# Algebraic doubt concerning orthogonal polynomials in physics research paper

I am not sure if this post should go into the math or physics stackexchange. I am sorry if it is misplaced and I'm obviously fine with moving it accordingly.

So I am currently reading this paper and I am having a hard time seeing the validity of equation (4) on page 4. I try to summarize what seems to be important here for convenient access despite the paywall, so there are:

• vectors $\vec{\sigma} = \lbrace \sigma_1, \sigma_2, \dots, \sigma_N\rbrace$ of length $N$. Depending on another parameter $M$, the components $\sigma_i$ of those vectors may be $$\sigma_i = \begin{cases} \pm m, \pm(m-1), \dots, \pm 1 & M = 2m\\ \pm m, \pm(m-1), \dots, \pm 1, 0 & M = 2m + 1 \end{cases}$$

• a scalar product for arbitrary functions of $\vec{\sigma}$, $f\left(\vec{\sigma}\right)$ and $g\left(\vec{\sigma}\right)$, is given by $$\langle f,g \rangle = \rho_N^0\mathrm{Tr}^{(N)}f\cdot g\tag{1}$$ with the trace operator $\mathrm{Tr}^{(N)} = \sum\limits_{\sigma_1}\sum\limits_{\sigma_2}\cdots\sum\limits_{\sigma_N}$ over the $M^N$ different vectors, and the normalization $\rho^0_N = M^{-N}$.

• a set of (real valued) polynomials $\Theta_n\left(\sigma_p\right)$ is defined as a function of the vector's components $\sigma_p$ by \begin{align} \Theta_n\left(\sigma_p\right) = \begin{cases} \Theta_{2s}\left(\sigma_p\right) = \sum\limits_{k=0}^{s} c_k^{(s)}\sigma_p^{2k}, & s = \begin{cases} 0, 1, \dots, m-1 & M = 2m\\ 0, 1, \dots, m-1, m & M = 2m+1 \end{cases} \\ \Theta_{2s+1}\left(\sigma_p\right) = \sum\limits_{k=0}^{s} d_k^{(s)}\sigma_p^{2k+1}, & s=0, 1, \dots, m-1 \end{cases}\tag{2} \end{align}

The coefficients of the polynomials in $(2)$ are chosen such that the scalar product according to $(1)$ yields \begin{align}\langle\Theta_n\left(\sigma_p\right), \Theta_{n^\prime}\left(\sigma_p\right)\rangle &= \rho^0_N\mathrm{Tr}^{(N)} \Theta_n\left(\sigma_p\right) \Theta_{n^\prime}\left(\sigma_p\right)\\ &= M^{-1}\sum\limits_{\sigma_p=-m}^{m} \Theta_n\left(\sigma_p\right) \Theta_{n^\prime}\left(\sigma_p\right) = \delta_{nn^\prime}\tag{3} \end{align} with the Kronecker delta $\delta_{nn^\prime} = \begin{cases}0 & n \neq n^\prime\\1 & n = n^\prime\end{cases}$.

I can see $(3)$ is some kind of orthogonality relation, the paper then however states

For any two lattice points $p$ and $p^\prime$ and $n \geq 1$, $n^\prime \geq 1$, eq. (3) generalizes to

$$\langle\Theta_n\left(\sigma_p\right), \Theta_{n^\prime}\left(\sigma_{p^\prime}\right)\rangle = M^{-2}\sum\limits_{\sigma_p=-m}^{m}\sum\limits_{\sigma_{p^\prime}=-m}^{m} \Theta_n\left(\sigma_p\right) \Theta_{n^\prime}\left(\sigma_{p^\prime}\right) = \delta_{nn^\prime}\delta_{pp^\prime}\tag{4}$$

Can someone show me how I can see $(4)$ is in fact a "generalization" of $(3)$?

I see how $(4)$ results in $(3)$ when considering the same point $p$: \begin{align} &M^{-2}\sum\limits_{\sigma_p=-m}^{m}\sum\limits_{\sigma_{p^\prime}=-m}^{m} \Theta_n\left(\sigma_p\right) \Theta_{n^\prime}\left(\sigma_p\right)\\ =& M^{-2}M\sum\limits_{\sigma_p=-m}^{m}\Theta_n\left(\sigma_p\right) \Theta_{n^\prime}\left(\sigma_p\right)\\ =& M^{-1}\sum\limits_{\sigma_p=-m}^{m}\Theta_n\left(\sigma_p\right) \Theta_{n^\prime}\left(\sigma_p\right) = \delta_{n,n^\prime}, \end{align} so in that sense $(4)$ seems to be some kind of "generalization" of $(3)$.

However, I do not get how I know the last equal sign in $(4)$ holds, i. e. how I know the second Kronecker delta $\delta_{p p^\prime}$ must go there. If I'm not mistaken, to understand this I'd need to be able to show the nested sum indeed sums up to zero when considering different points $p$ and $p^\prime$ when $(3)$ holds, but I am not able to do so.

EDIT: After the first answer, I feel as if I should clarify my reasoning a bit. The polynomials are fully defined once their coefficients have been determined from the orthogonality relation eq. (3). In fact, constructing the polynomials for $M=3$, \begin{align} \Theta_0\left(\sigma_p\right) &= 1\\ \Theta_1\left(\sigma_p\right) &= \sqrt{\frac{3}{2}}\sigma_p\\ \Theta_2\left(\sigma_p\right) &= \sqrt{2} - \frac{3}{\sqrt{2}}\sigma_p^2 \end{align} as given in the paper as an example, eq. (4) holds despite the polynomials being defined with the help of eq. (3) only. This is why I think eq. (3) somehow needs to be enough to guarantee eq. (4) holds but I fail to see how this is facilitated.

EDIT 2: After the answers I received I thought about my problem a little further and I unfortunately still can't fully comprehend the reasoning behind knowing the last equation in (4) holds. However, I think I can give a more precise statement of what my stumbling block actually is.

So, starting from the generalization of (3) in (4), I split up the summations into two partial sums: $$M^{-2}\sum\limits_{\sigma_p=-m}^{m}\sum\limits_{\sigma_{p^\prime}=-m}^{m} \Theta_n\left(\sigma_p\right) \Theta_{n^\prime}\left(\sigma_{p^\prime}\right) = M^{-2}\left( \sum\limits_{\sigma_p} \Theta_n\left(\sigma_p\right)\Theta_{n^\prime} \left(\sigma_p\right) + \sum\limits_{\sigma_{p^\prime}\neq \sigma_p} \Theta_n\left(\sigma_p\right) \Theta_{n^\prime}\left(\sigma_{p^\prime}\right) \right) .$$ The first partial sum contains only those summands with equal arguments for both polynomials, while the second only contains summands with different arguments for both polynomials.

E.g, for $M = 3$ as given in the paper, the first sum contains $3$ summands ($\sigma_p$ cycling through $-1, 0, 1$) and the second sum contains $6$ summands (cycling through $(-1,0), (-1,1), (0,-1), (0,1), (1,-1), (1,0)$) which together yield the same $9$ summands the original double sum in the generalization of (3) in (4) cycles through.

So, the first partial sum can be simplified easily using (3), see the annotation of its underbrace. However, the last equality in (4) implies the second partial sum evaluates to what is annotated in its underbrace below: \begin{align} & M^{-2}\left( \underbrace{\sum\limits_{\sigma_p} \Theta_n\left(\sigma_p\right)\Theta_{n^\prime} \left(\sigma_p\right)}_{=M\delta_{nn^\prime}\text{ from (3)}} + \underbrace{\sum\limits_{\sigma_{p^\prime}\neq \sigma_p} \Theta_n\left(\sigma_p\right) \Theta_{n^\prime}\left(\sigma_{p^\prime}\right)}_{=\left(M^2\delta_{pp^\prime}-M\right)\delta_{nn^\prime}} \right)\\ = & M^{-2}\left( M\delta_{nn^\prime} + \left(M^2\delta_{pp^\prime}-M\right)\delta_{nn^\prime} \right)\\ = & M^{-2}\left( M + M^2\delta_{pp^\prime}-M \right)\delta_{nn^\prime}\\ = & M^{-2}\left( M^2\delta_{pp^\prime} \right)\delta_{nn^\prime}\\ = & \delta_{pp^\prime}\delta_{nn^\prime} \end{align} So if I did not make a mistake here, my comprehension problem boils down to:

How is it, that $\sum\limits_{\sigma_{p^\prime}\neq \sigma_p} \Theta_n\left(\sigma_p\right) \Theta_{n^\prime}\left(\sigma_{p^\prime}\right) = \left(M^2\delta_{pp^\prime}-M\right)\delta_{nn^\prime}$ is guaranteed to hold?

I do not understand where this property of the polynomials comes from - as opposed to (3) because after all, (3) was used to define the polynomials, i.e. they are designed to fulfill (3). As I see things now, this additional property can't be due to (3) as (3), to my understanding, only makes a statement considering the same arguments for both polynomials in all summands. So where does this additional property come from?

Thank you very much in advance.

If you follow the paper you realize that the orthogonality of the basis $\Theta_n(\sigma_p)$ is defined in one site of the lattice, Eq. (3) is actually that. There is nothing that allows you to go from Eq. (3) to Eq. (4), except for the fact that physically makes sense that the single-site states you use to describe one site $p$ should be independent from states to describe another site $q$. So you impose this condition by hand
$$\langle \Theta_n(\sigma_p)| \Theta_n(\sigma_q)\rangle = \delta_{pq}$$
• I still don't get how $\langle \Theta_n\left(\sigma_p\right) , \Theta_n\left(\sigma_q\right) \rangle = \delta_{pq}$ could be imposed though. After all, the polynomials are fully defined after their coefficients are determined from eq. (3). So my reasoning still is that eq. (3) ought to be enough to guarantee eq. (4) holds. In fact, if you construct the polynomials for $M=3$ as given in the example, you find eq. (4) really holds without explicitly imposing anything in addition. I just keep failing to see how this is facilitated ... – JohnSmithOptional Nov 17 '16 at 6:38
• @JohnSmithOptional Let me give you another example, imagine you have two independent H atoms: $1$ and $2$, each atom is described in the basis $|a\rangle \equiv |lmn\rangle$, so the state $|a\rangle_1$ describes the first atom, and $|a \rangle_2$ the second one. We know that you can define the basis such that ${}_1\langle a | a'\rangle_1 = \delta_{aa'}$. Now, based on this information only, can you predict what is going to be the result of ${}_1\langle a | a'\rangle_2$? – caverac Nov 19 '16 at 13:25