# A fundamental confusion with the expression for path integral

Consider the path-integral $$\big\langle q_f\big|\exp\big(-\frac{iHT}{\hbar}\big)\big|q_i\big\rangle=\int Dq(t) \exp\Big[\frac{i}{\hbar}\int_0^Tdt(\frac{1}{2}m\dot{q}^2-V(q)\Big]$$ where $$\int Dq(t)=\lim_{N\to\infty}\Big(\frac{-2\pi imN}{T}\Big)^{N/2}\prod_{k=1}^{N-1}\int_{-\infty}^{+\infty}dq_k.$$

From the LHS it seems that the result should depend on $$q_i\equiv q_0$$ and $$q_f\equiv q_N$$. But why is that not clear form the expression of RHS? Where is the information on the RHS that I have held $$q_0=0$$ and $$q_N=50$$ but not $$q_0=50$$ and $$q_f=100$$?

• That information is there, it's just usually suppressed for brevity. You only integrate over paths with the desired endpoints. – knzhou Jan 5 '20 at 17:37
• @knzhou I know. But where is that on the RHS? I cannot understand. How would the expression change if $q_0=50$ and $q_N=100$ instead of $q_0=0$ and $q_N=50$? What/which factor would change on the RHS? – mithusengupta123 Jan 5 '20 at 17:39
• When you discretize the time derivative $\dot{q}$, it will couple $q_0 = q_i$ and $q_N = q_f$ to $q_1$ and $q_{N-1}$ respectively. Then dependence on those endpoints will remain after performing the integrals. – Seth Whitsitt Jan 5 '20 at 18:16

Although the existing answer is complete and there is nothing more to be said, I believe a new answer can at best clarify notation. So I will re-write the expression that the OP has to make the appearance of the boundary conditions manifest: Here I will call $$q_0=q_i$$, $$q_N=q_f$$ $$\int Dq(t) \exp\Big[\frac{i}{\hbar}\int_0^Tdt(\frac{1}{2}m\dot{q}^2-V(q)\Big]$$$$=\lim_{N\to\infty}\Big(\frac{-2\pi imN}{T}\Big)^{N/2}\prod_{k=1}^{N-1}\int_{-\infty}^{+\infty}dq_k.\exp\Big[\frac{i}{\hbar}\sum_{k=1}^N \epsilon\left(\frac{1}{2}m\left(\frac{q_k-q_{k-1}}{\epsilon}\right)^2-V(q_k)\right)\Big]$$ Here $$\epsilon=\frac{T}{N}$$. In particular, the boundary conditions always influences the time derivative term at the ends of the path integral and the potential term at one boundary(which one is a matter of convention, and is deemed unimportant in the large $$N$$ limit): $$=\lim_{N\to\infty}\Big(\frac{-2\pi imN}{T}\Big)^{N/2}\prod_{k=1}^{N-1}\int_{-\infty}^{+\infty}dq_k.\exp\Big[\frac{i}{\hbar}\sum_{k=2}^{N-1} \epsilon\left(\frac{1}{2}m\left(\frac{q_k-q_{k-1}}{\epsilon}\right)^2-V(q_k)\right)+\frac{i}{\hbar}\epsilon\left(\frac{1}{2}m\left(\frac{q_1-q_{0}}{\epsilon}\right)^2-V(q_1)+\frac{1}{2}m\left(\frac{q_N-q_{N-1}}{\epsilon}\right)^2-V(q_N)\right)\Big]$$ Where in the last line, I have separated the $$k=1,N$$ terms to explicitly see the boundary conditions in the path integral.
• I think, you meant "...$k=0,N$ terms to explicitly see..." . – mithusengupta123 Jan 18 '20 at 12:08
The boundary conditions are implicitly implied in the notation. One could e.g. make the boundary conditions more explicit by writing $$\big\langle q_f\big|\exp\big(-\frac{iHT}{\hbar}\big)\big|q_i\big\rangle~=~\int_{q(0)=q_i}^{q(T)=q_f} Dq(t) \exp\Big[\frac{i}{\hbar}\int_0^Tdt(\frac{1}{2}m\dot{q}^2-V(q)\Big].$$ Unlike the rigorous LHS written in the operator formalism, one should understand that the path integral on the RHS is still only a formal expression. In particular, there are operator ordering issues in the discretization procedure, that among other things affect the boundary conditions, cf. e.g. this Phys.SE post.
• But this is just a notation. How does the boundary cond. goes into the evaluation of the integral? See that in my expression $q_1,q_2,...q_{N-1}$ are all integrated from $-\infty$ to $+\infty$. – mithusengupta123 Jan 5 '20 at 17:42