The definition is: $${\bf{j}} = \frac{\hbar}{2mi} (\psi^* \nabla \psi - \psi \nabla \psi^*)$$ However: Where ever I have looked, the above "pops out of nowhere".

I was wondering how can I obtain some intuition about this, and/or, can it be derived from some related definition/s?


I know of two ways to derive this: the first is to take the time derivative of $|\psi|^2$, then use the Schrödinger and the continuity equations, and the second is to start with the Schrödinger lagrangian and find the Noether current. Indeed:

First way

The Schrödinger equation is $$\left( - \frac{\hbar^2}{2m} \nabla^2 + V(\vec{r}) \right) \psi(\vec{r},t) = i\hbar \frac{\partial \psi}{\partial t}(\vec{r},t)$$ and the continuity equation $$\vec{\nabla}\cdot \vec{J} + \frac{\partial \rho}{\partial t} = 0.$$ We have ($\rho \equiv \psi\psi^*$), $$\begin{split} \frac{\partial }{\partial t} \psi \psi^* &= \frac{1}{i\hbar} \left( - \frac{\hbar^2}{2m} \nabla^2 \psi + V\psi\right)\psi^* - \frac{1}{i\hbar} \left( - \frac{\hbar^2}{2m} \nabla^2 \psi^* + V\psi^*\right) \psi\\ &= - \frac{\hbar}{2mi} \vec{\nabla}\cdot \left( \psi^* \vec{\nabla} \psi - \psi \vec{\nabla} \psi^* \right) = - \vec{\nabla} \cdot \vec{J}.\end{split}$$

Second way

In the following $k$ denotes spatial and $\mu$ spacetime indices. The lagrangian is $$\mathcal{L} \equiv \frac{\hbar}{2m} \vec{\nabla} \psi^* \cdot \vec{\nabla} \psi - i \frac{\partial \psi}{\partial t} \psi^* + V \psi\psi^*.$$ This has a global U(1) symmetry, $\psi \to \psi e^{i a}$, hence $\delta \psi = i\psi$, and Noether's theorem yields: $$\begin{split} J^0 &= \frac{\partial \mathcal{L}} { \partial \partial_0\psi} \delta \psi = \psi\psi^* =\rho,\\ J^k &= \frac{\partial \mathcal{L}} {\partial \partial_k \psi} \delta \psi + \frac{\partial \mathcal{L}} {\partial \partial_k \psi^*} \delta \psi^* = \frac{\hbar}{2mi} ( \psi^* \partial^k \psi - \psi \partial^k \psi^*), \end{split}$$ and from $\partial_\mu J^\mu = 0$ the continuity equation follows.

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    $\begingroup$ Quick note: the proper spelling is Noether, without the umlaut (even in German). $\endgroup$ – Emilio Pisanty May 27 '14 at 13:11
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    $\begingroup$ @EmilioPisanty My German language teacher used to say that ö can be replaced by oe and vice versa. Is this wrong? $\endgroup$ – auxsvr May 27 '14 at 13:21
  • $\begingroup$ In normal German ö and oe are indeed interchangeable, with ö being preferable where it is available. Nevertheless, certain names have a fixed spelling, which is the case here. Don't over-umlaut the Germans themselves - if they don't use it, you shouldn't. $\endgroup$ – Emilio Pisanty May 27 '14 at 15:16
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    $\begingroup$ Ok, thanks. This page indicates that the form preferred by the family is Noether indeed, although both forms have been used officially. $\endgroup$ – auxsvr May 27 '14 at 15:28
  • $\begingroup$ I pity the fool who thinks he can out-umlaut the Germans. $\endgroup$ – Marty Green May 27 '14 at 15:49

An intuitive approach, which is less formal, is the conservation of charge: \begin{equation} \nabla \cdot\vec{j} + \partial_t \rho = 0 \end{equation} we know the "charge density" of a wave function is $|\psi|^2$, and we are left to figure out the correct $\vec{j}$ to conserve charge. This is, of course, the Noether current, but intuitively this is what the current is saying.


The way I would approach this problem is to integrate the probability distribution, $P(r,t)=|\psi(r,t)|^2$ over all space and see how this varies with time (take a time derivative). So using the bra ket formalism we have

$$ \frac{\partial}{\partial t} \langle\psi|\psi\rangle $$

Which can be expanded as

$$ \langle\frac{\partial \psi}{\partial t}|\psi\rangle + \langle\psi|\frac{\partial \psi}{\partial t}\rangle $$

The re-express $\frac{\partial \psi}{\partial t}$ and its complex conjugate $\frac{\partial \psi *}{\partial t}$ using the time dependent schrodinger equation in 3D. ie. $$ \frac{\partial}{\partial t} \psi(\mathbf{r},t) = -\frac{i}{ \hbar} \left [ \frac{-\hbar^2}{2m}\nabla^2 + V(\mathbf{r},t)\right ] \psi(\mathbf{r},t) $$

The result should follow using this method.

You may also find the divergence theorem useful in completing the argument.

$$ \iiint_\tau\left(\mathbf{\nabla}\cdot\mathbf{F}\right)\,dV=\iint_{\partial \tau} \mathbf{F}\cdot\mathbf{dS} . $$

I hope this helps, it should at least give you some idea of the motivation behind the definition of $\mathbf{j}$.


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