Indexing in Momentum Space

For the position-space representation $\varphi(\vec{n})$ of the field, with $1\leq n_{\mu}\leq N$ and $\mu=1,\ldots,d$, one can index the field into a single $N$-dimensional vector, using the single index $\iota$ instead of all the position coordinates, where

$$\iota= 1+(n_{1}-1)N^{0}+(n_{2}-1)N^{1}+(n_{3}-1)N^{2}+\ldots+(n_{d}-1)N^{d-1}.$$

Defined in this way the index $\iota$ runs from $1$ to $N^{d}$, thus enumerating all the sites of the lattice. There is also an algorithm to obtain the components of $\vec{n}$ back from the value of $\iota$, as we will discuss later. We would like to do the same for the momentum-space representation $\widetilde\varphi (\vec{k})$, with $k_{m}\leq n_{\mu}\leq k_{M}$, $\mu=1,\ldots,d$, where the extremes of the range are defined as

$$k_{m}=-(N-1)/2,\mbox{   }k_{M}=(N-1)/2,$$

for odd $N$, and as

$$k_{m}=-N/2+1,\mbox{   }k_{M}=N/2,$$

for even $N$. In order for the index value of $0$ to correspond to the zero mode $\vec{k}=\vec{0}$, we would like to use the index $\kappa$ given by

$$\kappa=k_{1}N^{0}+k_{2}N^{1}+k_{3}N^{2}+\ldots+k_{d}N^{d-1}.$$

In order to show that one can in fact do this, and to discover how to invert the relation, getting the components of $\vec{k}$ out of $\kappa$, we start with a version of the momentum-space index just like the position-space one,

$$\kappa'=1+(k_{1}-k_{m})N^{0}+(k_{2}-k_{m})N^{1}+(k_{3}-k_{m})N^{2} +\ldots+(k_{d}-k_{m})N^{d-1},$$

which ranges from $1$ to $N^{d}$ and where $0\leq(k_{\mu}-k_{m})\leq N-1$, $\mu=1,\ldots,d$, just like the position-space index. Now, starting from this index we have

$$\begin{eqnarray*} \kappa' & = & 1+k_{1}N^{0}+k_{2}N^{1}+k_{3}N^{2}+\ldots+k_{d}N... ...1} +1-k_{m}\frac{\displaystyle N^{d}-1}{\displaystyle N-1}, \\ \end{eqnarray*}$$

where $N^{d}-1$ is always divisible by $N-1$, of course. Therefore, considering the definition of $\kappa$, we have

$$\kappa'=\left(1-k_{m}\frac{\displaystyle N^{d}-1}{\displaystyle N-1}\right)+\kappa.$$

Since the two indices are related by an additive constant, either one can be used to index the modes. Since $\kappa'$ ranges from $1$ to $N^{d}$, we have for the extreme values of $\kappa$

$$(N^{d}-1)\frac{\displaystyle k_{m}}{\displaystyle N-1}=\kapp... ...)\left(1+\frac{\displaystyle k_{m}}{\displaystyle N-1}\right).$$

Note that $k_{m}$ is negative, so that the lower extreme is negative. For odd $N$ the extremes can be written explicitly as

$$-\frac{\displaystyle N^{d}-1}{\displaystyle 2}\leq\kappa\leq\frac{\displaystyle N^{d}-1}{\displaystyle 2},$$

and the range is therefore symmetrical around $0$. Note that since $N$ is odd so is $N^{d}$, and therefore $N^{d}-1$ is even and thus divisible by $2$. For even $N$ we get

$$-\frac{\displaystyle (N-2)(N^{d}-1)}{\displaystyle 2(N-1)}\l... ...appa\leq\frac{\displaystyle N(N^{d}-1)}{\displaystyle 2(N-1)},$$

so that in this case the range is not symmetrical, and there are $(N^{d}-1)/(N-1)$ more positive-index elements than negative-index elements. A little analysis will show that the integer divisions are exact in this case also, without any truncation: since $N^{d}-1$ is always divisible by $N-1$, we are left with $-(N-2)/2$ and $N/2$ to be considered, and since $N$ is even both $N$ and $N-2$ are divisible by $2$.

If we recall our organization of real and imaginary parts as independent coordinates, as described in the previous section, it is interesting to observe that one can verify that for odd $N$ the negative values of $\kappa$ correspond to imaginary parts, the value $0$ to the real mode, and the positive values to real parts. However, the same is not true for even $N$, due to the modes with one or more components equal to $N/2$, some of which have their imaginary parts at positive values of the index. Due to this, in general it will be necessary to implement an additional pair of indexing arrays in order to map real parts to the corresponding imaginary parts and vice-versa, whenever it becomes necessary to recover the two parts of one and the same mode.

It remains for us to discuss the inversion algorithm. The algorithm for extracting the position coordinates $\vec{n}$ from the position-space index $\iota$ works by integer division, with truncation, and in $d$ dimensions is given by

$$\begin{array}{rcl} \tau & = & \iota-1, n_{d} & = & 1+\tau... ..._{2}-1)N^{1}, n_{1} & = & 1+\tau/\!\!/N^{0}, \\ \end{array}$$

where $\tau$ is a temporary variable. It is therefore clear that we can use exactly the same algorithm for extracting the $d$ shifted momentum coordinates $k_{\mu}-k_{m}+1$ from the index $\kappa'$. Since $\kappa$ and $\kappa'$ are related by an additive constant, we can also use the algorithm for $\kappa$, so long as we start by adding to it the constant $1-\kappa_{m}$, and substitute $n_{\mu}-1$ by $k_{\mu}-k_{m}$ in the formulas, in order to get the components $k_{\mu}$ of $\vec{k}$, and therefore we have the algorithm

$$\begin{array}{rcl} \tau & = & \kappa-\kappa_{m}, k_{d} & ... ...-k_{m})N^{1}, k_{1} & = & k_{m}+\tau/\!\!/N^{0}. \end{array}$$

This completes the discussion of the indexing scheme in momentum space.