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After reviewing this question as well as this one, I am left with some confusion, mainly about the nature of complex and real representations of the Lorentz group and how we do physics. I understand that representations are 'borrowed' from the complexification, and so naturally we consider how the Lorentz group acts over complex vector spaces. But then whenever possible we seem, via the following prescription, to descend to the real form of the group/algebra:

$$v = v_1+iv_2 \quad| \quad v_i\in\mathbb{R}^n$$ $$g\in SO(1,3;\mathbb{R})$$ $$\rho_\mathbb{C}(g)\cdot v = \rho_\mathbb{R}(g)\cdot v_1+i \rho_\mathbb{R}(g)\cdot v_2$$

My question is why everywhere in physics we choose to 'drop down' and work over real vector spaces except the spinor rep $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$, when a real spinor rep does exist in the Majorana rep. Is this just convenience? Are there consequences if one chooses to work in a complex representation for, say, field strength tensors $(1,0)\oplus(0,1)$ (perhaps potentially unphysical/chiral field strength configurations)? Or is everything fine no matter what vector space we choose to work in?

Craig
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    I think I already said this in a previous question, but the reason a typical intro QFT course uses Dirac spinors is because that's what you need to describe QED, which is a very important example. Of course, a Dirac spinor is equivalent to a pair of Majorana spinors, and you can also have a theory with only one Majorana spinor, but QED isn't one of them. – knzhou Mar 31 '22 at 00:28
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    Similarly, there's nothing stopping you from using a complex field strength tensor. But just like how a complex scalar field is equivalent to two real scalar fields, a complex field strength is equivalent to two real field strengths. It would be useful in a world where there are two identical photon fields, but our world only has one photon field. – knzhou Mar 31 '22 at 00:29
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    The point is, the types of elementary fields emphasized in a QFT course are not dictated by pure mathematics -- they're determined by what is necessary to match experiment. – knzhou Mar 31 '22 at 00:32
  • I think it is clear now. Thank you – Craig Mar 31 '22 at 00:53

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