# Infinitesimal transformations that leave the action invariant

I have the Klein-Gordon Lagrangian for three scalar fields and I want to find three independent infinitesimal transformations that leave the action invariant.

I suppose that these three transformations are rotation, translation and Lorentz transformation. But, if this first step is right I have no idea how to calculate the conserved currents because at least for translation and Lorentz transformation only the coordinates change and not the fields. Do you have any hint/idea?

• Hint: field do change under active Lorentz transformations. The change in the field is equal to $\delta \varphi = \partial_{\mu} \varphi \cdot \delta x^{\mu}$ where $\delta x^{\mu}$ is the change in the coordinate. Hint 2: there isn't a lot of different things you can write down for a simple scalar field. Your currents can be constructed from the components of the stress-energy tensor. – Prof. Legolasov Apr 11 at 15:25
• Ok, thank you for your helpful hints. So, I suppose that the three transformations I considered are correct, right? – Vaso Karanasou Apr 11 at 15:32
• Geia sou Vaso, in D spacetime dimensions you can do D independent translations, D(D-1)/2 independent rotations (in the D-dimensional sense, so this is all Lorentz transformations which contain spatial rotations), and if you have conformal invariance you can also do 1 independent scale transformation and D independent so-called "special conformal transformations'', and these are only the spacetime symmetries. So, if D=4 there are 10 independent (spacetime) symmetry transformations (in absence of conformal symmetry) and 15 independent symmetry transformations in presence of conformal symmetry. – Wakabaloola Apr 11 at 17:20

This is fairly stardard QFT material: How to compute the Noether current for a general group of transformations that may involve both the fields and the coordinates \left\lbrace \begin{aligned} x^\mu\to {x'}^\mu &= x^\mu + \varepsilon\,\xi^\mu(x)\,,\\ \phi^i(x) \to {\phi'}^i(x') &= \phi^i(x) + \varepsilon\,t^i[\phi](x)\,. \end{aligned} \right. The operator $$t^i[\phi]$$ may be simply $$t^i_{\,j}\phi^j$$ but we can allow also for a more general functional dependence on $$\phi$$. The variation of the fields $$\delta_\varepsilon\phi$$ is (note the primes) $$\delta_\varepsilon\phi(x) \equiv {\phi'}^i(x) - \phi^i(x) = \varepsilon \,\big(t^i[\phi](x) - \xi^\mu\partial_\mu\phi^i(x)\big)\,.$$ Asking for invariance amounts to the requirement that the Lagrangian transforms as a total derivative, so one must have $$\delta_\varepsilon \mathcal{L} = \partial_\mu K_\varepsilon^\mu\,, \tag{1}\label{k}$$ for some $$K_\varepsilon^\mu$$. Define now the Noether current as $$J_\varepsilon^\mu(x) = K_\varepsilon^\mu(x) - \frac{\delta\mathcal{L}}{\delta\partial_\mu\phi^i}\delta_\varepsilon\phi^i(x)\,.\tag{2}\label{j}$$ This is a Noether current because it is conserved when the equations of motion are satisfied. Indeed $$\partial_\mu J^\mu_\varepsilon = \delta_\varepsilon\phi^i \frac{\delta\mathcal{L}}{\delta\phi^i} \approx 0\,.$$ The symbol $$\approx$$ means equality modulo stuff which vanishes when equations of motion are satisfied.
You can check for yourself that under Poincaré + dilatations the variations of the field are \begin{aligned} &\mbox{Translations}:\\ &\;\;\;\delta_\epsilon \phi^i(x) = - \epsilon^\mu \partial_\mu \phi^i(x)\,,\qquad &&\epsilon^\mu\;\mbox{constant vector}\\ &\mbox{Lorentz}:\\ &\;\;\;\delta_\omega \phi^i(x) = - \tfrac12 \omega_{\mu\nu}\big((\Sigma^{\mu\nu})^i_{\phantom{i}j}\phi^j(x) + (x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})\phi^i(x)\big)\,,\qquad &&\omega_{\mu\nu}\;\mbox{antisymm. matrix}\\ &\mbox{Dilatations}:\\ &\;\;\;\delta_\lambda\phi^i(x) = - \lambda \big(x^\mu\partial_\mu +\Delta_i\big)\phi^i(x)\,,\qquad &&\lambda\;\mbox{real constant} \end{aligned} The matrix $$(\Sigma^{\mu\nu})^i_{\phantom{i}j}$$ is the spin representation (e.g. $$\frac14[\gamma^\mu,\gamma^\nu]$$ for Dirac spinors, in your case it would just be zero because you have scalar fields) and $$\Delta$$ is the conformal dimension of $$\phi$$.
In order to compute the respective Noether currents you need to make the variation of the Lagrangian so as to compute $$K^\mu$$ as in \eqref{k} and then merely plug everything in \eqref{j}. The results will be of the form $$J_{\epsilon}^\mu = \epsilon_{\nu} T^{\mu\nu}\,,\qquad J_{\omega}^\mu = \omega_{\nu\rho} \mathcal{S}^{\mu\nu\rho}\,,\qquad J_{\lambda}^\mu = \lambda \mathcal{D}^{\mu}\,.$$ So you can extract the current you need by basically dropping the $$\epsilon_\mu,\,\omega_{\mu\nu}$$ and $$\lambda$$. Just remember to antisymmetrize the last two indices of $$\mathcal{S}^{\mu\nu\rho}$$ before removing $$\omega_{\nu\rho}$$.
• good answer, I would add that for a scalar in particular (which is what the question is about) we have by definition $t^i[\phi]=0$, that the $\Sigma^{\mu\nu}$ term is absent. You also have a typo in the case of Lorentz, you should erase the last factor, $\partial_\mu\phi^i$. Also, if you're going to add dilatations you might as well include special conformal transformations too, since the two really go hand in hand in most examples we know of. – Wakabaloola Apr 12 at 8:15
• And if $\Delta_i\neq0$ then $\phi^i$ does not transform as a scalar under dilatations (since in this case $t^i[\phi]\neq0$). – Wakabaloola Apr 12 at 8:38
• Most if the times by a scalar one means a spin scalar. In any CFT the only operator with $\Delta=0$ is the identity. – MannyC Apr 12 at 11:15
• Since this is meant to be pedagogical I wanted to emphasise the distinction, that the meaning of the word 'scalar' is always with respect to a specified symmetry. If you want to include dilatations as a symmetry (which is what your post does), one should make the distinction between scalars under the Poincare group and scalars under the full conformal group. Regarding your second comment, in dimensions other than $D=4$ there do of course exist non-trivial scalars under the full conformal group (e.g. in $D=2$ such a scalar is the standard bosonic matter field of worldsheet string theory). – Wakabaloola Apr 12 at 15:39