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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 $b$.

Qmechanic
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PPR
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  • 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 02 '13 at 17:15

1 Answers1

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Firstly, note that they postulate those commutation relations in the beginning of section 3.5 in order to show that they are wrong, which they demonstrate in the ensuing pages. The ultimate point is to show that one needs to impose anti-commutation relations on fermionic fields.

In fact, the correct relations are postulated in equation 3.96; \begin{align} \{\psi_a(\mathbf x), \psi_b^\dagger(\mathbf y)\} &= \delta^{(3)}(\mathbf x - \mathbf y) \delta_{ab} \end{align} You could then ask, are these equivalent to the anti-commutation relations of the mode operators that they write in (3.97)? Namely, \begin{align} \{a^r_\mathbf p, {a^s_\mathbf q}^\dagger\} = \{b^r_\mathbf p, {b^s_\mathbf q}^\dagger\} = (2\pi)^3\delta^{(3)}(\mathbf p - \mathbf q)\delta^{rs} \end{align} And the answer is yes.

To show that the second set implies the first, write the fields in their integral mode expansions, compute the anti-commutator of these integral expressions, and apply the anti-commutators between modes. To show that the first set implies the second, invert the integral expressions for the fields in terms of the modes to obtain integral expressions for the modes in terms of the fields, and do the analogous thing.

Main Point. The commutators/anti-commutators between fields are equivalent to the commutators/anti-commutators between modes.

joshphysics
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