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?
Answer
@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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