I stumbled upon this point several times, the latest beeing this question: Connection between conserved charge and the generator of a symmetry

I want to understand, why Quantum fields transform under symmetry transformations as

$(g\cdot\phi)(y) = T_g^{-1}\phi(y) T_g = e^{-tX}\phi(y) e^{tX} = \big[ 1 + t[X,\cdot]+\mathcal{O}(t^2)\big]\phi(y)$

Two ideas from me:

1) Fields are treated as operators in the Heisenberg picture, and instead of transforming the states with $ T_g |x>$, the states stay as they are and all operators, including the quantum fields transform as $ T_g^{-1}O T_g$

2) There is a good reason why quantum fields live in $T_e$, i.e. the tangent space at the identity of the group (=the Lie algebra), onto which the group acts with the adjoint action: $Ad_g(x)= T_g^{-1} X T_g$ $\forall X \in T_e$

Any ideas or reading tips would be awesome!


1 Answer 1


Your idea 1) is the right idea: It's just the law of transformation of matrices generalized from the transformation of matrices:

If we apply a general linear transformation $U : V \to V$ on a vector space, the matrices/operators on it transform as $M \mapsto U^\dagger MU$

For unitary operators $U^\dagger = U^{-1}$, so the transformation law becomes $M \mapsto U^{-1}MU$. Since $T_g$ is the representation of the group upon our space of states, the quantum fields as operators transform according to this law.

  • $\begingroup$ shouldn't $M$ transform as $$M \rightarrow UMU^{-1}$$so that $$M|\alpha \rangle \rightarrow (U M U^{-1}) U |\alpha \rangle = U M | \alpha \rangle$$? $\endgroup$
    – glS
    Oct 13, 2014 at 21:36
  • $\begingroup$ @glance: I'm always a bit confused by that myself, but that would only hold if we were doing a basis change of the space (then, the trafo would also be $U^{-1}$ from the start). Instead, when applying a symmetry, we either implement it on the states as $\lvert \psi \rangle \mapsto U\lvert \psi \rangle$ or on the operators by $ M \mapsto U^\dagger M U$. It's basically the same as switching between the Schrödinger and the Heisenberg picture. $\endgroup$
    – ACuriousMind
    Oct 13, 2014 at 21:44

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