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I hope this question makes sense.

Let $\phi$ be a quantized Dirac spinor with four components $\phi_{\alpha}$, $\alpha=1,2,3,4$. If we denote by $\pi$ the conjugate momentum, then they obey the canonical anti-commutation relations $$(CAR) \ \ \ \ \ [\phi_{\alpha}(x), \pi_{\beta}(y)]_+=\delta(x-y)\delta_{\alpha \beta}$$ in a distributional sense that we won't worry about in detail. (The normalization will also be ignored here.) Write $\phi=(\phi_+, \phi_-)$ for the decomposition into two Weyl spinors. If I put $$\psi(x)=\phi_+(x) v_0,$$ where $v_0$ is the vacuum, then in some non-relativistic limit, I've been told that $\psi(x)$ will go to $\psi^{pauli}(x)$, the wave function for a single Pauli electron. Thus, the position $x$ and momentum $p$ observables acting on $\psi^{pauli}$ satisfy the canonical commutation relations $$(CCR)\ \ \ \ \ \ \ [x_i,p_j]=\delta_{ij}$$ (again ignoring the correct normalization).

My question is: Can we derive this $(CCR)$ directly from the $(CAR)$ above? Intuitively, I would guess yes, since the position and momentum of the field should determine those quantities for the particles it creates. But how does this go precisely?

I might remark that for a novice like me, the expositions of the Dirac picture in standard textbooks look only loosely related to the QM picture. Hence, my confusion.

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    $\begingroup$ I'm not quite sure I understand what you mean: the field operator for a fermion satisfies an anti-commutation relation, irrespective of whether you describe the electron fully relativistically (4-component Dirac spinor) or not (2 -component Pauli spinor). $\endgroup$
    – Carlo Beenakker
    Oct 13, 2014 at 18:11
  • $\begingroup$ The commutation relation between $x$ and $p$ has nothing to do with what it acts on, be it $\psi^{\rm Pauli}$ or your friendly wave function from QM instead of QFT. It is just because $x$ is multiplication by $x$ and $p$ is $-i$ times $d/dx$. $\endgroup$
    – Abdelmalek Abdesselam
    Oct 13, 2014 at 19:28
  • $\begingroup$ Carlo Beenakker: Yes, I agree. But does this answer my question of deriving (CCR) for $x$ and $p$ from the (CAR) for the Dirac field? $\endgroup$
    – Minhyong Kim
    Oct 13, 2014 at 20:37
  • $\begingroup$ Abdelmalek Abdesselam: I agree with this as well. However, from a rather fundamentalist view, the representation of position and momentum you describe is a consequence of (CCR). $\endgroup$
    – Minhyong Kim
    Oct 13, 2014 at 20:40
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    $\begingroup$ Dear Jeff: OK, I think I've figured out how to write down a position and momentum operator just on single particle states for the Dirac field. I'll try to actually calculate before too long and see what happens. But I didn't shut up just yet so as to tell you not to waste your time explaining anything yet. $\endgroup$
    – Minhyong Kim
    Oct 18, 2014 at 0:36

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