Calculation of fluctuation corrections to the saddle point approximation

Question

From Statistical physics of fields by Mehran Kardar page 45 section 3.6

[...] the partition function including small fluctuations is $$\begin{align} Z \approx e^{-V\left(t/2 \cdot \overline m^2 +u \overline m^4\right)} &\cdot \int \mathcal{D}\phi_l(x)e^{-K/2 \int d^d x\left[(\nabla \phi_l)^2 + \phi_l^2/\xi_l^2\right]} \\ &\cdot \int \mathcal{D}\phi_t(x)e^{-K/2\int d^d x\left[(\nabla\phi_t)^2+\phi_t^2/\xi_t^2\right]}. \end{align}\tag{3.56}$$ Each of the Gaussian kernels is diagonalized by the Fourier transforms $$ \overline \phi(\mathbf{q})=\int d^d \mathbf{x}\ e^{ -i\mathbf{q}\cdot \mathbf{x}} \phi(\mathbf{x}) / \sqrt{V},$$ with the corresponding eigenvalues $K(\mathbf{q})= K(q^2+\xi^{-2})$. [...]

What is the expression of the integral for Z in terms of the Fourier transformed coordinate $ \phi(q)$? What I cannot understand is how does the $ D\phi(x)$ changes under the Fourier transform. Given that the Fourier transform originally was defined as a discrete sum do the no of modes remain conserved under the transformation ( if x was defined over N points there would be N modes of q) As the complex conjugate of $\phi(q) $ also gives the same integral as $\phi(q)$ ?

Answer 1

Consider

$$Z = \int D \phi(x) \, e^{ -K/2 \int d^d x[(\nabla \phi)^2 + \phi^2/ \xi^2]}.$$

We have $D\phi(x) = D\phi(q)$. One way to see this is that the Fourier transform applied twice returns to the original partition function. So $K^2 =1$, and since $K$ is the absolute value of a determinant we have $K=1$.

The Fourier transform of the action is derived by plugging in $\phi(x) = \int d^d q/(2\pi)^d \phi(q) e^{-iq\cdot x}$. In the continuum case, both have an infinite number of modes. In the discrete case, say on $x=1,\ldots,N$, $\phi(x) = \frac{1}{\sqrt{N}} \sum_{q=1}^N \phi(q)e^{-i \frac{2\pi q}{N} x}$, so there are indeed $N$ modes of $\phi(q)$. This is in keeping with $D\phi(x) = D\phi(q)$.