# Why is a bath necessary for nonconservative forces in quantum mechanics

In classical mechanics, it's straightforward to include nonconservative forces. For a particle in 1D, Liouville's equation becomes,

$$\partial_t \rho + \dot{q}\partial_q \rho + \dot{p}\partial_p \rho + \rho \partial_p Q=0,$$

where $Q$ is the nonconservative force.

In quantum mechanics, the standard approach seems to be to invent a "bath" that that supplies the forces to the system you're interested in. Is there an analogous approach to get nonconservative forces as with classical systems?

In quantum mechanics, the standard approach seems to be to invent a "bath" that that supplies the forces to the system you're interested in.

Yes, this is indeed the usual approach.

Is there an analogous approach to get nonconservative forces as with classical systems?

Yes, absolutely.

Fundamentally, every closed system is non-dissipative. We only see apparent dissipation when we observe a sub-part of the full system. For example, when an object rolling on the ground comes to rest, some of the object's kinetic energy has been transferred into heat and other forms of energy that we're not paying attention to.

The classical and quantum equations of motion are both fundamentally conservative. However, in some cases where "extra" degrees of freedom cause dissipation in the degrees of freedom we care about, it is possible to find "reduced equations of motion" that describe the motion of the degrees of freedom we care about. That's what's implicitly going on in the equation

$$\partial_t \rho + \dot{q}\partial_q \rho + \dot{p}\partial_p \rho + \rho \partial_p Q=0$$

in the original post. The $\rho \partial_p Q$ term represents the overall effect of forces from the environment/bath (or whatever you want to call it) on the main degree of freedom represented by $q$ and $p$.

## Caldeira-Leggett

A famous example of explicitly reducing the interaction between the main degrees of freedom and the environmental degrees of freedom is the Caldeira-Leggett model. In this model, a main system $S$ with $q$ and $p$ is coupled to a set $E$ of harmonic oscillators. The total system $S$ coupled to all the oscillators in $E$ conserves energy. However, It turns out that in the limit that the number of oscillators in $E$ goes to infinity, the motion of $S$ alone can be represented by a dissipative equation of motion similar to the one quoted above. In particular, $S$ experiences a damping force $F \propto -\dot{q}$.

## Classical and quantum

The Caldeira-Leggett model is normally written entirely in terms of Hamiltonian mechanics. It is, therefore, entirely classical. It can, of course, be applied in quantum systems wherein the Heisenberg equations of motion match the classical equations of motion, e.g. when $S$ is a damped oscillator. Things are a bit different when $S$ is a two level system (i.e. a spin) because the effect of damping is a little different.

## Some intuition about why an infinite set of non-dissipative elements looks dissipative

Suppose we send an electrical pulse down a transmission line of some fixed length. The pulse travels down the line, bounces off the end, and then comes back to us. Energy is conserved. Of course, for some amount of time, the energy is in the transmission line, and we asked "How much energy do we have on our end of the line?", then the answer would be "zero" during the time that the pulse is travelling. Of course, the energy eventually does return. Note that a finite length transmission line can be modeled as a discrete, infinite set of harmonic oscillators.

Now suppose that the line is infinitely long. The infinitely long line can be modeled as a continuously infinite set of harmonic oscillators. We send out the pulse, and it never comes back. So, from our restricted point of view, the energy is gone, and the continuously infinite set of oscillators acts like a source of damping that never gives the energy back.

• Why is it necessary to explicitly account for those leaky degrees of freedom? In classical mechanics you don't have to think as hard about where the energy lost to friction goes (although I suppose you can as per the Caldeira-Leggett model you mentioned)
– Ravi
Mar 30, 2018 at 0:17
• @Ravi There are two ways to answer: 1) You don't think about the leaky degrees of freedom because you already happen to have an equation of motion that works. For example, we all know that friction force is $F = -\mu F_\text{normal}$ because it works empirically. 2) You've already done the microscopic analysis (e.g. the Caldeira-Leggett model) so you know the answer and can just write down the reduced equations of motion. It's the same situation for quantum mechanics, by the way. There are lots of cases where we just write the reduced (dissipative) equations of motion. Mar 30, 2018 at 0:18
• 1) is a good point. For some reason I thought that because drag was so simple in CM, it should translate perfectly to QM. Thanks for thorough answer! I'll spend more time on the Caldeira-Leggett model
– Ravi
Mar 30, 2018 at 0:39
• @Ravi I actually wrote up a really detailed calculation that goes through the Caldeira Leggett model, with pictures etc. I can send it to you if you want. Ping me in the chat room to talk about it. Mar 30, 2018 at 0:40
• It looks like I don't have the reputation to use chat here :(. Would it be all right if I email you?
– Ravi
Mar 30, 2018 at 1:03

I'm not even sure this question even makes sense. Non conservative forces in Newtonian mechanics are macroscopic phenomena that are due to interactions at the quantum level. For example frictional forces are fundamentally interactions between atoms on the the frictional surface and the atoms of the thing that is sliding across it. Once you get to the quantum level you just have particles and the fundamental forces between them.

• A couple things: Whatever the origin of non conservative forces, they are a useful construct in classical mechanics. Since there is active research in dissipative quantum mechanics, it seems useful there too
– Ravi
Mar 29, 2018 at 21:51