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Recently, I've learned that the Clifford algebra can be regarded as the quantization of Grassmann algebra. This is shown from the following two papers by Berezin.

'Classical spin and Grassmann algebra'

http://www.jetpletters.ac.ru/ps/1476/article_22521.shtml

'Particle spin dynamics as the Grassmann variant of classical mechanics'

http://www.sciencedirect.com/science/article/pii/0003491677903359

I've also noticed that when doing classical Dirac fields, sometimes they are treated as complex-valued spinors but sometimes they are treated as Grassmann-valued spinors. They are treated as complex-valued spinors because they are the representation of the group $SL(2,\mathbb{C})$. But when we are dealing with the canonical and path-integral quantization of Dirac fields, we have to treat them as Grassmann-valued spinors.

If there were such a representation, does it make any sense to construct Grassmann-valued spin up and spin down states in non-relativistic quantum mechanics?

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when doing classical Dirac fields, sometimes they are treated as complex-valued spinors but sometimes they are treated as grassmann-valued spinors.

Dirac fields should always be treated as grassmann-valued, period.

While investigating a spinor field in isolation, folks can get away with treating it as if it is complex-valued, since the issue of ordering multiple spinor fields does not come into picture.

On the other hand, when it comes to spinor bilinears (and for that matter, any expression involving multiplying two or more spinor fields), one is obliged to treat spinor fields as grassmann-valued. Case in point: Majorana mass term (which is a spinor bilinear) is allowed only if spinor fields are grassmann-valued.

Another example: one should tread carefully with Fierz identities, which concern terms with four spinor fields. Unfortunately, the grassmann nature of spinor fields is often overlooked in such cases.

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  • $\begingroup$ The bilinears (and higher order $Spin$-symmetric combinations) are objects of usual representation theory. There's no need to think of the inputs as anything other than spinors over $\mathbb{C}$. $\endgroup$ – Ryan Thorngren Aug 31 '18 at 16:49

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