Just a quick question: Suppose I want to consider QED or YM in 4 dimensions we always say that the coupling constants are dimensionless and that the field then has a specific mass dimension. What happens if we change the dimensions we are considering? Do the mass dimensions of the fields then change or do the coupling constants become dimensionful?
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2 Answers
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Do the mass dimensions of the fields then change or do the coupling constants become dimensionful?
As checked by @Funzies, the mass dimension of the fields change, for instance the mass dimension for scalar bosonic fields is $[\phi]=\frac{d-2}{2}$. This is because you have always a kinetic term $(\partial \phi)^2$ in the Lagrangian, and that the Lagrangian hass mass dimension $d$, such as the mass dimension of the action is zero.
When you have in the Lagrangian , an interacting term of kind $\alpha ~\phi^p$, you must have $[\alpha]+ p [\phi]= d$, so finally $[\alpha] = d - \frac{p(d-2)}{2}$
For instance, in $d=4$ dimensions, a interacting term in $\alpha ~\phi^4$ has a dimensionless coupling constant $\alpha$. For other dimensions, this coupling constant is dimensionfull, for instance, in $d=6$ dimensions, $[\alpha]=-2$
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The action $S$ must always be dimensionless. In $d$ dimensions it looks like: $$ S = \int d^dx \mathcal{L}, $$ with $\mathcal{L}$ the Lagrange density. Recall that $[x] = -1$, this follows from $[x,p]=i$ (commutator with $\hbar := 1$), such that $[x] = -[p] = -1$, thus we must have: $[\mathcal{L}] = d$. A standard mass term would be $m^2\phi^2$, so the dimension of the field becomes $[\phi] = 1/2(d-2)$.
