# Derivation of probability density of isolated polymers

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} \end{align} where $$b_{i\alpha}$$ is the component of $$\mathbf{b}_i$$ in the $$\alpha$$ direction, for the lattice the polymer lies in.

He makes the statement, $$\frac{1}{z}\sum _{i=1}^z b_{i\alpha}b_{i\beta} = \frac{\delta _{\alpha\beta}b^2}{3}$$ where $$z$$ is the coordination number. I don't understand where this comes from.

What is the proof of this statement?

In this context Doi considers polymer as a random walk on a lattice, where $$\mathbf{b}=(b_{ix}, b_{iy}, b_{iz})$$ are the lattice vectors pointing from site $$i$$ of the lattice towards its nearest neighbors. The correctness of the equation above can be verified explictly for any specific lattice. E.g., let us consider a cubic lattice with $$z=6$$ and $$\mathbf{b}_1=(+b,0,0)\\ \mathbf{b}_2=(-b,0,0)\\ \mathbf{b}_3=(0,+b,0)\\ \mathbf{b}_4=(0,-b,0)\\ \mathbf{b}_5=(0,0,+b)\\ \mathbf{b}_6=(0,0, -b)$$ The veracity of the Doi's equations (1.6) and (1.7) can be immediately verified: $$\frac{1}{z}\sum_{i=1}^zb_{i\alpha}=0$$ is the sum of the columns of in the matrix above formed by the six vectors. $$\frac{1}{z}\sum_{i=1}^zb_{i\alpha}b_{i\beta}=\frac{\delta_{\alpha,\beta}b^2}{3}$$ is the sum of the products of the columns.