In canonical quantization the Dirac equation is a complex column matrix, while in path integral formulation it's Grassmann numbers.
Is there a formula to convert from complex matrix to Grassmann numbers and vice versa?
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2 Answers
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In canonical quantization the Dirac field is a column vector of operators that obey anticomutation relations such as $\{\psi,\psi^\dagger\}= \hbar \delta(x-y)$. Grassmann numbers are the classical limit of such operators that remain when we set $\hbar\to0$. The Dirac field is never a column vector of complex numbers, although we make use of such column vectors when expanding the field in terms of normal modes.
answered Aug 19, 2022 at 22:04
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$\begingroup$ Exactly. Just as we quantize by replacing (for example) a real-number-valued scalar field with a hermitian operator at each point $\phi^\dagger = \phi$, the quantization for fermions involves replacing a Grassman number with an operator. Alternatively, from what I understand, you could also just think of grassman numbers as just being an invention to make the path integral approach work for fermions as well, since you can't build it for just complex-number-valued fields as you could for bosons. $\endgroup$ Commented Aug 20, 2022 at 0:00
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$\begingroup$ Sort of like how we defined $i = \sqrt{-1}$ because it ended up being useful. Does it exist? Eh, who knows what that even really means, but we can use it and it's really useful! $\endgroup$ Commented Aug 20, 2022 at 0:01
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To tie to OP's terminology, a 4D Dirac spinor $\psi$ is a $4\times 1$ column matrix where each component is a field.
The components may be operator- and/or supernumber-valued depending on context.
When considering an equation that is linear in $\psi$ (such as, e.g. the Dirac equation), it becomes agnostic to whether we treat $\psi$ as Grassmann-odd or Grassmann-even.
$\psi$ is Grassmann-odd in an interacting fermionic theory.
For more information, see e.g. this and this related Phys.SE posts.
answered Aug 20, 2022 at 8:35