So the kernel for the Lagrangian of the 1D harmonic oscillator:

$$\mathcal{L} = \tfrac12 m \dot{x}^2 - \tfrac12 m \omega^2 x^2$$

is solvable (from the Wikipedia entry on Path Integral Formulation) this is:

$$K(x_f, t_f; x_i, t_i) = \left( \frac{m \omega}{2 \pi i \hbar \sin\omega T } \right)^\frac12 \exp{ \left( \frac{i}{\hbar} \tfrac12 m \omega \frac{ (x_i^2 + x_f^2) \cos \omega T - 2 x_i x_f }{ \sin \omega T } \right) }$$

where $T=t_f-t_i$.

  1. What about for a more general case for N dimensions: $$\mathcal{L} = \tfrac12 m \dot{x}^2 -\tfrac12 m x^T M x$$ where $M$ is a symmetric matrix? Is this known? One could guess by replacing $\omega$ with the matrix $\sqrt{M}$ and writing the trig functions as series. But I'm not certain that would be correct? And this wouldn't work for the first factor.

  2. Also as a bonus, does this Lagrangian have any physical meaning?

  • $\begingroup$ Why don't you solve for the propagator by the method described in Feynman-Hibbs for quadratic Lagrangians. The Lagrangian corresponds to N-coupled harmonic oscillators. $\endgroup$ – Sunyam Oct 16 '17 at 18:18

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