# How does a $\Theta$ function arise in this correlator?

I am currently reading the paper by Coleman on Symmetry breaking in 2d, which can be found here. On page 262 (4th page in the document), he is evaluating the following distribution:

$$F_{\mu}(k)=\int d^2x\ e^{i k x} \langle0|j_\mu(x)\phi(0)|0\rangle.\tag{11b}$$

And finds since $k^\mu F_\mu = 0$, that it must be $$\sigma k_\mu\delta(k^2) \Theta(k^0) + \epsilon_{\mu\nu}k^\nu \rho(k^2) \Theta(k^0).\tag{13}$$

For some number $\sigma$ and function $\rho$.

I cannot see where-ever the $\Theta(k^0)$ part is coming from, can somebody point me the right direction?

• Is it correct to assume there is a fermi-surface at $k^0$? In that case the $\Theta$ functions will keep track of that surface. – Mikael Fremling May 16 '16 at 15:38
• @MikaelFremling I don't think so, I'm not an expert but Fermi surfaces come from periodic lattices if I recall correctly, which is not the case here (its just an arbitrary QFT in 2d). – s.harp May 16 '16 at 16:14
• Post about same paper: math.stackexchange.com/q/1787656/11127 – Qmechanic May 17 '16 at 12:37
• @Qmechanic yes that is my post^ – s.harp May 17 '16 at 12:40

Conventionally in QFT, particles and antiparticles are defined with positive energy $k^0\geq 0$ only. (Recall that would-be negative energy states are reinterpreted as matter/antimatter of the opposite kind in order to make the vacuum stable.)
• Would attempting to decompose $j_\mu$ and $\phi$ in a way similar to the Källen Lehman representation be the right way to show that the negative modes are zero? – s.harp May 16 '16 at 16:12