Understanding Z2 Symmetry in Models

[touqra]

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

 

[shoehorn]

Are you talking about $\mathbb{Z}_2$?

 

[BenTheMan]

Z2 is usually a symmetry like something goes to - something.

So, for example, I can write this lagrangian:

$$\mathcal{L} = \frac{1}{2}\partial_{\mu} \phi \partial^{\mu} \phi+\lambda\phi^4$$

The Z2 symmetry is manifest---that is I can always take $$\phi$$ to $$-\phi$$ 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.

 

[xepma]

Even simple, $\mathbb{Z}_2$ refers to a group, namely that of additon mod 2. It only has two elements, the identity and element a, which satisfies a^2 = 1.

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 $\mathbb{Z}_2$ 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 $\mathbb{Z}_2$ operation. A simple case is the example given by BenTheMan. Other contexts include the Ising model, $\mathbb{Z}_2$ topological quantum field theory, orbifolds, etc.

 

[touqra]

Z2 is usually a symmetry like something goes to - something.

So, for example, I can write this lagrangian:

$$\mathcal{L} = \frac{1}{2}\partial_{\mu} \phi \partial^{\mu} \phi+\lambda\phi^4$$

The Z2 symmetry is manifest---that is I can always take $$\phi$$ to $$-\phi$$ 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.

What does it mean to have a $$-\phi$$ field ?

 

[BenTheMan]

What does it mean to have a $$-\phi$$ field ?

Hmm. I don't know. In the context of the example I gave you, it doesn't mean anything because of the symmetry. $$\phi$$ and $$-\phi$$ 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 $$-\phi$$.

 

[xepma]

In a classical model $-\phi(x)$ stands for the amplitude of the field. The amplitude can be all sorts of things, most prominent example being the displacement of an atom with respect to some mean lattice.

In that case a positive $\phi(x)$ is a displacement of the field in the positive direction at the point x, and a negative $\phi(x)$ 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 different states $\phi(x)$ and $-\phi(x)$ carry the same energy.

 

[robousy]

$\mathbb{Z}_2$ is useful in higher dimensional theories to project out unwanted zero modes for the photon.

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

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

The $\mathbb{Z}_2$ orbifolding is obtained after identifying y with -y.

The invariance of the interval $ds^2$ under the symmetry determines the transformation properties of all the fields:

$$g_{\mu\nu}(y)=g_{\mu\nu}(-y)$$

$$A_{\mu}(y)=-A_{\mu}(-y)$$

$$\phi(y)=\phi(-y)$$

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

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

:)