# Complex scalar field and contraction

1. Aug 20, 2011

### wod58

Hi guys,

If I use the definition of the scalar complex field as the combination of two scalar real fields, I can get

$$\phi (x) = \int \frac{d^3 p}{(2\pi )^3} \frac{1}{\sqrt{2p_0}} [ \hat a _{\vec{p}} e^{-ip.x} + \hat b _{\vec{p}}^{\dagger } e^{ip.x}]$$

which I can rewrite in terms of (like in Peskin & Schroeder)

$$\phi (x) = \phi ^{+} (x) + \phi ^{-} (x)$$

where $\langle 0|\phi ^{-} = 0$ and $\phi ^{+} |0\rangle = 0$.

My problem is: when you try to calculate the contraction of the field with itself

$$\text{\contraction}\{\phi (x)\}\{\phi (y)\} = \begin{cases} [\phi ^{+} (x), \phi ^{-} (y)] , & \text{if } x_0 > y_0 \\ [\phi ^{+} (y), \phi ^{-} (x)] , & \text{if } x_0 < y_0 \end{cases}$$

which is supposed to be the Feynman Propagator, you obtain it for a scalar real field, but for a scalar complex field as defined above, you obtain terms with $\hat a _{\vec{p}} \hat b _{\vec{p \prime}}^{\dagger }$. The operators commute, so the vacuum expectation of these terms would be 0.

I guess I'm wrong, but can someone see where? :)

2. Aug 21, 2011

### kurros

Hmm, I think the problem is that you are trying to contract the complex scalar field with itself, when I think you can actually only contract it with it's conjugate field (similarly to spin-1/2 fields, i.e. see page 116 of Peskin and Schroeder). I haven't been able to find a reference to back me up on this but I think it must be the case, for the very reason you have discovered.

I.e. in your definition of the contraction there should be a dagger on the second scalar field in the first commutators and on the first field in the second commutator -if you want them to be equal to the Feynman propagator- and the commutators you have written down are indeed zero.

It makes perfect sense now that I think about it more. The positive and negative frequency components of a complex scalar field are totally separate fields in some sense, so of course their commutators should vanish, in a free theory anyway...

Last edited: Aug 21, 2011
3. Aug 22, 2011

### wod58

Now that you say it, it seems pretty logical. Thanks for explanation. :)