# Mass dimension of coupling constant -- always an integer?

1. Feb 14, 2015

### guest1234

Just a simple question -- can the dimension of coupling constant be a rational number or should it always be an integer?

The question arose when I was trying to construct a Lagrangian with an interaction term involving two spin-1 particles and a fermion. The dimensions add up to 7/2, which leaves 1/2 for the coupling.

2. Feb 14, 2015

### Orodruin

Staff Emeritus
Any Lagrangian with an odd number of fermions will not be a Lorentz scalar. Therefore you will always end up with an even number of fermions leading to an integer mass dimension.

3. Feb 14, 2015

### guest1234

Thinking beyond SM, what are the physical consequences if the Lagrangian contains an interaction term with an odd number (i.e. only one) of fermions?

4. Feb 14, 2015

### Orodruin

Staff Emeritus
Going beyond the SM does not help (in fact, this was my assumption in my first post, in the SM there definitely are no such interactions), your Lagrangian will still not be a Lorentz scalar. You simply cannot do this within the confinements of QFT without breaking Lorentz invariance.

5. Feb 14, 2015

### guest1234

Well I was thinking whether it'd be pseudoscalar (that shouldn't be any big problem -- if we really want a parity violating theory). I just thought about gauge transformations, and yup, you were right -- the term wouldn't transform as a (pseudo)scalar but as a spinor. Adding scalar and spinor terms together doesn't make any sense, even mathematically.
Thanks anyways

Last edited: Feb 14, 2015
6. Feb 17, 2015

### vanhees71

If the Lagrangian has a term with an odd number of fermion fields, so will have the Hamiltonian, representing energy density. Then due to the canonical equal-time anticommutators for fermions, in general the commutator of the Hamiltonian at equal times in general won't commute, which violates the principle of microcausality, which is pretty bad, because the the time ordering appearing in the S matrix won't be Lorentz invariant anymore, and you'd loose causality (and perhaps unitarity) of the S matrix.

To keep the story short: It simply doesn't make much sense :-).