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Baker-Campbell-Hausdorff formula

  1. Oct 3, 2011 #1
    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
    [itex]U(\alpha_I)=\exp(i \alpha_I T^I).[/itex]
    Proof the Baker-Campbell-Hausdorff formula
    [itex]U(\alpha_I) U(\beta_I) = U(\gamma_I),[/itex]
    up to the third order.

    First Order
    [itex]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)[/itex]


    Second Order
    [itex]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[/itex]
    [itex]\Rightarrow \gamma^{(2)}_I = \alpha_I + \beta_I - \frac{1}{2} \alpha_I \beta_J [T^I, T^J][/itex]

    Third Order
    HOW TO CALCULATE THIS?

    THX for every help!!
     
  2. jcsd
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