0
$\begingroup$

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 $a$?

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $\phi$.

$\endgroup$
10
  • $\begingroup$ what does it mean to "defin"e as a scalar field. Isn't anything automatically a scalar field if it is a scalar function of spacetime? $\endgroup$ Feb 13, 2015 at 1:56
  • $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Feb 13, 2015 at 2:04
  • $\begingroup$ 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? $\endgroup$ Feb 13, 2015 at 2:07
  • 1
    $\begingroup$ @KyleLee: Indeed, $\phi$ is a scalar by assumption. $\endgroup$
    – ACuriousMind
    Feb 13, 2015 at 2:15
  • $\begingroup$ So what do I do for other $R$-valued function? How do I determine whether it's a scalar or not? $\endgroup$ Feb 13, 2015 at 2:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.