Understanding the definition of a path integral

Question

In the Book "Quantum Mechanics and Path Integrals" by Feynman & Hibbs the path integral is approximated (page 32 and following) by $$ K(b,a)\approx\int...\int\int\phi[x(t)]dx_1dx_2...dx_{N-1}\tag{2.20} $$ with $b=(x_b,t_b)$ and $a=(x_a,t_a)$ being the start and endpoints of the path and $$\phi[x(t)]=const\cdot e^{(i/\hbar)S[x(t)]}=const\cdot e^{(i/\hbar)\int_{t_a}^{t_b} L[x(t),v(t),t]dt}.\tag{2.15}$$ Now I dont quite get this approximation.

1. First of all I assume that the $dx_1dx_2...dx_{N-1}$ integrals have to be executed first and only after that the $dt$ integral in $\phi[x(t)]$ (or rather in $S[x(t)]$) should be executed. Is that right?
   

2. And the second thing is that I dont get the meaning behind the $dx_1dx_2...dx_{N-1}$ integrals itself (each is integrated from $-\infty$ to $\infty$ according to wikipedia). So in the book the path was divided into straight lines between $x_k$ and $x_{k+1}$ with equal length and $x_0=x_a$ and $x_N=x_b$. That's why I would have thought the integration would not go from $-\infty$ to $\infty$ but rather from $x_k$ to $x_{k+1}$. So the integral would then look sth. like this $$K(b,a)\approx\int_{x_{N-1}}^{x_N}...\int_{x_1}^{x_2}\int_{x_0}^{x_1}\phi[x(t)]dx_0dx_1...dx_{N-1} $$ Could someone explain to me in an easy way why that is not the case?

Answer 1

1. The action functional $S[t\mapsto x(t)]$ with its $dt$-integration is the continuum limit of a discretion $S[x_0,\ldots, x_N]$ that is implicitly implied on the RHS of eq. (2.20), so in eq. (2.20) there is strictly speaking not a $dt$-integration to perform, cf. OP's question. It only emerges in the continuum limit.

2. OP's proposal does not make sense as written as integration limits can only refer to later performed integrations, not the other way around. Anyway, Feynman's point is that we should sum over all histories, i.e. the integration variables should cover the whole target space, i.e. position space $\mathbb{R}$ from $-\infty$ to $\infty$.

References:

1. R.P. Feynman & A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965; chapters 2 + 3 + 4.

2. W. Dittrich & M. Reuter, Classical and Quantum Dynamics, 6th ed, 2020; chapters 19 + 20.

Answer 2

The whole idea of a path integral is based on the concept of a functional, i.e., a function of functions. Therefore the path integral is really in integration over functions in a functional space rather than the values of a variable. In Feynman's approximation, he breaks the functions over which is being integrated up into all the different function values at all the different points in these functions and then integrates over all those values, to ensure that the path integral actually integrates over all possible functions. That is why the individual $x$-integration runs from $-\infty$ to $\infty$. Therefore one would not be able to evaluate the $t$-integral first. That would require knowledge of the function $x(t)$, but the function can be anything. It is the "variable" over which the integration is performed. Hope this helps.

Answer 3

For question 2, the bounds should be from $-\infty$ to $\infty$. That is because at each time, we integrate over every possible position through the use of the identity

$1 = \int _{-\infty}^{\infty} |x \rangle \langle x | dx$

In terms of interpretation, at each point in time $t$, the position on the path can be any real number. So consecutive positions are not bounded by their neighbors.

For question 1, the way I think of it is to think of the $dx_k$ integrals as riemann sums. For each term in the result, you have a piecewise path $\bar{x}(t)$ with which the integral $dt$ can be evaluated. The limit as those riemann sums tend to integrals is the result. In that sense I think of doing the $dt$ integral first.