# A fluid-mechanical visualisation of the quantum-mechanical continuity equation

The concept of a probability current is useful in quantum mechanics for analysing quantum scattering and tunnelling phenomena, among other things. However, I have noticed that the same rather abstract and non-visual approach to introducing probability currents is repeated almost verbatim in every textbook (also see, e.g., this Wikipedia article). The standard approach essentially involves defining a probability current from the outset as

$\vec{j} = \frac{i \hbar}{2m}(\Psi \nabla \Psi^{*} - \Psi^{*} \nabla \Psi)$

and then using Schrödinger’s equation to show that this satisfies a fluid-like continuity equation of the form

$\frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} = 0$

with

$\rho \equiv \Psi^{*} \Psi$

In the present note I want to briefly explore a more intuitive and visual approach involving a model of the actual flow of a probability fluid’. I want to begin with a fluid-mechanical model and then obtain the standard expression for the quantum-mechanical continuity equation from this, rather than starting out with an abstract definition of the probability current and then showing that this satisfies a continuity equation. The essential problem one faces when trying to do this is that although in classical mechanics the position $\vec{r}(t)$ of a point particle and its velocity $\vec{v}(t) = d\vec{r}(t)/dt$ are well defined, this is not the case in conventional quantum mechanics. Quantum mechanics is done probabilistically, treating a particle as a wave packet such that the square of the amplitude of the corresponding wave function acts as a probability density which can be used to measure the probability that the particle will occupy a particular region of space at a particular time. It is not possible to say definitively where a particular particle will be at a particular time in quantum mechanics, which makes it difficult to apply the conventional deterministic equations of fluid mechanics.

A huge specialist literature on quantum hydrodynamics has in fact arisen which tries to circumvent this problem in a number of ways. A standard reference is Wyatt, R. E., 2005, Quantum Dynamics with Trajectories: Introduction to Quantum Hydrodynamics (Springer). The route that a large part of this literature has taken is intriguing because it is based on Bohmian mechanics, an approach to quantum mechanics developed by David Bohm in 1952 which is regarded by most mainstream physicists today as unconventional. The key feature of the Bohmian mechanics approach is that classical-like particle trajectories are possible. Using this approach one can obtain Newtonian-like equations of motion analogous to those in conventional fluid mechanics and this is how this particular literature seems to have chosen to treat quantum particle trajectories in a fluid-like way. Attempts have also been made to introduce mathematically equivalent approaches, but defined within conventional quantum mechanics (see, e.g., Brandt, S. et al, 1998, Quantile motion and tunneling, Physics Letters A, Volume 249, Issue 4, pp. 265-270).

In the present note I am not looking to solve any elaborate problems so I will simply consider a free quantum wave packet which is not acted upon by any forces and try to visualise probability currents and the quantum continuity equation in a fluid-like way by using the momentum vector operator $\hat{\vec{p}}$ to characterise the velocity of the particle. I will then show that the probability current obtained in this fluid-mechanical model is the same as the one defined abstractly in textbooks.

In quantum mechanics, calculations are done using operators to represent observables. Every possible observable that one might be interested in for the purposes of experiment has a corresponding operator which the mathematics of quantum mechanics can work on to produce predictions. The key operator for the purposes of the present note is the momentum vector operator

$\hat{\vec{p}} = -i \hbar \nabla$

which is the quantum mechanical analogue of the classical momentum vector

$\vec{p} = m \vec{v}$

The key idea for the present note is to regard the velocity vector of the quantum particle as being represented by the operator

$\frac{\hat{\vec{p}}}{m}$

by analogy with the classical velocity vector which can be obtained as

$\vec{v} = \frac{\vec{p}}{m}$

We will imagine the total probability mass

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \Psi^{*} \Psi \text{d}V = 1$

as a fluid in steady flow throughout the whole of space and obeying mass conservation. The fact that the flow is steady reflects the fact that there are no forces acting on the quantum particle in this model, so we must have

$\frac{\partial }{\partial t}\big[\frac{\hat{\vec{p}}}{m}\big] = 0$

The velocity can vary from point to point in the probability fluid but at any given point it cannot be varying over time.

In a classical fluid we have a mass density per unit volume $\rho$ and we regard the velocity vector $\vec{v}$ as a volumetric flow rate per unit area, i.e., the volume of fluid that would pass through a unit area per unit time. Then $\rho \vec{v}$ is the mass flow rate per unit area, i.e., the mass of fluid that would pass through a unit area per unit time. In quantum mechanics we can regard the probability mass density per unit volume $\rho \equiv \Psi^{*} \Psi$ as analogous to the mass density of a classical fluid. We can interpret $\frac{\hat{\vec{p}}}{m}$ as the volumetric flow rate per unit area, i.e., the volume of probability fluid that would pass through a unit area per unit time. When doing probability calculations with quantum mechanical operators we usually sandwich’ the operator between $\Psi^{*}$ and $\Psi$, so following that approach here we can define the probability current density as

$\Psi^{*} \frac{\hat{\vec{p}}}{m} \Psi$

This is to be interpreted as the probability mass flow rate per unit area, i.e., the amount of probability mass that would pass through a unit area per unit time, analogous to $\vec{j} = \rho \vec{v}$ in the classical case. To see how close the analogy is, suppose the quantum wave function is that of a plane wave

$\Psi(x, y, z, t) = A\mathrm{e}^{i(\vec{k} \cdot \vec{r} - \omega t)}$

Then

$\Psi^{*} \frac{\hat{\vec{p}}}{m} \Psi = A \mathrm{e}^{-i(\vec{k} \cdot \vec{r} - \omega t)} \frac{(-i \hbar)}{m} \nabla A \mathrm{e}^{i(\vec{k} \cdot \vec{r} - \omega t)}$

$= A \mathrm{e}^{-i(\vec{k} \cdot \vec{r} - \omega t)} \frac{(-i \hbar)}{m} A \ i \ \vec{k} \ \mathrm{e}^{i(\vec{k} \cdot \vec{r} - \omega t)}$

$= A^2 \frac{\hbar \vec{k}}{m}$

$= \rho \vec{v}$

which looks just like the mass flow rate in the classical case with $\rho = \Psi^{*} \Psi$ and $\vec{v} \equiv \frac{\hbar \vec{k}}{m}$. Note that in this example the probability current density formula we are using, namely $\Psi^{*} \frac{\hat{\vec{p}}}{m} \Psi$, turned out to be real-valued. Unfortunately this will not always be the case. Since the probability current vector must always be real-valued, the fluid-mechanical model in the present note will only be applicable in cases when this is true for the formula $\Psi^{*} \frac{\hat{\vec{p}}}{m} \Psi$.

As in classical fluid mechanics, a continuity equation can now be derived by considering the net outflow of probability mass from an infinitesimal fluid element of volume $\mathrm{d}V \equiv \mathrm{d} x \mathrm{d} y \mathrm{d} z$.

Considering only the $y$-component for the moment, we see from the diagram that on the left-hand side we have the probability mass flow rate coming into the volume element through the left-hand face. The mass flow rate coming out of the fluid element through the right-hand face can be approximated using a Taylor series expansion as being equal to the mass flow rate through the left-hand face plus a differential adjustment based on the gradient of the probability current density and the length $\mathrm{d} y$. The net probability mass flow rate in the $y$-direction is then obtained by subtracting the left-hand term from the right-hand term to get

$\frac{\partial }{\partial y} \big(\Psi^{*}\frac{\hat{\vec{p}}}{m} \Psi\big) \mathrm{d}V$

Using similar arguments for the $x$ and $z$-directions, the net mass flow rate out of the fluid element in all three directions is then

$\nabla \cdot \big(\Psi^{*}\frac{\hat{\vec{p}}}{m} \Psi \big) \mathrm{d} V$

Now, the probability mass inside the fluid element is $\rho \mathrm{d} V$ where $\rho = \Psi^{*} \Psi$ and if there is a net outflow of probability fluid this mass will be decreasing at the rate

$- \frac{\partial \rho}{\partial t} \mathrm{d} V$

Equating the two expressions and dividing through by the volume of the fluid element we get the equation of continuity

$\frac{\partial \rho}{\partial t} + \nabla \cdot \big(\Psi^{*}\frac{\hat{\vec{p}}}{m} \Psi \big) = 0$

What I want to do now is show that if we work out $\Psi^{*}\frac{\hat{\vec{p}}}{m} \Psi$ we will get the same formula for the probability current as the one usually given in quantum mechanics textbooks. We have

$\Psi^{*}\frac{\hat{\vec{p}}}{m} \Psi$

$= - \frac{i \hbar}{m} \Psi^{*} \nabla \Psi$

$= - \frac{i \hbar}{2m} \Psi^{*} \nabla \Psi - \frac{i \hbar}{2m} \Psi^{*} \nabla \Psi$

We now note that since the probability current density must be a real vector, the last two terms above must be real. Therefore they are not affected in any way if we take their complex conjugate. Taking the complex conjugate of the second term in the last equality we get

$\Psi^{*}\frac{\hat{\vec{p}}}{m} \Psi$

$= - \frac{i \hbar}{2m} \Psi^{*} \nabla \Psi + \frac{i \hbar}{2m} \Psi \nabla \Psi^{*}$

$= \frac{i \hbar}{2m}(\Psi \nabla \Psi^{*} - \Psi^{*} \nabla \Psi)$

$= \vec{j}$

This is exactly the expression for the probability current density that appears in textbooks, but rather than introducing it `out of nowhere’ at the beginning, we have obtained it naturally as a result of a fluid-mechanical model.