# Functional determinants

I wish to know what is the result of this Gaussian Functional Integral

$$Z[\chi] = \int[\mathcal{D}\phi]~e^{-i\int d^dx ~\phi^2\chi}$$ where $$\phi, \chi$$ are position dependent fields. Now, my question is whether

$$Z[\chi] = (\det\chi)^{-1/2} ?$$ But, since, $$\chi$$ is not an operator just a scalar field is $$Z[\chi] = \chi^{-1/2}$$ then? What is the correct answer?

• Well, are you interpreting your $\chi^{-1/2}$ as the product of its value at every argument, as the functional determinant asks you to? Why don't you schematize your functional integral in a space with just three spacetime points? – Cosmas Zachos Jun 23 at 13:13
• @CosmasZachos I am sorry I do not understand, can you please explain what do you mean by product of its value at every argument? – user44690 Jun 23 at 16:49
• You already got a sound answer. Test-drive it with a 3-point spacetime, consisting of $x_1;x_2;x_3$, as when you were taught what a functional integral really is.. Convert spacetime integrals to sums. Write down your functional integral as a triple standard integral. – Cosmas Zachos Jun 23 at 16:52
• @CosmasZachos Yes I agree it is a good answer, however, it would be helpful if you can comment a bit about its generalization to curved-spaces too? – user44690 Jun 23 at 16:54
• Ah... to quote JS Bach this would take some preparation... – Cosmas Zachos Jun 23 at 17:00

$$\det \chi$$ is a slight abuse of notation. You are actually computing the determinant of the multiplication operator $$m_\chi:\phi \mapsto\chi\phi$$. As @CosmasZachos pointed out in the comment, this is the product integral : $$\det m_\chi = \prod_{x\in\mathbb R^d} \chi(x)^{\text d^dx} = \exp\left(\int_{x\in\mathbb R^d}\ln(\chi(x))\text d^dx\right)$$