Educational notebook

Quantum Equilibrium Relaxation

Does any Bohmian initial distribution relax towards the Born distribution?

If the initial particle distribution is \(\rho = |\psi|^2\), equivariance says it remains matched to the Born density.

However, if the particles begin in a different distribution, the flow can stretch and fold that distribution in ways that resemble relaxation toward quantum equilibrium.

Mixing towards the Born distribution does not happen for eigenstates. In general a superposition of eigen states is required for the mixing to happen. Moreover, including the spin-current into the guiding law greatly facilitates the mixing process.

The Born Distribution and Equivariance

First let us demonstrate that if we start with a distribution which follows Borns rule, then it will stay that way. This is called quantum equilibrium. Only in this equilibrium will the Pilot wave theory give the same empirical predictions as the standard interpretation.

Try switching between different mixed states and different guiding laws. Notice how adding spin will increase general mixing.

Born equilibrium

The particles start from the quantum equilibrium. Change the state and guiding law.

Non-equilibrium starting distribution

This time we do not begin from the equilibrium. Let us pick for example an artificial uniform square distribution initially. Will it relax towards the Born distribution as time evolves?

Try it different wave states. You will see that the ground state and the second state are completely unable to mix the starting distribution. The third superposition state, however, manages to create some mixing, but it seems unclear weathers it is able to eventually reach the Born distribution.

It seems clear from these three states, that more complicated states (more superpositions) are better able to reach the quantum equilibrium.

Spin current increases mixing

This time we take into account the spin. The guiding wave is two component spinorial wave, Pauli wave. In the guiding law we add the spin current, which has the effect of driving particles around the density level curves.

This time even the second state mixes well and pretty fast. With spin-current added the relaxation into quantum equilibrium quickly starts to seem believable. Only the ground state, which is a pure eigen state, remains unmixing.

The question about quantum relaxation is interesting since, if the relaxation happens naturally, then we do not need to explain why the initial distribution (of a subsystem or even the whole universe) happens to already obey the Born distribution.

It also invites us to ask the question: "Can we somehow find violations of the equilibrium?" If we can, that would open a door to all kinds of new physics.

Equilibrium and equivariance

Quantum equilibrium is the special distribution \(\rho(\mathbf{x},t)=|\psi(\mathbf{x},t)|^2\). In Bohmian mechanics, particles move with the velocity field determined by the wave, so both the Born density and an ensemble of Bohmian positions satisfy the same continuity equation.

Equivariance \[ \rho(\mathbf{x},0)=|\psi(\mathbf{x},0)|^2 \quad\Longrightarrow\quad \rho(\mathbf{x},t)=|\psi(\mathbf{x},t)|^2 . \]

This is why Born seeding should remain visually matched to the wave density, apart from sampling noise and numerical error. It is also a useful diagnostic for the trajectory integrator.

A relaxation experiment

The Square and Uniform initializations deliberately violate quantum equilibrium. Under a time-dependent superposition, the velocity field can develop stretching, folding, and mixing. At coarse resolution this can make a non-equilibrium distribution look more like the Born distribution, even though the exact fine-grained distribution is simply transported.

Box eigenstate superposition \[ \psi(x,y,t) = \sum_k c_k \sin\!\left(\frac{n_{x,k}\pi x}{L_x}\right) \sin\!\left(\frac{n_{y,k}\pi y}{L_y}\right) e^{-iE_k t/\hbar}. \]

The preset states mix different \(x\) and \(y\) quantum numbers so that the density, phase, and trajectories are genuinely two-dimensional. The ground state is included as a stationary reference where the Born distribution does not move.

Guidance law

The applet supports the standard Schrodinger current and an optional Pauli spin-current term.

Schrodinger current \[ \dot{\mathbf{x}} = \frac{\mathbf{j}}{\rho}, \qquad \mathbf{j} = \frac{\hbar}{m} \operatorname{Im}\left(\psi^*\nabla\psi\right), \qquad \rho=|\psi|^2 . \]

2D Pauli spin-current option \[ \dot{\mathbf{x}}_{\mathrm{Pauli}} = \frac{\mathbf{j}}{\rho} + \frac{\hbar s}{m\rho} \left(\partial_y\rho,-\partial_x\rho\right). \]

Imagining Physics