0) + y^2 f_{yy}(\mathbf 0)\big) + \ldots$$ In index notation $$f(\mathbf r) = f(\mathbf 0) + r^i\big( \partial_i f\big)(\mathbf 0) + \frac{1}{2} r^j r^k \big(\partial_j \partial_k f\big)(\mathbf 0) + \ldots $$

In each term, the dummy indices are summed over. Strictly speaking, there is nothing wrong with reusing dummy indices in different terms (so we could write the second term as $r^i r^j\partial_i\partial_j$ without risk of ambiguity) but it's not a bad thing to avoid any potential confusion by not doing so.

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The set of all $n\times n$ Hermitian matrices is a real vector space of dimension $n^2$. Letting $t_a$ be a basis of this space for $a=1,\ldots, n^2$, one could write an arbitrary $n\times n$ Hermitian matrix $A$ as $$A = \sum_{a=1}^{n^2} \theta^a t_a$$ for some $\theta^a \in \mathbb R$. The matrix exponential $$U = \exp\left[i A\right] = \exp\left[i\sum_a \theta^a t_a\right]$$ is a unitary matrix whose power series expansion about the point $\boldsymbol \theta = \mathbf 0$ is $$U = \mathbb I + i \sum_{a=1}^{n^2} \theta^a t_a - \sum_{b=1}^{n^2}\sum_{c=1}^{n^2} \theta^b \theta^c t_a t_b + \ldots $$

The product $t_at_b$ is not generally Hermitian if $[t_a,t_b]\neq 0$. However, noting the symmetry in the sum we could define $$t_{bc} \equiv \frac{1}{2}(t_bt_c+t_ct_b)$$

and then write $$U = \mathbb I + i \sum_{a=1}^{n^2}\theta^a t_a - \sum_{b=1}^{n^2} \sum_{c=1}^{n^2} \theta^b \theta^c t_{bc} + \ldots $$ where now the $t_a$'s and the $t_{bc}$'s are all Hermitian (for each $a,b,c= 1,\ldots, n^2$).

It turns out that at least in a connected neighborhood of the identity element, all $n\times n$ unitary matrices can be written in this way. This is what Weinberg is doing.

Answer 2

A summation over those indices is intended. Since $a$ goes from $1$ to $Dim\big[Lie(T)\big]$ you can put the parameters into vector form and the generators as Lie-algebra valued vectors (ie, vectors having as components vectors of the Lie Algebra). Your summation can be intended as:

$\theta^at_a = \langle\vec{\theta}; \vec{t}\rangle$

Where $\langle ●$ ; $●\rangle$ is the inner product. Note that the result is not a scalar since $\vec{t}$ does not have numbers as components but it's a vector in $Lie(T)$.

Since $t_a$ is a vector it has no meaning to square it, so the second order term $t_b t_c$ is just the 2-tensor given by the (Lie) product of two generators. Denoting the product between the vectors as $\star$ you can understand the second term as:

$\big(\langle\vec{\theta}; \vec{t}\rangle\big)\star\big(\langle\vec{\theta}; \vec{t}\rangle\big)$

This is the Lie multiplication of the two vectors given by $ \langle\vec{\theta}; \vec{t}\rangle$ which in a Lie algebra it's again a vector of the algebra.