Basis Transformations in Configuration Space

The whole formalism of Fourier transformation can be applied in identical form both to the dimensionless field $\varphi$ and to the dimensionfull field $\phi$, as well as to any other quantity of the theory that is a function of position on the lattice, whatever the boundary conditions may be. In fact, the transformation to momentum space is a fundamental operation, not only in the classical theory but in the quantum theory as well, because, as we shall see in more detail later, it corresponds to a transformation from a representation in terms of point-like quantities, whose quantum expectation values cannot be observed directly, to a representation in terms of extended quantities, whole expectation values are associated in a more direct way to the quantities that are in fact physically observable.

Any operation that is linear on the fields and that involves an integration (in the continuum case) or sum (in the discrete case) over the whole lattice, as is the case for the Fourier transforms, may be understood as a matrix multiplication operation on a vector. This is true both for periodical and for fixed boundary conditions. The Fourier transformation itself, for periodical boundary conditions, which is given by

$$\widetilde\varphi (\vec{k})=\frac{1}{N^{d}}\sum_{\vec{n}} e^{\imath\frac{2\pi}{N}\vec{k}\cdot\vec{n}}\varphi(\vec{n}),$$

may be written as a matrix operation so long as we represent the positions $\vec{n}$ of the sites by means of the index $\iota$, as discussed in section 2.4, at the same time that we exchange the integer coordinates $\vec{k}$ of momentum space for another index $\kappa$ defined in an analogous way. Once we have “piled up” in this way both the $\vec{n}$ and the $\vec{k}$ coordinates into indices of vectors of dimension $N^{d}$, we may write the transformation as

$$\widetilde\varphi _{\kappa}=\sum_{\iota=1}^{N^{d}}F_{\kappa\iota}\varphi_{\iota} =F_{\kappa\iota}\varphi_{\iota},$$

where we will usually omit the explicit sum as we did here, hence adopting the summation convention for the matrix products. The equation above represents the matrix product of the matrix $\mathbb{F}$ of components $F_{\kappa\iota}$ by the vector $\varphi$ of components $\varphi_{\iota}$, resulting in another vector $\widetilde\varphi$ of components $\widetilde\varphi _{\kappa}$. In this language the Fourier transform of the field $\varphi_{\iota}$ is an $N^{d}$-vector $\widetilde\varphi _{\kappa}$ that may be written in matrix notation as $\widetilde\varphi =\mathbb{F}\varphi$.

From the point of view of configuration space, Fourier transformation is a simple change of basis in a vector space. We may consider the set of $N^{d}$ vectors $\widehat{\varphi}_{\iota}$, which are equal to $1$ at a particular site in position space and to $0$ at all the others, as a basis of the space of configurations, since any configuration $\varphi$ may be written as a linear combination of these basis versors. In the same way, the set of $N^{d}$ vectors $\widehat{\widetilde\varphi }_{\kappa}$, which are equal to $1$ at a particular mode in momentum space and to $0$ at all the others, also form a basis of the same space. This is so because, since the Fourier transformation exists for any configuration, is linear and invertible, any configuration may also be written as a linear combination of this other set of basis versors.

This represents in fact a simple decomposition of the configuration in terms of normal modes of oscillation in momentum space. The transformation of basis is represented by the matrix $F_{\kappa\iota}$, with the sites of coordinates $n_{\mu}$ represented by the index $\iota$ and the modes of coordinates $k_{\mu}$ by the corresponding index $\kappa$. In the case of periodical boundary conditions, according to the normalization convention defined before, this matrix is given by

$$F_{\kappa\iota}=\frac{1}{N^{d}}e^{\imath\frac{2\pi}{N}\vec{k}\cdot\vec{n}}.$$

It is easy to represent this explicitly in one dimension, where the index $\iota$ is simply the site coordinate $n=n_{1}$ and the index $\kappa$ the momentum coordinate $k=k_{1}$. In this case we simply get

$$F_{\kappa\iota}=F_{kn}=\frac{1}{N}e^{\imath 2\pi k n/N}.$$

Up to the normalization convention adopted, in any dimension $d$ the Fourier transformation is an unitary transformation and one can check that the transformation matrix is unitary, with a constant determinant, independent of the fields. In other words we have that $\mathbb{F}^{\dagger}\sim\mathbb{F}^{-1}$ up to the normalization convention. For the normalization that we adopt here we have in fact $\mathbb{F}^{-1}=N^{d}\mathbb{F}^{\dagger}$ and one can check (problem 2.9.2) that

$$N^{d}F^{\dagger}_{\iota'\kappa}F_{\kappa\iota}=I_{\iota'\iota},$$

where $I_{\iota'\iota}=\delta_{\iota'\iota}$ is the unit matrix. Again, this is just a consequence of the orthogonality and completeness relations satisfied by the mode functions. One can get the transformation to become truly unitary with the normalization

$$F^{\rm (u)}_{\kappa\iota}= \frac{1}{N^{d/2}}e^{\imath\frac{2\pi}{N}\vec{k}\cdot\vec{n}}.$$

As we shall see later on, in the quantum theory it will be necessary to consider the determinant of this transformation. Since the transformation is linear its determinant is a constant, in the sense that it does not depend on the fields. One can easily show in the one-dimensional case (problem 2.9.3) that, with our usual normalization, we have

  $$\det(\mathbb{F})=\left\{ \begin{array}{l} N^{-N/2}\;\imath^{(N+2)... ... \\ N^{-N/2}\;\imath^{(1-N)/2}\mbox{~~if $N$\ is odd,} \\ \end{array}\right.$$ (2.9.1)

while for the inverse transformation we have $\det(\mathbb{F}^{-1})=1/\det(\mathbb{F})$, naturally. For larger dimensions the calculation is more complex but the result still has the all-important property that it does not depend on the fields.

Problems

1. Write explicitly the matrix $\mathbb{F}$ in the one-dimensional case $d=1$ for the following lattice sizes: $N=2$, $N=4$ and $N=6$. Write also the matrix for the two-dimensional case $d=2$ with $N=2$.

2. If $\mathbb{F}$ is the Fourier matrix with our usual normalization and $\mathbb{I}$ is the unit matrix, show that $\mathbb{F}^{-1}=N^{d}\mathbb{F}^{\dagger}$, that is, show that

   
   

   $$N^{d}\mathbb{F}^{\dagger}\mathbb{F}=\mathbb{I}.$$
   
   

3. Calculate the determinant of $\mathbb{F}$ in the one-dimensional case, for an arbitrary $N$, obtaining the result given in equation (2.9.1).

4. ($\star$) Calculate the determinant of $\mathbb{F}$ in the two-dimensional case, for an arbitrary $N$.