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I am reading Srednicki's book on QFT and there's a thing I don't quite see in chapter 6 (Path integrals in QM)

equation (6.7) is

$\langle{}q^{''},t^{''}|q^{'},t^{'}\rangle=\int\prod_{k=1}^Ndq_k\prod_{j=0}^N\frac{dp_j}{2\pi}e^{ip_j(q_{j+1}-q_j)}e^{-iH(p_j,\bar{q_j})\delta t}$

where he says $\bar{q_j}=\frac{1}{2}(q_j+q_{j+1})$,$q_0=q'$, $q_{N+1}=q^{''}$ and takes the limit $\delta t\to{}0$ and he gets equation (6.8)

$\langle{}q^{''},t^{''}|q^{'},t^{'}\rangle=\int\mathcal{D}q\mathcal{D}p\cdot e^{i\int_{t'}^{t''}dt(p(t)\dot{q(t)}-H(p(t),q(t)))} $

My question is, where does the integral on the exponential come from?

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2 Answers 2

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A product of exponentials is equivalent to an exponential of a sum, e.g.

$$e^{A}e^{B}=e^{A+B}.$$

The limit of $\delta t$ going to zero turns the discrete sum into a continuous integral.

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The products of the exponentials form a sum in the exponent. And in the limit this sum becomes an integral

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