4
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Are you dealing with a four dimensional spacetime? $\endgroup$ Commented Jan 7, 2017 at 12:36

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.