# Non-locality of the Madelung equations?

As far as I understand, the Madelung equations $$\frac{\partial \rho}{\partial t} + \nabla \cdot \rho \mathbf{u} = 0 \\ \frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u} = -\frac{1}{m} \nabla (Q + V)$$ with $$Q = -\frac{\hbar^2}{2m} \frac{\nabla^2 \sqrt{\rho}}{\sqrt{\rho}}$$ can be obtained from the Schrödinger equation via $$\psi = \sqrt{\rho} e^{i \frac{S}{\hbar}}, \qquad \mathbf{u} = \frac{\nabla S}{m}$$ and are a completely equivalent formulation. Often, the “quantum potential” $$Q$$ is rewritten using a “quantum pressure tensor” $$\mathbf{p}_{\text{Q}} = -\left(\frac{\hbar}{2m}\right)^2 \rho \nabla \otimes \nabla \ln(\rho)$$ with $$\nabla Q = -\frac{m}{\rho} \nabla \cdot \mathbf{p}_{\text{Q}}.$$

Now, I have seen that this tensor is sometimes described as “non-local” – e. g. arXiv:1503.03869, arXiv:1705.05845. In these examples specifically, the only explanation for the term “non-local” is that the quantum pressure tensor depends of the gradient of the density. But I would say that the gradient is a local quantity, since it only depends on an infinitesimal environment of a single point, and $$\rho$$ itself is simply the familiar squared absolute value of the Schrödinger wave function, which evolves locally. In any case, surely the gradient $$\nabla \rho$$ is not any more or less local than, for example, $$\nabla V$$, which also appears in the equations?

I suspect that there might be some confusion stemming from the similarity of the Madelung equations to Bohmian mechanics, whose equations are indeed explicitly non-local. In fact, the only other reference to a non-local quantum pressure which I have found is the Wikipedia page on the Bohmian quantum potential:

### Relation to the Madelung pressure tensor

In the Madelung equations presented by Erwin Madelung in 1927, the non-local quantum pressure tensor has the same mathematical form as the quantum potential. The underlying theory is different in that the Bohm approach describes particle trajectories whereas the equations of Madelung quantum hydrodynamics are the Euler equations of a fluid that describe its averaged statistical characteristics.

However, I’m not sure whether this is trying to say that the Madelung quantum pressure tensor is, in fact, non-local (if so, again: why?) or simply that there is a (local) equivalent in the Madelung equations to Bohm’s non-local quantum potential.