Let $\phi(x)$ be a real classical scalar field which satisfies the Klein-Gordon equation; without loss of generality it can be written as:

$$ \phi(x) = \int \frac{d\vec k}{(2\pi)^{3/2} 2 \omega_k} \left \{ \tilde \phi (k) e^{-ikx} + \tilde \phi ^*(k) e^{ikx} \right \}_{\big|_{k^0=\omega(\vec k)}}$$

Since $\phi(x)$ is a scalar field I require that under Lorentz transformation

$$ x \rightarrow x'(x)=\Lambda x \; \Rightarrow \; \phi (x) \rightarrow \phi'(x'(x)) \overset{!}{=}\phi (x)$$

and applying that condition to the integral representation of $\phi (x)$ I deduce that even $\tilde \phi (k)$ has to be a scalar field: if

$$\begin{cases} x \rightarrow x'(x)=\Lambda x \\ k \rightarrow k'(k)=\Lambda k \end{cases}$$

the measure $\frac{d\vec k}{(2\pi)^{3/2} 2 \omega_k}$ and the exponential $e^{-ikx}$ are invariant and hence

$$ \phi'(x'(x)) = \phi (x) \Leftrightarrow \tilde \phi '(k'(k)) = \tilde \phi(k) \\$$

QUESTION: But when it comes to translation invariance how do things work?

Since $\phi(x)$ is a scalar field I require, as before, that:

$$ x \rightarrow x'(x)=x+a \; \Rightarrow \; \phi (x) \rightarrow \phi'(x'(x)) \overset{!}{=}\phi (x) $$

but in this case I find a strange transformation rule for $\tilde \phi(k)$: if

$$\begin{cases} x \rightarrow x'(x)= x + a \\ k \rightarrow k'(k)= k \end{cases}$$

the measure $\frac{d\vec k}{(2\pi)^{3/2} 2 \omega_k}$ is invariant but the exponential is not: $e^{-ikx} \rightarrow e^{-ik'x'}=e^{-ikx-ika} $ and hence I find:

$$ \phi'(x'(x)) = \phi (x) \Leftrightarrow \tilde \phi '(k'(k)) = \tilde \phi(k) e^{ika}\\$$

Is this transformation rule correct? what kind of field is $\tilde \phi(k)$ then?

  • $\begingroup$ Note that the Fourier transform of $f(x+a)$ is $\mathrm e^{ika}\tilde f(k)$ (cf. wikipedia). $\endgroup$ – AccidentalFourierTransform Oct 23 '17 at 17:37
  • $\begingroup$ @AccidentalFourierTransform do you think the transformation rule for $\tilde \phi(k)$ is correct then? do fields with that behavior (invariant under Lorents transformation but not under translation) have a particular name? Like, for example, pseudo-scalar are fields invariant under Lorentz transformation but not under parity $\endgroup$ – L.R. Oct 23 '17 at 22:22

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