I have a question about the dimensions of quantum fields. In natural units, the dimensions of bosonic fields (both scalar and vector) is 1. The dimension of spin=1/2 fermion fields is 3/2. This is all very good, but I have never read any explanation anywhere why we cannot have other types of fields where the dimension of the field is different from those mentioned above. For example, is there a reason why a field (of any spin) cannot have dimension= 1/2? For such a scalar field f, for example, there will be a term f^8 in the Lagrangian density with a dimensionless coupling constant in 4 space-time dimensions. For such a term, naive dimensional counting appears not to rule out renormalizibility outright. (In contrast, ordinary scalar fields of dimension=1 cannot have a renormalizable f^8 interaction.)(adsbygoogle = window.adsbygoogle || []).push({});

How about something even more exotic: why can't we have spin=1/2 fermion fields of dimension=1, instead of 3/2? Is there some simple argument or symmetry in QFT that forbids such exotics?

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# Dimensions of fields in QFT

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