# Is there a scalar field that is not a lorentz scalar if we begin with Lorentz invariant Lagrangian?

In Quantum Field Theory by Mark Srednicki chapter 3 and 4, he constructs Lorentz invariant theory for scalar field by assuming that the scalar field transforms by $$\tag1U(\Lambda)^{-1}\phi(x)U(\Lambda) = \phi(\Lambda^{-1}x)$$ and then he finds equation that is Lorentz invariant under transformation.

The equation that he finds is namely the Klein-Gordon equation, which for the real field produces $$\tag2 \phi(x) = \int d\tilde{k} a(\mathbf{k})e^{ikx} + a^\dagger(\mathbf{k})e^{-ikx}$$ as a solution.

He then defines $$\phi^+ =\int d\tilde{k} a(\mathbf{k})e^{ikx}$$ and$$\phi^- =\int d\tilde{k} a^\dagger(\mathbf{k})e^{-ikx}$$.

He then says, because $\phi$ transforms to $\phi(\Lambda^{-1}x)$ and $a(\mathbf{k})$ transforms to $a(\Lambda^{-1}\mathbf{k})$, $\phi^+$ also transforms to $\phi^+(\Lambda^{-1}x)$, and is thus a Lorentz scalar.

What I am confused about is, while I agree that $\phi^+$ transforms like that and is thus a Lorentz scalar, I don't understand his reason. $\phi^+$ does not transform in such a way because of $\phi$ or $a$, but because it's a scalar field, no? In this context, doesn't scalar field always transform in such fashion and is thus always a Lorentz scalar? Why does he say the transformation rule of $\phi^+$ comes from the transformation rule of $\phi$ and $a$?

He says that the transformation behaviour of $\phi^+$ comes from that of $\phi$ because you have not defined $\phi^+$ as "a scalar field", but as the integration of the annihilation modes of $\phi$, so the transformation behaviour of $\phi^+$ is by definition that of $a$, which in turn is that of $\phi$.
• @KyleLee: Scalar is a word describing the transformation behaviour of a thing - in this case, a scalar field is a field that transforms exactly as $\phi(x) \mapsto \phi(\Lambda^{-1}x)$ does. So, something that is a scalar function of spacetime is by definition something that transforms like this - being $\mathbb{R}$-valued is not sufficient for that - e.g. a component of a vector field is also just an $\mathbb{R}$-number, but is not a scalar. – ACuriousMind Feb 13 '15 at 2:04
• I see. I think I was confused with the mathematical definition of "scalar". So when I said scalar function, I meant a function that associates spacetime with a number. So if I have an arbitrary R-valued function of spacetime, how do I know if this is a scalar? As for $\phi$, didn't we $a$ $priori$ decide that it transforms as a scalar? So how do I determine the transformation rule for other arbitrary R-valued functions? – Quantization Feb 13 '15 at 2:07
• @KyleLee: Indeed, $\phi$ is a scalar by assumption. – ACuriousMind Feb 13 '15 at 2:15
• So what do I do for other $R$-valued function? How do I determine whether it's a scalar or not? – Quantization Feb 13 '15 at 2:16