Path integral for complex scalar field I am taking a QFT course which focuses on the path integral formulation. At a certain point, I was confused because we saw that, when integrating over complex Grassmann fields for fermions, we defined the complex conjugate as
$$(\theta\eta)^* = \eta^*\theta^*\tag{1}$$
and then said that we could treat $\theta$ and $\theta^*$ as independent variables, so we integrate over both variables. When I asked the lecturer about it, he said it was because complex conjugation is not uniquely defined for Grassmann variables, which means you can’t really obtain $\theta^*$ from $\theta$, so you have to integrate over both. However, let’s consider we are calculating path integrals for a complex scalar field $\phi$. Would we integrate only over $\phi$ or both $\phi$ and $\phi^*$? Somehow I have seen both options in different references (for example, Peskin and Schroeder integrate only over $\phi$ in section 9.6). In this case $\phi$ and $\phi^*$ are dependent on each other, so you should only have one integration measure, right? Also, how would the integration results such as Gaussian integrals change when considering complex fields?
 A: For simplicity, let's consider the standard complex integral (the path integral is just the limit of the product of many standard integrals).
A function defined on the complex plane $f(z)$ is the same as a function of two real variable $f(x,y)$, where $x+iy =z$. Therefore, integrating over the complex plane is the same as integrating over two real variables. Formally, we could write the measure $\text d^2 z= \text dx \text dy$ (with the $2$ exponent to remember that this is a $2$ dimensional integral).
However, it is often useful to take $(z,\bar z)$ as variables. At first, it might not be clear why this works. For differentiation, we can define $\partial = \frac12(\partial_x - i \partial_y)$ and $\bar{\partial} = \frac12(\partial_x + i \partial_y)$. Then you check that those differential operators satisfy the Leibniz rule as well as :
$$\partial z = \bar \partial \bar z = 1 \qquad \text{ and }\qquad \partial\bar z = \bar \partial z = 0$$
Therefore, everything happens as if $z$ and $\bar z$ where independent variables.
For integration, it goes the same way. We define $\text d z = \text d x + i \text dy$ and $\text d\bar z = \text dx - i \text dy$, and check that, up to normalization :
$$\text d^2 z = \text dx \text dy = \text d z \text d\bar z$$
Here again, we can act as if the two variables are independent.
Conclusion For a complex field (bosonic or fermionic), the path integral is an infinite product of $2$ dimensional complex integrals. Whether we write :
$$\mathcal D\phi = \prod_x \text d^2 \phi(x)$$
or
$$\mathcal D\phi\mathcal D\bar \phi = \prod_x \text d \phi(x) \prod_x \text d\bar \phi(x)$$
is purely a matter of conventions.
A: *

*In the same way that a Grassmann-even field can be real or complex valued, a Grassmann-odd field can be real or complex valued.


*Concerning P&S section 9.6, they consider a Grassmann-even complex field $\phi$. They write the path integral measure as ${\cal D}\phi$, while other authors would instead write the path integral measure as
${\cal D}\phi^{\ast}{\cal D}\phi$, but they really mean the same thing, cf. e.g. this Phys.SE post.


*Concerning whether to treat the complex conjugate field as independent, see e.g. this related Phys.SE post.


*Concerning OP's eq. (1) be aware that there exist different sign conventions in the literature for the complex conjugation of a product of Grassmann-odd variables
