In quantum field theory, it is crutial that two experiments can not effect each other at space-like seperation. Thus $[\mathcal{O}_1(x), \mathcal{O}_2(y)] = 0 $ if $(x-y)^2 < 0$.

For the Klein-Gordan field we now the equal times commutation relation $[\phi(x), \pi(y)] = i \delta^{(3)}(\mathbf{x} - \mathbf{y})$, i.e. the commutator is zero at spacelike separation. Any observable such as momentum, energy, or charge involves these operators and thus the measurements can not effect each other.

For the Dirac field however, we only have an anti commutation relation $\lbrace \psi_a(x), \bar{\psi}_b(y) \rbrace = \delta^{(3)}(\mathbf{x} - \mathbf{y}) \delta_{ab}$. In section 3.5 Peskin and Schroeder say that this is enough to ensure that any two Observables will commutate at space-like separation since any reasonable observable is made up of an even number of spinor fields. But I can not see how this is the case.

So could someone explain to me how the anti-commutation relation implies the commutation of observables?


1 Answer 1


I think it's easier to start from the general expression for a field operator $\phi$:

$$ \hat\phi_{A,B}(x) = \int \frac{\mathrm{d}^3 \mathbf{p}}{(2\pi)^3}\frac{1}{2E_{\mathbf{p}}} \sum_s \left [ \mathrm{e}^{-\mathrm{i}px}f_{A,B}(\mathbf{p},s)\hat a(\mathbf{p},s) + \mathrm{e}^{\mathrm{i}px}h_{A,B}(\mathbf{p},s)\hat a^\dagger(\mathbf{p},s) \right ]\bigg\vert_{p^0=E_\mathbf{p}},$$

where $A$ and $B$ are just labels for two different particles.

The bosonic or fermionic nature of the particles is reflected in the (anti-)commutation relationship between the creation and annihilator operators $a^\dagger$ and $a$.

Because operators $\mathcal{O}(x)$ are usually just a product of field operators, so that $\mathcal{O}(x) \propto \prod_i \hat\phi_i(x)$, requiring $[\mathcal{O}_1(x), \mathcal{O}_2(y)]=0$ is the same as requiring $[\hat\phi_A(x), \hat\phi_B(y)]=0$.

For bosons, you have that: $$ [\hat a(\mathbf{p},s), \hat a^\dagger(\mathbf{p}',s')] = 2E_\mathbf{p}(2\pi)^3 \delta^{(3)}(\mathbf{p}-\mathbf{p}')\delta_{s,s'},$$ while for fermions, you have that: $$ \{\hat a(\mathbf{p},s), \hat a^\dagger(\mathbf{p}',s')\} = 2E_\mathbf{p}(2\pi)^3 \delta^{(3)}(\mathbf{p}-\mathbf{p}')\delta_{s,s'}.$$

Then you can show that this results in: $$ \text{For bosons: }[\hat\phi_A(x), \hat\phi_B(y)] \propto 1 - (-1)^{2s} \stackrel{!}{=} 0 \implies s \text{ is an integer.}$$ $$ \text{For fermions: }[\hat\phi_A(x), \hat\phi_B(y)] \propto 1 + (-1)^{2s} \stackrel{!}{=} 0 \implies s \text{ is a half-integer.}$$

The maths is done in Weinberg's The quantum theory of fields volume I, in the chapter called General causal fields.

  • $\begingroup$ I'm not sure this is quite right. Following Sterman p167, the fermionic field commutator doesn't vanish at space-like separations. However, fermionic fields aren't observables; the spinors are double-valued representations, so under a rotation by $2\pi$ one has in your notation $\phi_A(x) \rightarrow -\phi_A(x)$. Rather, commutators of observables such as currents of fermionic fields $j^\mu(x) \equiv \bar \phi_A(x)\gamma^\mu\phi_A(x)$ do correctly vanish for space-like separations, utilizing the anti-commutation relations between the fields (Sterman exercise 6.7). $\endgroup$ Nov 1, 2021 at 16:19

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