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Basically, I wonder what Chromoelectric dipole moment is and how it is formed...

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- Thread starter Sehwook Lee
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Basically, I wonder what Chromoelectric dipole moment is and how it is formed...

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ChrisVer

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[itex] H_{EDM} = - \vec{d}_f \cdot \vec{E}[/itex]

For a spin-1/2 particle, this corresponds to the effective lagrangian density:

[itex] \mathcal{L}_{EDM} = \frac{-i}{2} d_f F_{\mu \nu} \bar{\psi}_f \sigma^{\mu \nu} \gamma^5 \psi_f [/itex]

you can have a similar term if you look at quarks [itex]\mathcal{q}_r[/itex], by replacing the electromagnetism with strong ints:

[itex] \mathcal{L}_{CEDM} = \frac{-i}{2} d_r^{CEDM} G_{\mu \nu}^a \bar{\mathcal{q}}_r \sigma^{\mu \nu} \gamma^5 \frac{\lambda^a}{2} \mathcal{q}_r [/itex]

and that's how it arises... In general both EDMs and CEDMs generate P and T (so CP) violations.

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