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
\[ C = \frac{1}{2\pi} \sqrt{\frac{2 \pi m}{i \Delta t}} = \sqrt{\frac{m}{2\pi i \Delta t}} \]