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The usual way to compute such a path integral is by writing the fields (in your case: the paths) as "classical configuration" (the straight line) plus "quantum fluctuations". So if you write your paths as $\gamma(\tau) = a + \tau b + \hat\gamma(\tau)$ (with $\tau\in[0,1]$ a parameter describing the path), then $\hat\gamma$ will be the "quantum fluctuation" ...

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It is the space of field configurations, if you conceive of a field as a (possibly vector-valued) $p$-form on the boundary, which, as a section of an exterior algebra naturally carries a vector space structure. Think about it - the classical fields are always $\mathbb{R}^n$ or $\mathbb{C}^n$ valued (with subtleties because they may not only transform under ...

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I can give an outline of the arguments that show how to arrive at the path integral representation of the expression $\text{tr}\, e^{-i \hat{H}t}$. For simplicity I set $\hbar=1$ and only consider the 1-particle case. The multi-particle case is technically more demanding but conceptionally the same. So the Hamiltonian I consider is simply $$\hat{H}=h ... 2 Second quantisation is named only to differentiate it from first quantisation, but essentially the process is the same in each case: take a physical quantity and upgrade it to an operator. In the case of first quantisation, this refers to taking position and momentum and upgrading them to operators:$$ [\hat{x},\hat{p}]=i $$The main reason you do this, is ... 3 There are several points: The first is that for usual self-adjoint Hamiltonians of the form H=-\Delta +V(x), with a common densely defined domain (and I am being very pedantic here mathematically, you may just ignore that remark) the limit process is well defined and it gives a meaning to the formal expression \int_{q_a}^{q_b} ... -1 ...He derives the path integral and shows it to be:$$\int_{q_a}^{q_b}\mathcal{D}p\mathcal{D}q\exp\{\frac{i}{\hbar}\int_{t_a}^{t_b} \mathcal{L}(p, q)\}$$This is clear to me. He then likens it to a discrete sum$$\sum_\limits{\text{paths}}\exp\left(\frac{iS}{\hbar}\right) where $S$ is the action functional of a particular path. Now, ...

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No, the easy way to do this is to not use that formula -- which incidentally is part of a completely different formulation of quantum mechanics. The $q$s and $\dot{q}$s up inside the path integral are classical c-number valued paths, whose interference will give the quantum content in correlation functions. The problem you're being asked to solve seems to be ...

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