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

which is invariant under

[itex]\phi^{i}[/itex] --> R[itex]^{ij}[/itex] [itex]\phi^{j}[/itex],

or, the Orthogonal Group O(N).

To show this, I've been using:

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

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

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!