Derive the propagator for a free particle from first principles


 * 1) Expand \(\langle q_F|q_I \rangle\): \(\langle q_F|q_I \rangle = \langle q_F|e^{-i\hat{\mathcal{H}}t/\hbar}|q_I \rangle\)
 * 2) Insert the Hamiltonian for a free particle: \(\langle q_F|q_I \rangle = \langle q_F|e^{-i\hat{p}^2t/(2m \hbar)}|q_I \rangle\)
 * 3) Insert the identity operator to the right of the operator: \(\langle q_F|q_I \rangle = \langle q_F|e^{-i\hat{p}^2t/(2m \hbar)}\mathbb{I}_p|q_I \rangle\)
 * 4) Use the completeness of states for Hilbert space for the momentum operator \( (\mathbb{I}=\int \frac{dp}{2\pi\hbar}|p\rangle\langle p|) \) to substitute out the identity operator: \(\langle q_F|q_I \rangle = \int \frac{dp}{2\pi\hbar} \langle q_F|e^{-i\hat{p}^2t/(2m \hbar)}|p\rangle\langle p|q_I \rangle\)
 * 5) Solve the operator \(\hat{p}\): \(\langle q_F|q_I \rangle = \int \frac{dp}{2\pi\hbar} e^{-ip^2t/(2m \hbar) + iq_Fp/\hbar - ipq_I/\hbar}\)
 * 6) Simplify: \(\langle q_F|q_I \rangle = \int \frac{dp}{2\pi\hbar} \exp{\left[\frac{-it}{2m\hbar} \left(p^2 - \frac{2m}{t}\left(q_F - q_I\right)\right)\right]}\)
 * 7) Factorise: \(\langle q_F|q_I \rangle = \int \frac{dp}{2\pi\hbar} \exp{\left[\frac{-it}{2m\hbar} \left(p - \frac{m}{t}\left(q_F - q_I\right)\right)^2 + \frac{it}{2m\hbar}\frac{m^2}{t^2}\left(q_F - q_I\right)^2\right]}\)
 * 8) Take term without \(p\) outside the integral: \(\langle q_F|q_I \rangle = \exp\left[\frac{it}{2m\hbar}\frac{m^2}{t^2}\left(q_F - q_I\right)^2\right]\int \frac{dp}{2\pi\hbar} \exp{\left[\frac{-it}{2m\hbar} \left(p - \frac{m}{t}\left(q_F - q_I\right)\right)^2\right]}\)
 * 9) Solve the integral: \(\langle q_F|q_I \rangle = \exp\left[\frac{it}{2m\hbar}\frac{m^2}{t^2}\left(q_F - q_I\right)^2\right] \frac{1}{2\pi\hbar} \sqrt{\frac{2m\hbar\pi}{it}}\)
 * 10) Simplify: \(\langle q_F|q_I \rangle = \sqrt{\frac{m}{2it\pi\hbar}} \exp\left[\frac{im}{2t\hbar}\left(q_F - q_I\right)^2\right] \)