Composing a few transformations

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SUMMARY

The discussion focuses on the simplification of a sequence of infinitesimal transformations represented by the expression $$U = e^{i\varepsilon K_{\mu}}e^{i\varepsilon K_{\nu}}e^{-i\varepsilon K_{\mu}}e^{-i\varepsilon K_{\nu}}$$ involving generators ##K_{\mu}## and ##K_{\nu}##. The user initially derived an expression that included an extra term, leading to confusion regarding the expected result quoted in the textbook. The correct simplification should yield $$U = I + \varepsilon^2 (K_{\nu} K_{\mu} - K_{\mu} K_{\nu}) + \mathcal{O}(\varepsilon^3)$$ without additional terms, emphasizing the importance of accurately expanding the transformations to the second order in ##\varepsilon##.

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I messed up somewhere, but don't know why! We consider this sequence of infinitesimal transformations,$$U = e^{i\varepsilon K_{\mu}}e^{i\varepsilon K_{\nu}}e^{-i\varepsilon K_{\mu}}e^{-i\varepsilon K_{\nu}}$$with ##K_{\mu}## and ##K_{\nu}## being two generators. I said, this simplifies to$$\begin{align*}

e^{i\varepsilon K_{\mu}}e^{i\varepsilon K_{\nu}}e^{-i\varepsilon K_{\mu}}e^{-i\varepsilon K_{\nu}} &= (I+i\varepsilon K_{\mu})(I+i\varepsilon K_{\nu})(I-i\varepsilon K_{\mu})(I-i\varepsilon K_{\nu}) + \mathcal{O}(\varepsilon^5) \\

&= \left[I + i\varepsilon(K_{\mu} + K_{\nu}) - \varepsilon^2 K_{\mu} K_{\nu} \right]\left[I - i\varepsilon(K_{\mu} + K_{\nu}) - \varepsilon^2 K_{\mu} K_{\nu} \right] + \mathcal{O}(\varepsilon^5) \\

&= I+ \varepsilon^2 (K_{\mu}^2 + K_{\mu} K_{\nu} + K_{\nu} K_{\mu} + K_{\nu}^2) -2 \varepsilon^2 K_{\mu} K_{\nu} + \mathcal{O}(\varepsilon^3) \\
&= I + \varepsilon^2 (K_{\nu} K_{\mu} - K_{\mu} K_{\nu}) + \varepsilon^2(K_{\mu}^2 + K_{\nu}^2) + \mathcal{O}(\varepsilon^3)

\end{align*}$$but the textbook only quotes$$U = I + \varepsilon^2 (K_{\nu} K_{\mu} - K_{\mu} K_{\nu}) + \mathcal{O}(\varepsilon^3)$$I wondered why I ended up with an extra term? Thank you.
 
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If you have an answer to second order ##\epsilon^2## then you ought to have quadratic terms in your original expansion.
 
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PeroK said:
If you have an answer to second order ##\epsilon^2## then you ought to have quadratic terms in your original expansion.

Ohhh, you're right! Guess I'll get started on expanding all that out, then...
 

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