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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 $q_f=100$?

Qmechanic
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Solidification
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    That information is there, it's just usually suppressed for brevity. You only integrate over paths with the desired endpoints. – knzhou Jan 05 '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? – Solidification Jan 05 '20 at 17:39
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    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 05 '20 at 18:16

2 Answers2

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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.

Qmechanic
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  • 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 $+\infty$. – Solidification Jan 05 '20 at 17:42
  • @mithusengupta123: but then you're just jumping between a bunch islands. For instance, gauge orbits can't suddenly jump while you're jumping between the islands so you have to impose boundary conditions. – Cinaed Simson Jan 16 '20 at 23:50
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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.

The wisdom in Qmechanic's answer indicates, correctly, that the continuum representation of the path integral, is entirely formal and has little tangible meaning. This limiting expression is the closest one can get to explicitly writing down a path integral. And in this expression, the boundary terms are apparent.

Anonjohn
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