# Global symmetry of an N-component Klein-Gordon theory?

1. Dec 5, 2009

### weejee

The Lagrangian is given by,
$$\sum_{a=1}^N \left[(\partial^{\mu}\phi_{a}^{\ast})(\partial_{\mu}\phi_{a})-m^{2}\phi_{a}^{\ast}\phi_{a}\right]$$.

Is the symmetry SO(2N), SU(N) or U(N)?

It seemed quite obvious to me and some of my friends that such theory has an SO(2N) symmetry. If we view these N copies of complex K-G fields as 2N copies of real K-G fields, the Lagrangian is invariant under any rotation in the 2N dimensional space. It also seems that there should be N(N-1)/2, which is the number of generators in the SO(2N) group, conserved currents for this theory.

However, I have faced some objections to my claim. What they say is that the actual symmetry is SU(N). The reasoning for this claim was that the real and imaginary parts of these complex K-G fields cannot be considered independent, since they are related by causality. If we allowed an arbitrary SO(2N) rotation, particles and antiparticles would mix each other and the causality would be violated.

What makes me uncomfortable about this statement is that, first, I haven't been able to see any convincing formal development, rather than some hand waving arguments, for it, second, it would mean that one-component complex K-G theory doesn't have a U(1) symmetry.

If somebody said that the symmetry is U(N) due to causality, I would be less unhappy since we can save the 1-component K-G theory from not having even a U(1) symmetry.

I would be very grateful if any of you could clarify this issue, and if this causality argument is right, show me some formal elaboration to it. (or let me know where I can find it)

2. Dec 5, 2009

### ansgar

We dont you just try to implement the action of these symmetries and see if the Lagrangian is unchanged? Consider also the infinitesimal action of such symmetry transformations.

3. Dec 5, 2009

### weejee

It is clearly unchanged. The problem is that some say that we should also consider "causality", rather than just the invariance of the Lagrangian, which I haven't been able to understand.

4. Dec 5, 2009

### ansgar

why should we? I have never read that / been taught that - and I have studied 8 or 9 QFT textbooks. As you know, causality is due to existence of antiparticles, see QFT book by peskin chapter 2.

5. Dec 5, 2009

### Avodyne

It's always hard to understand incorrect arguments! :)