This is from Maggiore's "A Modern Introduction to Field Theory", Page 15.

I have a Lie group with matrix generators

$$ T^{a}_{R}$$

Where $a$ takes values from 1 to the dimension of the Lie group.

I want to show that $$ \alpha_{a} \beta_{b}[T^{a}_{R}, T^{b}_{R} ] = i \gamma_{c}(\alpha, \beta) T^{c}_{R} $$

(Einstein summation is implied.)

Where $$ \gamma_{c} = - 2 (\delta_{c}(\alpha, \beta) - \alpha_c - \beta_{c}).$$

In the matrix representation,

$$ \exp(i \alpha_{a} T^{a}_{R}) \exp(i \beta_{b} T^{b}_{R})= \exp(i \delta_{c}(\alpha, \beta) T^{c}_{R}) .$$

If I expand the exponentials (up to quadratic order) and take the natural logarithm of both sides, I have

$$ i \delta_{a} T^{a}_{R} = \log \Big(1 + i(\alpha_a + \beta_a)T^{a}_{R} - \frac{1}{2}(\alpha_a T^{a}_{R})^2 - \frac{1}{2}(\beta_a T^{a}_{R})^2 - \alpha_{a} \beta_{b} T^{a}_{R} T^{b}_{R}\Big).$$

I used the fact that $\log(1 + x) \approxeq x - \frac{x^2}{2}$ to get (up to quadratic T)

$$ i \delta_{a} T^{a}_{R} = i(\alpha_a + \beta_a)T^{a}_{R} - \frac{1}{2}(\alpha_a T^{a}_{R})^2 - \frac{1}{2}(\beta_a T^{a}_{R})^2 - \alpha_{a} \beta_{b} T^{a}_{R} T^{b}_{R} + \frac{(\alpha_{a} + \beta_{a})(\alpha_{b} + \beta_{b}) T^{a}_{R} T^{b}_{R}}{2}$$

Which implies that

$$ - 2i (\delta_{a} - \alpha_{a} - \beta_{a})T^{a}_{R} = (\alpha_a T^{a}_{R})^{2} + (\beta_{a} T^{a}_{R})^2 - (\alpha_{a} + \beta_{a})(\alpha_{b} + \beta_{b})T^{a}_{R} T^{b}_{R}.$$

The left hand side is just as required. However, I cannot get the commutator. The only surviving term after cancellation on the right hand side is

$$ -(\alpha_a \beta_b + \beta_a \alpha_b) T^{a}_{R} T^{b}_{R}.$$

Is it possible to get the commutator from here? Where have I gone wrong?


In the second last step I missed a term: $2 \alpha_{a} \beta_{b} T^{a}_{R} T^{b}_{R}$. The RHS reads

$$ (\alpha_{a} \beta_{b} - \beta_{a} \alpha_{b}) T^{a}_{R} T^{b}_{R}$$

In the second term, we relabel $a\leftrightarrow b$ to get the commutator $\alpha_{a}\beta_{b} [T^{a}_{R}, T^{b}_{R} ]$.

Thank you Mane.andrea for pointing out my error.

  • $\begingroup$ In your last equation there should be a minus between the two terms, you lost a $2\alpha_a \beta_b T^a_RT^b_R$ from the third to last equation to the second to last equation (I think). Then relabel $a\leftrightarrow b$ in the second term of the last equation and there you go. $\endgroup$ – MannyC Feb 13 at 13:59

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