Derivation of probability density of isolated polymers

Question

(Crossposted on physics SE) I am reading Introduction to Polymer Physics by Doi, and in his proof for the probability distribution for ideal polymers of length $N$ and end-to-end vector $\mathbf{R}$, he does the following: \begin{align} P(\mathbf{R}-\mathbf{b}_i,N-1)=P(\mathbf{R},N)-\frac{\partial P}{\partial N}-\frac{\partial P}{\partial R_{\alpha}}b_{i\alpha}+\frac{1}{2}\cdot \frac{\partial ^2 P}{\partial R_{\alpha}\partial R_{\beta}}b_{i\alpha}b_{i\beta}\label{1}\tag{1} \end{align} where $b_{i\alpha}$ is the component of $\mathbf{b}_i$ in the $\alpha$ direction, $R_{\alpha}$ is a component of $\mathbf{R}$,for the lattice the polymer lies in. We are using the Einstein convention for summation in the above equation.

He makes the statement, $$\frac{1}{z}\sum _{i=1}^z b_{i\alpha}b_{i\beta} = \frac{\delta _{\alpha\beta}b^2}{3}\label{2}\tag{2}$$ where $z$ is the coordination number, $\delta _{ij}$ is the Kronecker delta, $b$ is the length of the lattice edge, so $|\mathbf{b}_i| = b$. I don't understand where this comes from.

What is the proof of this statement?

Since I have a problem with the Taylor expansion, I am going to write down what I am not understanding. \begin{align} P(\mathbf{R}-\mathbf{b}) = P([R_x-b_x, R_y-b_y]^T)\approx P([R_x,R_y]^T)-\frac{\partial P}{\partial R_x} b_x-\frac{\partial P}{\partial R_y}b_y+\frac{1}{2}\cdot \left(\frac{\partial ^2 P}{\partial R_x^2}b_x^2+\frac{\partial ^2 P}{\partial R_y^2}b_y^2 + 2\frac{\partial ^2 P}{\partial R_xR_y}b_xb_y \right) \end{align} This was derived using the standard higher-dimensional Taylor expansion formula for $f(\mathbf{x})$ where \begin{align} f(\mathbf{x}) = f(\mathbf{a}) + Df(\mathbf{x})(\mathbf{x}-\mathbf{a})^T+\frac{1}{2}(\mathbf{x}-\mathbf{a}) Hf(\mathbf{x})(\mathbf{x}-\mathbf{a})^T \end{align} where $H$ is the Hessian of $f$.

Answer 1

For context to future readers, Doi starts with a model of an $N$ unit polymer formed by a random walk along a uniform grid of lattice length $b$.

It seems to be implicitly assumed in the described derivation that the lattice used is $b\mathbb{Z}^3$, that is the uniform 3D grid. Note there are other types of lattices for which Eq.$\,(\ref{2})$ is not true. Also note the factor of $\frac{1}{3}$ comes from the dimension, so this expression is also not true for the 2D grid they show as an example.

For the 3D uniform grid, the result arises fairly simply. We have $z=6$ corresponding to the vectors $(\pm \mathbf{b}_1,\pm \mathbf{b}_2,\pm \mathbf{b}_3)$. For simplicity, we can say these just align with the usual Cartesian axes $(\pm x,\pm y,\pm z)$, since Eq.$\,(\ref{2})$ should not be dependent on the orientation of the grid. Clearly if $\alpha\neq\beta$, the sum equals zero, as all of the $\mathbf{b}_i$ vectors only point along one Cartesian direction. For $\alpha=\beta$, only $2$ of the $6$ vectors (i.e. $\frac{1}{3}$) will have this component be nonzero and for these vectors the product is just $b^2$. So for any Cartesian, the sum will result in $\frac{1}{3}zb^2$, which when divided by $z$ gives the expression from Eq.$\,(\ref{2})$.