The Lorentz groups act on the scalar fields as: $\phi'(x)=\phi(\Lambda^{-1} x)$

The conditions for an action of a group on a set are that the identity does nothing and that $(g_1g_2)s=g_1(g_2s)$. This second condition is not fulfilled because of the inverse on $\Lambda$. What is then the action of the Lorentz group on the scalar fields?

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    $\begingroup$ Would you mind writing out more carefully why the second condition isn't fulfilled? $\endgroup$ – user1504 Dec 29 '12 at 13:11
  • $\begingroup$ $(\Lambda_1 \Lambda_2)^{-1}=\Lambda_2^{-1} \Lambda_1^{-1}$ which is not $\Lambda_1^{-1} \Lambda_2^{-1}$ $\endgroup$ – inquisitor Dec 29 '12 at 13:20
  • $\begingroup$ It's OK, nothing else is needed. $\endgroup$ – Vladimir Kalitvianski Dec 29 '12 at 14:01

Denote by $g_1\phi$ the field transformed by the action of $\Lambda_1$ : $$(g_1\phi)(x) = \phi(\Lambda_1^{-1}(x))$$ Similarly $g_2$ has action $$(g_2\psi)(x) = \psi(\Lambda_2^{-1}(x))$$ Substitute $g_1\phi$ for $\psi$ $$(g_2g_1\phi)(x) = (g_1\phi)(\Lambda_2^{-1}(x)) = \phi(\Lambda_1^{-1}\Lambda_2^{-1}(x)) = \phi((\Lambda_2\Lambda_1)^{-1}(x)) $$ So the group action looks correct.

  • $\begingroup$ +1, you forgot the subscripts on the first two lines, though it's not really unclear :) $\endgroup$ – kηives Dec 29 '12 at 18:37
  • $\begingroup$ Thank you, just to write it on the original notation: $(g_2g_1)\phi(x)=\phi((\Lambda_2 \Lambda_1)^{-1} x)=\phi(\Lambda_1^{-1}\Lambda_2^{-1} x)$, and $g_2(g_1\phi)(x)=g_1\phi(\Lambda_2^{-1} x)= \phi(\Lambda_1^{-1}\Lambda_2^{-1} x)$ $\endgroup$ – inquisitor Dec 29 '12 at 21:20

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