From Shankar's QM book pg. 166:

The continuity equation for probability density in QM is $$\frac{\partial P(\vec{r},t)}{\partial t}=-\nabla \cdot \vec{j}(\vec{r},t),$$ where $P=\psi^*\psi$ is the probability density and $\vec{j}$ is the probability current density given by$$\vec{j}=\frac{\hbar}{2mi}(\psi^*\nabla\psi-\psi\nabla\psi^*).$$

Integrating over all space, we get $$\frac{d}{dt}\int P(\vec{r},t) d^3\vec{r}=-\int \nabla\cdot\vec{j}(\vec{r},t)d^3\vec{r}=-\int_{S_\infty}\vec{j}\cdot d\vec{S}$$ where the last integral is over the sphere at infinity.

It was said that for wave functions which are normalizable to unity, $r^{3/2}\psi\rightarrow0$ as $r\rightarrow \infty$ in order that $\int\psi^*\psi r^2drd\Omega$ is bounded, and the surface integral of $\vec{j}$ on $S_\infty$ vanishes.

Why is the author considering the expressions $r^{3/2}\psi$ and $\int\psi^*\psi r^2drd\Omega$? What do these expressions represent?

My understanding is that $\psi\rightarrow0$ as $r\rightarrow \infty$ for normalizable wavefunctions, hence $\vec{j}$ vanishes and the integral $\int_{S_\infty} \vec{j} \cdot d\vec{S}$ vanishes.

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    $\begingroup$ This is integration in cylindrical coordinates: $dxdydz=r^2drd\thetad\phi$. $\endgroup$ May 5 at 6:59

1 Answer 1


The integral $$\int \psi^*\psi\ r^2\ dr\ d\Omega$$ is just the same as integral $$\int P(\mathbf{r},t) d^3\mathbf{r}$$ representing the total probability of finding the particle anywhere. Remember that $P(\mathbf{r},t)=\psi^*\psi$ is the probability density, and $d^3\mathbf{r}=r^2\ dr\ d\Omega$ is the volume element (a slice with thickness $dr$ and area $r^2\ d\Omega$).

The requirement $\lim_{r\to\infty}r^{3/2}\psi=0$ makes sure that in the integral $$\int \psi^*\psi\ r^2\ dr\ d\Omega$$ the integrand $\psi^*\psi\ r^2$ stays smaller than $O(\frac{1}{r})$, and thus the integral from $r=0$ to $r=\infty$ doesn't get infinitely large.


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