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


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

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


1 Answer 1


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 $$\bigg(\frac{-i}{\hbar}\bigg)^2\int^{t_f}_{t_i}\,dt_2\int_{t_i}^{t_2}\,dt_1\int_{-\infty}^\infty dx_2\int_{-\infty}^\infty dx_1\, K_0(x_ft_f,x_2t_2)V(x_2,t_2)\times K_0(x_2t_2,x_1t_1)V(x_1,t_1)K_0(x_1t_1;x_it_i)$$

  • $\begingroup$ How did you get $j_2$ and $j_1$? The original quation 5.14 says $V^2$? $\endgroup$ Commented Jul 24, 2022 at 7:10
  • $\begingroup$ I mean, in 5.14, when you do taylor expansion on something like $e^V$, you get $1+V+V^2/2+...$ $\endgroup$ Commented Jul 24, 2022 at 7:19
  • $\begingroup$ 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 $\sum_{j_2>j_1}b_{j_2}b_{j_1}\prod_{j=1}^{n}{e^{a_j}}$. $\endgroup$
    – Andy Chen
    Commented Jul 24, 2022 at 7:20
  • $\begingroup$ 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 $+\infty$. $\endgroup$
    – Andy Chen
    Commented Jul 24, 2022 at 7:23
  • $\begingroup$ This is what I mean "The two $V$'s do not have to occur at equal time". Hope this answer your question. $\endgroup$
    – Andy Chen
    Commented Jul 24, 2022 at 7:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.