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## Main Question or Discussion Point

Can someone explain what's Z2 symmetry ? Is it necessary to have it in a model, even SM ?

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Can someone explain what's Z2 symmetry ? Is it necessary to have it in a model, even SM ?

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Are you talking about [itex]\mathbb{Z}_2[/itex]?

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So, for example, I can write this lagrangian:

[tex]\mathcal{L} = \frac{1}{2}\partial_{\mu} \phi \partial^{\mu} \phi+\lambda\phi^4[/tex]

The Z2 symmetry is manifest---that is I can always take [tex]\phi[/tex] to [tex]-\phi[/tex] and get the same lagrangian back.

As far as necessarily needing it for anything, I don't know, but I don't suspect there's anything particularly deep about it.

- #4

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You can also view this as the numbers 0 and 1 with "group multiplication" identified with addition modulo 2. See for yourself:

0+0 = 0

0+1 = 1+ 0 = 1

1+1 = 0 (mod 2)

In the context of physics the [itex]\mathbb{Z}_2[/itex] symmetry usually refers to the fact that we are dealing with some system which, among a lot more stuff, contains an invariance with respect to some [itex]\mathbb{Z}_2[/itex] operation. A simple case is the example given by BenTheMan. Other contexts include the Ising model, [itex]\mathbb{Z}_2[/itex] topological quantum field theory, orbifolds, etc.

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What does it mean to have a [tex]-\phi[/tex] field ?

So, for example, I can write this lagrangian:

[tex]\mathcal{L} = \frac{1}{2}\partial_{\mu} \phi \partial^{\mu} \phi+\lambda\phi^4[/tex]

The Z2 symmetry is manifest---that is I can always take [tex]\phi[/tex] to [tex]-\phi[/tex] and get the same lagrangian back.

As far as necessarily needing it for anything, I don't know, but I don't suspect there's anything particularly deep about it.

- #6

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Hmm. I don't know. In the context of the example I gave you, it doesn't mean anything because of the symmetry. [tex]\phi[/tex] and [tex]-\phi[/tex] give the same lagrangian, so the equations of motion are unchanged---that is, the physics is exactly the same, so (in some sense) there IS no meaning to [tex]-\phi[/tex].What does it mean to have a [tex]-\phi[/tex] field ?

- #7

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In that case a postive [itex]\phi(x)[/itex] is a displacement of the field in the positive direction at the point x, and a negative [itex]\phi(x)[/itex] is one in the negative direction. (think of it as a rubber sheet stretched out, with bumps here and there).

The fact that we have a symmetry means that the

- #8

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You can see this quite nicely in the Kaluza-Klein lagrangian where the metric is parameterized on an [itex]S^1[/itex] compactification:

[tex]ds^2=\phi^{-1/3}(g_{\mu\nu}+A_\mu A_\nu\phi)dx^\mu dx^\nu+\phi^{2/3}A_\mu dx^\mu dy+\phi^{2/3}dy^2[/tex]

The [itex]\mathbb{Z}_2[/itex] orbifolding is obtained after identifying y with -y.

The invariance of the interval [itex]ds^2[/itex] under the symmetry determines the transformation properties of all the fields:

[tex]g_{\mu\nu}(y)=g_{\mu\nu}(-y)[/tex]

[tex]A_{\mu}(y)=-A_{\mu}(-y)[/tex]

[tex]\phi(y)=\phi(-y)[/tex]

Since the field [itex]A_\mu[/itex] is odd under this symmetry it cannot have a zero mode, eg when you write the fields out like:

[tex]A_\mu(\vec{x},y)=\sum_{n=-\infty}^\infty A(\vec{x})e^{iny/r}[/tex]

:)

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