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The anti-commutation relation between the components of a fermion field $\psi$ is given by $$[\psi _\alpha(x),\psi_\beta^\dagger(y)]_+=\delta_{\alpha\beta}\delta^{(3)}(\textbf{x}-\textbf{y}).$$

  1. In case of two different and independent fermion fields, should I impose commutation or anticommutation between them?

  2. If we continue to use anticommutation, how should the RHS change for two different fermion fields $\psi^1$ and $\psi^2$? $$[\psi^1 _\alpha(x),\psi_\beta^{2\dagger}(y)]_+=?$$ $$[\psi^1 _\alpha(x),\psi_\beta^2(y)]_+=?$$ $$[\psi^{1\dagger} _\alpha(x),\psi_\beta^{2\dagger}(y)]_+=?$$

SRS
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    @Qmechanic Curiously, the question is answered in the question itself: the non-specified $\alpha$ and $\beta$ can be used for different fermion fields. If the OP had understood her notation, she would not have asked ... – FraSchelle Dec 01 '16 at 13:29
  • @FraSchelle $\alpha,\beta$ are not fermion type label. They are spinor indices for a given fermion field $\psi$. – SRS Dec 01 '16 at 21:00
  • @SRS that's exactly FraSchelle's point, I believe. Spinor indices label different fermionic degrees of freedom just as $(1)$ and $(2)$ do! – Prof. Legolasov Dec 02 '16 at 05:17
  • @SolenodonParadoxus Isn't $\alpha,\beta$ label the indices of the elements/entries of the column vector $\psi$ and elements of the row vectors of $\bar{\psi}$? When I say, $\psi^1$ and $\psi^2$ I talk about two different fields, say electron and muon. – SRS Dec 02 '16 at 08:05
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    @SRS different entries of the column vector $\psi$ are also different quantum fields. We only write them together because they form a Lorentz multiplet (that is, the Lorentz transformation for one of the entries depend on the values of all other entries). Btw I upvoted your question, I think it is perfectly reasonable to ask, especially if you are confused. – Prof. Legolasov Dec 02 '16 at 11:16
  • @SolenodonParadoxus $\alpha$ and $\beta$ are not defined in your question. You are thus free to define them the way you want, the rule will be the same. You can even suppose $\alpha = {\alpha_{1},\alpha_{2}, ...}$ if you wish, and $\delta_{\alpha,\beta}=\delta_{\alpha_{1},\beta_{1}}\delta_{\alpha_{2},\beta_{2}}...$ In short, the indices can be used in any way you want in order to distinguish the two fermion fields (spin up, spin down, brother and sister, blue and green, whatever and co, ... ). So the answer to your question is already there, in the Kronecker's notation :-) – FraSchelle Dec 02 '16 at 12:35
  • @FraSchelle the only thing that remains a mystery to me is why you address your comment to me when I clearly agree (you probably meant to address it to OP). – Prof. Legolasov Dec 02 '16 at 13:05
  • @SolenodonParadoxus Because you were part of the discussion, and the OP is automatically informed of a new comment. So I informed both of you of this comment. Sorry for the possible spam indeed. – FraSchelle Dec 05 '16 at 11:40
  • @FraSchelle np, its just that I thought you misinterpreted my position in the argument, sorry :) – Prof. Legolasov Dec 05 '16 at 11:56