According to the Copenhagen interpretation of quantum mechanics, particles do not have locations until they are observed. Another interpretation, due to Louis de Broglie [4] and
David Bohm [2, 3], provides an alternative view in which particles have precise positions at all times.
Let $q(t)$ denote the configuration of a physical system, i.e. particles, $Q$ be its configuration space and $\psi(q,t)$ be the wave function underlying such a system. According to this theory, both wave and particles actually
exist as separate entities. We characterise the time evolution of a physical system according to a set of two equations: the Schrödinger equation
\begin{equation} \label{eq:schrodinger}
i\hbar \pdv{\psi}{t} = H\psi,
\end{equation}
where $H$ is the Hamiltonian operator, and the guidance equation
\begin{equation}
\dv{q}{t} = v^\psi(q),
\end{equation}
where $v^\psi = J/\rho$ is a velocity field that can be obtained from the Schrödinger equation itself through the quantum probability density $\rho = \inner{\psi}{\psi}$ and the quantum probability current $J$ [5].
For a non-relativistic system of $N$ particles, each of mass $m_i$, moving subject to a potential $V(r,t)$, the Hamiltonian writes as
\begin{equation}
H = \sum_{i=1}^N \frac{p_i^2}{2m_i} + V = - \sum_{i=1}^N \frac{\hbar^2}{2m_i} \nabla_i^2 + V,
\end{equation}
where $p_i = -i\hbar\grad_i$ deblog the momentum ascribed to the coordinates of the $i$-th particle, respectively. Similarly, the guidance equation becomes
\begin{equation} \label{eq:guidance}
\dv{r_i}{t} = \frac{1}{m_i} \, \Re{\left( \frac{\inner{\psi}{p_i\psi}}{\inner{\psi}{\psi}} \right)} = \frac{\hbar}{m_i} \, \Im{\left( \frac{\inner{\psi}{\grad_i{\psi}}}{\inner{\psi}{\psi}} \right)}.
\end{equation}
An immediate consequence of \eqref{eq:guidance} is that quantum trajectories do not cross each other: if they had, two particles would have shared the same point, making the wave function multi-valued at that point.
The double-slit experiment
One advantage of this theory is that it resolves the dilemma of wave-particle duality in the double-slit experiment. In this experiment, a source sends a stream of particles, e.g. electrons, towards a pair of narrow slits. After the passage
through the slits, the particles reach a photographic plate, leaving a black spot. After a while, these spots start to form an interference pattern on the screen, as if the particles were (or behaved as) waves. Note that this happens even if we
send one particle at a time.
Our model is as follows. Since the Schrödinger equation is linear, we assume that the incident wave function comes out of the slits as a superposition of two Gaussian wave packets, which then propagate forward. We can use other wave packets, e.g.
Airy functions, which propagate freely without envelope dispersion [1], but the end result does not change substantially. Hence, for the $j$-th slit, our problem reads as
\begin{equation}
\begin{cases}
i\hbar \pdv{\psi_j}{t} = -\frac{\hbar^2}{2m} \left( \pdvn{\psi_j}{x}{2} + \pdvn{\psi_j}{y}{2} \right), & (x, t) \in \RR \times [0, \infty], \\
\psi_j(x, y, 0) = \exp{\left[ -\frac{(y-y_j)^2}{2\sigma^2} + ikx \right]}, & t = 0, \\
\psi_j(x, y, t) \to 0, & x, y \to \pm\infty.
\end{cases}
\end{equation}
Separation of variables and a standard application of the Fourier transform give that
\begin{equation}
\psi_j(x,y,t) = \frac{\sigma}{\sigma_t} \, \exp{\left[ -\frac{(y-y_j)^2}{2\sigma_t^2} + i \left(kx - \frac{\hbar k^2}{2m}t\right) \right]},
\end{equation}
with
\begin{equation}
\sigma_t := \sqrt{\sigma^2 + i\frac{\hbar}{m}t} \,.
\end{equation}
Hence, we substitute $\psi = \psi_1 + \psi_2$ into the guiding equation to obtain $x = kt$ and
\begin{equation}
\dv{y}{t} = \Im{\left( \frac{1}{\psi} \pdv{\psi}{y} \right)} = \Im{\left[ \frac{(y_1 - y) \, \psi_1 + (y_2 - y) \, \psi_2}{\sigma_t^2 \, (\psi_1 + \psi_2)} \right]},
\end{equation}
which we solve using the classic 4-th order Runge-Kutta scheme from RungeKutta.jl whilst setting $\hbar = m = k = 1$ and $\sigma = 0.225$:
using RungeKutta
problem = DoubleSlit2D(σ = 0.225)
solver = RK4(h = 1e-3)
solution = solve(problem, solver)
We plot our result in Figure 1. As it turns out, the slits cause the guiding wave to form an interference pattern, which in turn determines the possible trajectories of
each particle.
Figure 1. Left: particle trajectories for the double-slit experiment. Right: (non-normalised) quantum equilibrium distribution at the photographic plate.