Derive the propagator for a free particle from the Feynman path integral


 * 1) Apply the Lagrangian form of the Feynman path integral to the \(\langle q_F|q_I\rangle\): \(\langle q_F|q_I\rangle = \int{\left( \prod \limits^{N-1}_{n-1} dq_n\right)\left( \prod \limits^{N}_{n-1} \sqrt{\frac{m}{i\delta t2\pi\hbar}}\right)\,\exp{\left[\frac{im}{2\hbar\delta t}\sum \limits^{N}_{n=1}\left(q_{n} - q_{n-1}\right)^2\right]}}\)
 * 2) Solve the first integral term.
 * 3) Write out the sum in the integral explicitly: \(\langle q_F|q_I\rangle = \int{\left( \prod \limits^{N-1}_{n-1} dq_n\right)\left( \prod \limits^{N}_{n-1} \sqrt{\frac{m}{i\delta t2\pi\hbar}}\right)\,\exp{\left[\frac{im}{2\hbar\delta t}\left(\left(q_1 - \frac{q_0 + q_2}{2}\right)^2 + \frac{\left(q_{2} - q_{0}\right)^2}{2} + \left(q_{3} - q_{2}\right)^2 \cdots \right)\right]}}\)
 * 4) Set \(\left(q_1 - \frac{q_0 + q_2}{2}\right)^2 = \bar{q}_1\) and integrate over \(\bar{q}_1\): \(\langle q_F|q_I\rangle = \int{\left( \prod \limits^{N-1}_{n-2} dq_n\right)\left( \prod \limits^{N}_{n-1} \sqrt{\frac{m}{i\delta t2\pi\hbar}}\right)\,\sqrt{\frac{i\delta t2\pi\hbar}{m}}\frac{1}{\sqrt{2}}\exp{\left[\frac{im}{2\hbar\delta t}\left(\frac{\left(q_{2} - q_{0}\right)^2}{2} + \left(q_{3} - q_{2}\right)^2 \cdots \right)\right]}}\)
 * 5) Repeat for each integral term, bringing down a term of \(\sqrt{\frac{i\delta t2\pi\hbar}{m}}\) and \(\frac{n}{n+1}\) each time: \(\langle q_F|q_I\rangle = \left( \prod \limits^{N}_{n-1} \sqrt{\frac{m}{i\delta t2\pi\hbar}}\right)\,\left(\sqrt{\frac{i\delta t2\pi\hbar}{m}}\right)^{N-1}\frac{1}{\sqrt{N}}\exp{\left[\frac{im}{2\hbar\delta t}\left(q_I - q_F \right)^2\right]}\)
 * 6) Simplify: \(\langle q_F|q_I \rangle = \sqrt{\frac{m}{2it\pi\hbar}} \exp\left[\frac{im}{2t\hbar}\left(q_F - q_I\right)^2\right] \)