# Free particle and harmonic oscillator coupled

I'm currently playing with a toy model given by the Lagrangian $$L=\frac{m\dot{x}^2}{2}+\frac{m\dot{y}^2}{2}+\frac{1}{2}m\omega^2x^2+x y,$$ which is basically a free particle (described by $y(t)$) and a harmonic oscillator (described by $x(t)$) coupled in a simple way.

My question is if you guys know any system that could correspond to such a Lagrangian? Basically this is a free particle that is forced to oscillate by a harmonic oscillator that pulls it...

• Is there a factor in front of the $xy$ term? Otherwise a problem with dimensions. – jim Apr 7 '16 at 14:04
• This is very similar to the Lagrange an of an electron in the beam of a free electron laser – Lewis Miller Apr 7 '16 at 14:07
• Dimensionally its incorrect. – Anubhav Goel Apr 7 '16 at 14:17
• This question (v1) seems like a list question. – Qmechanic Apr 7 '16 at 15:21
• @jim I just encountered this Lagrangian in a course on path integration, where it was a toy model to solve analytically. It might be that in a physical system there can be a factor in front of it (depending on what units you use I guess ;-) )! – Nick Apr 8 '16 at 9:45

First of all I guess that what you wrote is the Hamiltonian and not the Lagrangian of the system and $\dot{x}$ stays for $p_x$ and $\dot{y}$ stays for $p_y$.
You can decouple the problem redefining $$(X,Y)^t = R(x,y)^t$$ for a suitable $R\in O(2)$ diagonalizing the symmetric matrix in the potential part of your Hamiltonian. This way you see the final Hamiltonian is $$\left(\frac{p^2_X}{2m} + \lambda_+ X^2\right) + \left(\frac{p^2_Y}{2m} + \lambda_- Y^2\right)$$ where $\lambda_\pm$ are the eigenvalues of the above symmetric matrix.
In the considered case (up to the dimensional problem already stressed) you find that $\lambda_+ \lambda_-<0$ (because the determinant of the symmetric matrix is negative nomatter the sign in front of $xy$).