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What is the meaning and significance, for example, of the following commutators $[\phi,P_\mu]=i\partial_\mu\phi$ and $[\psi,J_{\mu\nu}]=(i(x_\mu\partial_\nu-x_\nu\partial_\mu)+\frac{1}{2}\sigma_{\mu\nu})\psi$? Here, we see that the commutator of a field with a generator gives an explicit representation of the generator acting on that field.

What can we conclude from these commutator relations?

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  • $\begingroup$ Those relations are nothing but that the "infinitesimal" representation of the adjoint action of $P_{\mu}$ and $J_{\mu\nu}$ onto the fields as translations and rotations, respectively. $\endgroup$ – gented Nov 30 '16 at 19:47
  • $\begingroup$ @GennaroTedesco What do you mean by adjoint action? $\endgroup$ – SRS Nov 30 '16 at 19:50
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    $\begingroup$ The group of transformations, say, translations, for example, act on the fields as $U_a^*\phi(f)U_a = \phi(f_a)$ with $f_a(x)= f(x-a)$, where here $U_a$ is a finite translation generated as $U_a = e^{iP_{\mu}a^{\mu}}$. $\endgroup$ – gented Nov 30 '16 at 19:53
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Representations on the Hilbert space $|\psi\rangle\mapsto\mathcal R_g|\psi\rangle$ can be lifted to the space of linear operators by the adjoint action $\mathcal O\mapsto \mathcal R_g\mathcal O \mathcal R_g^{-1}$, or on the level of the Lie algebra $\mathcal O\mapsto [\mathcal R_a,\mathcal O]$, where $a$ is a Lie algebra element. The space of linear operators can thus be decomposed into direct sums of irreducible representations of the Lie algebra (just like the Hilbert space can), under the adjoint action $[\mathcal R_a, \cdot]$.

The meaning of your commutation relations is the following: operators satisfying the commutation relations will transform in a representation of your group/Lie algebra and have well-defined spin, scaling dimension (for conformal symmetry) etc. Thus it makes sense to organize your theory around these operators, if the group is a symmetry group. You can in principle create states in the Hilbert space using these operators, that will tranform in a well defined representation of the symmetry group. Any operator commuting with your symmetry group (observables for example), will map states to other states within their irreducible representation, and not mix with other representations. Thus it's matrix elements between different representations will be zero. This is one of the reasons why this decomposition is useful.

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