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Howard Georgi in his book on Lie algebras mentions a very interesting formula

$$\dfrac{\partial}{\partial a_b}e^{ia_aX_a} = \int_{0}^{1} ds \ e^{isa_aX_a}(iX_b)e^{i(1-s)a_cX_c}.$$

How can one use this formula to derive the Baker Hausdorff formulas and in consequence say calculate a general rotation matrix, $U(\theta) = e^{i\theta_a X_a}$, for spin 1 representation of $\mathrm{SU(2)}$?

EEEB
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    This seems to be a pure math question. – ACuriousMind Oct 15 '15 at 22:13
  • More specifically which Baker-Hausdorff formulas? – Qmechanic Oct 15 '15 at 22:19
  • Say I want to calculate $e^{i(\theta_x J_x + \theta_y J_y + \theta_z J_z)}$ –  Oct 15 '15 at 22:24
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    I do not know, it seems to me a very involved procedure if your goal is to compute that exponential. An explicit formula for $e^{i \theta \sum_{k=1} \sigma_k n_k}$ is instead given by $\cos(\theta)I + i\sin(\theta)\sum_{k=1} ^3\sigma_k n_k$, where $(n_1,n_2,n_3) \in \mathbb R^3$ is any unit vector. – V. Moretti Oct 17 '15 at 11:15

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You really don't need to go that painful way. It pays to collect your $\theta_a \equiv \theta~ n_a$, where $n_a$ are the components of a unit 3-vector $\hat{\boldsymbol{n}}$.

As a consequence, for spin 1, that is, the triplet representation of SU(2), the celebrated Rodrigues formula gives $$ e^{i\theta(\hat{\boldsymbol{n}}\cdot\boldsymbol{J})}=I_{3}+i(\boldsymbol{\hat {n}}\cdot\boldsymbol{J})\sin{\theta}+(\hat{\boldsymbol{n}}\cdot\boldsymbol{J})^{2}(\cos\theta-1)~. $$ The reason is that the eigenvalues of $\hat{\boldsymbol{n}}\cdot\boldsymbol{J}$ are the eigenvalues of $J_{3}=\mathrm{diag}(1,0,-1)$, and the Cayley-Hamilton theorem (dictating this matrix satisfies its own characteristic equation) reduces any higher power of it to a linear combination of the identity and the first two powers.

It is then straightforward to obtain this expression.

(You might find the compact formula for any irrep in Curtright et al, 2014.)

Cosmas Zachos
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