I am playing around with a Lagrangian of type $$L = \frac{1}{2}m\dot{q}^2 - \frac{1}{2}mw^2q^2 - \alpha q \dot{q}.$$ If I solve the classical equations of motion using the Euler-Lagrange, I end up with the same equations of motion as the Harmonic Oscillator because each of the new terms cancel each other out. If I formulate a Hamiltonian however, I end up with the form $$H = \frac{1}{2m}p^2 + \frac{\alpha}{m}qp + \frac{1}{2}\left(\frac{\alpha^2}{m} + mw^2 \right)q^2$$ Whose equations of motion are different from the Lagrangian method. I am quite convinced that this is a non-conservative system, since there is explicit time dependence in the Hamiltonian on the generalised momentum, however, I am unable to prove this rigorously.

Beyond this, I am confused as to whether the Hamiltonian is applicable to be fit into the Schrodinger Equation due to its non-conservative nature since it doesen't resemble the notion of total energy. Do I have to use the evolution of the density operator to be able to model this system quantum mechanically? Can I interpret the term as both damping and an added potential? I am just confused on how to proceed with this.

Note: I tried interpreting it as a damped qho and using the methods from https://doi.org/10.1103/PhysRevE.92.062927, although I am not sure if that's a simplification of the system.

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    $\begingroup$ Hint: $\alpha$ plays the role of a uniform $B$-field, cf. en.wikipedia.org/wiki/Landau_quantization There is no damping. $\endgroup$
    – Qmechanic
    Commented Feb 1, 2022 at 13:46
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    $\begingroup$ The fact that you can write down a Hamiltonian already shows that it is conservative. It is basically the same Hamiltonian you get solving Landau level problem in the Landau gauge. $\endgroup$
    – Meng Cheng
    Commented Feb 1, 2022 at 14:23
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    $\begingroup$ there is explicit time dependence in the Hamiltonian on the generalised momentum Where is the explicit time dependence? Perhaps you means something else? $\endgroup$
    – Roger V.
    Commented Feb 1, 2022 at 14:52
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    $\begingroup$ Yeah turns out to be the Landau Quantization problem after all. Thanks for your guidance. $\endgroup$
    – Carpe
    Commented Feb 3, 2022 at 16:30

1 Answer 1


The correct equation for a dissipative oscillator is actually $$\ddot{q}(t) +\gamma \dot{q}(t)+\omega_0^2 q(t)=\xi(t),$$ where $\xi(t)$ is the noise term, whose spectral density is related to the damping coefficient via the fluctuation-dissipation theorem (See also Einstein relation). Without such a noise term the equation is unphysical, as it breaks the fluctuation-dissipation theorem - in this sense introduction of an additional term in the Lagrangian is just a mathematical trick - the Lagrangian does not correspond necessarily to a real physical system, which puts in cause the very application of the least action principle and the Euler-Lagrange equations.

Even with the noise term the equation stull remains an ad-hoc way of introducting the dissipation into the oscillator. There are equivalent quantum methods, such as dissipative Schrödinger equation or quantum Langevin equations, but they are not popular and plagged by various issues, such as violating normalization etc.

A more principled way is coupling the system (oscillator in this case) to a bath, which itself is very often taken to be a bath of harmonic oscillators with continuous spectrum. One then considers joint Hamiltonian evolution of the system and the bath, whereas the dissipation (and irreversibility) is then obtained in the thermodynamic limit, as the system never returning to its initial state due to the energy lost somewhere in the bath. There exist multiple methods for doing so: Caldeira-Legget approach, Lindbad master equation, Keldysh Green's function method, etc. Neither of them is particularly simple, but this is how it is usually done.


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