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Why is the derivative necessary to connect left and right-hand spinors?

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I am studying Weyl and Dirac spinors. Suppose we have two Weyl fermions $\eta, \chi$ transforming under $(1/2,0)$ representation of the Lorentz group. I learned that to construct Lorentz invariant term in the Lagrangian, we have terms like $i\eta^\dagger\bar\sigma^\mu\partial_\mu\chi$ or $i\eta\sigma^\mu\partial_\mu\chi^\dagger$. I understand that $\sigma^\mu$ maps left-handed spinors to right-handed and it is Lorentz invariant, therefore sandwiched between the two spinors, but I am still not quite sure how can I justify that $\partial^\mu$ is needed, and we don't need higher-derivative terms?

Also, why for a vector field $A_\mu$, the possible interaction is similar to the derivative term: $i\eta^\dagger\bar\sigma^\mu A^\mu\chi$$i\eta^\dagger\bar\sigma^\mu A_\mu\chi$ ?

Thanks for the help!

I am studying Weyl and Dirac spinors. Suppose we have two Weyl fermions $\eta, \chi$ transforming under $(1/2,0)$ representation of the Lorentz group. I learned that to construct Lorentz invariant term in the Lagrangian, we have terms like $i\eta^\dagger\bar\sigma^\mu\partial_\mu\chi$ or $i\eta\sigma^\mu\partial_\mu\chi^\dagger$. I understand that $\sigma^\mu$ maps left-handed spinors to right-handed and it is Lorentz invariant, therefore sandwiched between the two spinors, but I am still not quite sure how can I justify that $\partial^\mu$ is needed, and we don't need higher-derivative terms?

Also, why for a vector field $A_\mu$, the possible interaction is similar to the derivative term: $i\eta^\dagger\bar\sigma^\mu A^\mu\chi$ ?

Thanks for the help!

I am studying Weyl and Dirac spinors. Suppose we have two Weyl fermions $\eta, \chi$ transforming under $(1/2,0)$ representation of the Lorentz group. I learned that to construct Lorentz invariant term in the Lagrangian, we have terms like $i\eta^\dagger\bar\sigma^\mu\partial_\mu\chi$ or $i\eta\sigma^\mu\partial_\mu\chi^\dagger$. I understand that $\sigma^\mu$ maps left-handed spinors to right-handed and it is Lorentz invariant, therefore sandwiched between the two spinors, but I am still not quite sure how can I justify that $\partial^\mu$ is needed, and we don't need higher-derivative terms?

Also, why for a vector field $A_\mu$, the possible interaction is similar to the derivative term: $i\eta^\dagger\bar\sigma^\mu A_\mu\chi$ ?

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IGY
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  • 5
  • 13

Why is the derivative necessary to connect left and right-hand spinors

I am studying Weyl and Dirac spinors. Suppose we have two Weyl fermions $\eta, \chi$ transforming under $(1/2,0)$ representation of the Lorentz group. I learned that to construct Lorentz invariant term in the Lagrangian, we have terms like $i\eta^\dagger\bar\sigma^\mu\partial_\mu\chi$ or $i\eta\sigma^\mu\partial_\mu\chi^\dagger$. I understand that $\sigma^\mu$ maps left-handed spinors to right-handed and it is Lorentz invariant, therefore sandwiched between the two spinors, but I am still not quite sure how can I justify that $\partial^\mu$ is needed, and we don't need higher-derivative terms?

Also, why for a vector field $A_\mu$, the possible interaction is similar to the derivative term: $i\eta^\dagger\bar\sigma^\mu A^\mu\chi$ ?

Thanks for the help!