In Peskin & Schroeder page 56, after introducing anti commutation relations for the fields instead of commutation relations (in order to fix the negative energy problem as well as to have proper causality conditions stemming from particles propagating from y->x and anti-particles propagating from x->y), it is stated that even though now we have anti-commutation relations, any two observables evaluated outside of the lightcone should still commute (and not anti commute) and thus our theory still preserves causality, in the sense that if two observables commute then measuring one does not affect the measurement of the other. The reason they give for this is that any observable is ultimately going to be composed out of an even number of fields.

I have tried to put this claim into mathematical sense, by doing the following:

  1. Assume $g(x-y,x-y)<0$ where $g$ is our metric.
  2. Thus we know that $<0|\{\Psi(x),\,\bar{\Psi}(y)\}|0>=0$ from our previous calculations.
  3. Define $O_1(x):=<0|\bar{\Psi}(x)A_1(x)\Psi(x)|0>$ and $O_2(x):=<0|\bar{\Psi}(x)A_2(x)\Psi(x)|0>$. $A_1$ and $A_2$ are two differential / spinor operators which are thus far arbitrary.
  4. Try to show that $[O_1(x),\,O_2(y)]=0$, because $g(x-y,x-y)<0$.

I have failed at showing 4. assuming 2. (I have tried to somehow make linear combinations of the commutator as anti-commutators of the fields and it didn't work out).

My question is:

  • Is this the most general formulation of the claim that one could make? If not, how would you better formulate it?
  • Once we have the formulation we want, how to prove it assuming 1. thus 2.?
  • $\begingroup$ Fermionic fields are not observable but only its current as in the answer of Trimok. Your definition of $O_i(x)$ is no operator, so the commutator does not make sense!. $\endgroup$
    – Marcel
    Commented Dec 16, 2013 at 8:45

1 Answer 1


In fact, you have $\{\psi^a(x), \bar \psi_b(y)\} = 0$, as an operator, for a space-like interval $(x-y)^2 <0$ (stricly speaking, this is a distribution, for instance, at $x_0=y_0$, this is the distribution $\delta^a_b \delta^3(\vec x-\vec y))$, together with relations $\{\psi^a(x), \psi_b(y)\} =\{\bar \psi^a(x), \bar \psi_b(y)\} = 0$

Now, if you look at a observable quantity, like the current for instance, you have :

$J_\mu(x) J_\nu(y) = (\gamma_\mu)^b_a (\gamma_\nu)^c_d \bar\psi_b(x)\psi^a(x) \bar \psi_c(y)\psi^d(y)$

Now to get $J_\nu(y) J_\mu(x)$, you have to put the $y$-dependent quantities on the left, so you have $2$ times $2$ minus sign, so finally you have a plus sign, which means that $J_\mu(x) J_\nu(y) = J_\nu(y) J_\mu(x)$, for a space-like interval $(x-y)^2<0$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.