Linear Sigma Model Invariance Under O(N)


by dm4b
Tags: invariance, linear, model, sigma
dm4b
dm4b is offline
#1
Jan19-13, 12:48 PM
P: 315
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.
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andrien
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#2
Jan20-13, 07:50 AM
P: 987
ψ'j2=(Rjkψk)(Rjlψl)=δklψkψljlψjψlj2
dm4b
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#3
Jan20-13, 10:58 AM
P: 315
Quote Quote by andrien View Post
ψ'j2=(Rjkψk)(Rjlψl)=δklψkψljlψjψlj2
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
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#4
Jan20-13, 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.
dm4b
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#5
Jan20-13, 03:26 PM
P: 315
Quote Quote by Einj View Post
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!


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