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The decay constant $f$ of a hadron with momentum $P$ or its in-hadron condensate $\kappa$ (See this paper, Eq. (8)) are given by terms like

$ \begin{align*} f_\pi P^\mu &\sim \langle 0|J_\text{axial}^\mu (0) |\pi(P)\rangle,\tag{1}\\ \kappa_\pi &\sim \langle 0|J_\text{pseudoscalar} (0) |\pi(P)\rangle.\tag{2} \end{align*} $

Why do we use the vacuum $|0\rangle$ and the pion $|\pi\rangle$ together? How would I interpret these terms if we would replace the vacuum with the pion state?

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    $\begingroup$ Hi. Isn't what you have here a correlation between the vacuum and the pion states, which in turn are products of spontaneous symmetry breaking? Doesn't that mean then that what the relation (1) shows you is the fact that the degeneracy of the vacuum is lifted and a pion field is produced cause of the symmetry breaking? $\endgroup$ Commented Mar 18, 2018 at 13:11
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    $\begingroup$ physics.stackexchange.com/questions/252744/… $\endgroup$ Commented Mar 18, 2018 at 13:30

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Are you asking about the meaning of these expressions or their relevance to physics? I'll address the former and mention the latter in brief in closing.

Injection of a particular axial current operator onto a pion line gives rise to its decay. This is represented by the matrix element $\langle 0|J^{\mu}_{\text{axial}}|\pi(P)\rangle,$ that is to say the operator valued insertion acts on an initial pion state leaving the vacuum in the final state with respect to the total number of initial state hadrons.

Replacing $\langle 0|$ with another pion state $\langle \pi(P’)|$, say, would imply the pion is existing in both the initial and final state. Matrix elements of this kind are common in e.g elastic scatterings where the operator insertion is not responsible for the destruction or decay of the initial state. See e.g the upper electron line in the canonical DIS set up, where the familiar QED vector insertion accounts for interaction of electron with the photon, changing only its momentum.

With regards to the utility of these terms, and perhaps to be viewed as an aside, the matrix element in question is useful for predicting the pion decay constant to some loop order using its convenient proportionality to it. The contributing diagram topologies are encoded in the axial current. This is all useful within chiral perturbation theory.

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  • $\begingroup$ "Matrix elements of this kind are common in e.g elastic scatterings where the operator insertion is not responsible for the destruction or decay of the initial state." -- I think this answers my question. Could you please elaborate on this or the "vector insertion" you mentioned? $\endgroup$
    – ersbygre1
    Commented Mar 18, 2018 at 16:29
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    $\begingroup$ In DIS you have a Leptonic and hadronic tensor parametrisation of the matrix element. The former is of the form $\langle k’ | j^{\mu} | k \rangle$ , which stands for interaction of an initial state electron of momentum $k$ with a photon (via the QED operator vertex $\bar{\psi} \gamma^{\mu} \psi$) to give an electron of momentum $k’$ in the final state. This current does not destroy the electron but only changes its momentum. In this sense I meant an elastic collision as electron in initial/final state and it’s a vector insertion because current operator transf. as vector under Lorentz transf. $\endgroup$
    – CAF
    Commented Mar 18, 2018 at 18:04
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@CAF answered your question nicely, but I'd like to illustrate a point on the generic answer to your question.

The matrix element of the type $ \langle \pi (P')|J ^\mu (x) |\pi(P)\rangle$ for generic current J is a form factor derived in current algebra, and typically couples to a gauge boson, depending on the current, so a photon for an e.m. current, a W for a weak current (with which your axial one overlaps leading to charged pion decay, etc.

Its particular value is a subtle nonperturbative QCD expression depending on chiral symmetry breaking, which is why this type of current algebra phenomenology is useful: people parameterize their ignorance segregating it into few unknown matrix elements, like the ones you wrote, estimating those, and then correlating everything else to them.

However, the iron rule in current algebra is respect of all the symmetries of the problem, about which more confidence is at hand. Typically, the parity of the axial current of your (1) is unconventional (+) and that of the pion is (-), so the product has parity (-), like the vector momentum on the left-hand side. In the prototypical low energy effective theory, the σ-model, $$ J_A^{a~\mu} \sim f_\pi \partial^\mu \pi^a+ \frac{\pi^a}{f_\pi} \pi\cdot \partial^\mu \pi+... ~, $$ where you are invited to check the parities and isospin of each term. The crucial point is there are only odd terms of the pion field in the expansion (a ready manifestation of the parity and G-parity preservation in the strong interactions). Sandwiching this between the vacuum and a pion state yields the iconic eqn (1) of charged pion decay.

As a consequence, the extended two-pion amps of the type (1), vanish, $$ \langle \pi (P')|J_A ^\mu (x) |\pi(P)\rangle =0, $$ and so need no interpretation--If they involved vector currents, instead, it would be a different matter altogether.

Similarly, your extension of (2), $\langle \pi (P')|J_P (x) |\pi(P)\rangle =0$, since the pseudoscalar operator is the interpolating field for the pions, $\bar{\psi} \vec{\tau} \gamma_5 \psi \leftrightarrow \vec{\pi}$.

To obtain form factors, one uses two of (1) or (2), in insertions of complete sets of states, which, however, are dominated by the vacuum state, on account of the lowness of the energy involved.

  • In your reference, their (1),(2),(3),(13), amount to the cornerstone Dashen's theorem, a crowning achievement of current algebra, manifest much more painlessly and conceptually directly in the original proof. The quantity $\langle 0| [[Q^a_A,H(0)],Q^b_A] |0\rangle $ amounts to $\langle \pi^a(0)| \Delta_{\chi SB}H| \pi^b(0)\rangle \sim m_\pi^{2~ab} f^2_\pi$ in the pion picture by above, while in the quark picture the double commutator with the explicit symmetry breaking quark mass term yields $\sim m_q \langle 0|\bar{\psi}\psi|0\rangle \delta^{ab}$. (Giving both quarks the same mass for the sake of simplicity. Note this condensate is the parity-preserving scalar, not the pseudoscalar you were considering! The two axial charges had their abnormal parity cancel each other's.)

  • The takeaway is the almost magical property of chiral symmetry breaking in QCD that the square of the almost-goldston pseudoscalars is a linear function of quark masses. Profound types were marveling at this in the late 60s, before QCD.

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  • $\begingroup$ Thank you for this detailed answer! I do not understand what you mean with "...since the pseudoscalar operator is the interpolating field for the pions...". Could you please explain this part? $\endgroup$
    – ersbygre1
    Commented Mar 18, 2018 at 21:03
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    $\begingroup$ I'd rather do it here than in the answer: it is the simplest bilinear of the quark field operators with the quantum numbers (Lorentz, parity, isospin...) of the pion (which, however, is the least appropriate hadron ever to represent as a bound state of a quark and an antiquark in a quark model!). So you might think of it behaving in many processes as a pion.... $\endgroup$ Commented Mar 18, 2018 at 21:06

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