Path Integrals 1: Feynman's Derivation

Consider the Hilbert space $\mathcal{H} = L^2(\mathbb{R})$ with Lebesgue measure and a Hamiltonian $H = T(k) + V(x)$ (a sum of kinetic and potential energy). Then the quantum hamiltonian $\hat{H}$ acts as

$$\begin{align} (\hat{H}\psi)(y) &= \frac{1}{2\pi} \int e^{ik(y-x)} T(k) \psi(x) dx dk + V(y)\psi(y) \\ &= \frac{1}{2\pi}\int e^{ik(y-x)} H(k,x) \psi(x) dx dk \\ \end{align}$$

Now consider the Schrodinger equation
$$\frac{\partial \psi}{\partial t} = i \hat{H} \psi.$$

We would like to obtain a formula for the solution operator $U_t = e^{-i \hat{H} t}$. Let us consider its Schwartz kernel $\langle{y}|U_t|{x}\rangle$. Let $N$ be a large integer so that $\Delta t = t/N$ is "small". Then we can write

$$U_t = U_{\Delta t}^N,$$

Now consider a single term:

$$\begin{align}  \langle{y}|U_{\Delta t}|x\rangle &\simeq \langle{y}|1-i\Delta{t}\hat{H}|x\rangle \\ &= \delta(y-x) -i \Delta t \langle x|\hat{H}|y\rangle \\ &= \frac{1}{2\pi}\int e^{ik(y-x)}( 1 - i \Delta t H(k,x)) dk \\ &\simeq \frac{1}{2\pi}\int e^{ik(y-x) - i \Delta t H(k,x)} dk\\ \end{align}$$
Now we have (taking $x_0 = x$ and $x_N = y$)
$$\begin{align} \langle y|U_t|x\rangle &= \int dx_1 \cdots dx_{N-1} \\ & \ \times \langle x_N|U_{\Delta t}|x_{N-1}\rangle \cdots \langle x_1|U_{\Delta t}|x_0\rangle \\ &= \frac{1}{(2\pi)^N} \int dx_1 \cdots dx_{N-1} dk_0 \cdots dk_{N-1} \\ & \ \times \ e^{ik_N(x_N-x_{N-1}) - i \Delta t H(k_{N-1},x_{N-1})} \cdots e^{-ik_N(x_1-x_0) - i \Delta t H(k_0,x_0)} \\ &= \frac{1}{(2\pi)^N} \int dx_1 \cdots dx_{N-1} dk_0 \cdots dk_{N-1} \\ & \ \times \ \exp \sum_{j=0}^{N-1} ik_j(x_{j+1} - x_j) - i \Delta t H(k_j,x_j) \end{align}$$
Note this kind of integral seems like it should have a natural interpretation in symplectic or contact geometry, since the expression $dxdk/(2\pi)^N$ is very nealy the Liouville measure. This is the most general form of the path integral.

Now assume that $H(k,x) = k^2/2m + V(x)$. Then the $k$-dependent terms have the form
$$\int e^{ik(y-x) - i\Delta t k^2/2m}.$$
Complete the square
$$\begin{align}  ik(y-x) -i\Delta t k^2/2m &= -\frac{i \Delta t}{2m} (k^2 - \frac{2m}{\Delta t}k(y-x)) \\ &= -\frac{i \Delta t}{2m} (k^2 - \frac{2m}{\Delta t}k(y-x) + \frac{m^2}{\Delta t^2}(y-x)^2 -\frac{m^2}{\Delta t^2}(y-x)^2) \\ &= -\frac{i \Delta t}{2m}(k - \frac{m}{\Delta t}(y-x))^2 + \frac{i m}{2 \Delta t}(y-x)^2 \end{align}$$
Now, using that
$$\int e^{-ak^2} = \sqrt{\frac{\pi}{a}}$$
We have
$$\int e^{-\frac{i \Delta t}{2m}(k-\frac{m}{\Delta t}(y-x))^2} = \sqrt{\frac{2\pi m}{i \Delta t}}$$
Putting it altogether, we get the more familiar version of the path integral,
$$\begin{align} \langle y|U_t|x\rangle &\simeq C^N \int dx_1 \cdots dx_{N-1} \\ & \ \times \ \exp i\sum_{j=0}^{N-1} \frac{m}{2\Delta t}(x_{j+1}-x_j)^2 - V(x_j) \Delta t \end{align}$$
where
$$C = \frac{1}{2\pi} \sqrt{\frac{2 \pi m}{i \Delta t}} = \sqrt{\frac{m}{2\pi i \Delta t}}$$