# Linear scale transformations for bosons and fermions

In the book from Coleman: The Aspects of Symmetry, p. 70; linear scale transformations or dilations are defined as $$x \rightarrow e^\alpha x$$ with $\alpha$ being a real number. The fields change as $$\phi(x) \rightarrow e^{\alpha d} \phi (e^{\alpha x})$$ which yields an infinitesimal transformation $$\delta \mathbf{\phi} = (d + x^\lambda \partial_\lambda ) \mathbf \phi$$where $d$ is a matrix. (It is clearer as $\delta \phi_i = (d_{ij} + x^\lambda \partial_\lambda \delta_{ij} ) \phi_j$, where $\delta_{ij}$ is the Kronecker delta.)

Now, there is a statement I have not been able to prove:

For a large class of theories (including all renormalizable field theories) these transformations are symmetries, if all non-dimensionless coupling constants (including the masses) are set equal to zero, and if $d$ is chosen to be a matrix that multiplies all Bose fields by one and all Fermi fields by $\frac{3}{2}$.

Based on this quote, I have tried using the lagrangian $\mathcal L = \partial_\mu \phi_i \partial^\mu \phi_i$ which is the Klein Gordon lagrangian with $m=0$. Computing $\mathcal L[\phi + \delta \phi] - L[\phi]$ gives the variation of the action which should be 0 up to a total derivative when $d_{ij} = 1 \cdot \delta_{ij}$. However I find terms with extra derivatives that do not cancel. I would like to know how to find the 1 and 3/2 for bosons and fermions. I guess this is a general result and there is no need to pick a specific lagrangian.

• Are you dealing with a four dimensional spacetime? Commented Jan 7, 2017 at 12:36

For a massless scalar field, $$I[\phi] :=\int g^{\mu\nu}\frac{\partial \phi(x)}{\partial x^\mu}\frac{\partial \phi(x)}{\partial x^\nu} dx^0dx^1 dx^2dx^3$$ where $g = diag(-1,1,1,1)$.
If replacing $\phi \to \phi_\lambda$ where $$\phi_\lambda(x) := \lambda \phi(\lambda x) \quad \mbox{for \lambda >0}$$ (evidently $\lambda = e^\alpha$), we have $$I[\phi_\lambda]=\int g^{\mu\nu}\frac{\partial \phi(\lambda x)}{\partial x^\mu}\frac{\partial \phi(\lambda x)}{\partial x^\nu} \lambda^2 dx^0dx^1dx^2dx^3 \:.$$ That is $$I[\phi_\lambda]=\int g^{\mu\nu}\frac{\partial \phi(\lambda x)}{\partial \lambda x^\mu}\frac{\partial \phi(\lambda x)}{\partial \lambda x^\nu} \lambda^4 dx^0dx^1dx^2 dx^3 = \int g^{\mu\nu}\frac{\partial \phi(\lambda x)}{\partial \lambda x^\mu}\frac{\partial \phi(\lambda x)}{\partial \lambda x^\nu} d \lambda x^0 d \lambda x^1 d\lambda x^2 d \lambda x^3 = \int g^{\mu\nu}\frac{\partial \phi( y)}{\partial y^\nu}\frac{\partial \phi(y)}{\partial y^\nu} dy^0dy^1dy^2dy^3 = I[\phi]\:.$$ thus $$I[\phi_\lambda]= I[\phi]\:.$$ The difference with spinors is that only one derivative enters the action and this explains the different power. I leave you the simple computations.