I am having trouble understanding how people parameterise hadronic matrix elements in terms of form factors. For example the decay of a pseudoscalar B to a vector meson: $$ \langle V(k, \eta) | \bar q \gamma^\mu \gamma_5 b | B(p) \rangle = i \eta^*_\nu \bigg[ g^{\mu \nu} (m_V+m_B) A_1 - \frac{(p+k)^\mu q^\nu}{m_B + m_V} A_2 - \frac{2m_V q^\mu q^\nu}{q^2} (A_3 - A_0) \bigg] \qquad (1) $$ $$ \langle V(k, \eta) | \bar q \gamma^\mu b | B(p) \rangle = \frac{2 V}{m_B + m_V} \epsilon^{\mu \nu \rho \sigma} \eta_\nu^* p_\rho k_\sigma \qquad (2) $$ where $A_{0 \rightarrow 3}$ and $V$ are form factors that depend on $q^2 = (p-k)^2$.

What leads us to write down the matrix element in this form?

  • Why the complicated form for (1), why not just $g^{\mu \nu}A + p^\mu p^\nu B + k^\mu p^\mu C \qquad (k\cdot \eta = 0)$?
  • Indeed why are there 4 (not 3) form factors for (1) but only 1 for (2)?
  • Why are we allowed an $\epsilon$ tensor in (2) but not in (1)? I know this is something to do with parity (B being $0^-$ and $\gamma^\mu \gamma_5$ being axial or something) but what exactly is the argument we make?

Many thanks.

  • $\begingroup$ What is $q^\mu$ in the RHS of (1)? $\endgroup$
    – Prahar
    Commented Jul 7, 2021 at 11:49
  • $\begingroup$ In the above $ q^\mu = p^\mu - k^\mu $. $\endgroup$ Commented Jul 7, 2021 at 13:45

1 Answer 1


In (1) you have exactly 3 form factors $A_1$, $A_2$ and $A_3 - A_0$. The complicated form used here will probably become clear later (or should already be clear based on earlier discussions). Strictly speaking, the exact form that the author choses to work with is arbitrary.

In (2), the LHS is not parity invariant (since $B$ is a pseudoscalar). It follows that the tensor structure on the RHS must also not be parity invariant and there is only one such structure we can write.

  • $\begingroup$ Thank you for your answer! So the choice (1) is basically just for convenience, rather than some deep reason. Regarding (2), a vector meson like a $\rho$ has negative parity like the B ($J^P = 1^-$ vs $0^-$), and under parity $\bar q \gamma^\mu b \rightarrow \bar q \gamma_\mu b $. Under parity we get two minus signs total? One from the $|B \rangle$, one from the $| V \rangle$ and none from the $\gamma^\mu$ bit? How does this lead to the LHS not being parity invariant? $\endgroup$ Commented Jul 7, 2021 at 18:00

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