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You can get a vector field from a pair of spinor fields with $A_\mu(x)=\psi(x) \gamma_\mu \overline{\psi}(x)$. Using this fact could you define a space-time vector in terms of Grasman numbers?

Say you had 16 Grassmann numbers and their conjugates. If you defined a coordinate:

$$ x_\mu \equiv \Theta ^\alpha \Gamma^{\alpha\beta}_\mu \overline{\Theta}^\beta\tag{1}$$

And then fields depended on this variable. $\phi(x)$. Could you get a consistent field theory? You would still have $[x_\mu,x_\nu]=0$ but the only difference I can see is that no function of $x$ could have powers greater than 16.

How would this effect physics? It would be the same as normal physics except for the strange rule:

$$|x|^{17}=0\tag{2}$$

Or in other words the coefficients of the fields $\phi(x)$ would be zero after the 17th term. Would this be disastrous?! You wouldn't be able to have functions like $e^{-x^2}$ as they would terminate after the 17th term. Unless you might define a coordinate $y$ as:

$$ x_\mu \equiv e^{-|y|^2}y_\mu\tag{3}$$

and take the $y$ as the space-time coordinates.

But on the other hand there are stranger algebras such as non-commutative algebra.

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  • $\begingroup$ Interesting question.. you can definitely create a a mapping to the complexes and the contraction of a GV with a spinor, $\theta_\alpha \psi^\alpha$. So, in that sense, yea, but I see what you're saying.. $\endgroup$
    – InertialObserver
    Commented Jul 16, 2018 at 22:48
  • $\begingroup$ Interesting background : motls.blogspot.com/2011/11/celebrating-grassmann-numbers.html $\endgroup$
    – user198207
    Commented Jul 16, 2018 at 22:49

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Besides the problem with the 17th power, the first problem is that the soul of a supernumber [like one of OP's $x$-variables] is an indeterminate, i.e. a placeholder. Unlike a body indeterminate, it is not possible to assign a number with a value to a soul. In supermathematics a value can only be achieved by integrating out Grassmann-odd indeterminates. It does not make sense to measure a soul-valued output in an actual physical experiment, cf. e.g. this Phys.SE post.

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  • $\begingroup$ But when you have fields such as $\phi(x)$ the $x^n$ are kind of placeholders for the coefficients of the expansion of $\phi$. And a wavefunction of $\phi$ doesn't contain $x$ at all: $\Psi[\phi]$. So it calls into question how vital it is than $x$ are real numbers. Also, if this were the case how would non-commutative geometry work? $\endgroup$
    – user84158
    Commented Jul 17, 2018 at 14:31
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic ♦
    Commented Jul 17, 2018 at 14:48

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