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Complex Scalar Field in Terms of Two Independent Real Fields

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ghotra
#1
Oct22-05, 10:33 PM
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,

[tex] \phi = \frac{1}{\sqrt{2}} \left(\phi_1 + i \phi_2) [/tex]

where [itex]\phi_1[/itex] and [itex]\phi_2[/itex] are real.

When deriving,

[tex] [\phi(\vec{x},t),\dot{\phi}(\vec{x}',t)] = 0 [/tex]

I get terms like the following:

[tex][\phi_1(\vec{x},t),\dot{\phi}_2(\vec{x}',t)][/tex]

which I need to vanish. It makes sense to me that they should vanish, but how do I show this?
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ghotra
#2
Oct23-05, 01:03 AM
P: 53
Hmm...I think that we just take that as the quantization condition. That is,

[tex]
[\phi_r(\vec{x},t),\pi_s(\vec{x}{\,}',t}] = i \delta^3(\vec{x}-\vec{x}{\,}')\delta_{rs}
[/tex]

Is this correct?
SpaceTiger
#3
Oct23-05, 10:15 AM
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Quote Quote by ghotra
Hmm...I think that we just take that as the quantization condition. That is,

[tex]
[\phi_r(\vec{x},t),\pi_s(\vec{x}{\,}',t}] = i \delta^3(\vec{x}-\vec{x}{\,}')\delta_{rs}
[/tex]

Is this correct?
Since [itex]\phi_1[/itex] and [itex]\phi_2[/itex] are independent, they'll only be canonically conjugate with their own momenta (the [itex]\delta_{rs}[/itex] on the left). Your equation just states that in combination with the usual commutation relation of the real scalar field.

snooper007
#4
Oct25-05, 08:51 PM
P: 36
Complex Scalar Field in Terms of Two Independent Real Fields

[tex]\phi_1[/tex] and [tex]\phi_2[/tex]
are independent fields, so
[tex][\phi_1, \dot{\phi}_2][/tex]=0
dextercioby
#5
Oct26-05, 05:33 AM
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P: 11,928
What is the Poisson bracket between the classical fields ? If you know that, you can canonically quantize using Dirac's rule.

Daniel.


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