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In many texts the anti-commutation relations for fermions are given as

$$\{ \bar{\psi}^\alpha (\vec{x}), \psi^\beta(\vec{y}) \} = \delta^{\alpha\beta} \delta(\vec{x} - \vec{y})$$ $$\{ \psi^\alpha (\vec{x}), \psi^\beta(\vec{y}) \} = 0$$ $$\{ \bar{\psi}^\alpha (\vec{x}), \bar{\psi}^\beta(\vec{y}) \} = 0.$$

(See for example eq. (3.102) in Peskin & Schroeder page 58.)

It is also commonly stated that in the classical limit $\hbar \to 0$ the fermions behave as Grassmann numbers. However, I struggle to see this because there is no $\hbar$ in the above equation. I've tried to find it using dimensional analysis, but unfortunately I can't find it.

I have found other questions on this site which do have an $\hbar$ in this expression, but my textbooks don't have it so I'm a bit confused.

My attempt: Starting from the dimensionful Dirac equation, I now that the Lagrangian contains a term $m c^2 \bar{\psi} \psi$. Knowing the units of the following quantities: $$[S] = [\hbar ] = J s = kg \, m^2 \, s^{-1}$$ $$[dt d^3\vec{x} ] = s m^3$$ $$[m c^2] = kg \, m^2 \, s^{-2}$$ I therefore deduce that $[\bar{\psi} \psi ] = m^{-3}$. Looking at the first anti-commutation relation above, the only thing giving units on the right side of the equality is the $\delta^{(3)}$ term, which has units of $[d^3 k] = m^{-3}$ so the sides appear to me to be in agreement.

However, this would mean that this does not have Grassmann variables as a classical limit so I must have done something wrong.

tBuLi
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  • So what is the physics unit of Grassmann variables in your definition? Is it different from real numbers (i.e. dimensionless)? – MadMax Jul 31 '19 at 15:04
  • I would say the dimensions will stay the same, right? All that happens is shrinking the non-anticommuting part down to zero. – tBuLi Jul 31 '19 at 15:12
  • If Grassmann number is as dimensionless as the real number, then why is it an issue for being a Grassmann variable? – MadMax Jul 31 '19 at 15:35
  • Your dimensional analysis is wrong. The Lagrangian contains a term $m{\color{red}{c^2}}\bar\psi\psi$, not $m{\color{red}{c}}\bar\psi\psi$. – AccidentalFourierTransform Aug 01 '19 at 13:01
  • Thanks, that would indeed explain the missing factor of $c$. Indeed in other articles this is mentioned, so the original wiki article I used was not a good starting point. However, this still leaves the $\hbar$ since the extra $c$ does not contain any $kg$. ( I have updated my question to include your comment) – tBuLi Aug 01 '19 at 13:16

1 Answers1

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Peskin & Schroeder write on p. xix that they work in "God-given" units $\hbar=1=c$. So one should first reinstate $\hbar$ on the RHS of the canonical anticommutation relations (CAR). The CAR may be viewed as a first principle similar to the CCR. Next take the classical limit $\hbar\to 0$. And voila! The fermions become Grassmann numbers, which anticommute.

See also e.g. this related Phys.SE post.

Qmechanic
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  • Thank you for your answer. However, this doesn't quite answer my question because I'm asking why it's there in the first place. According to my dimensional analysis, the units seem correct without including $\hbar$. So I'm trying to see where I went wrong. The only references to this I find are online sources such as the one I linked and which you kindly repeated, but none of my textbooks have it and I can't seem to reinstate it myself. Your help is much appreciated! :) – tBuLi Aug 01 '19 at 09:24