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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 $\psi^2$? $$[\psi^1 _\alpha(x),\psi_\beta^{2\dagger}(y)]_+=?$$ $$[\psi^1 _\alpha(x),\psi_\beta^2(y)]_+=?$$ $$[\psi^{1\dagger} _\alpha(x),\psi_\beta^{2\dagger}(y)]_+=?$$

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    $\begingroup$ 1. anticommutation. 2. zero. Related: physics.stackexchange.com/q/17893/2451 $\endgroup$
    – Qmechanic ♦
    Commented Nov 30, 2016 at 11:22
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    $\begingroup$ @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 ... $\endgroup$
    – FraSchelle
    Commented Dec 1, 2016 at 13:29
  • $\begingroup$ @FraSchelle $\alpha,\beta$ are not fermion type label. They are spinor indices for a given fermion field $\psi$. $\endgroup$
    – SRS
    Commented Dec 1, 2016 at 21:00
  • $\begingroup$ @SRS that's exactly FraSchelle's point, I believe. Spinor indices label different fermionic degrees of freedom just as $(1)$ and $(2)$ do! $\endgroup$
    – Prof. Legolasov
    Commented Dec 2, 2016 at 5:17
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    $\begingroup$ @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. $\endgroup$
    – Prof. Legolasov
    Commented Dec 2, 2016 at 11:16

2 Answers 2

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We use anticommutation relations:

$$[\psi^1 _\alpha(x),\psi_\beta^{2\dagger}(y)]_+=0$$ $$[\psi^1 _\alpha(x),\psi_\beta^2(y)]_+=0$$ $$[\psi^{1\dagger} _\alpha(x),\psi_\beta^{2\dagger}(y)]_+=0$$

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    $\begingroup$ Please put some explanation together with your answer $\endgroup$
    – fffred
    Commented Dec 1, 2016 at 9:53
  • $\begingroup$ @fffred It is a postulate of quantization ala Fermi, I don't know what you want me to explain. $\endgroup$
    – Prof. Legolasov
    Commented Dec 1, 2016 at 9:54
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    $\begingroup$ Well, maybe what you just commented would be a good explanation :) $\endgroup$
    – fffred
    Commented Dec 1, 2016 at 9:55
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As far as I remember, it is also possible to choose the commutation relations for different fermions. However, traditionally, anti-commutation relations are chosen for creation and annihilation operators of different fermions and commutation relations for creation and annihilation operators of fermions and bosons or different bosons. EDIT:(12/1/2016) See, e.g., https://arxiv.org/abs/1312.0831

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