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 \[Q = -\frac{\hbar^2}{2 m} \frac{(\sqrt{\rho})''}{\sqrt{\rho}} = -\frac{\hbar^2}{2 m} \left( \frac{\rho''}{2 \rho} - \frac{(\rho')^2}{4 \rho^2} \right) \,.\]