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.

  • $\begingroup$ 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 $\endgroup$
    – d_b
    Dec 19, 2019 at 22:54
  • $\begingroup$ 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. $\endgroup$ Dec 20, 2019 at 1:52
  • $\begingroup$ @ Guthrie Douglas Prentice I include the missing bracket $\endgroup$ Dec 20, 2019 at 2:24
  • $\begingroup$ @Claudio Sapinski Thank you. $\endgroup$ Dec 20, 2019 at 3:00
  • $\begingroup$ Closely related. $\endgroup$ Dec 20, 2019 at 17:14

1 Answer 1


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.

  • $\begingroup$ J has really dimension of speed if h is not squared. But that is the case, because I had divided only the left side by h. Now the expression for J is correct in my post. $\endgroup$ Dec 20, 2019 at 16:51
  • $\begingroup$ Aaaah, the curse of copy-and-paste!!!!!!!!!!! $\endgroup$
    – hyportnex
    Dec 20, 2019 at 17:21

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