Complex Scalar Field in Terms of Two Independent Real Fields

[ghotra]

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?

 

[ghotra]

Hmm...I think that we just take that as the quantization condition. That is,

<br /> [\phi_r(\vec{x},t),\pi_s(\vec{x}{\,}&#039;,t}] = i \delta^3(\vec{x}-\vec{x}{\,}&#039;)\delta_{rs}<br />

Is this correct?

 

[SpaceTiger]

Hmm...I think that we just take that as the quantization condition. That is,

<br /> [\phi_r(\vec{x},t),\pi_s(\vec{x}{\,}&#039;,t}] = i \delta^3(\vec{x}-\vec{x}{\,}&#039;)\delta_{rs}<br />

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.

 

[snooper007]

\phi_1 and \phi_2
are independent fields, so
[\phi_1, \dot{\phi}_2]=0

 

[dextercioby]

What is the Poisson bracket between the classical fields ? If you know that, you can canonically quantize using Dirac's rule.

Daniel.