# Are stationary quantum states attractors?

While explaining the quantum adiabatic theorem recently, I appealed to a thermodynamic analogy: when slowly contracting the walls containing a classical gas, the relaxation timescale can be taken to be fast, so the system remains always in thermal equilibrium. By analogy, the quantum system remains always in a stationary state.

The hidden assumption here is that stationary states (eigenstates of the Hamiltonian) are in some sense attractors of the system: that a system in some energy-superposition will, over time, relax to a single energy eigenstate.

Is this actually true? It seems like maybe it isn't, since for example coherent states of the harmonic oscillator are stable (?). Maybe it's true in neighbourhoods around the stationary states only?

• I think your question is whether there is some mapping from "thermodynamic system goes to equilibrium" to "quantum system goes to stationary energy eigenstate." Is that correct? If so, this is not quite right but is getting at something interesting. I encourage you to read about the Eigenstate Thermalization Hypothesis (including a recent SE question: physics.stackexchange.com/questions/213733/…) Nov 12, 2015 at 2:15
• @Rococo Yes, this is the essence of the question. Thanks - I'll read about it. I'm in gravity so quantum mechanics is more of a hobby :P.
– AGML
Nov 12, 2015 at 16:51

If the system is not interacting with some "environment" and the dynamics is purely hamiltonian, superposition states will not relax to stationary states, even if they are "close" to some such stationary state. Instead they evolve on "orbits". Say, for instance, that $\psi_n$, $\psi_m$ are eigenstates (stationary states) of hamiltonian $H$, $$H\psi_n = E_n \psi_n\\ H\psi_m = E_m \psi_m$$ A state $\psi$ "close" to $\psi_n$, of the form $$\psi = A(\psi_n + \epsilon \psi_m)$$ where $A$ is a normalization constant and $|\epsilon|<< 1$ is a small "overlap" parameter, will evolve in time according to $$\psi(t) = A\left(e^{-\frac{i}{\hbar}E_nt}\psi_n + \epsilon e^{-\frac{i}{\hbar}E_mt}\psi_m \right) =\\ = A e^{-\frac{i}{\hbar}E_nt}\left(\psi_n + \epsilon e^{-\frac{i}{\hbar}(E_m-E_n)t}\psi_m\right) =\\ = A(t)\left(\psi_n + \epsilon(t)\psi_m \right)$$
where $|A(t)| = |A|$ and $|\epsilon(t)| = \epsilon$. Obviously $\psi(t)$ is of the same superposition type and is still "in the neighborhood" of $\psi_n$, but will never decay to $\psi_n$.

Even in the presence of a perturbation $V$, $\psi$ will not "decay" to a stationary state unless it happens to be itself a stationary state of the perturbed hamiltonian $H+V$. Otherwise, it will evolve on an "orbit" determined now by $H+V$ instead of $H$.

What the adiabatic theorem points out is that if a perturbation $V$ is turned on infinitely slowly in time, then a stationary state of $H$, say $\psi_n$, will evolve into a stationary state of $H+V$, say $\Psi_n$. This adiabatic switching-on is equivalent not so much to a continuous fast relaxation from "non-equilibrium" to "equilibrium", but rather to a continuous adjusting of an "orbit". The difference may seem subtle when dealing with large many-body systems, but looking at it this way keeps things consistent regardless of system size.

The "relaxation to equilibrium" picture is more adequate when the system interacts with an environment and its local dynamics is no longer hamiltonian, but dissipative.

• @urdv It looks like you appealed to the time-independent Schrodinger equation in your calculation - is this correct? Does 'small initial deviation equals small final deviation' hold for time-dependent Hamiltonians, or for non-Schrodinger quantum theories?
– AGML
Nov 12, 2015 at 17:14
• Correct, I assumed a time-independent Hamiltonian for a non-interacting system as reference. And no, small deviations need not stay small for arbitrary time dependent Hamiltonians. Think electromagnetic interaction. Same for non-Schroedinger dynamics. Generally 'small deviations stay small' is taken to mean a 'small enough' perturbation relative to the time-independent reference dynamics, but the detailed stability math may be more involved.
– udrv
Nov 12, 2015 at 22:13