Physics Asked on June 18, 2021
I’m reading the book "Symmetry and the Standard Model" by Matthew Robinson. On page 80 of the book the author asserts that given a representation $D(g)$ of a group $g$, we can expand around the identity element as:
$$D_n(g(0 + delta alpha_i)) = mathbb{I} + delta alpha_i frac{partial D_n (g(alpha_i))}{partial alpha_i}Bigr|_{alpha_i=0} + …$$
And we can define the generators of this representations as
$$X_i = -i frac{partial D_n}{partial alpha_i}Bigr|_{alpha_i=0}$$
Then the author derives the generators for $SO(2)$ arguing that rotations in the plane leave the scalar product $v^Tv$ invariant and for some generator of SO(2) we have that
$$v rightarrow R(theta) v = exp(i theta X) v$$
So expanding up to first order we get:
$$v^T e^{i theta X^T}e^{i theta X}v = v^T(1 + itheta X^T + i theta X) v = v^Tv + v^Ti theta(X+X^T)v$$
So we conclude that $X = – X^T$ for the dot product to vanish, asserting that $X$ is given by:
$$X= frac{1}{i} left[ begin {array}{cc} 0&1 -1&0end {array}
right]
$$
But why does the generator assumes this form? In particular, why are the digonal elements zero? Why can’t they any other number? Since this will still product an antisymmetric 2×2 matrix.
Why are the digonal elements zero?
You said it yourself that $X = -X^T$. Along the diagonal elements this means that
$$ X_{ii} = -X_{ii} qquad text{(no sum on $i$)} $$
So the only solution for the diagonal entries is that they are zero.
Correct answer by InertialObserver on June 18, 2021
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