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I'm currently studying Goldberger-Treiman relation from the book by S. Coleman (Aspects of Symmetry, chapter 2) in which, working in the framework of a not better precised "weak interaction Hamiltonian", at one point he considers the following matrix element (I'm omitting numerical constants and kinematic factors):

$$ \langle 0 |A^\mu | \pi(p) \rangle = ie^{-ip \cdot x}p^\mu F_\pi \tag{2.2.7}$$

where $A^\mu$ stands for the hadronic axial current [maybe a certain contribution to it] and of course $|\pi\rangle$ is the one particle pion state. Then he takes the derivative of this expression, writing

$$ \langle 0 |\partial_\mu A^\mu | \pi(p) \rangle = e^{-ip \cdot x}m_\pi^2 F_\pi . \tag{2.2.8}$$

Now, the question is: why does it look like Coleman is taking the derivative outside the bra-ket? More explicitly: why is he computing the derivative as

$$\partial_\mu \langle 0 |A^\mu | \pi(p) \rangle ~ ?$$

My concern is that if I apply LSZ formula in (2.2.7), I receive a $T$-order product which, if I'm correct, taking the 4-derivative of the new matrix element, would yield a Ward-Identity of the form:

$$ \partial_\mu \langle T(J^\mu O[\phi]) \rangle = \langle T(\partial_\mu J^\mu O[\phi]) \rangle + C.T. \tag{3}$$

where "$C.T.$" stands for contact terms and $O[\phi]$ is a generic composite operator, that satisfies LSZ hypothesis. If I could ignore these terms then I would understand the possibility of taking the 4-derivative inside the $T$-order product. I have come up with 3 possibilities:

  1. I should regard this fact as a particular case, due to the specific form of $A^\mu$ (the expression of which is to me quite a mystery, just by reading these notes).

  2. This is a general fact: if a matrix element is not LSZ-reduced (i.e. if there is no explicit $T$-order in it), I can take the 4-derivative inside the $T$-order product (which would make no sense to me, since it's the same expression rewritten into two different ways).

  3. Equation (3) is wrong - which could be the case since rarely I've found an expression for Ward Identity for a non-conserved current (one example is from Coleman itself: see for example his eq. (3.2.5) in chapter 3) that I'm suspecting it isn't a real thing. In this case the possibility would restrict just to the first point of this list.

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3 Answers 3

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OP asks:

Why does it look like Coleman is taking the derivative outside the bra-ket?

Because that's what he's doing. This means that the implicit time-order in eq. (2.2.8) is the covariant time-order $T_{\rm cov}$, cf. e.g. my Phys.SE answer here.

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I believe that the contact terms that are generated from applying the derivative to the time ordered product do not have the correct pole structure or number of poles to give a contribution to the S matrix for processes where PCAC is used. See this stack exchange answer: Contact terms in Dyson-Schwinger equation can be ignored?

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There is only one spacetime point dependence, so the time-ordering is trivial and does not generate any additional term from differentiating any possible step functions in this case. That is, for one-point expectation values derivative outside or inside the expectation values are the same. Namely: $$T[\phi(x)] = \phi(x),$$ (there is nothing to order). If we write the expectation value you have above $$\partial_\mu \langle 0 | T[A^\mu(x)] | \pi_p\rangle = \partial_\mu \langle 0 | A^\mu(x) | \pi_p\rangle = \langle 0 | \partial_\mu A^\mu(x) | \pi_p\rangle $$

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  • $\begingroup$ I don't think I understand what you mean when you're saying "there is only one spacetime point dependence". If I used LSZ I would get two spacetime points (of which only one, the one carried by the current, has the same dependence of the derivative; but this is the general case that holds for every Green Function involved in Ward Identity, isn't it?). May you give a clarification about that? $\endgroup$
    – Boreanaz
    Commented Jan 31, 2022 at 20:41

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