Generating functional for scalar field vs Dirac field

Consider the generating functional for the free Dirac field.

$$Z_0[J,\overline J] := \int \mathcal D\overline \psi \mathcal D \psi \ \exp{(i \overline \psi (i\gamma^\mu{\partial_\mu} - m_0) \psi + iJ\overline\psi + i\overline J\psi)} \,.$$

Contrast it with the generating functional for the complex scalar field.

$$Z_0[J,J^*] := \int \mathcal D\phi\ \exp{(-\frac{i}{2} \phi^*(\partial^2 + m_0^2) \phi + iJ\phi^* + iJ^*\phi)} \,.$$

Compare the measures in the two cases.

Why don't we have an integral over $\mathcal D\phi^*\mathcal D\phi$ for the scalar field, just like we had for an integral over $\mathcal D\overline \psi \mathcal D \psi$ for fermions?

• Given that the path integral measure exists only heuristically to begin with, do you have any reason to believe these measures actually are different in a meaningful way and this is not just a notational inconsistency? – ACuriousMind Jul 19 '17 at 8:41
• @ACuriousMind If the scalar field $\phi$ is real, then the generating functional above reduces to the following. $$Z_0 [J] = \int \mathcal D\phi \int\mathcal D\phi \exp(-\frac{i}{2} \phi(\partial^2 + m_0^2) \phi + 2iJ\phi) \\ \propto \int \mathcal D\phi \exp \Big(-\frac{i}{2}\int d^4x\int d^4y\ J(-x) \Delta_F(x-y)J(y) \Big)$$ This is not good, right? We could hide the extra $\int \mathcal D\phi$ in an overall proportionality constant, but that sounds like a dirty trick to me. Is it really notational inconsistency? Isn't the well-posedness of these measures important due to anomalies? – Nanashi No Gombe Jul 19 '17 at 13:42
• When you examine the measures for anomalies like in the Fujikawa method, you work with some regularised form for which your "measures" here are merely shorthand notation. As long as you know what that shorthand stands for, it really doesn't matter how you write it, does it? – ACuriousMind Jul 19 '17 at 13:54