Using the "Dirac Prescription", we can preserve the format of a differential equation in its QM form. If we define the canonical variables s.t. they have the same commutation relations times $i$ as the Poisson brackets, Heisenberg and Hamiltons equations will have the same form.

This gives us a very simple way to quantise a boson field: Just set the commutation relations in the right way, and Heisberg's equation will just be the wave equation. From this point everything will follow: Heisenberg equation will be Lorentz invariant if the Lagrangian is, the Noether current is conserved in the QM version of the theory. Birds will sing.

my question is: What about the Dirac field?

Since we are defining the anti-commutator instead of the commutator, the same logic do not apply. Anti-commutator cannot (as far as I can tell) act as a derivative on a formal series like the commutator.

So why are we justified in quantising the Dirac Field like that?

(If you can, I would appreciate justifications that don't have anything to do with Feynman integrals)


1 Answer 1


TL;DR: The same logic still applies to Grassmann-odd variables with pertinent sign factors.

  1. The Dirac prescription $\{\cdot,\cdot\}_{PB}\to [\cdot,\cdot]/i\hbar$ in principle also work for supervariables (up to higher-order quantum corrections, cf. the deformation quantization paradigm). The relevant (graded) bracket structures $\{\cdot,\cdot\}_{PB}$ and $[\cdot,\cdot]$ are a super-Poisson bracket and a supercommutator, respectively.

  2. The Schrödinger representation$^1$
    $$ \hat{q}~=~q , \qquad \hat{p}~=~\frac{\hbar}{i}\frac{\partial_R}{\partial q}~=~(-1)^{|q|}\frac{\hbar}{i}\frac{\partial_L}{\partial q}, \qquad[\hat{q},\hat{p}]~=~i\hbar{\bf 1},\tag{1}$$ also works for supervariables & supercommutators.

  3. For further details, see e.g. this, this & this related Phys.SE posts.


$^1$ Here the subscript $R$ & $L$ denote left & right derivatives, and $|q|$ denotes the Grassmann-parity of $q$. NB: Apart from the explicitly written sign factor in eq. (1), there may be additional sign factors in accordance with the Koszul sign rule/convention, stemming from permuting Grassmann-odd objects.

  • $\begingroup$ Thank you very much for your answer! I have read it multiple times, but still cannot understand how this might solve the problem. 1. To get Dirac's equation, I never have to assume that the field is a Grassmann-valued-field. 2. In addition to that, I can write Classical Hamiltonians that give Heisenberg equation =! Hamilton equation (for example, $H=P$, or $H=P_1P_2$). Could you help? $\endgroup$ Commented Feb 5, 2019 at 20:49
  • $\begingroup$ 1. This is essentially asked here. $\endgroup$
    – Qmechanic
    Commented Feb 5, 2019 at 22:03

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