Question

For each of the diagrams shown in equation 8.20, write down the corresponding formula in terms of $f-$functions, and explain why the symmetry factor gives the correct overall coefficient.

Short answer

The formula for the figures are:

$\left(a\right)=\left(\frac{1}{2}\right)\left(N\right)(N-1)\left(\frac{1}{V^2}\right)\int d^3{r}_1{d}^3{r}_2{f}_{12}$

$\left(b\right)=\left(\frac{1}{2}\right)\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{13}$

$\left(c\right)=\left(\frac{1}{8}\right)\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34}$

$\left(d\right)=\left(\frac{1}{6}\right)\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}{f}_{13}$

$\left(e\right)=\left(\frac{1}{6}\right)\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{23}{f}_{24}$

$\left(f\right)=\left(\frac{1}{4}\right)\left(N\right)(N-1)(N-2)(N-3)(N-4)\left(\frac{1}{V^5}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{34}{f}_{45}$

$\left(g\right)=\left(\frac{1}{48}\right)\left(N\right)(N-1)(N-2)(N-3)(N-4)(N-5)\left(\frac{1}{V^6}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{f}_{12}{f}_{34}{f}_{56}$

$\left(h\right)=\left(\frac{1}{6}\right)\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{13}{f}_{24}$

Step-by-step solution

Step 1. The rules which helps in converting pictures into the formula:

(1) For each dot the equation is written as $i\text{as}\left(\frac{1}{V}\right)\int d^3{r}_i$, than for 1 dot multiply it by $N$, for 2 multiply by $N-1$and for 3 dot multiply by $N-2$and so on.

(2) The factor $f_{i{j}_{}}$is written for the line which connects the $i$and $j$dots.

(3) Divide the equation by the symmetric factor of the diagram. Symmetric factor is defined as the permutation of dots that does not alter the figure.

Step 2. The diagrams given in the equation

Figure : The representation of give diagrams in numbers.

Step 3. The representation of give diagrams in numbers. 

(a) Equation for 2 dots:

$\left(\frac{1}{V^2}\right)\int d^3{r}_1{d}^3{r}_2$

Multiplying the equation with $N,N-1$for 2 dots

$\left(N\right)(N-1)\left(\frac{1}{V^2}\right)\int d^3{r}_1{d}^3{r}_2$

The factor for two dots is$f_{12}$

$\left(N\right)(N-1)\left(\frac{1}{V^2}\right)\int d^3{r}_1{d}^3{r}_2{f}_{12}$

The symmetry factor for the two dots is 2 because the two dots is arranged in two ways.

Thus, the formula for two dots in the figure becomes,

$\left(a\right)=\left(\frac{1}{2}\right)\left(N\right)(N-1)\left(\frac{1}{V^2}\right)\int d^3{r}_1{d}^3{r}_2{f}_{12}$

(b) Equation for 3 dots:

$\left(\frac{1}{V^3}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3$

Multiplying with $N,N-1,N-2$for 3 dots

$\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3$

The factor for two dots is $f_{12},f_{13}$

$\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{13}$

The symmetry factor for the three dots is 2 because the first and third dot is arranged in two ways.

Thus, the formula for three dots in the figure becomes,

$\left(b\right)=\left(\frac{1}{2}\right)\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{13}$.

Step 4. Equation for (c) and (d) part.

(c) Equation for 4 dots:

$\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

Multiplying by $N,N-1,N-2,N-3$for 4 dots,

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

The factors for 4 dots are $f_{12}and{f}_{34}$,

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34}$,

The symmetry factor 4 dots is 8

Thus, the formula for 4 dots become:

$\left(c\right)=\left(\frac{1}{8}\right)\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34}$

(d) Equation for 3 dots:

$\left(\frac{1}{V^3}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3$

Multiplying with $N,N-1,N-2$for three dots,

$\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3$

The factor for three dots are $f_{12},f_{23},f_{13}$,

$\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}{f}_{13}$

The symmetry factor for 3 dots in the figure is 6, because of the addition of third line in dots

Thus, the formula for the figure is,

$\left(d\right)=\left(\frac{1}{6}\right)\left(N\right)(N-1)(N-2)\left(\frac{1}{V^3}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}{f}_{13}$

Step 5. Equation for (e) and (f) part

(e) Equation for 4 dots:

$\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

Multiplying with $N,N-1,N-2,N-3$for 4 dots,

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

The factor for 4 dots for the figure is,$f_{12},f_{23},\text{and}{f}_{24}$

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{23}{f}_{24}$

The symmetry factor for the figure is 6, attained by putting dot 2 in center and arrange the other dots. So this is arranged in 6 ways.

Thus, the formula for figure (e) is:

$\left(e\right)=\left(\frac{1}{6}\right)\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{23}{f}_{24}$

(f) Equation for 5 dots:

$\left(\frac{1}{V^5}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5$

Multiplying with $N,N-1,N-2,N-3,N-4$for 5 Dots,

$\left(N\right)(N-1)(N-2)(N-3)(N-4)\left(\frac{1}{V^5}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5$

The factor for 5 dots is $f_{12}{f}_{34}{f}_{45}$.

$\left(N\right)(N-1)(N-2)(N-3)(N-4)\left(\frac{1}{V^5}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{34}{f}_{45}$

The symmetry factor for the figure is 4, by putting dot number 4 in center, the other number is arranged in 4 ways.

Thus, the formula for figure (f) is

$\left(f\right)=\left(\frac{1}{4}\right)\left(N\right)(N-1)(N-2)(N-3)(N-4)\left(\frac{1}{V^5}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{34}{f}_{45}\text{.}$

Step 6. Equation for (g) and (h) part

(g) Equation for 6 dots:

$\left(\frac{1}{V^6}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6$

Multiplying by $N,N-1,N-2,N-3,N-4,N-5$for 6 dots,

$\left(N\right)(N-1)(N-2)(N-3)(N-4)(N-5)\left(\frac{1}{V^6}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6$

The factors for 6 dots are $f_{12}{f}_{34}{f}_{56}$,

$\left(N\right)(N-1)(N-2)(N-3)(N-4)(N-5)\left(\frac{1}{V^6}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{f}_{12}{f}_{34}{f}_{56}$

The symmetry factor for 6 dots is 48.

Thus, the formula for 6 dots become:

$\left(g\right)=\left(\frac{1}{48}\right)\left(N\right)(N-1)(N-2)(N-3)(N-4)(N-5)\left(\frac{1}{V^6}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{f}_{12}{f}_{34}{f}_{56}$

(h) Equation for 4 dots:

$\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

Multiplying by $N,N-1,N-2,N-3$for 4 dots,

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4$

The factors for figure are $f_{12}{f}_{13}{f}_{24}$,

$\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{13}{f}_{24}$

The symmetry factor for the figure is 6, attained by putting one dot fixed and arrange the other dots. So this is arranged in 6 ways.

Thus the formula for the figure:

$\left(h\right)=\left(\frac{1}{6}\right)\left(N\right)(N-1)(N-2)(N-3)\left(\frac{1}{V^4}\right)\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{13}{f}_{24}$