Derivation of probability density of isolated polymers

Question

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?

Answer 1

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.

The equations can be generalized to an arbitrary monatomic lattice, using the point symmetries thereof. However, for the purposes of understanding the argument in the book, the square lattice is sufficient.