Probability in Pilot Wave Theory

Pilot-wave theory assigns definite configurations guided by a wavefunction. The motion can be deterministic while preparation outcomes are statistically distributed. Its probability problem is to explain why configurations are governed by the quantum-equilibrium measure $|\psi|^2$, and why accessible records inherit Born statistics.

Required background. Born rule: Translate a specified state and measurement into outcome probabilities.

Equivariance of the squared wavefunction

For a spinless particle with Hamiltonian $-\hbar^2\nabla^2/(2m)+V$, the guiding velocity is

$$\mathbf v^\psi= \frac{\hbar}{m}\operatorname{Im}\frac{\nabla\psi}{\psi},$$

away from nodes. Schrödinger evolution implies

$$\partial_t|\psi|^2+\nabla\cdot(|\psi|^2\mathbf v^\psi)=0.$$

An ensemble transported by this velocity obeys the same continuity equation. If its initial density is $\rho_0=|\psi_0|^2$, it remains $\rho_t=|\psi_t|^2$ whenever the evolution and boundary conditions are well defined. This property is equivariance.

The pilot-wave interpretation page introduces the dynamics and its physical interpretation. Here the important point is that preservation of equilibrium does not prove that every possible initial distribution starts in equilibrium.

Conditional subsystem probabilities

Write the universal configuration as $(X,Y)$, where $X$ is the subsystem and $Y$ its environment. Given the universal equilibrium measure, ordinary conditional probability yields

$$\Pr(X\in dx\mid Y=y) = \frac{|\Psi(x,y)|^2\,dx} {\int dx'\,|\Psi(x',y)|^2}.$$

When the environment configuration lies in the support of one effectively disjoint branch, the conditional wavefunction is proportional to that branch’s subsystem wavefunction. This gives the familiar $|\psi(x)|^2$ distribution for a prepared subsystem.

The measure assumption and effective-branch approximation are visible in this derivation. Neither should be hidden behind the phrase “determinism produces randomness.”

A real wavefunction need not imply measured zero momentum

For a stationary nondegenerate state with a real spatial wavefunction, the guiding velocity vanishes. Yet a momentum-measurement apparatus can produce a nontrivial distribution of pointer positions. This is not a contradiction: the operator called momentum summarizes outcomes of a particular interaction, not necessarily a nondisturbing readout of a pre-existing value $m\dot Q$.

For example, release a one-dimensional Gaussian with initial position variance $\sigma_0^2$ and real wavefunction

$$\psi_0(x)=(2\pi\sigma_0^2)^{-1/4}e^{-x^2/(4\sigma_0^2)}.$$

Free evolution gives a spreading packet and trajectories

$$v(x,t)=\frac{xt}{\tau^2+t^2},\qquad Q(t)=Q_0\sqrt{1+(t/\tau)^2},\qquad \tau=\frac{2m\sigma_0^2}{\hbar}.$$

Every trajectory starts with zero velocity. Nevertheless, the momentum inferred from a distant detection, $p_{\mathrm{TOF}}=\lim_{t\to\infty}mQ(t)/t=mQ_0/\tau$, has equilibrium variance

$$\operatorname{Var}(p_{\mathrm{TOF}}) =\frac{m^2\sigma_0^2}{\tau^2} =\frac{\hbar^2}{4\sigma_0^2}.$$

This agrees with the Fourier momentum distribution of the initial state. Releasing the packet changes its subsequent motion; the time-of-flight result does not reveal the particle’s initial guiding velocity.

Typicality versus dynamical relaxation

Typicality regards Born-like subsystem records as typical with respect to a universal equilibrium measure. Relaxation arguments ask whether nonequilibrium ensembles develop coarse-grained equilibrium under suitable mixing. These are different strategies.

Fine-grained deterministic transport preserves information about an initial nonequilibrium ratio $\rho/|\psi|^2$. It cannot by itself establish universal convergence. The distinction matters when assessing proposals for observable nonequilibrium deviations.

Exercises

Assume both $\rho$ and $|\psi|^2$ obey the same continuity equation. Show that $f=\rho/|\psi|^2$ is constant along trajectories where the denominator is nonzero.

Solution

For a density $g$, the material derivative satisfies $(\partial_t+\mathbf v\cdot\nabla)g=-g\nabla\cdot\mathbf v$. Applying the quotient rule to $\rho/g$ cancels the two divergence terms. Hence $(\partial_t+\mathbf v\cdot\nabla)f=0$. Equilibrium $f=1$ is preserved; arbitrary $f$ is transported, not erased.

References

• D. Bohm, “A suggested interpretation of the quantum theory in terms of ‘hidden’ variables. I,” Physical Review 85, 166–179 (1952), doi:10.1103/PhysRev.85.166.
• D. Dürr, S. Goldstein, and N. Zanghì, “Quantum equilibrium and the origin of absolute uncertainty,” Journal of Statistical Physics 67, 843–907 (1992), author manuscript.