Linear Sigma Model Invariance Under O(N)

[dm4b]

In addition to my Faddeev-Popov Trick thread, I'm still tying up a few other loose ends before going into Part III of Peskin and Schroeder.

I was able to show that the other Lagrangians introduced thus far are indeed invariant under the transformations given. But, I am hung up on what I think should probably be the easiest - the linear sigma model from page 349, Chapter 11:

L_{LSM} = (1/2) ( \partial_{\mu} \phi^{i} )^2 + (1/2)\mu^2 ( \phi^{i} )^2 - (\lambda/4!) ( \phi^{i} )^4

which is invariant under

\phi^{i} --> R^{ij} \phi^{j},

or, the Orthogonal Group O(N).

To show this, I've been using:

\phi^{j} ^2 --> R^{ij} R^{ik} \phi^{j} \phi^{k}
= \delta^{j}_{k} \phi^{j} \phi^{k}
= \phi^{j} ^2

but, I guess I haven't convinced myself. Seems contrived (with the indices)

Any help/clarification would be greatly appreciated.

 

[andrien]

ψ'_j²=(R_jkψ_k)(R_jlψ_l)=δ_klψ_kψ_l=δ_jlψ_jψ_l=ψ_j²

 

[dm4b]

ψ'_j²=(R_jkψ_k)(R_jlψ_l)=δ_klψ_kψ_l=δ_jlψ_jψ_l=ψ_j²

Thanks andrien.

Looks like that's exactly what I have above in the OP, so I guess you're confirming that's correct.

Don't know why it still leaves me uneasy. I'll probably work out some explicit examples next, as that usually clears things up.

 

[Einj]

You can see it in a "vectorial way". O(N) are rotations and you know that these kind of transformations leave the value of the square of the vector unchanged. That's the same thing.

 

[dm4b]

You can see it in a "vectorial way". O(N) are rotations and you know that these kind of transformations leave the value of the square of the vector unchanged. That's the same thing.

Thanks Einj. I totally get it in a conceptual way like that.

It was just the notation with the math. Wasn't quite sure I had it right!