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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?

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    $\begingroup$ 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? $\endgroup$ Jun 23, 2021 at 13:13
  • $\begingroup$ @CosmasZachos I am sorry I do not understand, can you please explain what do you mean by product of its value at every argument? $\endgroup$
    – user44690
    Jun 23, 2021 at 16:49
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    $\begingroup$ 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. $\endgroup$ Jun 23, 2021 at 16:52
  • $\begingroup$ @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? $\endgroup$
    – user44690
    Jun 23, 2021 at 16:54
  • $\begingroup$ Ah... to quote JS Bach this would take some preparation... $\endgroup$ Jun 23, 2021 at 17:00

1 Answer 1

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$\det \chi$ 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 : $$\det m_\chi = \prod_{x\in\mathbb R^d} \chi(x)^{\text d^dx} = \exp\left(\int_{x\in\mathbb R^d}\ln(\chi(x))\text d^dx\right) \tag 1$$

Edit

The multiplication operator is diagonal, as it can be written as an integral kernel proportional to $\delta(x-y)$ : $$m_\chi(x)\phi(x) = \int_{\mathbb R^d} \text d^dy \ m_\chi(x)\delta^{(d)}(x-y)\phi(y)$$

Then, the determinant is just the product of the diagonal entries, which gives equation $(1)$ above.

Another way to see this, more rigorously, is to discretize space on a finite lattice $\Lambda$ of size $L^d$ and lattice step $a \to 0$ and take the limit $a\to 0, L\to \infty$.

Then, a scalar field is just an element of $\mathbb R^{L^d}$ and a multiplication operator is a diagonal $L^d \times L^d$ matrix. Therefore : $$\det m_\chi |_{\Lambda,a} = \prod_{x\in \Lambda}\chi(x)$$ To get a regular limit, we take it to the power $a$, to get : \begin{align} \det m_\chi &= \lim_{a\to 0,\Lambda \to \infty}\det m_\chi|_{\Lambda,a } \\ &= \lim \exp\left(a\sum_{x\in\Lambda}\ln \chi(x) \right) \\ &= \exp\left(\int_{x\in\mathbb R^d}\ln\chi(x)\text d^dx\right) \end{align}

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  • $\begingroup$ Hi, it would be very helpful if you provided some references about this. Can you also explain if this can be generalized to curved-spaces or manifolds? $\endgroup$
    – user44690
    Jun 23, 2021 at 16:50
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    $\begingroup$ This is equation 14.25 here, and equation 14.20 and 14.21 are also of interest $\endgroup$ Jun 23, 2021 at 16:56
  • $\begingroup$ Hi! Can you please explain how did you get the first equality in your answer? $\endgroup$
    – user44690
    Apr 26 at 7:15

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