# How to understand vector $J$ as probability current from Schroedinger equation?

Multiplying the Schroedinger equation by $$\psi^*$$:
\begin{align} \psi^*[i\hbar\frac{\partial\psi}{\partial t}] = \psi^*[-\frac{\hbar^2}{2m}\nabla^2 + V]\psi \end{align} Its complex conjugate by $$\psi$$: \begin{align} \psi[-i\hbar\frac{\partial\psi^{\ast}}{\partial t}]= \psi[-\frac{\hbar^2}{2m}\nabla^2 + V]\psi^* \end{align}

and subtracting:

\begin{align}i\hbar[\psi^*\frac{\partial\psi}{\partial t} + \psi\frac{\partial\psi^{\ast}}{\partial t}] = \psi^*(-\frac{\hbar^2}{2m}\nabla^2\psi) + \psi^*V\psi\; -\; \psi(-\frac{\hbar^2}{2m}\nabla^2\psi^*) - \psi V\psi^*\end{align}

The potential terms cancel, and dividing by $$\hbar$$ and multiplying by $$-i$$:

\begin{align}[\psi^*\frac{\partial\psi}{\partial t} + \psi\frac{\partial\psi^{\ast}}{\partial t}] = i\frac{\hbar}{2m}[\psi^*\nabla^2\psi\; -\; \psi\nabla^2\psi^*]\end{align}

The left side is the derivative of the product $$\psi^*\psi$$, what is by definition the density of probability $$\rho$$:

\begin{align}\frac{\partial\rho}{\partial t} = \frac{\partial}{\partial t}[\psi^*\psi ] = i\frac{\hbar}{2m}[\psi^*\nabla^2\psi - \psi\nabla^2\psi^*]\end{align}

But now things are not so clear: creating the vectors $$\nabla\psi$$ and $$\nabla\psi^*$$, the right side can be expressed as the divergence of a vector:

\begin{align}\nabla\cdot[\psi^*\nabla\psi - \psi\nabla\psi^*] = \nabla\psi^*\cdot\nabla\psi + \psi^*\nabla^2\psi - \nabla\psi\cdot\nabla\psi^* - \psi\nabla^2\psi^* = \psi^*\nabla^2\psi - \psi\nabla^2\psi^*\end{align}

Of course if we call $$-i\hbar/(2m)[\psi^*\nabla\psi - \psi\nabla\psi^*] = \mathbf{J}$$, the expression becomes similar to the continuity equation:

\begin{align}\frac{\partial\rho}{\partial t} = i\frac{\hbar}{2m}\nabla\cdot[\psi^*\nabla\psi - \psi\nabla\psi^*] => \frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf{J} = 0\end{align}

I understand the idea: if the probability decreases in any region, it must flow outside, so that the total probability is always one. But how such a flow relates to the vector difference in the brackets? I mean: the left side is a known subject, the square of the wave function is associated to the probability, but it is not the case of the right side.

• The probability current can be derived from Noether's theorem applied to infinitesimal $U(1)$ variations of the Schrödinger action. See physics.stackexchange.com/q/279635 – d_b Dec 19 '19 at 22:54
• There is a missing closed bracket in line 3 of this derivation, after "and subtracting:". Could the closed bracket be added in the correct place so I can follow this properly? Thanks. – Guthrie Douglas Prentice Dec 20 '19 at 1:52
• @ Guthrie Douglas Prentice I include the missing bracket – Claudio Saspinski Dec 20 '19 at 2:24
• @Claudio Sapinski Thank you. – Guthrie Douglas Prentice Dec 20 '19 at 3:00
• Closely related. – Cosmas Zachos Dec 20 '19 at 17:14

Let $$\psi = |\psi|e^{\mathfrak{j} \alpha}$$, then $$\ln \left( \frac{\psi}{\psi^*} \right)=2\mathfrak{j}\alpha$$ and $$\\ 2\mathfrak{j} \nabla \alpha = \nabla \ln \left( \frac{\psi}{\psi*} \right)=\nabla \ln\psi - \nabla \ln\psi^* \\=\frac{1}{\psi}\nabla \psi - \frac{1}{\psi^*}\nabla \psi^*.$$ Now multiply both sides with $$\rho=\psi\psi^*$$ $$\psi\psi^*2\mathfrak{j} \nabla \alpha = \psi\psi^* \left(\frac{1}{\psi}\nabla \psi - \frac{1}{\psi^*}\nabla \psi^* \right)\\ = \psi^* \nabla \psi -\psi \nabla\psi^*.$$ But since $$-\mathfrak{j}\hbar/(2m)[\psi^*\nabla\psi - \psi\nabla\psi^*] = \mathbf{J}$$ we also have $$\mathbf{J} = -\mathfrak{j}\frac{\hbar}{2m}\psi\psi^*2\mathfrak{j} \nabla \alpha \\= \rho \frac{\hbar}{m}\nabla\alpha = \rho \mathbf{v}$$
Here I defined a "velocity" $$\mathbf{v} = \frac{\hbar}{m} \nabla \alpha$$ with which the probability density $$\rho$$ is "convected". Notice that $$\mathbf{v}$$ is proportional with the gradient of the equiphase surfaces $$\alpha$$, in other words the velocity field comprise the orthogonal trajectories, rays, of these surfaces.