In undergraduate courses the introduction to Hamiltonian mechanics usually starts from a Newtonian view point. One has equations of motions of the form (not sure if it is ok to use covariant notation for forces, but I will do it anyway):

$F_\mu = -\frac{\partial V}{\partial x^\mu}$

Then assuming certain properties of the potential (e.g. that it is independent of the velocity coordinates) one can show that it can be represented by Hamiltonian mechanics.

Now my question is if the systems that do not fulfill these conditions (e.g. dissipative systems, two energy conserving examples are given @JohnSidles answer to this question). Can such systems be quantized in any meaningful sense?

What I am really trying to achieve with this question is to understand better what quantization actually is. We usually have a Hamiltonian system and replace the Poisson bracket by commutation or anti-commutation relations and promote the functions to operators. But is this necessary for quantization or is there an underlying principle that can also be applied to other things?

  • 1
    $\begingroup$ You can certainly approach it from the Lagrangian point of view; but a quick search found another method for you: Quantization of non-Hamiltonian and Dissipative Systems. I only read the abstract. $\endgroup$ Apr 14 '16 at 2:22
  • $\begingroup$ I can answer this but I want to make sure you're being fair about what you're asking. You say you're generally interested in how to quantize a system whose classical version does not admit a Hamiltonian description. However, the title of the post specifically mentions only dissipative systems. I can give you a great answer about dissipative systems, but I want to do that only if that really qualifies as an answer to the post. If you really only want to know about dissipative systems, I think you should edit the text to reflect that, i.e. make it more focused. $\endgroup$
    – DanielSank
    Apr 14 '16 at 7:27
  • $\begingroup$ @Peter Diehr: Note that Lagrangian and Hamiltonian formulations traditionally go hand in hands, cf. my Phys.SE answer here. (I should specify that this is in the context of theories with a variational principle.) $\endgroup$
    – Qmechanic
    Apr 15 '16 at 14:57
  • $\begingroup$ @Qmechanic: I'm very familiar with the Lagrangian for classical mechanics, for which it is fairly easy to take into account dissipative forces. But this doesn't transfer to the classical Hamiltonian, which is usually required to be conservative. I learned Lagrangian theory, originally, from the book by Lanczos, The Variational Principles of Mechanics, and only later the text book version of Goldstein. $\endgroup$ Apr 15 '16 at 16:44
  • $\begingroup$ Yes, there is also a non-variational generalization of Lagrangian formulation, where the forces do not necessarily have potentials, e.g. the case of Rayleigh dissipation function. However, OP's question seems to be essentially about non-variational theories (as opposed to variational theories) rather than an issue of being Lagrangian vs. Hamiltonian. (Here I'm assuming OP asks for fully quantized theories, not just effective/non-equilibrium methods/descriptions of dissipative systems coupled to an environments/bath, which is a huge topic in itself.) $\endgroup$
    – Qmechanic
    Apr 15 '16 at 17:04

Dissipative quantum mechanics does not preserve the pureness of a state, hence must be formulated in terms of density operators.

Conservative dynamics is classically described by a conservative dynamics obtained through an action principle. The quantum version is given on the level of density operators by the von Neumann equation expressing the derivative of the density operator as a commutator with the Hamiltonian.

Dissipative autonomous systems are classically described by modifying the conservative dynamics through adding dissipative terms. Similarly, the quantum version adds to the von Neumann equation dissipative terms of double commutator type. Most prominent is the Lindblad equation, extensively used in dissipative quantum optics.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.