Consider a quantum particle of mass $$m$$ moving along a line, the $$x$$ axis, in the presence of a potential $$V(x,t)$$. The time-evolution of the particle's wave function $$\Psi(x,t)$$ is governed by the Schrödinger equation, $i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2 m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi \,.$ The Schrödinger equation can be written in a hydrodynamic form, known as the Madelung equations: $\frac{\partial \rho}{\partial t} + \frac{\partial (\rho v)}{\partial x} = 0 \,,$ $\frac{\partial v}{\partial t} + v \frac{\partial v}{\partial x} + \frac{1}{m} \frac{\partial (V + Q)}{\partial x} = 0 \,.$ Here, $$\rho(x,t)$$ is the probability density associated with $$\Psi$$, $\rho = |\Psi|^2 \,,$ $$v(x,t)$$ is the velocity field associated with the flow of $$\rho$$, $\rho v = \frac{\hbar}{m} \operatorname{Im} \left\{ \Psi^* \frac{\partial \Psi}{\partial x} \right\} \,,$ and $$Q(x,t)$$ is the Bohm quantum potential, $Q = -\frac{\hbar^2}{2 m} \frac{1}{\sqrt{\rho}} \frac{\partial^2 \sqrt{\rho}}{\partial x^2} \,.$
Proof: Let us represent the wave function in polar form, $\Psi = \sqrt{\rho} \exp \left( i \frac{S}{\hbar} \right) \,,$ where $$S(x,t)$$ is a real-valued function. Then, denoting $$\frac{\partial}{\partial t}$$ by $$\dot{}$$ and $$\frac{\partial}{\partial x}$$ by $$\, {}'$$, we have $\dot{\Psi} = \left( \frac{\dot{\rho}}{2 \rho} + i \frac{\dot{S}}{\hbar} \right) \Psi$ and $\Psi' = \left( \frac{\rho'}{2 \rho} + i \frac{S'}{\hbar} \right) \Psi \,,$ $\Psi'' = \left[ \frac{\rho''}{2 \rho} - \frac{(\rho')^2}{2 \rho^2} + i \frac{S''}{\hbar} + \left( \frac{\rho'}{2 \rho} + i \frac{S'}{\hbar} \right)^2 \right] \Psi \,,$ or $\Psi'' = \left( \frac{\rho''}{2 \rho} - \frac{(\rho')^2}{4 \rho^2} - \frac{(S')^2}{\hbar^2} + i \frac{S''}{\hbar} + i \frac{\rho' S'}{\hbar \rho} \right) \Psi \,.$ Then, we substitute these expressions into the Schrödinger equation, divide both sides by $$\Psi$$, and analyze the real and imaginary parts of the resulting equation.
• For the imaginary part, we have $i \hbar \frac{\dot{\rho}}{2 \rho} = -\frac{\hbar^2}{2 m} \left( i \frac{S''}{\hbar} + i \frac{\rho' S'}{\hbar \rho} \right) \,,$ or $\dot{\rho} = -\frac{(\rho S')'}{m} \,.$ Since $v = \frac{\hbar}{m \rho} \operatorname{Im} \Psi^* \Psi' = \frac{\hbar}{m |\Psi|^2} \operatorname{Im} \left( \frac{\rho'}{2 \rho} + i \frac{S'}{\hbar} \right) |\Psi|^2 = \frac{S'}{m} \,,$ we see that this equation can be written as $\dot{\rho} + (\rho v)' = 0 \,,$ which is nothing but the continuity equation, or the first Madelung equation.
• For the real part, we have $i \hbar \left( i \frac{\dot{S}}{\hbar} \right) = -\frac{\hbar^2}{2 m} \left( \frac{\rho''}{2 \rho} - \frac{(\rho')^2}{4 \rho^2} - \frac{(S')^2}{\hbar^2} \right) + V \,,$ or $\dot{S} + \frac{(S')^2}{2 m} + V - \frac{\hbar^2}{2 m} \left( \frac{\rho''}{2 \rho} - \frac{(\rho')^2}{4 \rho^2} \right) = 0 \,.$ Differentiating the last equation with respect to $$x$$, and making use of the fact that $$S' = m v$$, we obtain $\dot{v} + \frac{(v^2)'}{2} + \frac{1}{m} \left[ V' - \frac{\hbar^2}{2 m} \left( \frac{\rho''}{2 \rho} - \frac{(\rho')^2}{4 \rho^2} \right)' \right] = 0 \,.$ The last equation coincides with the second Madelung equation, provided the quantum potential $$Q$$ can be written as $Q = -\frac{\hbar^2}{2 m} \left( \frac{\rho''}{2 \rho} - \frac{(\rho')^2}{4 \rho^2} \right) \,.$ This is indeed the case: since $(\sqrt{\rho})'' = \left( \frac{\rho'}{2 \sqrt{\rho}} \right)' = \frac{\rho''}{2 \sqrt{\rho}} - \frac{(\rho')^2}{4 \rho^{3/2}} \,,$ the quantum potential reads $Q = -\frac{\hbar^2}{2 m} \frac{(\sqrt{\rho})''}{\sqrt{\rho}} = -\frac{\hbar^2}{2 m} \left( \frac{\rho''}{2 \rho} - \frac{(\rho')^2}{4 \rho^2} \right) \,.$