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In Peskin and Schroeder p. 669 it is argued that the axial current can be parametrized between the vacuum and an on-shell pion state as:

$$<0|j^{\mu 5}(x)|\pi^b(p)>=-ip^\mu f_\pi \delta^{ab}e^{-ipx}$$

This is then described as a parametrization of the amplitude for the axial current to create a pion state from the vacuum. This interpretation puzzles me: isn't it the role of the creation operators of a given theory to do this? Does the above expression correspond to an actual physical process or is it just part of an amplitude in a Feynman diagram for an actual process?

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    $\begingroup$ A related Phys.SE post by OP about the axial current: physics.stackexchange.com/q/16877/2451 $\endgroup$
    – Qmechanic ♦
    Commented Nov 14, 2012 at 10:32
  • $\begingroup$ Well if $\langle0|j|\pi\rangle$ is not zero, then clearly the states $|\pi\rangle$ and the state $j|0\rangle$ overlap. After all, $j$ is some expression in terms of some creation operators too. If your Hamiltonian involves the current $j$ any state which involves in time might apriori connect to $\pi$. The book motivates this kind of language from the beginning, see e.g. (2.42), isn't that the exact same situation? I think the discussion on page 32 is similar. $\endgroup$
    – Nikolaj-K
    Commented Nov 14, 2012 at 10:57
  • $\begingroup$ @Nick : In (2.42) the operator is the field operator. I am used to think of $\phi(x)$ acting on the vacuum as creating a particle at $x$. But I was suprised to see it attached to the current operator. Is this language valid whenever an operator has a non-zero matrix element between the vacuum and a particle state? $\endgroup$
    – Whelp
    Commented Nov 14, 2012 at 19:29
  • $\begingroup$ @Whelp: Go back and search for the definition of the current. You will eventually trance it back to an expression involving the operators you already know, plus maybe some derivatives and symmetrizations. Just take e.g. at look at the plugging together of spinors and a gamma matrix $j^\mu\propto\bar \psi \gamma^\mu\psi$ to form a bosonic operator, just so $\psi$ can be coupled to another bosonic field $A$ in $j^\mu A_\mu$. $\endgroup$
    – Nikolaj-K
    Commented Nov 14, 2012 at 20:19

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