# Path integral derivation of Dyson's series

I'm studying Ryder's quantum field theory book, in chapter 5 section 2 Ryder tries to derive Dyson's series using Feynman's path integral.

On page 160 we have the following definition of Feynman's path integral:

$$\tag{5.14} \langle q_ft_f|q_it_i\rangle=\lim_{n\rightarrow\infty}\bigg(\frac{m}{i\hbar\tau}\bigg)^{(n+1)/2}\int\prod_{1}^n\,dq_j\exp\bigg\{\frac{i\tau}{\hbar}\sum_0^n\bigg[\frac{m}{2}\bigg(\frac{q_{j+1}-q_j}{\tau}\bigg)^2-V\bigg(\frac{q_j+q_{j+1}}{2}\bigg)\bigg]\bigg\}$$

the book then does the Taylor expansion

$$\tag{5.16}\exp\bigg[\frac{-i}{\hbar}\int_{t_i}^{t_f}V(x,t)\,dt\bigg]=1-\frac{i}{\hbar}\int_{t_i}^{t_f}V(x,t)\,dt-\frac{1}{2!\hbar^2}\bigg[\int_{t_i}^{t_f}V(x,t)\,dt\bigg]^2+...$$

The first term gives the free propogator $$K_0(x_ft_f,x_it_i)$$, the second term gives first order perturbation $$\tag{5.20} K_1(x_ft_f,x_it_i)=-\frac{i}{\hbar}\int_{\infty}^{-\infty}\,dt\int_{-\infty}^\infty \,dx \,K_0(x_ft_f,xt)V(x,t)K_0(xt,x_it_i)$$

I get these two terms. But how did we get the 2nd order perturbation?

$$\tag{5.22} K_2(x_ft_f,x_it_i)=\bigg(\frac{-i}{\hbar}\bigg)^2\int^\infty_{-\infty}\,dt_1\int_{-\infty}^\infty\,dt_1\int_{-\infty}^\infty dx_1\int_{-\infty}^\infty dx_1\, K_0(x_ft_f,x_2t_2)V(x_2,t_2)\times K_0(x_2t_2,x_1t_1)V(x_1,t_1)K_0(x_1t_1;x_it_i).$$

I mean, if we use equation 5.14, the two $$V$$'s should occur at equal time, why shouldn't it something like

$$K_2(x_ft_f,x_it_i)=?-\frac{1}{2!\hbar}\int_{\infty}^{-\infty}\,dt\int_{-\infty}^\infty \,dx \,K_0(x_ft_f,xt)V(x,t)^2K_0(xt,x_it_i)$$

The two $$V$$'s do not have to occur at equal time. Note that in Eq. 5.14, the second order perturbation in $$V$$ is \begin{align} \lim_{n\rightarrow\infty}{\bigg(\frac{m}{2\pi i\hbar\tau}\bigg)^{(n+1)/2}\sum_{j_2>j_1}} & {\int{\prod_{j_2}^n\,dq_j\exp\bigg\{\frac{i\tau}{\hbar}\sum_{j=j_2}^{n}\bigg[\frac{m}{2}\bigg(\frac{q_{j+1}-q_j}{\tau}\bigg)^2\bigg]\bigg\}}\bigg({-i\tau \over \hbar}V\bigg(\frac{q_{j_2}+q_{j_2+1}}{2}\bigg)\bigg)} \\ & {\int{\prod_{j_1}^{j_2-1}\,dq_j\exp\bigg\{\frac{i\tau}{\hbar}\sum_{j=j_1}^{j_2-1}\bigg[\frac{m}{2}\bigg(\frac{q_{j+1}-q_j}{\tau}\bigg)^2\bigg]\bigg\}}\bigg({-i\tau \over \hbar}V\bigg(\frac{q_{j_1}+q_{j_1+1}}{2}\bigg)\bigg)} \\ & {\int{\prod_{1}^{j_1-1}\,dq_j\exp\bigg\{\frac{i\tau}{\hbar}\sum_{j=1}^{j_1-1}\bigg[\frac{m}{2}\bigg(\frac{q_{j+1}-q_j}{\tau}\bigg)^2\bigg]\bigg\}}} \end{align} This is just $$\bigg(\frac{-i}{\hbar}\bigg)^2\int^{t_f}_{t_i}\,dt_2\int_{t_i}^{t_2}\,dt_1\int_{-\infty}^\infty dx_2\int_{-\infty}^\infty dx_1\, K_0(x_ft_f,x_2t_2)V(x_2,t_2)\times K_0(x_2t_2,x_1t_1)V(x_1,t_1)K_0(x_1t_1;x_it_i)$$
• How did you get $j_2$ and $j_1$? The original quation 5.14 says $V^2$? Commented Jul 24, 2022 at 7:10
• I mean, in 5.14, when you do taylor expansion on something like $e^V$, you get $1+V+V^2/2+...$ Commented Jul 24, 2022 at 7:19
• I use Eq. 5.14 and expand the term of $\text{exp}(-{i\tau \over \hbar}V)$. The second order of the expansion gives $j_1$ and $j_2$ naturally. Consider some product $\prod_{j=1}^{n}{e^{a_j+b_j}}=\prod_{j=1}^{n}{e^{a_j}e^{b_j}}$. And the second order term in $b_j$'s is ${1 \over 2!}\sum_{j_2,j_1}b_{j_2}b_{j_1}\prod_{j=1}^{n}{e^{a_j}}$. If $n$ is large, we can just take $\sum_{j_2>j_1}b_{j_2}b_{j_1}\prod_{j=1}^{n}{e^{a_j}}$. Commented Jul 24, 2022 at 7:20
• The idea here is that perturbation in $V$ can come from different time frames, and the contribution of the second order term from the same time frame $V({q_j+q_{j+1} \over 2})^2$ is, in fact, small when $n$ goes to $+\infty$. Commented Jul 24, 2022 at 7:23
• This is what I mean "The two $V$'s do not have to occur at equal time". Hope this answer your question. Commented Jul 24, 2022 at 7:25