91影视

We need to find all the numerical coefficients give precisely the correct symmetry factors for the disconnected diagrams.

Lets first write out the expression $8.23$in mathematical form (without diagrams):

$Z_c=\exp \left[\frac{1}{2}\frac{N\left(N鈭�1\right)}{V^2}鈭�d^3{r}_1{d}^3{r}_2{f}_{12}+\frac{1}{2}\frac{N\left(N鈭�1\right)\left(N鈭�2\right)}{V^3}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}\right]$

Now we can do the approximation $N鈮�N-1鈮�N-2鈥�.$. Then we need to expand 's $Z_c$exp function to the third power:

$$\begin{gathered} Z_c鈮�\exp \left[\frac{1}{2}\frac{N^2}{V^2}鈭�d^3{r}_1{d}^3{r}_2{f}_{12}+\frac{1}{2}\frac{N^3}{V^3}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}\right] \\ \exp \left(位\right)鈭�1+位+\frac{1}{2}{位}^2+\frac{1}{6}{位}^3+鈥� \\ {Z}_c=1+\frac{1}{2}\frac{N^2}{V^2}鈭�d^3{r}_1{d}^3{r}_2{f}_{12}+\frac{1}{2}\frac{N^3}{V^3}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23} \\ +\frac{1}{8}\frac{N^4}{V^4}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34}+\frac{1}{8}\frac{N^6}{V^6}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6鈰�f_{12}{f}_{23}{f}_{45}{f}_{56} \\ +\frac{1}{4}\frac{N^5}{V^5}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{14}{f}_{45}+鈥� \end{gathered}$$

To examine those integrals we have to add one more degree of integrals comming from the ${位}^3$of the. This would give:

$$\begin{gathered} Z_c=1+\frac{1}{2}\frac{N^2}{V^2}鈭�d^3{r}_1{d}^3{r}_2{f}_{12}+\frac{1}{2}\frac{N^3}{V^3}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23} \\ +\frac{1}{8}\frac{N^4}{V^4}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34}+\frac{1}{8}\frac{N^6}{V^6}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6鈰�f_{12}{f}_{23}{f}_{45}{f}_{56} \\ +\frac{1}{4}\frac{N^5}{V^5}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{34}{f}_{45}+ \\ \frac{1}{48}\frac{N^6}{V^6}鈭�d^3{r}_1{d}^3{r}_2{d}^3{V}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{f}_{12}{f}_{34}{f}_{56} \\ +\frac{1}{16}\frac{N^7}{V^7}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{d}^3{r}_7{f}_{12}{f}_{34}{f}_{56}{f}_{67} \\ +\frac{1}{48}\frac{N^9}{V^9}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{d}^3{r}_7{d}^3{r}_8{d}^3{r}_9{f}_{12}{f}_{23}{f}_{45}{f}_{56}{f}_{78}{f}_{89} \end{gathered}$$

We can see that in front of every integral asymmetry factor arises, we can check that on the example of this last integral:

$+\frac{1}{48}\frac{N^9}{V^9}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{d}^3{r}_7{d}^3{r}_8{d}^3{r}_9{f}_{12}{f}_{23}{f}_{45}{f}_{56}{f}_{78}{f}_{89}$

If we draw it's diagram and look for dots that can change the place we get exactly $48$permutations:

Another example can be integral:

$\frac{1}{16}\frac{N^8}{V^8}鈭�d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{d}^3{r}_7{d}^3{r}_8{f}_{12}{f}_{34}{f}_{45}{f}_{67}{f}_{78}$

We can see that counting permutations, which is counting the dots with the same role in the integral and possible place where they can be interchanged, is exactly what we get from Taylor's expansion of $\$Z,c\$.$By doing the expansion, the symmetry factors come out naturally.