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On generators of $SO(2)$

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.

One Answer

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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