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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?

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2 Answers 2

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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.

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  • $\begingroup$ Thank you very much. So, if we discretized the integral $\int d^2z \ f(z,\overline{z})$, we would just sum over one set of variables $z$, right? So, for each point in the integral $z_i$, we would just substitute $f(z,\overline{z})\to f(z_{i},\overline{(z_{i})})$. $\endgroup$
    – Marcosko
    Feb 22, 2022 at 18:08
  • $\begingroup$ With the points $(z_i)$ filling the complex plane (or the relevant domain), yes. $\endgroup$ Feb 22, 2022 at 18:21
  • $\begingroup$ Understood. I guess I was confused about the notation, then. Thank you! $\endgroup$
    – Marcosko
    Feb 22, 2022 at 18:33
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  1. 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.

  2. 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.

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

  4. 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.

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  • $\begingroup$ Thank you very much for your answer and the links. I can see that the two fields are independent with respect to the derivative, which is useful for the classical equations of motion. The integral part is what I have a little more trouble visualizing, because it seems like we are summing over double the sets of fields in one case vs the other. $\endgroup$
    – Marcosko
    Feb 22, 2022 at 18:12

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