
#1
Jan1913, 12:48 PM

P: 315

In addition to my FaddeevPopov 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. 



#2
Jan2013, 07:50 AM

P: 987

ψ'_{j}^{2}=(R_{jk}ψ_{k})(R_{jl}ψ_{l})=δ_{kl}ψ_{k}ψ_{l}=δ_{jl}ψ_{j}ψ_{l}=ψ_{j}^{2}




#3
Jan2013, 10:58 AM

P: 315

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. 



#4
Jan2013, 11:49 AM

P: 235

Linear Sigma Model Invariance Under O(N)
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.




#5
Jan2013, 03:26 PM

P: 315

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


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