1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I added something to my answer that you might want to read. $\endgroup$
    – Ryan Unger
    Dec 5, 2014 at 23:06

1 Answer 1

3
$\begingroup$

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 $U(1)\cong SO(2)$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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