# Quantizing the Dirac Field: which commutation relations are more fundamental?

When quantizing a system, what is the more (physically) fundamental commutation relation, $[q,p]$ or $[a,a^\dagger]$? (or are they completely equivalent?)

For instance, in Peskin & Schroeder's QFT, section 3.5, when trying to quantize the Dirac field, they first say what commutation relation they expect to get for $[\Psi(\vec{x}),\Psi^\dagger(\vec{y})]$ (where $i\Psi^\dagger$ is the conjugate momentum to $\Psi$), in analogy to the Klein-Gordon field, then they postulate a commutation relation between $[a^r_{\vec{p}},a^{s\dagger}_{\vec{q}}]$ etc., and then verify that they indeed get what they expected for $[\Psi(\vec{x}),\Psi^\dagger(\vec{y})]$.

Why did we need to postulate the value of $[a^r_{\vec{p}},a^{s\dagger}_{\vec{q}}]$? Couldn't we have just computed it off of $[\Psi(\vec{x}),\Psi^\dagger(\vec{y})]$? (which, by expecting to get it, we could have just as well already postulated it).

I suppose that would entail explicitly writing something like: $$a_{\vec{p}}^r = \frac{1}{\sqrt{2E_{\vec{p}}}} u^{r\dagger}(\vec{p})\int_{\mathbb{R}^3}d^3\vec{x}\,e^{-i\vec{p}\cdot\vec{x}}\Psi(\vec{x})$$ and a similar expression for $b$.

• Commutative relations between operators refer to the postulate of positively definite energy levels of Dirac particles. Commutative relations for Dirac spinors refer to fermions. We may formulate Dirac theory in terms both of them, because each of them may lead to another one. – Andrew McAddams Dec 2 '13 at 17:15