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Not sure if this is a path integral. Never finished QFT, but I remember stuff like this. I have a density function like,

$$ P(z) \mathcal{D}z \propto \exp \left[ - \iint_0^T ds dt ~ \lambda(s, t) ~ z(s) z(t) \right] \mathcal{D}z $$

and I want to get $\mathbb{E}[z(s)z(t)]$, so I was planning to get the partition function and take variations of $\lambda$.

Summing over Fourier terms $ Z_n = \frac{1}{T} \int_0^T dt ~ z(t) \exp\left[ -i \frac{2\pi n t}{T} \right] $, $$\begin{align} \sum_{n,m=-N}^{N} Z_n Z_m \Lambda_{nm} &= \frac{1}{T^2} \iint_0^T ds dt ~ z(s) z(t) \sum_{n,m=-N}^{N} \exp\left[ -i \frac{2\pi (n t + m s)}{T} \right] \Lambda_{nm} \\ \lambda(s, t) &= \frac{1}{T^2} \sum_{n,m=-N}^{N} \exp\left[ -i \frac{2\pi (n t + m s)}{T} \right] \Lambda_{nm} \\ \Lambda_{nm} &= \iint ds dt ~ \lambda(s, t) ~ \exp\left[ i \frac{2\pi (n t + m s)}{T} \right] \end{align}$$

So then?

$$\begin{align} P(z) ~ \mathcal{D}z &\propto \exp \left[ - \sum_{n,m=-N}^{N} Z_n Z_m \Lambda_{nm} \right] \frac{d^Nz}{\mathcal{D}z} \mathcal{D}z \end{align}$$

I need to figure out the "Jacobian" $\frac{d^Nz}{\mathcal{D}z}$, but I don't even know how to define it.

How do I get the Jacobian here?

  • Quick hint: If one discretizes $\int_{0}^{T} dt$-integral and do change of variables (in the functional integral) to "Fourier space", the transformation matrix ("Discrete Fourier Transformation" matrix) is unitary with unit determinant. – Sunyam Nov 10 '20 at 06:10
  • @Sunyam Perhaps I am misunderstanding, but I'm not doing a Fourier transform. I am putting my integral into a Fourier series. The transform matrix of a discrete Fourier Series transform would be non-square and ill-determined. –  Nov 10 '20 at 06:15
  • I get it now. Haven't done this in a while. Forgot the limit $N \rightarrow \infty$ then the making the integral discrete makes sense. And yeah it's unitary, so $J=1$, my favorite number. –  Nov 10 '20 at 15:51

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