# Complex Scalar Field in Terms of Two Independent Real Fields

 P: 53 I am working with a complex scalar field written in terms of two independent real scalar fields and trying to derive the commutator relations. So, $$\phi = \frac{1}{\sqrt{2}} \left(\phi_1 + i \phi_2)$$ where $\phi_1$ and $\phi_2$ are real. When deriving, $$[\phi(\vec{x},t),\dot{\phi}(\vec{x}',t)] = 0$$ I get terms like the following: $$[\phi_1(\vec{x},t),\dot{\phi}_2(\vec{x}',t)]$$ which I need to vanish. It makes sense to me that they should vanish, but how do I show this?
 P: 53 Hmm...I think that we just take that as the quantization condition. That is, $$[\phi_r(\vec{x},t),\pi_s(\vec{x}{\,}',t}] = i \delta^3(\vec{x}-\vec{x}{\,}')\delta_{rs}$$ Is this correct?
Emeritus
 Quote by ghotra Hmm...I think that we just take that as the quantization condition. That is, $$[\phi_r(\vec{x},t),\pi_s(\vec{x}{\,}',t}] = i \delta^3(\vec{x}-\vec{x}{\,}')\delta_{rs}$$ Is this correct?
Since $\phi_1$ and $\phi_2$ are independent, they'll only be canonically conjugate with their own momenta (the $\delta_{rs}$ on the left). Your equation just states that in combination with the usual commutation relation of the real scalar field.
 P: 36 Complex Scalar Field in Terms of Two Independent Real Fields $$\phi_1$$ and $$\phi_2$$ are independent fields, so $$[\phi_1, \dot{\phi}_2]$$=0