Baker-Campbell-Hausdorff formula

[Bresden]

Hi everybody!

I have a problem with the Baker-Campbell-Hausdorf formula in third order approximation. I hope there is anyone who can help me through this calculation!

1. The Problem Statement
All Lie group elements can be written as
$U(\alpha_I)=\exp(i \alpha_I T^I).$
Proof the Baker-Campbell-Hausdorff formula
$U(\alpha_I) U(\beta_I) = U(\gamma_I),$
up to the third order.

First Order
$U(\alpha_I) U(\beta_I) = (\mathbb{1} + i \alpha_I T^I)(\mathbb{1} + i \beta_I T^I) = \mathbb{1} + i(\alpha_I + \beta_I)T^I \Rightarrow \gamma^{(1)}_I = (\alpha_I + \beta_I)$


Second Order
$U(\alpha_I) U(\beta_I) = (\mathbb{1} + i \alpha_I T^I - \frac{1}{2} \alpha_I \alpha_J T^I T^J)(\mathbb{1} + i \beta_I T^I - \frac{1}{2} \beta_I \beta_J T^I T^J) = \mathbb{1} + i(\alpha_I + \beta_I)T^I - \frac{1}{2} \left( \alpha_I \alpha_J + \beta_I \beta_J + 2 \alpha_I \beta_J \right)T^I T^J = \mathbb{1} + i(\alpha_I + \beta_I)T^I - \frac{1}{2} \left[ (\alpha_I + \beta_I)(\alpha_J + \beta_J)T^I T^J + \alpha_I \beta_J[T^I,T^J] \right] \Rightarrow \gamma^{(1)}_I = \alpha_I + \beta_I$
$\Rightarrow \gamma^{(2)}_I = \alpha_I + \beta_I - \frac{1}{2} \alpha_I \beta_J [T^I, T^J]$

Third Order
HOW TO CALCULATE THIS?

THX for every help!

 

[NateD]

</code>The third order of the Baker-Campbell-Hausdorff formula is:U(\alpha_I) U(\beta_I) = (\mathbb{1} + i \alpha_I T^I - \frac{1}{2} \alpha_I \alpha_J T^I T^J + \frac{1}{6} \alpha_I \alpha_J \alpha_K T^I T^J T^K)(\mathbb{1} + i \beta_I T^I - \frac{1}{2} \beta_I \beta_J T^I T^J + \frac{1}{6} \beta_I \beta_J \beta_K T^I T^J T^K)= \mathbb{1} + i(\alpha_I + \beta_I)T^I - \frac{1}{2} \left[ (\alpha_I + \beta_I)(\alpha_J + \beta_J)T^I T^J + \alpha_I \beta_J[T^I,T^J] \right] + \frac{1}{6} \left[(\alpha_I + \beta_I)(\alpha_J + \beta_J)(\alpha_K + \beta_K)T^I T^J T^K + \alpha_I \beta_J \beta_K [T^I, T^J, T^K] + \alpha_I \beta_J \alpha_K [T^I, T^K, T^J] + \alpha_I \alpha_J \beta_K [T^J, T^K, T^I] \right] \Rightarrow \gamma^{(1)}_I = \alpha_I + \beta_I \Rightarrow \gamma^{(2)}_I = \alpha_I + \beta_I - \frac{1}{2} \alpha_I \beta_J [T^I, T^J]\Rightarrow \gamma^{(3)}_I = \alpha_I + \beta_I - \frac{1}{2} \alpha_I \beta_J [T