Functional measure variable change

Question

On Peskin & Schroeder's QFT, page 285, the book introduces the functional quantization of scalar fields.

We can replace the variables $\phi(x)$ defined on a continuum of points by variables $\phi(x_i)$ defined at the points $x_i$ of a square lattice. Let the lattice spacing be $\epsilon$, the four space-time volume $L^4$, and define $$\mathcal{D} \phi=\prod_i d \phi\left(x_i\right) \tag{9.20}$$ up to an irrelevant overall constant.

The field values $\phi(x_i)$ can be represented by a discrete Fourier series $$\phi\left(x_i\right)=\frac{1}{V} \sum_n e^{-i k_n \cdot x_i} \phi\left(k_n\right) \tag{9.21} $$ where $k_n^\mu=2\pi n^\mu/L$, with $n^\mu$ an integer. Since $\phi(x)$ is real, we have $\phi^*(k)=\phi(-k)$. Then the book consider the real and imaginary parts of the $\phi(k_n)$ with $k_n^0$ as independent variables. Then they have following unitary transformation $$ \mathcal{D} \phi(x)=\prod_{k_n^0>0} d \operatorname{Re} \phi\left(k_n\right) d \operatorname{Im} \phi\left(k_n\right) $$ I am troubled for why the Jacobian determinant equal to one only for $k_n^0>0$ case?

The Jacobian determinant ($J$) should look like $$J=\left| \begin{array}{ccccc} \frac{\partial \phi(x_1)}{\partial \text{Re} \phi(k_1)} & \frac{\partial \phi(x_1)}{\partial \text{Im} \phi(k_1)} & \frac{\partial \phi(x_1)}{\partial \text{Re} \phi(k_2)} & \frac{\partial \phi(x_1)}{\partial \text{Im} \phi(k_2)} & \ldots \\ \frac{\partial \phi(x_2)}{\partial \text{Re} \phi(k_1)} & \frac{\partial \phi(x_2)}{\partial \text{Im} \phi(k_1)} & \frac{\partial \phi(x_2)}{\partial \text{Re} \phi(k_2)} & \frac{\partial \phi(x_2)}{\partial \text{Im} \phi(k_2)} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}\right| $$

Here is my detailed questions.

1. How can we make sure that the number of $x_i$ equal to $k_i$? If so, $k_n^0>0$ makes sense, since we have $\text{Re}$ and $\text{Im}$.

2. Can we satisfy the orthogonal relation if we only consider $k_n^0>0$ case? Would this be the reason for $J=1$?

Answer 1

Question 1

First of all, a Jacobian is not a transformation, but a measure of how a transformation affects the volume or area elements. For example, if you have a function $f$ that maps $x$ to $y$, the Jacobian of $f$ is the ratio of the area element $dy$ to the area element $dx$. The Jacobian can be computed using partial derivatives and determinants.

A Lorentz transformation is a special kind of transformation that preserves the interval between two events in Minkowski spacetime. It can be represented by a $4*4$ matrix that satisfies some conditions. The Jacobian of a Lorentz transformation is always 1, meaning that it does not change the spacetime volume element $dtdxdydz$ ($d^4x$).

A Fourier transform is another kind of transformation that changes the basis of a function space. For example, if you have a function $f(x)$ defined on the real line, you can express it as a linear combination of periodic functions $f~(k)$ using the Fourier transform1. The Fourier transform can be seen as a change of coordinates from $x$ to $k$, where $k$ is the wavenumber or frequency. The Jacobian of the Fouriertransform is $\frac{1}{√(2π)}$, meaning that it scales down the area element $dk$ by a constant factor.

Now, to answer your specific question, you need to understand how the functional quantization of scalar fields works. Basically, it is a way of defining a quantum theory for scalar fields using path integrals and generating functionals. The idea is to replace the variables $\phi(x)$ defined on a continuum of points by variables $\phi\left(x_i\right)$ defined at the points $x_i$ of a square lattice. Then, you can write the correlation functions and Feynman rules for the scalar field theory using these discrete variables.

The Fourier transform comes in when you want to express the discrete variables $\phi\left(x_i\right)$ in terms of their wavenumber spectrum $\phi(k_n)$. This is useful because it simplifies some calculations and allows you to use Wick’s theorem and Feynman diagrams. The Fourier transform here is just a linear transformation between two bases: the point functions and the periodic functions.

The Jacobian determinant comes in when you want to change the measure of integration from $D\phi(x)$ to $D\phi(k)$. This is because when you change variables in an integral, you need to multiply by the absolute value of the Jacobian determinant to preserve the value of the integral1. In this case, the Jacobian determinant is just the product of all the partial derivatives $\frac{\partial\phi\left(x_i\right)}{\partial\phi(k_n)}$, where $i$ and $n$ are related by $x_i = \frac{k_nL}{(2π^3)}$.

The reason why you only consider $k_0^n > 0$ is because you are dealing with real scalar fields, which satisfy $\phi^*(k) = \phi(−k)$. This means that you only need half of the spectrum to describe the field completely. The other half is redundant and can be obtained by complex conjugation3. Therefore, you only need to integrate over $k_0^n > 0$ (or $k_0^n < 0$) and multiply by 2 to account for both signs.

Question 2

The orthogonal relation of the Fourier transform states that if $f(x)$ and $g(x)$ are two functions defined on the real line, then their Fourier transforms f~(k) and g~(k) satisfy the following equation:

$(f~(k), g~(k)) = (f(x), g(x))$

where $(f, g)$ denotes the inner product of two functions, defined as

$(f, g) = \int f(x)g(x)dx$

This relation means that the Fourier transform preserves the inner product of two functions, or equivalently, their norm and angle. It also implies that if $f(x)$ and $g(x)$ are orthogonal, meaning that $(f, g) = 0$, then so are $f~(k)$ and $g~(k)$.

Now, in your case, you are dealing with discrete variables $\phi(x_i)$ and $\phi(k_n)$, which are related by the discrete Fourier transform. The discrete Fourier transform can be seen as a matrix multiplication of a vector $\phi(x_i)$ by a matrix $F$ whose entries are complex exponentials. The matrix $F$ is called the Fourier matrix and it has some special properties. One of them is that it is unitary, meaning that its inverse is equal to its conjugate transpose. This implies that F preserves the inner product of two vectors, or equivalently, their norm and angle. It also implies that if $\phi(x_i)$ and $\psi(x_i)$ are orthogonal, meaning that their dot product is zero, then so are $F\phi(k_n)$ and $F\psi(k_n)$.