I have the following Lagrangian (density) for bosons

$$L = \partial_{\mu} \phi^i \partial^{\mu}\phi^i+ m^2\phi^i \phi^i$$

and I am trying to understand why this Lagrangian is invariant under global internal $SO(n)$ transformations. I have the following transformation $$\phi'^i = R_{\,\,j}^i\phi^j$$where $R_{\,\,j}^i = \delta_{\,\, j}^i-\epsilon r_{\,\,j}^i$ for $\epsilon <<1$. Of course $r_{\,\,j}^i=-r_{\,\,i}^j$. So what I am trying is to substitute the transformed fields into the Lagrangian. Still, after doing so I do not see how the extra terms cancel out so that I get the original Lagrangian! So, why is this Lagrangian $SO(n)$ invariant?

P.S. I can see this if $\phi$ is complex, but here I take it to be real.

  • $\begingroup$ I added something to my answer that you might want to read. $\endgroup$
    – Ryan Unger
    Dec 5, 2014 at 23:06

1 Answer 1


Stack all the $\phi^i$s into a column "vector" $\vec\phi$. The mass term $m^2\vec\phi\cdot\vec\phi$ is obviously invariant by $R^{-1}=R^T$. The same with the kinetic term $(\partial_\mu\vec\phi)\cdot(\partial^\mu\vec\phi)$ because $\partial_\mu R=0$. It is $SO(n)$ invariant because I take it $i$ runs over $n$ values. Thus your $r^i_j$ generates $SO(n)$.

EDIT: Not sure if OP will read this, but I'd like to clarify something. The complex case for two bosonic fields amounts to setting $\sqrt{2}\Phi\equiv \phi^1+i\phi^2$ and writing the Lagrangian as $$\mathcal{L}=\partial_\mu\Phi^*\partial^\mu\Phi+m^2|\Phi|^2$$ Alternatively we could write $$\vec\phi= \begin{pmatrix} \phi^1 \\ \phi^2 \end{pmatrix}$$ Above I showed that this is invariant under (in this case) $SO(2)$. However, the $\Phi$ Lagrangian also enjoys a $U(1)$ symmetry: $\Phi\longrightarrow e^{i\theta}\Phi$ and $\Phi^*\longrightarrow e^{-i\theta}\Phi^*$. So this theory actually has both $SO(2)$ and $U(1)$ symmetry. Mathematically this is because $U(1)\cong SO(2)$.


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