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.
